digital-signal-processing-objective.docx

1.Which of the following is an method for implementing an FIR system?
a) Direct form
b) Cascade form
c) Lattice structure
d) All of the mentioned
Answer: d
Explanation: There are several structures for implementing an FIR system, beginning with the
simplest structure, called the direct form. There are several other methods like cascade form
realization, frequency sampling realization and lattice realization which are used for
implementing and FIR system.
2. How many memory locations are used for storage of the output point of a sequence of
length M in direct form realization?
a) M+1
b) M
c) M-1
d) none of the mentioned
Answer: c
Explanation: The direct form realization follows immediately from the non-recursive difference
equation given by y(n)=∑M−1k=0bkx(n−k).
We observe that this structure requires M-1 memory locations for storing the M-1 previous
inputs.
3. What is the general systemfunction of an FIR system?
a) ∑M−1k=0bkx(n−k)
b) ∑Mk=0bkz−k
c) ∑M−1k=0bkz−k
d) None of the mentioned
Answer: c
Explanation: We know that the difference equation of an FIR system is given by
y(n)=∑M−1k=0bkx(n−k).=>h(n)=bk=>∑M−1k=0bkz−k.

4. Which of the following is an method for implementing an FIR system?
a) Direct form
b) Cascade form
c) Lattice structure
Answer: d
Explanation: There are several structures for implementing an FIR system, beginning with the
simplest structure, called the direct form. There are several other methods like cascade form
realization, frequency sampling realization and lattice realization which are used for
implementing and FIR system.
5. The realization of FIR filter by frequency sampling realization can be viewed as cascade
of how many filters?
a) Two
b) Three
c) Four
Answer: a
Explanation: In frequency sampling realization, the system function H(z) is characterized by the
set of frequency samples {H(k+ α)} instead of {h(n)}. We view this FIR filter realization as a
cascade of two filters. One is an all-zero or a comb filter and the other consists of parallel bank
of single pole filters with resonant frequencies.

6. Which of the following filters have a cascade realization as shown below?
a) IIR filter
b) Comb filter
c) High pass filter
d) FIR filter
Answer: d
Explanation: The system function of the FIR filter according to the frequency sampling
realization is given by the equation
H(z)=1M(1−z−Mej2πα)∑M−1k=0H(k+α)1−ej2π(k+α)Mz−1
The above system function can be represented in the cascade form as shown in the above block
diagram.
13. Where does the poles of the systemfunction of the second filter locate?
a) ej2π(k+α)M
b) ej2π(k+α)/M
c) ej2π(k-α)/M
d) ejπ(k+α)/M
Answer: b
Explanation: The system function of the second filter in the cascade of an FIR realization by
frequency sampling method is given by
H2(z)=∑M−1k=0H(k+α)1−ej2π(k+α)Mz−1.We obtain the poles of the above system function by
equating the denominator of the above equation to zero.

11. The zeros of the systemfunction of comb filter are located at
a) Inside unit circle
b) On unit circle
c) Outside unit circle
Answer: b
Explanation: The system function of the comb filter is given by the equation
H1(z)=1M(1−z−Mej2πα)
Its zeros are located at equally spaced points on the unit circle at zk=ej2π(k+α)/M k=0,1,2….M-1
12. By combining two pairs of poles to form a fourth order filter section, by what factor we
have reduced the number of multiplications?
a) 25%
b) 30%
c) 40%
d) 50%
Answer: d
Explanation: We have to do 3 multiplications for every second order equation. So, we have to
do 6 multiplications if we combine two second order equations and we have to perform 3
multiplications by directly calculating the fourth order equation. Thus the number of
multiplications are reduced by a factor of 50%.
13.What is the value of the coefficient α2(1) in the case of FIR filter represented in direct
form structure with m=2 in terms of K1 and K2?
a) K1(K2)
b) K1(1-K2)
c) K1(1+K2)
Answer: c
Explanation: The equation for the output of an FIR filter represented in the direct form structure
is given as y(n)=x(n)+ α2(1)x(n-1)+ α2(2)x(n-2).The output from the double stage lattice

structure is given by the equation,f2(n)= x(n)+K2(1+K2)x(n-1)+K2x(n-2)
By comparing the coefficients of both the equations, we get α2(1)= K1(1+K2).
14. Which of the following is true for the given signal flow graph?
a) Two pole system
b) Two zero system
c) Two pole and two zero system
Answer: c
Explanation: The equivalent filter structure of the given signal flow graph in the direct form-II
is given by as
Thus from the above structure, the system has two zeros and two poles.
15. What are the nodes that replace the adders in the signal flow graphs?
a) Source node
b) Sink node
c) Branch node
d) Summing node
Answer: d
Explanation: Summing node is the node which is used in the signal flow graph which replaces
the adder in the structure of a filter.
16. If we reverse the directions of all branch transmittances and interchange the input and
output in the flow graph, then the resulting structure is called as
a) Direct form-I
b) Transposed form

c) Direct form-II
Answer: b
Explanation: According to the transposition or flow-graph reversal theorem, if we reverse the
directions of all branch transmittances and interchange the input and output in the flow graph,
then the system remains unchanged. The resulting structure is known as transposed structure or
transposed form.
17. What does the structure given below represents?
a) Direct form-I
b) Regular Direct form-II
c) Transposed direct form-II
Answer: c
Explanation: The structure given in the question is the transposed direct form-II
structure of a two pole and two zero IIR system.

18. The structure shown below is known as
a) Parallel form structure
b) Cascade structure
c) Direct form
Answer: a
Explanation: From the given figure, it consists of a parallel bank of single pole filters and thus it
is called as parallel form structure.
19. If M and N are the orders of numerator and denominator of rational systemfunction
respectively, then how many memory locations are required in direct form-I realization of
that IIR filter?
a) M+N+1
b) M+N
c) M+N-1
d) M+N-2
Answer: a
Explanation: From the direct form-I realization of the IIR filter, if M and N are the orders of
numerator and denominator of rational system function respectively, then M+N+1 memory
locations are required.
20. If M and N are the orders of numerator and denominator of rational systemfunction
respectively, then how many additions are required in direct form-I realization of that IIR
filter?
a) M+N-1
b) M+N

c) M+N+1
d) M+N+2
Answer: b
Explanation: From the direct form-I realization of the IIR filter, if M and N are the orders of
numerator and denominator of rational system function respectively, then M+N additions are
required.
21) The interface between an analog signal and a digital processor is
a. D/A converter
b. A/D converter
c. Modulator
d. Demodulator
ANSWER: (b) A/D converter
22) The speechsignal is obtained after
a. Analog to digital conversion
b. Digital to analog conversion
c. Modulation
d. Quantization
ANSWER: (b) Digital to analog conversion
23) Telegraph signals are examples of
a. Digital signals
b. Analog signals
c. Impulse signals
d. Pulse train
ANSWER: (a) Digital signals

24) As compared to the analog systems, the digital processing of signals allow
1) Programmable operations
2) Flexibility in the system design
3) Cheaper systems
4) More reliability
a. 1, 2 and 3 are correct
b. 1 and 2 are correct
c. 1, 2 and 4 are correct
d. All the four are correct
ANSWER: (d) All the four are correct
25) The Nyquist theorem for sampling
1) Relates the conditions in time domain and frequency domain
2) Helps in quantization
3) Limits the bandwidth requirement
4) Gives the spectrum of the signal
c. 1 and 3 are correct
ANSWER: (c) 1 and 3 are correct
26) Roll-off factor is
a. The bandwidth occupied beyond the Nyquist Bandwidth of the filter
b. The performance of the filter or device
c. Aliasing effect
d. None of the above
ANSWER: (a) The bandwidth occupied beyond the Nyquist Bandwidth of the filter

27) A discrete time signal may be
1) Samples of a continuous signal
2) A time series which is a domain of integers
3) Time series of sequence of quantities
4) Amplitude modulated wave
ANSWER: (a) 1, 2 and 3 are correct
28) The discrete impulse function is defined by
a. δ(n) = 1, n ≥ 0= 0, n ≠ 1
b. δ(n) = 1, n = 0= 0, n ≠ 1
c. δ(n) = 1, n ≤ 0= 0, n ≠ 1
d. δ(n) = 1, n ≤ 0= 0, n ≥ 1
ANSWER: (b) δ(n) = 1, n = 0= 0, n ≠ 1
29) DTFT is the representation of
a. Periodic Discrete time signals
b. Aperiodic Discrete time signals
c. Aperiodic continuous signals
d. Periodic continuous signals
ANSWER:(b) Aperiodic Discrete time signals

30) The transforming relations performed by DTFT are
1) Linearity
2) Modulation
3) Shifting
4) Convolution
31) The DFT is preferred for
1) Its ability to determine the frequency component of the signal
2) Removal of noise
3) Filter design
4) Quantization of signal
32) Frequency selectivity characteristics of DFT refers to
a. Ability to resolve different frequency components from input signal
b. Ability to translate into frequency domain
c. Ability to convert into discrete signal
ANSWER: (a) Ability to resolve different frequency components from input signal

33) The Cooley–Tukey algorithm of FFT is a
a. Divide and conquer algorithm
b. Divide and rule algorithm
c. Split and rule algorithm
d. Split and combine algorithm
ANSWER: (a) Divide and conquer algorithm
34) FFT may be used to calculate
1) DFT
2) IDFT
3) Direct Z transform
4) In direct Z transform
ANSWER: (b) 1 and 2 are correct
35) DIT algorithm divides the sequence into
a. Positive and negative values
b. Even and odd samples
c. Upper higher and lower spectrum
d. Small and large samples
ANSWER: (b) Even and odd samples
36) The computational procedure for Decimation in frequency algorithm takes
a. Log2 N stages
b. 2Log2 N stages
c. Log2 N2 stages
d. Log2 N/2 stages

ANSWER:(a) Log2 N stages
37) The transformations are required for
1) Analysis in time or frequency domain
2) Quantization
3) Easier operations
4) Modulation
38) The s plane and z plane are related as
a. z = esT
b. z = e2sT
c. z = 2esT
d. z = esT/2
ANSWER: (a) z = eSt
39) The similarity between the Fourier transform and the z transform is that
a. Both convert frequency spectrum domain to discrete time domain
b. Both convert discrete time domain to frequency spectrum domain
c. Both convert analog signal to digital signal
d. Both convert digital signal to analog signal
ANSWER: (b) Both convert discrete time domain to frequency spectrum domain

40) The ROC of a systemis the
a. range of z for which the z transform converges
b. range of frequency for which the z transform exists
c. range of frequency for which the signal gets transmitted
d. range in which the signal is free of noise
ANSWER: (a) range of z for which the z transform converges
41) The several ways to perform an inverse Z transform are
1) Direct computation
2) Long division
3) Partial fraction expansion with table lookup
4) Direct inversion
42) The anti causal sequences have components in the left hand sequences.
a. Positive
b. Negative
c. Both a and b
ANSWER: (a) Positive
43) For an expanded power series method, the coefficients represent
a. Inverse sequence values
b. Original sequence values
c. Negative values only
d. Positive values only
ANSWER: (a) Inverse sequence values

44) The region of convergence of x/ (1+2x+x2) is
a. 0
b. 1
c. Negative
d. Positive
ANSWER: (b) 1
45) The IIR filter designing involves
a. Designing of analog filter in analog domain and transforming into digital domain
b. Designing of digital filter in analog domain and transforming into digital domain
c. Designing of analog filter in digital domain and transforming into analog domain
d. Designing of digital filter in digital domain and transforming into analog domain
ANSWER: (b) Designing of digital filter in analog domain and transforming into digital
domain
46) For a system function H(s) to be stable
a. The zeros lie in left half of the s plane
b. The zeros lie in right half of the s plane
c. The poles lie in left half of the s plane
d. The poles lie in right half of the s plane
ANSWER: (c) The poles lie in left half of the s plane
47) IIR filter design by approximation of derivatives has the limitations
1) Used only for transforming analog high pass filters
2) Used for band pass filters having smaller resonant frequencies
3) Used only for transforming analog low pass filters
4) Used for band pass filters having high resonant frequencies

48) The filter that may not be realized by approximation of derivatives techniques are
1) Band pass filters
2) High pass filters
3) Low pass filters
4) Band reject filters
49) In direct form for realisation of IIR filters,
1) Denominator coefficients are the multipliers in the feed forward paths
2) Multipliers in the feedback paths are the positives of the denominator coefficients
3) Numerator coefficients are the multipliers in the feed forward paths
4) Multipliers in the feedback paths are the negatives of the denominator coefficients
ANSWER:(c) 3 and 4 are correct

50) The direct form II for realisation involves
1) The realisation of transfer function into two parts
2) Realisation after fraction
3) Product of two transfer functions
4) Addition of two transfer functions
51) The cascade realisation of IIR systems involves
1) The transfer function broken into product of transfer functions
2) The transfer function divided into addition of transfer functions
3) Factoring the numerator and denominator polynomials
4) Derivatives of the transfer functions
ANSWER:(b) 1 and 3 are correct
52) The advantage of using the cascade form of realisation is
1) It has same number of poles and zeros as that of individual components
2) The number of poles is the product of poles of individual components
3) The number of zeros is the product of poles of individual components
4) Over all transfer function may be determined

53) Which among the following represent/s the characteristic/s of an ideal filter?
a. Constant gain in passband
b. Zero gain in stop band
c. Linear Phase Response
d. All of the above
ANSWER: (d) All of the above
54) FIR filters
A. are non-recursive
B. do not adopt any feedback
C. are recursive
D. use feedback
a. A & B
b. C & D
c. A & D
d. B & C
ANSWER:(a) A & B
55) In tapped delay line filter, the tapped line is also known as
a. Pick-on node
b. Pick-off node
c. Pick-up node
d. Pick-down node
ANSWER:(b) Pick-off node

56) How is the sensitivity of filter coefficient quantization for FIR filters?
a. Low
b. Moderate
c. High
d. Unpredictable
ANSWER: (a) Low
57) Decimation is a process in which the sampling rate is .
a. enhanced
b. stable
c. reduced
d. unpredictable
ANSWER:(c) reduced
58) Anti-imaging filter with cut-off frequency ωc = π/ I is specifically used
upsampling process for the removal of unwanted images.
a. Before
b. At the time of
c. After
d. All of the above
ANSWER: (c) After
59) Which units are generally involved in Multiply and Accumulate (MAC)?
a. Adder
b. Multiplier
c. Accumulator
d. All of the above
ANSWER: (d) All of the above

60) In DSP processors, which among the following maintains the track of addresses of
input data as well as the coefficients stored in data and program memories?
a. Data Address Generators (DAGs)
b. Program sequences
c. Barrel Shifter
d. MAC
ANSWER: (a) Data Address Generators (DAGs)
61) FIR filters
A. are non-recursive
B. do not adopt any feedback
C. are recursive
D. use feedback
ANSWER: A&B
62. If x(n) and X(k) are an N-point DFT pair, then x(n+N)=x(n).
a) True
b) False
Answer: a
Explanation: We know that the expression for an DFT is given as
X(k)=∑N−1n=0x(n)e−j2πkn/N
Now take x(n)=x(n+N)=>X1(k)=∑N−1n=0x(n+N)e−j2πkn/N
Let n+N=l=>X1(k)=∑0l=Nx(l)e−j2πkl/N=X(k)
Therefore, we got x(n)=x(n+N)

63. If x(n) and X(k) are an N-point DFT pair, then X(k+N)=?
a) X(-k)
b) -X(k)
c) X(k)
Answer: c
Explanation: We know that
x(n)=1N∑N−1k=0x(k)ej2πkn/N
Let X(k)=X(k+N)=>x1(n)=1N∑N−1k=0X(k+N)ej2πkn/N=x(n)
Therefore, we have X(k)=X(k+N)
64. If X1(k) and X2(k) are the N-point DFTs of X1(n) and x2(n) respectively, then what is the
N-point DFT of x(n)=ax1(n)+bx2(n)?
a) X1(ak)+X2(bk)
b) aX1(k)+bX2(k)
c) eakX1(k)+ebkX2(k)
Answer: b
Explanation: We know that, the DFT of a signal x(n) is given by the expression
Given x(n)=ax1(n)+bx2(n)
=>X(k)= ∑N−1n=0(ax1(n)+bx2(n))e−j2πkn/N=a∑N−1n=0x1(n)e−j2πkn/N+b∑N−1n=0x2(n)
65. If x(n) is a complex valued sequence given by x(n)=xR(n)+jxI(n), then what is the DFT of
xR(n)?
a) ∑Nn=0xR(n)cos2πknN+xI(n)sin2πknN
b) ∑Nn=0xR(n)cos2πknN−xI(n)sin2πknN
c) ∑N−1n=0xR(n)cos2πknN−xI(n)sin2πknN
d) ∑N−1n=0xR(n)cos2πknN+xI(n)sin2πknN
Answer: d
Explanation: Given x(n)=xR(n)+jxI(n)=>xR(n)=1/2(x(n)+x*(n))

Substitute the above equation in the DFT expression
Thus we get, XR(k)=∑N−1n=0xR(n)cos2πknN+xI(n)sin2πknN
66. If x(n) is a real sequence and X(k) is its N-point DFT, then which of the following is
true?
a) X(N-k)=X(-k)
b) X(N-k)=X*(k)
c) X(-k)=X*(k)
Answer: d
Now X(N-k)=∑N−1n=0x(n)e−j2π(N−k)n/N=X*(k)=X(-k)
Therefore,
X(N-k)=X*(k)=X(-k)
67. If x(n) is real and even, then what is the DFT of x(n)?
a) ∑N−1n=0x(n)sin2πknN
b) ∑N−1n=0x(n)cos2πknN
c) -j∑N−1n=0x(n)sin2πknN
Answer: b
Explanation: Given x(n) is real and even, that is x(n)=x(N-n)
We know that XI(k)=0. Hence the DFT reduces to
X(k)=∑N−1n=0x(n)cos2πknN ;0 ≤ k ≤ N-1
68. If x(n) is real and odd, then what is the IDFT of the given sequence?
a) j1N∑N−1k=0x(k)sin2πknN
b) 1N∑N−1k=0x(k)cos2πknN
c) −j1N∑N−1k=0x(k)sin2πknN
Answer: a
Explanation: If x(n) is real and odd, that is x(n)=-x(N-n), then XR(k)=0. Hence X(k) is purely

imaginary and odd. Since XR(k) reduces to zero, the IDFT reduces to
x(n)=j1N∑N−1k=0x(k)sin2πknN
69. If X1(n), x2(n) and x3(m) are three sequences each of length N whose DFTs are given as
X1(k), X2(k) and X3(k) respectively and X3(k)=X1(k).X2(k), then what is the expression for
x3(m)?
a) ∑N−1n=0x1(n)x2(m+n)
b) ∑N−1n=0x1(n)x2(m−n)
c) ∑N−1n=0x1(n)x2(m−n)N
d) ∑N−1n=0x1(n)x2(m+n)N
Answer: c
Explanation: If X1(n), x2(n) and x3(m) are three sequences each of length N whose DFTs are
given as X1(k), x2(k) and X3(k) respectively and X3(k)=X1(k).X2(k), then according to the
multiplication property of DFT we have x3(m) is the circular convolution of X1(n) and x2(n).
That is x3(m) = ∑N−1n=0x1(n)x2(m−n)N.
70. What is the circular convolution of the sequences X1(n)={2,1,2,1} and x2(n)={1,2,3,4}?
a) {14,14,16,16}
b) {16,16,14,14}
c) {2,3,6,4}
d) {14,16,14,16}
Answer: d
Explanation: We know that the circular convolution of two sequences is given by the expression
x(m)= ∑N−1n=0x1(n)x2(m−n)N
For m=0, x2((-n))4={1,4,3,2}
For m=1, x2((1-n))4={2,1,4,3}
For m=2, x2((2-n))4={3,2,1,4}
For m=3, x2((3-n))4={4,3,2,1}
Now we get x(m)={14,16,14,16}.

71. What is the circular convolution of the sequences X1(n)={2,1,2,1} and x2(n)={1,2,3,4},
find using the DFT and IDFT concepts?
a) {16,16,14,14}
b) {14,16,14,16}
c) {14,14,16,16}
Answer: b
Explanation: Given X1(n)={2,1,2,1}=>X1(k)=[6,0,2,0]
Given x2(n)={1,2,3,4}=>X2(k)=[10,-2+j2,-2,-2-j2]
when we multiply both DFTs we obtain the product
X(k)=X1(k).X2(k)=[60,0,-4,0]
By applying the IDFT to the above sequence, we get
x(n)={14,16,14,16}.
72. If X(k) is the N-point DFT of a sequence x(n), then circular time shift property is that
N-point DFT of x((n-l))N is X(k)e-j2πkl/N.
a) True
b) False
Answer: a
Explanation: According to the circular time shift property of a sequence, If X(k) is the N-point
DFT of a sequence x(n), then the N-pint DFT of x((n-l))N is X(k)e-j2πkl/N.
73. If X(k) is the N-point DFT of a sequence x(n), then what is the DFT of x*(n)?
a) X(N-k)
b) X*(k)
c) X*(N-k)
Answer: c
Explanation: According to the complex conjugate property of DFT, we have if X(k) is the N-
point DFT of a sequence x(n), then what is the DFT of x*(n) is X*(N-k).

74. By means of the DFT and IDFT, determine the response of the FIR filter with impulse
response h(n)={1,2,3} to the input sequence x(n)={1,2,2,1}?
a) {1,4,11,9,8,3}
b) {1,4,9,11,8,3}
c) {1,4,9,11,3,8}
d) {1,4,9,3,8,11}
Answer: b
Explanation: The input sequence has a length N=4 and impulse response has a length M=3. So,
the response must have a length of 6(4+3-1).
We know that, Y(k)=X(k).H(k)
Thus we obtain Y(k)={36,-14.07-j17.48,j4,0.07+j0.515,0,0.07-j0.515,-j4,-14.07+j17.48}
By applying IDFT to the above sequence, we get y(n)={1,4,9,11,8,3,0,0}
Thus the output of the system is {1,4,9,11,8,3}.
75.What is the sequence y(n) that results from the use of four point DFTs if the impulse
response is h(n)={1,2,3} and the input sequence x(n)={1,2,2,1}?
a) {9,9,7,11}
b) {1,4,9,11,8,3}
c) {7,9,7,11}
d) {9,7,9,11}
Answer: d
Explanation: The four point DFT of h(n) is H(k)=1+2e-jkπ/2+3 e-jkπ (k=0,1,2,3)
Hence H(0)=6, H(1)=-2-j2, H(3)=2, H(4)=-2+j2
The four point DFT of x(n) is X(k)= 1+2e-jkπ/2+2 e-jkπ+3e-3jkπ/2(k=0,1,2,3)
Hence X(0)=6, X(1)=-1-j, X(2)=0, X(3)=-1+j
The product of these two four point DFTs is
Ŷ(0)=36, Ŷ(1)=j4, Ŷ(2)=0, Ŷ(3)=-j4
The four point IDFT yields ŷ(n)={9,7,9,11}
We can verify as follows
We know that from the previous question y(n)={1,4,9,11,8,3}
ŷ(0)=y(0)+y(4)=9
ŷ(1)=y(1)+y(5)=7

ŷ(2)=y(2)=9
ŷ(3)=y(3)=11.
76. If the signal to be analyzed is an analog signal, we would pass it through an anti-aliasing
filter with B as the bandwidth of the filtered signal and then the signal is sampled at a rate
a) Fs ≤ 2B
b) Fs ≤ B
c) Fs ≥ 2B
d) Fs = 2B
Answer: c
Explanation: The filtered signal is sampled at a rate of Fs≥ 2B, where B is the bandwidth of the
filtered signal to prevent aliasing.
77. What is the highest frequency that is contained in the sampled signal?
a) 2Fs
b) Fs/2
c) Fs
Answer: b
Explanation: We know that, after passing the signal through anti-aliasing filter, the filtered
signal is sampled at a rate of Fs≥ 2B=>B≤ Fs/2.Thus the maximum frequency of the sampled
signal is Fs/2.
78. If {x(n)} is the signal to be analyzed, limiting the duration of the sequence to L samples,
in the interval 0≤ n≤ L-1, is equivalent to multiplying {x(n)} by?
a) Kaiser window
b) Hamming window
c) Hanning window
d) Rectangular window
Answer: d
Explanation: The equation of the rectangular window w(n) is given as

w(n)=1, 0≤ n≤ L-1=0, otherwise
Thus, we can limit the duration of the signal x(n) to L samples by multiplying it with a
rectangular window of length L.
79. What is the Fourier transform of rectangular window of length L?
a) sin(ωL2)sin(ω2)ejω(L+1)/2
b) sin(ωL2)sin(ω2)ejω(L−1)/2
c) sin(ωL2)sin(ω2)e−jω(L−1)/2
Answer: c
Explanation: We know that the equation for the rectangular window w(n) is given as
w(n)=1, 0≤ n≤ L-1=0, otherwise
We know that the Fourier transform of a signal x(n) is given as
X(ω)=∑∞n=−∞x(n)e−jωn=>W(ω)=∑L−1n=0e−jωn=sin(ωL2)sin(ω2)e−jω(L−1)/2
80. If x(n)=cosω0n and W(ω) is the Fourier transform of the rectangular signal w(n), then
what is the Fourier transform of the signal x(n).w(n)?
a) 1/2[W(ω-ω0)- W(ω+ω0)]
b) 1/2[W(ω-ω0)+ W(ω+ω0)]
c) [W(ω-ω0)+ W(ω+ω0)]
d) [W(ω-ω0)- W(ω+ω0)]
Answer: b
Explanation: According to the exponential properties of Fourier transform, we get
Fourier transform of x(n).w(n)= 1/2[W(ω-ω0)+ W(ω+ω0)]
81. Which of the following is the advantage of Hanning window over rectangular window?
a) More side lobes
b) Less side lobes
c) More width of main lobe
Answer: b
Explanation: The Hanning window has less side lobes and the leakage is less in this windowing
technique.

82. Which of the following is the disadvantage of Hanning window over rectangular
window?
a) More side lobes
b) Less side lobes
c) More width of main lobe
Answer: c
Explanation: In the magnitude response of the signal windowed using Hanning window, the
width of the main lobe is more which is the disadvantage of this technique over rectangular
windowing technique.
83.Which of the following is true regarding the number of computations required to
compute an N-point DFT?
a) N2 complex multiplications and N(N-1) complex additions
b) N2 complex additions and N(N-1) complex multiplications
c) N2 complex multiplications and N(N+1) complex additions
d) N2 complex additions and N(N+1) complex multiplications
Answer: a
Explanation: The formula for calculating N point DFT is given as
From the formula given at every step of computing we are performing N complex multiplications
and N-1 complex additions. So, in a total to perform N-point DFT we perform N2 complex
multiplications and N(N-1) complex additions.
84. Which of the following is true regarding the number of computations required to
compute DFT at any one value of ‘k’?
a) 4N-2 real multiplications and 4N real additions
b) 4N real multiplications and 4N-4 real additions
c) 4N-2 real multiplications and 4N+2 real additions
d) 4N real multiplications and 4N-2 real additions
Answer: d

N
N
N
X(k)=∑N−1n=0x(n)e−j2πkn/N From the formula given at every step of computing we are
performing N complex multiplications and N-1 complex additions. So, it requires 4N real
multiplications and 4N-2 real additions for any value of ‘k’ to compute DFT of the sequence.
85. WN
k+N/2
=?
a) W k
b) -W k
c) W -k
Answer: b
Explanation: According to the symmetry property, we get WNk+N/2=-WN
k.
86. What is the real part of the N point DFT XR(k) of a complex valued sequence x(n)?
a) ∑N−1n=0[xR(n)cos2πknN–xI(n)sin2πknN]
b) ∑N−1n=0[xR(n)sin2πknN+xI(n)cos2πknN]
c) ∑N−1n=0[xR(n)cos2πknN+xI(n)sin2πknN]
Answer: c
Explanation: For a complex valued sequence x(n) of N points, the DFT may be expressed as
XR(k)=∑N−1n=0[xR(n)cos2πknN+xI(n)sin2πknN]
87. The computation of XR(k) for a complex valued x(n) of N points requires
a) 2N2 evaluations of trigonometric functions
b) 4N2 real multiplications
c) 4N(N-1) real additions
Answer: d
Explanation: The expression for XR(k) is given as
XR(k)=∑N−1n=0[xR(n)cos2πknN+xI(n)sin2πknN]
So, from the equation we can tell that the computation of XR(k) requires 2N2 evaluations of
trigonometric functions, 4N2 real multiplications and 4N(N-1) real additions.

N
L
N
= W
88. If the arrangement is of the form in which the first row consists of the first M elements
of x(n), the second row consists of the next M elements of x(n), and so on, then which of the
following mapping represents the above arrangement?
a) n=l+mL
b) n=Ml+m
c) n=ML+l
Answer: b
Explanation: If we consider the mapping n=Ml+m, then it leads to an arrangement in which the
first row consists of the first M elements of x(n), the second row consists of the next M elements
of x(n), and so on.
89. If N=LM, then what is the value of W mqL?
a) WM
mq
b) W mq
c) W mq
Answer: a
Explanation: We know that if N=LM, then W mqL mq= W
N N/L M
mq
.
90. How many complex multiplications are performed in computing the N-point DFT of a
sequence using divide-and-conquer method if N=LM?
a) N(L+M+2)
b) N(L+M-2)
c) N(L+M-1)
d) N(L+M+1)
Answer: d
Explanation: The expression for N point DFT is given as
X(p,q)=∑L−1l=0{WlqN[∑M−1m=0x(l,m)WmqM]}WlpL
The first step involves L DFTs, each of M points. Hence this step requires LM2 complex
multiplications, second require LM and finally third requires ML2. So, Total complex
multiplications = N(L+M+1).

91. How many complex additions are performed in computing the N-point DFT of a
sequence using divide-and-conquer method if N=LM?
a) N(L+M+2)
b) N(L+M-2)
c) N(L+M-1)
d) N(L+M+1)
Answer: b
Explanation: The expression for N point DFT is given as
X(p,q)=∑L−1l=0{WlqN[∑M−1m=0x(l,m)WmqM]}WlpL
The first step involves L DFTs, each of M points. Hence this step requires LM(M-1) complex
additions, second step do not require any additions and finally third step requires ML(L-1)
complex additions. So, Total number of complex additions=N(L+M-2).
92. Which is the correct order of the following steps to be done in one of the algorithm of
divide and conquer method?
i) Store the signal column wise
ii) Compute the M-point DFT of each row
iii) Multiply the resulting array by the phase factors WN
lq.
iv) Compute the L-point DFT of each column.
v) Read the result array row wise.
a) i-ii-iv-iii-v
b) i-iii-ii-iv-v
c) i-ii-iii-iv-v
d) i-iv-iii-ii-v
Answer: c
Explanation: According to one of the algorithm describing the divide and conquer method, if
we store the signal in column wise, then compute the M-point DFT of each row and multiply the
resulting array by the phase factors WN
lq and then compute the L-point DFT of each column and
read the result row wise.

N 2
N 2
N 2
93. If we split the N point data sequence into two N/2 point data sequences f1(n) and f2(n)
corresponding to the even numbered and odd numbered samples of x(n) and F1(k) and
F2(k) are the N/2 point DFTs of f1(k) and f2(k) respectively, then what is the N/2 point DFT
X(k) of x(n)?
a) F1(k)+F2(k)
b) F1(k)-W k F (k)
c) F1(k)+W k F (k)
Answer: c
Explanation: From the question, it is given that
f1(n)=x(2n)
f2(n)=x(2n+1), n=0,1,2…N/2-1
X(k)=∑N−1n=0x(n)WknN, k=0,1,2..N-1
=∑nevenx(n)WknN+∑noddx(n)WknN
=∑(N2)−1m=0x(2m)W2kmN+∑(N2)−1m=0x(2m+1)Wk(2m+1)N
=∑(N2)−1m=0f1(m)WkmN/2+WkN∑(N/2)−1m=0f2(m)Wkm(N2)
X(k)=F1(k)+ W k F (k).
94. If X(k) is the N/2 point DFT of the sequence x(n), then what is the value of X(k+N/2)?
a) F1(k)+F2(k)
b) F1(k)-WN
k F2(k)
c) F1(k)+WN
k F2(k)
Answer: b
Explanation: We know that, X(k) = F1(k)+WN
k F2(k)
We know that F1(k) and F2(k) are periodic, with period N/2, we have F1(k+N/2) = F1(k) and
F2(k+N/2)= F2(k). In addition, the factor WN
k+N/2 = -WN
k.
Thus we get, X(k+N/2)= F1(k)- WN
k F2(k).

N
95. How many complex multiplications are required to compute X(k)?
a) N(N+1)
b) N(N-1)/2
c) N2/2
d) N(N+1)/2
Answer: d
Explanation: We observe that the direct computation of F1(k) requires (N/2)2 complex
multiplications. The same applies to the computation of F2(k). Furthermore, there are N/2
additional complex multiplications required to compute W k. Hence it requires N(N+1)/2
complex multiplications to compute X(k).
96. The total number of complex multiplications required to compute N point DFT by
radix-2 FFT is?
a) (N/2)log2N
b) Nlog2N
c) (N/2)logN
Answer: a
Explanation: The decimation of the data sequence should be repeated again and again until the
resulting sequences are reduced to one point sequences. For N=2v, this decimation can be
performed v=log2N times. Thus the total number of complex multiplications is reduced to
(N/2)log2N.
97. The total number of complex additions required to compute N point DFT by radix-2
FFT is?
a) (N/2)log2N
b) Nlog2N
c) (N/2)logN
Answer: b

performed v=log2N times. Thus the total number of complex additions is reduced to Nlog2N.
98. The following butterfly diagram is used in the computation of
a) Decimation-in-time FFT
b) Decimation-in-frequency FFT
c) All of the mentioned
Answer: a
Explanation: The above given diagram is the basic butterfly computation in the decimation-in-
time FFT algorithm.
99. For a decimation-in-time FFT algorithm, which of the following is true?
a) Both input and output are in order
b) Both input and output are shuffled
c) Input is shuffled and output is in order
d) Input is in order and output is shuffled
Answer: c
Explanation: In decimation-in-time FFT algorithm, the input is taken in bit reversal order and the
output is obtained in the order.

100. The following butterfly diagram is used in the computation of
a) Decimation-in-time FFT
b) Decimation-in-frequency FFT
Answer: b
Explanation: The above given diagram is the basic butterfly computation in the decimation-in-
frequency FFT algorithm.
101. For a decimation-in-time FFT algorithm, which of the following is true?
a) Both input and output are in order
b) Both input and output are shuffled
c) Input is shuffled and output is in order
d) Input is in order and output is shuffled
Answer: d

Explanation: In decimation-in-frequency FFT algorithm, the input is taken in order and the
output is obtained in the bit reversal order.
102. If x1(n) and x2(n) are two real valued sequences of length N, and let x(n) be a complex
valued sequence defined as x(n)=x1(n)+jx2(n), 0≤n≤N-1, then what is the value of x1(n)?
a) x(n)−x∗(n)2
b) x(n)+x∗(n)2
c) x(n)−x∗(n)2j
d) x(n)+x∗(n)2j
Answer: b
Explanation: Given x(n)=x1(n)+jx2(n)=>x*(n)= x1(n)-jx2(n)
Upon adding the above two equations, we get x1(n)=x(n)+x∗(n)2.
103. If x1(n) and x2(n) are two real valued sequences of length N, and let x(n) be a complex
valued sequence defined as x(n)=x1(n)+jx2(n), 0≤ n≤ N-1, then what is the value of x2(n)?
a) x(n)−x∗(n)2
b) x(n)+x∗(n)2
c) x(n)+x∗(n)2j
d) x(n)−x∗(n)2j
Answer: d
Explanation: Given x(n)=x1(n)+jx2(n)=>x*(n) = x1(n)-jx2(n)
Upon subtracting the above two equations, we get x2(n)=x(n)−x∗(n)2j.
104. If X(k) is the DFT of x(n) which is defined as x(n)=x1(n)+jx2(n), 0≤ n≤ N-1, then what is
the DFT of x1(n)?
a) 12[X∗(k)+X∗(N−k)]
b) 12[X∗(k)−X∗(N−k)]
c) 12j[X∗(k)−X∗(N−k)]
d) 12j[X∗(k)+X∗(N−k)]
Answer: a
Explanation: We know that if x(n)=x1(n)+jx2(n) then x1(n)=x(n)+x∗(n)2
On applying DFT on both sides of the above equation, we get

X1(k)=12DFT[x(n)]+DFT[x∗(n)]
We know that if X(k) is the DFT of x(n), the DFT[x*(n)]=X*(N-k=>X1(k)=12[X∗(k)+X∗(N−k)].
105. If X(k) is the DFT of x(n) which is defined as x(n)=x1(n)+jx2(n), 0≤ n≤ N-1, then what is
the DFT of x1(n)?
a) 12[X∗(k)+X∗(N−k)]
b) 12[X∗(k)−X∗(N−k)]
c) 12j[X∗(k)−X∗(N−k)]
d) 12j[X∗(k)+X∗(N−k)]
Answer: c
Explanation: We know that if x(n)=x1(n)+jx2(n) then x2(n)=x(n)−x∗(n)2j.
On applying DFT on both sides of the above equation, we get
X2(k)=12jDFT[x(n)]−DFT[x∗(n)]
We know that if X(k) is the DFT of x(n), the DFT[x*(n)]=X*(N-
k)=>X2(k)=12j[X∗(k)−X∗(N−k)].
106. If g(n) is a real valued sequence of 2N points and x1(n)=g(2n) and x2(n)=g(2n+1), then
what is the value of G(k), k=0,1,2…N-1?
a) X1(k)-W2
kNX2(k)
b) X1(k)+W2
kNX2(k)
c) X1(k)+W2
kX2(k)
d) X1(k)-W2
kX2(k)
Answer: b
Explanation: Given g(n) is a real valued 2N point sequence. The 2N point sequence is divided
into two N point sequences x1(n) and x2(n)
Let x(n)=x1(n)+jx2(n)=> X1(k)=12[X∗(k)+X∗(N−k)] and X2(k)=12j[X∗(k)−X∗(N−k)]
We know that g(n)=x1(n)+x2(n)=>G(k)=X1(k)+W2
kNX2(k), k=0,1,2…N-1.

107. If g(n) is a real valued sequence of 2N points and x1(n)=g(2n) and x2(n)=g(2n+1), then
what is the value of G(k), k=N,N-1,…2N-1?
a) X1(k)-W2
kX2(k)
b) X1(k)+W2
kNX2(k)
c) X1(k)+W2
kX2(k)
d) X1(k)-W2
kNX2(k)
Answer: d
Explanation: Given g(n) is a real valued 2N point sequence. The 2N point sequence is divided
into two N point sequences x1(n) and x2(n)
Let x(n)=x1(n)+jx2(n)
=> X1(k)=12[X∗(k)+X∗(N−k)] and X2(k)=12j[X∗(k)−X∗(N−k)]
We know that g(n)=x1(n)+x2(n)
=>G(k)=X1(k)-W 2
kNX2(k), k=N,N-1,…2N-1.
108. How many complex multiplications are need to be performed for each FFT algorithm?
a) (N/2)logN
b) Nlog2N
c) (N/2)log2N
Answer: c
performed v=log2N times. Thus the total number of complex multiplications is reduced to
(N/2)log2N.
109. How many complex additions are required to be performed in linear filtering of a
sequence using FFT algorithm?
a) (N/2)logN
b) 2Nlog2N
c) (N/2)log2N
d) Nlog2N

Answer: b
Explanation: The number of additions to be performed in FFT are Nlog2N. But in linear
filtering of a sequence, we calculate DFT which requires Nlog2N complex additions and IDFT
requires Nlog2N complex additions. So, the total number of complex additions to be performed
in linear filtering of a sequence using FFT algorithm is 2Nlog2N.
110. How many complex multiplication are required per output data point?
a) [(N/2)logN]/L
b) [Nlog22N]/L
c) [(N/2)log2N]/L
Answer: b
Explanation: In the overlap add method, the N-point data block consists of L new data points
and additional M-1 zeros and the number of complex multiplications required in FFT algorithm
are (N/2)log2N. So, the number of complex multiplications per output data point is [Nlog22N]/L.
111. Which of the following is a frequency domain specification?
a) 0 ≥ 20 log|H(jΩ)|
b) 20 log|H(jΩ)| ≥ KP
c) 20 log|H(jΩ)| ≤ KS
Answer: d
Explanation: We are required to design a low pass Butterworth filter to meet the following
frequency domain specifications.
KP ≤ 20 log|H(jΩ)| ≤ 0
and 20 log|H(jΩ)| ≤ KS.
112. What is the value of gain at the pass band frequency, i.e., what is the value of KP?
a) -10 log[1−(ΩPΩC)2N]
b) -10 log[1+(ΩPΩC)2N]
c) 10 log[1−(ΩPΩC)2N]
d) 10 log[1+(ΩPΩC)2N]

Answer: b
Explanation: We know that the formula for gain is K = 20 log|H(jΩ)|
We know that
|H(jΩ)|=1(1+(ΩΩC)2N√
By applying 20log on both sides of above equation, we get
K = 20 log|H(jΩ)|=−20[log[1+(ΩΩC)2N]]1/2
= -10 log[1+(ΩΩC)2N]
We know that K= KP at Ω=ΩP
=> KP=-10 log[1+(ΩPΩC)2N].
113. What is the value of gain at the stop band frequency, i.e., what is the value of KS?
a) -10 log[1+(ΩSΩC)2N]
b) -10 log[1−(ΩSΩC)2N]
c) 10 log[1−(ΩSΩC)2N]
d) 10 log[1+(ΩSΩC)2N]
Answer: a
Explanation: We know that the formula for gain is
K = 20 log|H(jΩ)|
We know that
|H(jΩ)|=1(1+(ΩΩC)2N√
By applying 20log on both sides of above equation, we get
K = 20 log|H(jΩ)|=−20[log[1+(ΩΩC)2N]]1/2
= -10 log[1+(ΩΩC)2N]
We know that K= KS at Ω=ΩS
=> KS=-10 log[1+(ΩSΩC)2N].
114. Which of the following equation is True?
a) [ΩPΩC]2N=10−KP/10+1
b) [ΩPΩC]2N=10KP/10+1
c) [ΩPΩC]2N=10−KP/10−1

Answer: c
Explanation: We know that,
KP=-10 log[1+(ΩPΩC)2N]
=>[ΩPΩC]2N=10−KP10−1
115. Which of the following equation is True?
a) [ΩSΩC]2N=10−KS/10+1
b) [ΩSΩC]2N=10KS/10+1
c) [ΩSΩC]2N=10−KS/10−1
Answer: b
KP=-10 log[1+(ΩSΩC)2N]
=>[ΩSΩC]2N=10−KS10−1
116. What is the order N of the low pass Butterworth filter in terms of KP and KS?
a) log[(10KP10−1)/(10Ks10−1)]2log(ΩPΩS)
b) log[(10KP10+1)/(10Ks10+1)]2log(ΩPΩS)
c) log[(10−KP10+1)/(10−Ks10+1)]2log(ΩPΩS)
d) log[(10−KP10−1)/(10−Ks10−1)]2log(ΩPΩS)
Answer: d
Explanation:
We know that, [ΩPΩC]2N=10−KP/10−1 and [ΩPΩC]2N=10−KS/10−1.
By dividing the above two equations, we get
=> [ΩP/ΩS]2N=(10−KS/10−1)(10−KP/10−1)
By taking log in both sides, we get
=> N=log[(10−KP10−1)/(10−Ks10−1)]2log(ΩPΩS).

117. What is the expression for cutoff frequency in terms of pass band gain?
a) ΩP(10−KP/10−1)1/2N
b) ΩP(10−KP/10+1)1/2N
c) ΩP(10KP/10−1)1/2N
Answer: a
[ΩPΩC]2N=10−KP/10−1
=> ΩC=ΩP(10−KP/10−1)1/2N.
118. What is the expression for cutoff frequency in terms of stop band gain?
a) ΩS(10−KS/10−1)1/2N
b) ΩS(10−KS/10+1)1/2N
c) ΩS(10KS/10−1)1/2N
Answer: c
[ΩSΩC]2N=10−KS/10−1
=> ΩC=ΩS(10−KS/10−1)1/2N.
119. What is the lowest order of the Butterworth filter with a pass band gain KP=-1 dB at
ΩP=4 rad/sec and stop band attenuation greater than or equal to 20dB at ΩS = 8 rad/sec?
a) 4
b) 5
c) 6
d) 3
Answer: b
Explanation: We know that the equation for the order of the Butterworth filter is given as
N=log[(10−KP/10−1)/(10−Ks/10−1)]2log(ΩPΩS)
From the given question,
KP=-1 dB, ΩP= 4 rad/sec, KS=-20 dB and ΩS= 8 rad/sec
Upon substituting the values in the above equation, we get

N=4.289
Rounding off to the next largest integer, we get N=5
120.Which of the following is done to convert a continuous time signal into discrete time
signal?
a) Modulating
b) Sampling
c) Differentiating
d) Integrating
Answer: b
Explanation: A discrete time signal can be obtained from a continuous time signal by replacing
t by nT, where T is the reciprocal of the sampling rate or time interval between the adjacent
values. This procedure is known as sampling.
121. The evenpart of a signal x(t) is?
a) x(t)+x(-t)
b) x(t)-x(-t)
c) (1/2)*(x(t)+x(-t))
d) (1/2)*(x(t)-x(-t))
Answer: c
Explanation: Let x(t)=xe(t)+xo(t)
=>x(-t)=xe(-t)-xo(-t)
By adding the above two equations, we get
xe(t)=(1/2)*(x(t)+x(-t)).
122. Which of the following is the odd component of the signal x(t)=e(jt)?
a) cost
b) j*sint
c) j*cost
d) sint

Answer: b
Explanation: Let x(t)=e(jt)
Now, xo(t)=(1/2)*(x(t)-x(-t))
=(1/2)*(e(jt) – e(-jt))
=(1/2)*(cost+jsint-cost+jsint)
=(1/2)*(2jsint)
=j*sint.
123. For a continuous time signal x(t) to be periodic with a period T, then x(t+mT) should
be equal to
a) x(-t)
b) x(mT)
c) x(mt)
d) x(t)
Answer: d
Explanation: If a signal x(t) is said to be periodic with period T, then x(t+mT)=x(t) for all t and
any integer m.
124. Let x1(t) and x2(t) be periodic signals with fundamental periods T1 and T2
respectively. Which of the following must be a rational number for x(t)=x1(t)+x2(t) to be
periodic?
a) T1+T2
b) T1-T2
c) T1/T2
d) T1*T2
Answer: c
Explanation: Let T be the period of the signal x(t)
=>x(t+T)=x1(t+mT1)+x2(t+nT2)
Thus, we must have
mT1=nT2=T
=>(T1/T2)=(k/m)= a rational number.

125. Let x1(t) and x2(t) be periodic signals with fundamental periods T1 and T2 respectively.
Then the fundamental period of x(t)=x1(t)+x2(t) is?
a) LCM of T1 and T2
b) HCF of T1and T2
c) Product of T1 and T2
d) Ratio of T1 to T2
Answer: a
Explanation: For the sum of x1(t) and x2(t) to be periodic the ratio of their periods should be a
rational number, then the fundamental period is the LCM of T1 and T2.
126. All energy signals will have an average power of
a) Infinite
b) Zero
c) Positive
d) Cannot be calculated
Answer: b
Explanation: For any energy signal, the average power should be equal to 0 i.e., P=0.
127. x(t) or x(n) is defined to be an energy signal, if and only if the total energy content of
the signal is a
a) Finite quantity
b) Infinite
c) Zero
Answer: a
Explanation: The energy signal should have a total energy value that lies between 0 and infinity.
128. What is the period of cos2t+sin3t?
a) pi
b) 2*pi
c) 3*pi
d) 4*pi

Answer: b
Explanation: Period of cos2t=(2*pi)/2=pi
Period of sin3t=(2*pi)/3
LCM of pi and (2*pi)/3 is 2*
129. Which of the following is common independent variable for speechsignal, EEG and
ECG?
a) Time
b) Spatial coordinates
c) Pressure
Answer: a
Explanation: Speech, EEG and ECG signals are the examples of information-bearing signals
that evolve as functions of a single independent variable, namely, time.
130. Which of the following conditions made digital signal processing more advantageous
over analog signal processing?
a) Flexibility
b) Accuracy
c) Storage
Answer: d
Explanation: Digital programmable system allows flexibility in reconfiguring the DSP
operations by just changing the program, as the digital signal is in the form of 1 and 0’s it is
more accurate and it can be stored in magnetic tapes.
131. Which property does y(t)=x(1-t) exhibit?
a) Time scaling
b) Time shifting
c) Reflecting
d) Time shifting and reflecting

Answer: d
Explanation: First the signal x(t) is shifted by 1 to get x(1+t) and it is reflected to get x(1-t). So,
it exhibits both time shifting and reflecting properties.
132. If x(n)=(0,1,2,3,3,0,0,0) then x(2n) is?
a) (0,2,4,6,6,0,0,0)
b) (0,1,2,3,3,0,0,0)
c) (0,2,3,0,0,0,0,0)
Answer: c
Explanation: Substitute n=0,1,2… in x(2n) and obtain the values from the given x(n).
133. If x(n)=(0,0,1,2,3,4,0,0) then x(n-2) is?
a) (0,0,2,4,6,8,0,0)
b) (0,0,1,2,3,4,0,0)
c) (1,2,3,4,0,0,0,0)
d) (0,0,0,0,1,2,3,4)
Answer: d
Explanation: The signal x(n) is shifted right by 2.
134. If x(n)=(0,0,1,1,1,1,1,0) then x(3n+1) is?
a) (0,1,0,0,0,0,0,0)
b) (0,0,1,1,1,1,0,0)
c) (1,1,0,0,0,0,0,0)
Answer: a
Explanation: First shift the given signal left by 1 and then time scale the obtained signal by 3.

135. If a signal x(t) is processed through a systemto obtain the signal (x(t)2), then the
systemis said to be
a) Linear
b) Non-linear
c) Exponential
Answer: b
Explanation: Let the input signal be ‘t’. Then the output signal after passing through the system
is y=t2 which is the equation of a parabola. So, the system is non-linear.
136. What are the important block(s) required to process an input analog signal to get an
output analog signal?
a) A/D converter
b) Digital signal processor
c) D/A converter
Answer: d
Explanation: The input analog signal is converted into digital using A/D converter and passed
through DSP and then converted back to analog using a D/A converter.
137. Which of the following block is not required in digital processing of a RADAR signal?
a) A/D converter
b) D/A converter
c) DSP
Answer: b
Explanation: In the digital processing of the radar signal, the information extracted from the
radar signal, such as the position of the aircraft and its speed, may simply be printed on a paper.
So, there is no need of an D/A converter in this case.

138. Which of the following wave is known as “amplitude modulated wave” of x(t)?
a) C.x(t) (where C is a constant)
b) x(t)+y(t)
c) x(t).y(t)
d) dx(t)/dt
Answer: c
Explanation: The multiplicative operation is often encountered in analog communication, where
an audio frequency signal is multiplied by a high frequency sinusoid known as carrier. The
resulting signal is known as “amplitude modulated wave”.
139. What is the physical device that performs an operation on the signal?
a) Signal source
b) System
c) Medium
Answer: b
Explanation: A system is a physical device which performs the operation on the signal and
modifies the input signal.
140. 1. Resolve the sequence into a sum of weighted impulse sequences.
a) 2δ(n)+4δ(n-1)+3δ(n-3)
b) 2δ(n+1)+4δ(n)+3δ(n-2)
c) 2δ(n)+4δ(n-1)+3δ(n-2)
Answer: b
Explanation: We know that, x(n)δ(n-k)=x(k)δ(n-k)
x(-1)=2=2δ(n+1)
x(0)=4=4δ(n)
x(2)=3=3δ(n-2)
Therefore, x(n)= 2δ(n+1)+4δ(n)+3δ(n-2).

141. The formula y(n)=∑∞k=−∞x(k)h(n−k) that gives the response y(n) of the LTI system
as the function of the input signal x(n) and the unit sample response h(n) is known as
a) Convolution sum
b) Convolution product
c) Convolution Difference
Answer: a
Explanation: The input x(n) is convoluted with the impulse response h(n) to yield the output
y(n). As we are summing the different values, we call it as Convolution sum.
142. What is the order of the four operations that are needed to be done on h(k) in order to
convolute x(k) and h(k)?
Step-1:Folding
Step-2:Multiplication with x(k)
Step-3:Shifting
Step-4:Summation
a) 1-2-3-4
b) 1-2-4-3
c) 2-1-3-4
d) 1-3-2-4
Answer: d
Explanation: First the signal h(k) is folded to get h(-k). Then it is shifted by n to get h(n-k).
Then it is multiplied by x(k) and then summed over -∞ to ∞.
143. The impulse response of a LTI systemis h(n)={1,1,1}. What is the response of the
signal to the input x(n)={1,2,3}?
a) {1,3,6,3,1}
b) {1,2,3,2,1}
c) {1,3,6,5,3}
d) {1,1,1,0,0}

Answer: c
Explanation: Let y(n)=x(n)*h(n)(‘*’ symbol indicates convolution symbol)
From the formula of convolution we get,
y(0)=x(0)h(0)=1.1=1
y(1)=x(0)h(1)+x(1)h(0)=1.1+2.1=3
y(2)=x(0)h(2)+x(1)h(1)+x(2)h(0)=1.1+2.1+3.1=6
y(3)=x(1)h(2)+x(2)h(1)=2.1+3.1=5
y(4)=x(2)h(2)=3.1=3
Therefore, y(n)=x(n)*h(n)={1,3,6,5,3}.
144. Determine the output y(n) of a LTI systemwith impulse response h(n)=anu(n),
|a|<1with the input sequence x(n)=u(n).
a) 1−an+11−a
b) 1−an−11−a
c) 1+an+11+a
Answer: a
Explanation: Now fold the signal x(n) and shift it by one unit at a time and sum as follows
y(0)=x(0)h(0)=1
y(1)=h(0)x(1)+h(1)x(0)=1.1+a.1=1+a
y(2)=h(0)x(2)+h(1)x(1)+h(2)x(0)=1.1+a.1+a2.1=1+a+a2
Similarly, y(n)=1+a+a2+….an=1−an+11−a.
145. Determine the impulse response for the cascade of two LTI systems having impulse
responses h1(n)=(12)2 u(n) and h2(n)=(14)2 u(n).
a) (12)n[2−(12)n], n<0
b) (12)n[2−(12)n], n>0
c) (12)n[2+(12)n], n<0
d) (12)n[2+(12)n], n>0
Answer: b
Explanation: Let h2(n) be shifted and folded.
so, h(k)=h1(n)*h2(n)=∑∞k=−∞h1(k)h2(n−k)

For k<0, h1(n)= h2(n)=0 since the unit step function is defined only on the right hand side.
Therefore, h(k)=(12)k(14)n−k
=>h(n)=∑nk=0(12)k(14)n−k
=(14)n∑nk=0(2)k
=(14)n.(2n+1−1)
=(12)n[2−(12)n],n>0
146.An LTI systemis said to be causal if and only if?
a) Impulse response is non-zero for positive values of n
b) Impulse response is zero for positive values of n
c) Impulse response is non-zero for negative values of n
d) Impulse response is zero for negative values of n
Answer: d
Explanation: Let us consider a LTI system having an output at time n=n0 given by the
convolution formula
y(n)=∑∞k=−∞h(k)x(n0−k)
We split the summation into two intervals.
=>y(n)=∑−1k=−∞h(k)x(n0−k)+∑∞k=0h(k)x(n0−k)
=(h(0)x(n0)+h(1)x(n0 -1)+h(2)x(n0-2)+….)+(h(-1)x(n0+1)+h(-2)x(n0+2)+…)
As per the definition of the causality, the output should depend only on the present and past
values of the input. So, the coefficients of the terms x(n0+1), x(n0+2)…. should be equal to zero.
that is, h(n)=0 for n<0 .
147. x(n)*δ(n-n0)=?
a) x(n+n0)
b) x(n-n0)
c) x(-n-n0)
d) x(-n+n0)
Answer: b
Explanation: x(n)*δ(n-n0)=∑∞k=−∞x(k)δ(n−k−n0)
=x(k)|k=n-n0
=x(n-n0)

148. 1. If x(n) is a discrete-time signal, then the value of x(n) at non integer value of ‘n’ is?
a) Zero
b) Positive
c) Negative
d) Not defined
Answer: d
Explanation: For a discrete time signal, the value of x(n) exists only at integral values of n. So,
for a non- integer value of ‘n’ the value of x(n) does not exist.
149. The discrete time function defined as u(n)=n for n≥0;u(n)=0 for n<0 is an
a) Unit sample signal
b) Unit step signal
c) Unit ramp signal
Answer: c
Explanation: When we plot the graph for the given function, we get a straight line passing
through origin with a unit positive slope. So, the function is called a unit ramp signal.
150. The phase function of a discrete time signal x(n)=an, where a=r.ejθ is?
a) tan(nθ)
b) nθ
c) tan-1(nθ)
Answer: b
Explanation: Given x(n)=an=(r.ejθ)n = rn.ejnθ
=>x(n)=rn.(cosnθ+jsinnθ)
Phase function is tan-1(cosnθ/sinnθ)=tan-1(tan nθ)=nθ.

151. The signal given by the equation ∑∞n=−∞|x(n)|2 is known as
a) Energy signal
b) Power signal
c) Work done signal
Answer: a
Explanation: We have used the magnitude-squared values of x(n), so that our definition applies
to complex-valued as well as real-valued signals. If the energy of the signal is finite i.e., 0<E<∞
then the given signal is known as Energy signal.
152. x(n)*δ(n-k)=?
a) x(n)
b) x(k)
c) x(k)*δ(n-k)
d) x(k)*δ(k)
Answer: c
Explanation: The given signal is defined only when n=k by the definition of delta function. So,
x(n)*δ(n-k)= x(k)*δ(n-k).
153. A real valued signal x(n) is called as anti-symmetric if
a) x(n)=x(-n)
b) x(n)=-x(-n)
c) x(n)=-x(n)
Answer: b
Explanation: According to the definition of anti-symmetric signal, the signal x(n) should be
symmetric over origin. So, for the signal x(n) to be symmetric, it should satisfy the condition
x(n)=-x(-n).

154. The odd part of a signal x(t) is?
a) x(t)+x(-t)
b) x(t)-x(-t)
c) (1/2)*(x(t)+x(-t))
d) (1/2)*(x(t)-x(-t))
Answer: d
Explanation: Let x(t)=xe(t)+xo(t)
=>x(-t)=xe(-t)-xo(-t)
By subtracting the above two equations, we get
xo(t)=(1/2)*(x(t)-x(-t)).
155. Time scaling operation is also known as
a) Down-sampling
b) Up-sampling
c) Sampling
Answer: a
Explanation: If the signal x(n) was originally obtained by sampling a signal xa(t), then
x(n)=xa(nT). Now, y(n)=x(2n)(say)=xa(2nT). Hence the time scaling operation is equivalent to
changing the sampling rate from 1/T to 1/2T, that is to decrease the rate by a factor of 2. So, time
scaling is also called as down-sampling.
156. What is the condition for a signal x(n)=Brn where r=eαT to be called as an decaying
exponential signal?
a) 0<r<∞
b) 0<r<1
c) r>1
d) r<0
Answer: b
Explanation: When the value of ‘r’ lies between 0 and 1 then the value of x(n) goes on
decreasing exponentially with increase in value of ‘n’. So, the signal is called as decaying
exponential signal.

157. The function given by the equation x(n)=1, for n=0; x(n)=0, for n≠0 is a
a) Step function
b) Ramp function
c) Triangular function
d) Impulse function
Answer: d
Explanation: According to the definition of the impulse function, it is defined only at n=0 and is
not defined elsewhere which is as per the signal given.
158. The output signal when a signal x(n)=(0,1,2,3) is processedthrough an ‘Identical’
systemis?
a) (3,2,1,0)
b) (1,2,3,0)
c) (0,1,2,3)
Answer: c
Explanation: An identical system is a system whose output is same as the input, that is it does
not perform any operation on the input and transmits it.
159. If a signal x(n) is passedthrough a systemto get an output signal of y(n)=x(n+1), then
the signal is said to be
a) Delayed
b) Advanced
c) No operation
Answer: d
Explanation: For example, the value of the output at the time n=0 is y(0)=x(1), that is the
system is advanced by one unit.

160. If the output of the systemis y(n)=∑nk=−∞x(y) with an input of x(n) then the system
will work as
a) Accumulator
b) Adder
c) Subtractor
d) Multiplier
Answer: a
Explanation: From the equation given, y(n)=x(n)+x(n-1)+x(n-2)+…. .This system calculates the
running sum of all the past input values till the present time. So, it acts as an accumulator.
161. What is the output y(n) when a signal x(n)=n*u(n)is passedthrough a accumulator
systemunder the conditions that it is initially relaxed?
a) n2+n+12
b) n(n+1)2
c) n2+n+22
Answer: b
Explanation: Given that the system is initially relaxed, that is y(-1)=0
According to the equation of the accumulator,
y(n)=∑nk=−∞x(n)
=∑−1k=−∞x(n)+∑nk=0x(n)
=y(−1)+∑nk=0n∗u(n)
=0+∑nk=0n(since u(n)=1 in 0 to n)
=n(n+1)2
162. The block denoted as follows is known as
a) Delay block
b) Advance block
c) Multiplier block
d) Adder block

Answer: a
Explanation: If the function to this block is x(n) then the output from the block will be x(n-1).
So, the block is called as delay block or delay element.
163. The output signal when a signal x(n)=(0,1,2,3) is processedthrough an ‘Delay’ system
is?
a) (3,2,1,0)
b) (1,2,3,0)
c) (0,1,2,3)
Answer: b
Explanation: An delay system is a system whose output is same as the input, but after a delay.
164. The systemdescribed by the input-output equation y(n)=nx(n)+bx3(n) is a
a) Static system
b) Dynamic system
c) Identical system
Answer: a
Explanation: Since the output of the system y(n) depends only on the present value of the input
x(n) but not on the past or the future values of the input, the system is called as static or memory-
less system.
165. If the output of the systemof the systemat any ‘n’ depends only the present or the past
values of the inputs then the systemis said to be
a) Linear
b) Non-Linear
c) Causal
d) Non-causal
Answer: c
Explanation: A system is said to be causal if the output of the system is defined as the function
shown below

y(n)=F[x(n),x(n-1),x(n-2),…] So, according to the conditions given in the question, the system is
a causal system.
166. If a systemdo not have a bounded output for bounded input, then the systemis said to
be
a) Causal
b) Non-causal
c) Stable
d) Non-stable
Answer: d
Explanation: An arbitrary relaxed system is said to be BIBO stable if it has a bounded output
for every value in the bounded input. So, the system given in the question is a Non-stable system.
167. Which of the following parameters are required to calculate the correlation between
the signals x(n) and y(n)?
a) Time delay
b) Attenuation factor
c) Noise signal
Answer: d
Explanation: Let us consider x(n) be the input reference signal and y(n) be the reflected signal.
Now, the relation between the two signals is given as y(n)=αx(n-D)+w(n)
Where α-attenuation factor representing the signal loss in the round-trip transmission of the
signal x(n)
D-time delay between the time of projection of signal and the reflected back signal
w(n)-noise signal generated in the electronic parts in the front end of the receiver.

168. The cross correlation of two real finite energy sequences x(n) and y(n) is given as
a) rxy(l)=∑∞n=−∞x(n)y(n−l) where l=0,±1,±2,…
b) rxy(l)=∑∞n=0x(n)y(n−l) where l=0,±1,±2,…
c) rxy(l)=∑∞n=−∞x(n)y(n−l) where -∞<l<∞
Answer: a
Explanation: If any two signals x(n) and y(n) are real and finite energy signals, then the
correlation between the two signals is known as cross correlation and its equation is given as
rxy(l)=∑∞n=−∞x(n)y(n−l) where l=0,±1,±2,…
169. Which of the following relation is true?
a) rxy(l)= rxy(-l)
b) rxy(l)= ryx(l)
c) rxy(l)= ryx(-l)
Answer: c
Explanation: we know that, the correlation of two signals x(n) and y(n) is
rxy(l)=∑∞n=−∞x(n)y(n−l)
If we change the roles of x(n) and y(n),
we get ryx(l)=∑∞n=−∞y(n)x(n−l)
Which is equivalent to
ryx(l)=∑∞n=−∞x(n)y(n+l) => ryx(−l)=∑∞n=−∞x(n)y(n−l)
Therefore, we get rxy(l)= ryx(-l).
170. What is the cross correlation sequence of the following sequences?
x(n)={….0,0,2,-1,3,7,1,2,-3,0,0….}
y(n)={….0,0,1,-1,2,-2,4,1,-2,5,0,0….}
a) {10,9,19,36,-14,33,0,7,13,-18,16,7,5,-3}
b) {10,-9,19,36,-14,33,0,7,13,-18,16,-7,5,-3}
c) {10,9,19,36,14,33,0,-7,13,-18,16,-7,5,-3}
d) {10,-9,19,36,-14,33,0,-7,13,18,16,7,5,-3}

Answer: b
Explanation: We know that rxy(l)=∑∞n=−∞x(n)y(n−l) where l=0,±1,±2,…
At l=0, rxy(0)=∑∞n=−∞x(n)y(n)=7
For l>0, we simply shift y(n) to the right relative to x(n) by ‘l’ units, compute the product
sequence and finally, sum over all the values of product sequence.
We get rxy(1)=13, rxy(2)=-18, rxy(3)=16, rxy(4)=-7, rxy(5)=5, rxy(6)=-3, rxy(l)=0 for l≥7
similarly for l<0, shift y(n) to left relative to x(n) We get rxy(-1)=0, rxy(-2)=33, rxy(-3)=-14, rxy(-
4)=36, rxy(-5)=19, rxy(-6)=-9, rxy(-7)=10, rxy(l)=0 for l≤-8
So, the sequence rxy(l)= {10,-9,19,36,-14,33,0,7,13,-18,16,-7,5,-3}
171. Which of the following is the auto correlation of x(n)?
a) rxy(l)=x(l)*x(-l)
b) rxy(l)=x(l)*x(l)
c) rxy(l)=x(l)+x(-l)
Answer: a
Explanation: We know that, the correlation of two signals x(n) and y(n) is
rxy(l)=∑∞n=−∞x(n)y(n−1)
Let x(n)=y(n) => rxx(l)=∑∞n=−∞x(n)x(n−l) = x(l)*x(-l)
172. What is the energy sequence of the signal ax(n)+by(n-l) in terms of cross correlation
and auto correlation sequences?
a) a2rxx(0)+b2ryy(0)+2abrxy(0)
b) a2rxx(0)+b2ryy(0)-2abrxy(0)
c) a2rxx(0)+b2ryy(0)+2abrxy(1)
d) a2rxx(0)+b2ryy(0)-2abrxy(1)
Answer: c
Explanation: The energy signal of the signal ax(n)+by(n-l) is
∑∞n=−∞[ax(n)+by(n−l)]2
= a2∑∞n=−∞x2(n)+b2∑∞n=−∞y2(n−l)+2ab∑∞n=−∞x(n)y(n−l)
= a2rxx(0)+b2ryy(0)+2abrxy(l)

173. What is the relation between cross correlation and auto correlation?
a) |rxy(l)|=rxx(0).ryy(0)−−−−−−−−−−√
b) |rxy(l)|≥rxx(0).ryy(0)−−−−−−−−−−√
c) |rxy(l)|≠rxx(0).ryy(0)−−−−−−−−−−√
d) |rxy(l)|≤rxx(0).ryy(0)−−−−−−−−−−√
Answer: d
Explanation: We know that, a2rxx(0)+b2ryy(0)+2abrxy(l) ≥0
=> (a/b)2rxx(0)+ryy(0)+2(a/b)rxy(l) ≥0
Since the quadratic is nonnegative, it follows that the discriminate of this quadratic must be non
positive, that is 4[r2
xy(l)- rxx(0) ryy(0)] ≤0 => |rxy(l)|≤rxx(0).ryy(0)−−−−−−−−−−√.
174. The normalized auto correlation ρxx(l) is defined as
a) rxx(l)rxx(0)
b) –rxx(l)rxx(0)
c) rxx(l)rxy(0)
Answer: a
Explanation: If the signal involved in auto correlation is scaled, the shape of auto correlation
does not change, only the amplitudes of auto correlation sequence are scaled accordingly. Since
scaling is unimportant, it is often desirable, in practice, to normalize the auto correlation
sequence to the range from -1 to 1. In the case of auto correlation sequence, we can simply
divide by rxx (0). Thus the normalized auto correlation sequence is defined as ρxx(l)=rxx(l)rxx(0).
175.What is the auto correlation of the sequence x(n)=anu(n), 0<a<l?
a) 11−a2 al (l≥0)
b) 11−a2 a-l (l<0)
c) 11−a2 a|l|(-∞<l<∞)
Answer: d
Explanation:
rxx(l)=∑∞n=−∞x(n)x(n−l)
For l≥0, rxx(l)=∑∞n=lx(n)x(n−l)

=∑∞n=lanan−l
=a−l∑∞n=la2n
=11−a2al(l≥0)
For l<0, rxx(l)=∑∞n=0x(n)x(n−l)
=∑∞n=0anan−l
=a−l∑∞n=0a2n
=11−a2a−l
So, rxx(l)=11−a2a|l| (-∞<l<∞)
176. Which of the following relation is true?
a) ryx(l)=h(l)*ryy(l)
b) rxy(l)=h(l)*rxx(l)
c) ryx(l)=h(l)*rxx(l)
Answer: c
Explanation: Let x(n) be the input signal and y(n) be the output signal with impulse response
h(n).
We know that y(n)=x(n)*h(n)=∑∞k=−∞x(k)h(n−k)
The cross correlation between the input signal and output signal is
ryx(l)=y(l)*x(-l)=h(l)*[x(l)*x(-l)]=h(l)*rxx(l).
177. If x(n) is the input signal of a systemwith impulse response h(n) and y(n) is the output
signal, then the auto correlation of the signal y(n) is?
a) rxx(l)*rhh(l)
b) rhh(l)*rxx(l)
c) rxy(l)*rhh(l)
d) ryx(l)*rhh(l)
Answer: b
Explanation: ryy(l)=y(l)*y(-l)
=[h(l)*x(l)]*[h(-l)*x(-l)]
=[h(l)*h(-l)]*[x(l)*x(-l)]
=rhh(l)*rxx(l).

178. Which of the following method is used to find the inverse z-transform of a signal?
a) Counter integration
b) Expansion into a series of terms
c) Partial fraction expansion
Answer: d
Explanation: All the methods mentioned above can be used to calculate the inverse z-transform
of the given signal.
179. What is the inverse z-transform of X(z)=11−1.5z−1+0.5z−2 if ROC is |z|>1?
a) {1,3/2,7/4,15/8,31/16,….}
b) {1,2/3,4/7,8/15,16/31,….}
c) {1/2,3/4,7/8,15/16,31/32,….}
Answer: a
Explanation: Since the ROC is the exterior circle, we expect x(n) to be a causal signal. Thus we
seek a power series expansion in negative powers of ‘z’. By dividing the numerator of X(z) by its
denominator, we obtain the power series
X(z)=11−1.5z−1+0.5z−2=1+32z−1+74z−2+158z−3+3116z−4+…
So, we obtain x(n)= {1,3/2,7/4,15/8,31/16,….}.
180. What is the inverse z-transform of X(z)=11−1.5z−1+0.5z−2 if ROC is |z| < 0.5?
a) {….62,30,14,6,2}
b) {…..62,30,14,6,2,0,0}
c) {0,0,2,6,14,30,62…..}
d) {2,6,14,30,62…..}
Answer: b
Explanation: In this case the ROC is the interior of a circle. Consequently, the signal x(n) is anti
causal. To obtain a power series expansion in positive powers of z, we perform the long division
in the following way:

Thus X(z)=11−1.5z−1+0.5z−2=2z2+6z3+14z4+30z5+62z6+… In this case x(n)=0 for n≥0.Thus
we obtain x(n)= {…..62,30,14,6,2,0,0}
181. What is the inverse z-transform of X(z)=log(1+az-1) |z|>|a|?
a) x(n)=(-1)n+1 a−nn, n≥1; x(n)=0, n≤0
b) x(n)=(-1)n-1 a−nn, n≥1; x(n)=0, n≤0
c) x(n)=(-1)n+1 a−nn, n≥1; x(n)=0, n≤0
Answer: c
Explanation: Using the power series expansion for log(1+x), with |x|<1, we have
X(z)=∑∞n=1(−1)n+1anz−nn
Thus
x(n)=(-1)n+1 ann, n≥1
=0, n≤0.
182. What is the proper fraction and polynomial form of the improper rational transform
X(z)=1+3z−1+116z−2+13z−31+56z−1+16z−2?
a) 1+2z-1+16z−11+56z−1+16z−2
b) 1-2z-1+16z−11+56z−1+16z−2
c) 1+2z-1+13z−11+56z−1+16z−2
d) 1+2z-1–16z−11+56z−1+16z−2
Answer: a
Explanation: First, we note that we should reduce the numerator so that the terms z-2 and z-3 are
eliminated. Thus we should carry out the long division with these two polynomials written in the
reverse order. We stop the division when the order of the remainder becomes z-1. Then we obtain
X(z)=1+2z-1+16z−11+56z−1+16z−2.

183. What is the partial fraction expansion of the proper function X(z)=11−1.5z−1+0.5z−2?
a) 2zz−1−zz+0.5
b) 2zz−1+zz−0.5
c) 2zz−1+zz+0.5
d) 2zz−1−zz−0.5
Answer: d
Explanation: First we eliminate the negative powers of z by multiplying both numerator and
denominator by z2.
Thus we obtain X(z)=z2z2−1.5z+0.5
The poles of X(z) are p1=1 and p2=0.5. Consequently, the expansion will be
X(z)z=z(z−1)(z−0.5)=2(z−1)–1(z−0.5)
(obtained by applying partial fractions)
=>X(z)=2z(z−1)−z(z−0.5).
184. What is the partial fraction expansion of X(z)=1+z−11−z−1+0.5z−2?
a) z(0.5−1.5j)z−0.5−0.5j–z(0.5+1.5j)z−0.5+0.5j
b) z(0.5−1.5j)z−0.5−0.5j+z(0.5+1.5j)z−0.5+0.5j
c) z(0.5+1.5j)z−0.5−0.5j–z(0.5−1.5j)z−0.5+0.5j
d) z(0.5+1.5j)z−0.5−0.5j+z(0.5−1.5j)z−0.5+0.5j
Answer: b
Explanation: To eliminate the negative powers of z, we multiply both numerator and
denominator by z2. Thus,
X(z)=z(z+1)z−2−z+0.5
The poles of X(z) are complex conjugates p1=0.5+0.5j and p2=0.5-0.5j
Consequently the expansion will be
X(z)= z(0.5−1.5j)z−0.5−0.5j+z(0.5+1.5j)z−0.5+0.5j.

185. What is the partial fraction expansion of X(z)=1(1+z−1)(1−z−1)2?
a) z4(z+1)+3z4(z−1)+z2(z+1)2
b) z4(z+1)+3z4(z−1)–z2(z+1)2
c) z4(z+1)+3z4(z−1)+z2(z−1)2
d) z4(z+1)+z4(z−1)+z2(z+1)2
Answer: c
Explanation: First we express X(z) in terms of positive powers of z, in the form
X(z)=z3(z+1)(z−1)2
X(z) has a simple pole at z=-1 and a double pole at z=1. In such a case the approximate partial
fraction expansion is
X(z)z=z2(z+1)(z−1)2=Az+1+Bz−1+C(z−1)2
On simplifying, we get the values of A, B and C as 1/4, 3/4 and 1/2 respectively.
Therefore, we get z4(z+1)+3z4(z−1)+z2(z−1)2.
186. What is the inverse z-transform of X(z)=11−1.5z−1+0.5z−2 if ROC is |z|>1?
a) (2-0.5n)u(n)
b) (2+0.5n)u(n)
c) (2n-0.5n)u(n)
Answer: a
Explanation: The partial fraction expansion for the given X(z) is
X(z)=2zz−1−zz−0.5
In case when ROC is |z|>1, the signal x(n) is causal and both the terms in the above equation are
causal terms. Thus, when we apply inverse z-transform to the above equation, we get
x(n)=2(1)nu(n)-(0.5)nu(n)=(2-0.5n)u(n).
187. What is the inverse z-transform of X(z)=11−1.5z−1+0.5z−2 if ROC is |z|<0.5?
a) [-2-0.5n]u(n)
b) [-2+0.5n]u(n)
c) [-2+0.5n]u(-n-1)
d) [-2-0.5n]u(-n-1)

Answer: c
Explanation: The partial fraction expansion for the given X(z) is
X(z)=2zz−1−zz−0.5
In case when ROC is |z|<0.5, the signal is anti causal. Thus both the terms in the above equation
are anti causal terms. So, if we apply inverse z-transform to the above equation we get
x(n)= [-2+0.5n]u(-n-1).
188. What is the inverse z-transform of X(z)=11−1.5z−1+0.5z−2 if ROC is 0.5<|z|<1?
a) -2u(-n-1)+(0.5)nu(n)
b) -2u(-n-1)-(0.5)nu(n)
c) -2u(-n-1)+(0.5)nu(-n-1)
d) 2u(n)+(0.5)nu(-n-1)
Answer: b
Explanation: The partial fraction expansion of the given X(z) is
X(z)=2zz−1−zz−0.5
In this case ROC is 0.5<|z|<1 is a ring, which implies that the signal is two sided. Thus one of the
signal corresponds to a causal signal and the other corresponds to an anti causal signal.
Obviously, the ROC given is the overlapping of the regions |z|>0.5 and |z|<1. Hence the pole
p2=0.5 provides the causal part and the pole p1=1 provides the anti causal part. SO, if we apply
the inverse z-transform we get
x(n)= -2u(-n-1)-(0.5)nu(n).
189. What is the causal signal x(n) having the z-transform X(z)=1(1+z−1)(1−z−1)2?
a) [1/4(-1)n+3/4-n/2]u(n)
b) [1/4(-1)n+3/4-n/2]u(-n-1)
c) [1/4+3/4(-1)n-n/2]u(n)
d) [1/4(-1)n+3/4+n/2]u(n)
Answer: d
Explanation: The partial fraction expansion of X(z) is X(z)=z4(z+1)+3z4(z−1)+z2(z−1)2
When we apply the inverse z-transform for the above equation, we get
x(n)=[1/4(-1)n+3/4+n/2]u(n).

190. Which of the following justifies the linearity property of z-transform?[x(n)↔X(z)].
a) x(n)+y(n) ↔ X(z)Y(z)
b) x(n)+y(n) ↔ X(z)+Y(z)
c) x(n)y(n) ↔ X(z)+Y(z)
d) x(n)y(n) ↔ X(z)Y(z)
Answer: b
Explanation: According to the linearity property of z-transform, if X(z) and Y(z) are the z-
transforms of x(n) and y(n) respectively then, the z-transform of x(n)+y(n) is X(z)+Y(z).
191. What is the z-transform of the signal x(n)=[3(2n)-4(3n)]u(n)?
a) 31−2z−1−41−3z−1
b) 31−2z−1−41+3z−1
c) 31−2z−41−3z
Answer: a
Explanation: Let us divide the given x(n) into x1(n)=3(2n)u(n) and x2(n)= 4(3n)u(n)
and x(n)=x1(n)-x2(n)
From the definition of z-transform X1(z)=31−2z−1 and X2(z)=41−3z−1
So, from the linearity property of z-transform
X(z)=X1(z)-X2(z)
=> X(z)=31−2z−1−41−3z−1.
192. What is the z-transform of the signal x(n)=sin(jω0n)u(n)?
a) z−1sinω01+2z−1cosω0+z−2
b) z−1sinω01−2z−1cosω0−z−2
c) z−1cosω01−2z−1cosω0+z−2
d) z−1sinω01−2z−1cosω0+z−2
Answer: d
Explanation: By Euler’s identity, the given signal x(n) can be written as
x(n) = sin(jω0n)u(n)=12j[ejω0nu(n)−e−jω0nu(n)]
Thus X(z)=12j[11−ejω0z−1−11−e−jω0z−1]

On simplification, we obtain
=> z−1sinω01−2z−1cosω0+z−2.
193. According to Time shifting property of z-transform, if X(z) is the z-transform of x(n)
then what is the z-transform of x(n-k)?
a) zkX(z)
b) z-kX(z)
c) X(z-k)
d) X(z+k)
Answer: b
Explanation: According to the definition of Z-transform
X(z)=∑∞n=−∞x(n)z−n
=>Z{x(n-k)}=X1(z)=∑∞n=−∞x(n−k)z−n
Let n-k=l
=> X1(z)=∑∞l=−∞x(l)z−l−k=z−k.∑∞l=−∞x(l)z−l=z−kX(z)
194. What is the z-transform of the signal defined as x(n)=u(n)-u(n-N)?
a) 1+zN1+z−1
b) 1−zN1+z−1
c) 1+z−N1+z−1
d) 1−z−N1−z−1
Answer: d
Explanation:
We know that Zu(n)=11−z−1
And by the time shifting property, we have Z{x(n-k)}=z-k.Z{x(n)}
=>Z{u(n-N)}=z−N.11−z−1
=>Z{u(n)-u(n-N)}=1−z−N1−z−1.

195. If X(z) is the z-transform of the signal x(n) then what is the z-transform of anx(n)?
a) X(az)
b) X(az-1)
c) X(a-1z)
Answer: c
Explanation: We know that from the definition of z-transform
Z{anx(n)}=∑∞n=−∞anx(n)z−n=∑∞n=−∞x(n)(a−1z)−n=X(a−1z).
196. If the ROC of X(z) is r1<|z|<r2, then what is the ROC of X(a-1z)?
a) |a|r1<|z|<|a|r2
b) |a|r1>|z|>|a|r2
c) |a|r1<|z|>|a|r2
d) |a|r1>|z|<|a|r2
Answer: a
Explanation: Given ROC of X(z) is r1<|z|<r2
Then ROC of X(a-1z) will be given by r1<|a-1z |<r2=|a|r1<|z|<|a|r2
197. What is the z-transform of the signal x(n)=an(sinω0n)u(n)?
a)az−1sinω01+2az−1cosω0+a2z−2
b)az−1sinω01−2az−1cosω0−a2z−2
c)(az)−1cosω01−2az−1cosω0+a2z−2
d)az−1sinω01−2az−1cosω0+a2z−2
Answer: d
Explanation:
we know that by the linearity property of z-transform of sin(ω0n)u(n) is
X(z)=z−1sinω01−2z−1cosω0+z−2
Now by the scaling in the z-domain property, we have z-transform of an (sin(ω0n))u(n) as
X(az)=az−1sinω01−2az−1cosω0+a2z−2

198. If X(z) is the z-transform of the signal x(n), then what is the z-transform of the signal
x(-n)?
a) X(-z)
b) X(z-1)
c) X-1(z)
Answer: b
Explanation: From the definition of z-transform, we have
Z{x(-n)}=∑∞n=−∞x(−n)z−n=∑∞n=−∞x(−n)(z−1)−(−n)=X(z−1)
199. X(z) is the z-transform of the signal x(n), then what is the z-transform of the signal
nx(n)?
a) −zdX(z)dz
b) zdX(z)dz
c) −z−1dX(z)dz
d) z−1dX(z)dz
Answer: a
Explanation:
From the definition of z-transform, we have
X(z)=∑∞n=−∞x(n)z−n
On differentiating both sides, we have
dX(z)dz=∑∞n=−∞(−n)x(n)z−n−1=−z−1∑∞n=−∞nx(n)z−n=−z−1Z{nx(n)}
Therefore, we get −zdX(z)dz = Z{nx(n)}.
200. What is the z-transform of the signal x(n)=nanu(n)?
a) (az)−1(1−(az)−1)2
b) az−1(1−(az)−1)2
c) az−1(1−az−1)2
d) az−1(1+az−1)2

Answer: c
Explanation:
We know that Z{anu(n)}=11−az−1=X(z) (say)
Now the z-transform of nanu(n)=−zdX(z)dz=az−1(1−az−1)2
201. Sampling rate conversion by the rational factor I/D is accomplished by what
connection of interpolator and decimator?
a) Parallel
b) Cascade
c) Convolution
Answer: b
Explanation: A sampling rate conversion by the rational factor I/D is accomplished by
cascading an interpolator with a decimator.
202. Which of the following has to be performed in sampling rate conversion by rational
factor?
a) Interpolation
b) Decimation
c) Either interpolation or decimation
Answer: a
Explanation: We emphasize that the importance of performing the interpolation first and
decimation second, is to preserve the desired spectral characteristics of x(n).
203. Which of the following operation is performed by the blocks given the figure below?

a) Sampling rate conversion by a factor I
b) Sampling rate conversion by a factor D
c) Sampling rate conversion by a factor D/I
d) Sampling rate conversion by a factor I/D
Answer: d
Explanation: In the diagram given, a interpolator is in cascade with a decimator which together
performs the action of sampling rate conversion by a factor I/D.
204. The Nth root of unity WN is given as
a) ej2πN
b) e-j2πN
c) e-j2π/N
d) ej2π/N
Answer: c
Explanation: We know that the Discrete Fourier transform of a signal x(n) is given as
X(k)=∑N−1n=0x(n)e−j2πkn/N=∑N−1n=0x(n)WknN
Thus we get Nth rot of unity WN= e-j2π/N
205. Which of the following is true regarding the number of computations requires to
compute an N-point DFT?
a) N2 complex multiplications and N(N-1) complex additions
b) N2 complex additions and N(N-1) complex multiplications
c) N2 complex multiplications and N(N+1) complex additions
d) N2 complex additions and N(N+1) complex multiplications
Answer: a
From the formula given at every step of computing we are performing N complex multiplications
and N-1 complex additions. So, in a total to perform N-point DFT we perform N2 complex
multiplications and N(N-1) complex additions.

N
206. Which of the following is true?
a) W∗ N=1NWN−1
b) WN−1=1NWN∗
c) WN−1=WN∗
Answer: b
Explanation: If XN represents the N point DFT of the sequence xN in the matrix form, then we
know that XN = WN.xN
By pre-multiplying both sides by W -1, we get
xN=WN-1.XN
But we know that the inverse DFT of XN is defined as
xN=1/N*XN
Thus by comparing the above two equations we get
WN-1=1/N WN*
207. What is the DFT of the four point sequence x(n)={0,1,2,3}?
a) {6,-2+2j-2,-2-2j}
b) {6,-2-2j,2,-2+2j}
c) {6,-2+2j,-2,-2-2j}
d) {6,-2-2j,-2,-2+2j}
Answer: c
Explanation: The first step is to determine the matrix W4. By exploiting the periodicity property
of W4 and the symmetry property
Wk+N/2N=−WNk
208. If X(k) is the N point DFT of a sequence whose Fourier series coefficients is given by
ck, then which of the following is true?
a) X(k)=Nck
b) X(k)=ck/N
c) X(k)=N/ck

4 x
Answer: a
Explanation: The Fourier series coefficients are given by the expression
ck=1N∑N−1n=0x(n)e−j2πkn/N=1NX(k)=>X(k)=Nck
209. What is the DFT of the four point sequence x(n)={0,1,2,3}?
a) {6,-2+2j-2,-2-2j}
b) {6,-2-2j,2,-2+2j}
c) {6,-2-2j,-2,-2+2j}
d) {6,-2+2j,-2,-2-2j}
Answer: d
Explanation: Given x(n)={0,1,2,3}
We know that the 4-point DFT of the above given sequence is given by the expression
In this case N=4
=>X(0)=6, X(1)=-2+2j, X(2)=-2, X(3)=-2-2j.
210. If W 100=W 200, then what is the value of x?
a) 2
b) 4
c) 8
d) 16
Answer: c
Explanation: We know that according to the periodicity and symmetry property,
100/4=200/x=>x=8.
211. Which of the following is the first method proposed for design of FIR filters?
a) Chebyshev approximation
b) Frequency sampling method
c) Windowing technique
Answer: c
Explanation: The design method based on the use of windows to truncate the impulse response

h(n) and obtaining the desired spectral shaping, was the first method proposed for designing
linear phase FIR filters.
212. The lack of precise control of cutoff frequencies is a disadvantage of which of the
following designs?
a) Window design
b) Chebyshev approximation
c) Frequency sampling
Answer: a
Explanation: The major disadvantage of the window design method is the lack of precise
control of the critical frequencies.
213. The values of cutoff frequencies in general depend on which of the following?
a) Type of the window
b) Length of the window
c) Type & Length of the window
Answer: d
Explanation: The values of the cutoff frequencies of a filter in general by windowing technique
depend on the type of the filter and the length of the filter.
214. In frequency sampling method, transition band is a multiple of which of the following?
a) π/M
b) 2π/M
c) π/2M
d) 2πM
Answer: b
Explanation: In the frequency sampling technique, the transition band is a multiple of 2π/M.

215.Which of the following values can a frequency response take in frequency sampling
technique?
a) Zero
b) One
c) Zero or One
Answer: c
Explanation: The attractive feature of the frequency sampling design is that the frequency
response can take either zero or one at all frequencies, except in the transition band.
216. Which of the following technique is more preferable for design of linear phase FIR
filter?
a) Window design
b) Chebyshev approximation
c) Frequency sampling
Answer: b
Explanation: The chebyshev approximation method provides total control of the filter
specifications, and as a consequence, it is usually preferable over the other two methods.
217Which of the following is the correct expression for transition band Δf?
a) (ωp– ωs)/2π
b) (ωp+ωs)/2π
c) (ωp.ωs)/2π
d) (ωs– ωp)/2π
Answer: d
Explanation: The expression for Δf i.e., for the transition band is given as
Δf=(ωs-ωp)/2π.

N
N
218. What is the transform that is suitable for evaluating the z-transform of a set of data on
a variety of contours in the z-plane?
a) Goertzel Algorithm
b) Fast Fourier transform
c) Chirp-z transform
Answer: c
Explanation: Chirp-z transform algorithm is suitable for evaluating the z-transform of a set of
data on a variety of contours in the z-plane. This algorithm is also formulated as a linear filtering
of a set of input data. As a consequence, the FFT algorithm can be used to compute the Chirp-z
transform.
219. According to Goertzel Algorithm, if the computation of DFT is expressedas a linear
filtering operation, then which of the following is true?
a) yk(n)=∑Nm=0x(m)W−k(n−m)N
b) yk(n)=∑N+1m=0x(m)W−k(n−m)N
c) yk(n)=∑N−1m=0x(m)W−k(n+m)N
d) yk(n)=∑N−1m=0x(m)W−k(n−m)N
Answer: d
Explanation: Since W -kN = 1, multiply the DFT by this factor. Thus
X(k)=W -kN∑N−1m=0x(m)W−kmN=∑N−1m=0x(m)W−k(N−m)N
The above equation is in the form of a convolution. Indeed, we can define a sequence yk(n) as
yk(n)=∑N−1m=0x(m)W−k(n−m)N
220. If yk(n) is the convolution of the finite duration input sequence x(n) of length N, then
what is the impulse response of the filter?
a) WN-kn
b) WN-knu(n)
c) WNkn u(n)

N
N k
k
Answer: b
Explanation: We know that yk(n)=∑N−1m=0x(m)W−k(n−m)N
The above equation is of the form yk(n)=x(n)*hk(n)
Thus we obtain, hk(n)= WN-kn u(n).
221. What is the systemfunction of the filter with impulse response hk(n)?
a) 11−W−kNz−1
b) 11+W−kNz−1
c) 11−WkNz−1
d) 11+WkNz−1
Answer: a
Explanation: We know that hk(n)= W -kn u(n)
On applying z-transform on both sides, we get
Hk(z)=11−W−kNz−1
222. What is the expression to compute yk(n) recursively?
a) yk(n)=W -ky (n+1)+x(n)
N k
b) yk(n)=W -ky (n-1)+x(n)
c) yk(n)=WNkyk(n+1)+x(n)
Answer: b
Explanation: We know that hk(n)=W -kn u(n)=y (n)/x(n)
=> yk(n)=W
N k
-ky (n-1)+x(n).
223. What is the equation to compute the values of the z-transform of x(n) at a set of points
{zk}?
a) ∑N−1n=0x(n)znk, k=0,1,2…L-1
b) ∑N−1n=0x(n)z−n−k, k=0,1,2…L-1
c) ∑N−1n=0x(n)z−nk, k=0,1,2…L-1
Answer: c
Explanation: According to the Chirp-z transform algorithm, if we wish to compute the values of
N

N k
the z-transform of x(n) at a set of points {zk}. Then,
X(zk)=∑N−1n=0x(n)z−nk, k=0,1,2…L-1
224. If the contour is a circle of radius r and the zk are N equally spaced points, then what
is the value of zk?
a) re-j2πkn/N
b) rejπkn/N
c) rej2πkn
d) rej2πkn/N
Answer: d
Explanation: We know that, if the contour is a circle of radius r and the zk are N equally spaced
points, then what is the value of zk is given by rej2πkn/N
225. How many multiplications are required to calculate X(k) by chirp-z transform if x(n)
is of length N?
a) N-1
b) N
c) N+1
Answer: c
Explanation: We know that yk(n)=W -ky (n-1)+x(n).Each iteration requires one multiplication
and two additions. Consequently, for a real input sequence x(n), this algorithm requires N+1 real
multiplications to yield not only X(k) but also, due to symmetry, the value of X(N-k).
226. If the contour on which the z-transform is evaluated is as shown below, then which of
the given condition is true?

a) R0>1
b) R0<1
c) R0=1
Answer: a
Explanation: From the definition of chirp z-transform, we know that V=R0ejθ.
If R0>1, then the contour which is used to calculate z-transform is as shown below.
227. How many complex multiplications are need to be performed to calculate chirp z-
transform?(M=N+L-1)
a) log2M
b) Mlog2M
c) (M-1)log2M
d) Mlog2(M-1)
Answer: b
Explanation: Since we will compute the convolution via the FFT, let us consider the circular
convolution of the N point sequence g(n) with an M point section of h(n) where M>N. In such a
case, we know that the first N-1 points contain aliasing and that the remaining M-N+1 points are
identical to the result that would be obtained from a linear convolution of h(n) with g(n). In view
of this, we should select a DFT of size M=L+N-1. Thus the total number of complex
multiplications to be performed are Mlog2M.

228. In IIR Filter design by the Bilinear Transformation, the Bilinear Transformation is a
mapping from
a) Z-plane to S-plane
b) S-plane to Z-plane
c) S-plane to J-plane
d) J-plane to Z-plane
Answer: b
Explanation: From the equation,
S=2T(1−z−11+z−1) it is clear that transformation occurs from s-plane to z-plane
229.The approximation of the integral in y(t) = ∫tt0y′(τ)dt+y(t0) by the Trapezoidal formula
at t = nT and t0=nT-T yields equation?
a) y(nT) = T2[y‘(nT)+y‘(T−nT)]+y(nT−T)
b) y(nT) = T2[y‘(nT)+y‘(nT−T)]+y(nT−T)
c) y(nT) = T2[y‘(nT)+y‘(T−nT)]+y(T−nT)
d) y(nT) = T2[y‘(nT)+y‘(nT−T)]+y(T−nT)
Answer: b
Explanation: By integrating the equation,
y(t) = ∫tt0y‘(τ)dt+y(t0) at t=nT and t0=nT-T we get equation,
y(nT) = T2[y‘(nT)+y‘(nT−T)]+y(nT−T).
230. We use y{‘}(nT)=-ay(nT)+bx(nT) to substitute for the derivative in y(nT)
= T2[y‘(nT)+y‘(nT−T)]+y(nT−T) and thus obtain a difference equation for the equivalent
discrete-time system. With y(n) = y(nT) and x(n) = x(nT), we obtain the result as of the
following?
a) (1+aT2)Y(z)−(1−aT2)y(n−1)=bT2[x(n)+x(n−1)]
b) (1+aTn)Y(z)−(1−aTn)y(n−1)=bTn[x(n)+x(n−1)]
c) (1+aT2)Y(z)+(1−aT2)y(n−1)=bT2(x(n)−x(n−1))
d) (1+aT2)Y(z)+(1−aT2)y(n−1)=bT2(x(n)+x(n+1))

Answer: a
Explanation: When we substitute the given equation in the derivative of other we get the
resultant required equation.
231. The z-transform of below difference equation is?
(1+aT2)Y(z)−(1−aT2)y(n−1)=bT2[x(n)+x(n−1)]
a) (1+aT2)Y(z)−(1−aT2)z−1Y(z)=bT2(1+z−1)X(z)
b) (1+aTn)Y(z)−(1−aT2)z−1Y(z)=bTn(1+z−1)X(z)
c) (1+aT2)Y(z)+(1−aTn)z−1Y(z)=bT2(1+z−1)X(z)
d) (1+aT2)Y(z)−(1+aT2)z−1Y(z)=bT2(1+z−1)X(z)
Answer: a
Explanation: By performing the z-transform of the given equation, we get the required
output/equation.
232. What is the systemfunction of the equivalent digital filter? H(z) = Y(z)/X(z) = ?
a) (bT2)(1+z−1)1+aT2−(1−aT2)z−1
b) (bT2)(1−z−1)1+aT2−(1+aT2)z−1
c) b2T(1−z−11+z−1+a)
d) (bT2)(1−z−1)1+aT2−(1+aT2)z−1 & b2T(1−z−11+z−1+a)
Answer: d
Explanation: As we considered analog linear filter with system function H(s) = b/s+a
Hence, we got an equivalent system function
where, s = 2T(1−z−11+z−1).
233. In the Bilinear Transformation mapping, which of the following are correct?
a) All points in the LHP of s are mapped inside the unit circle in the z-plane
b) All points in the RHP of s are mapped outside the unit circle in the z-plane
c) All points in the LHP & RHP of s are mapped inside & outside the unit circle in the z-plane
Answer: c
Explanation: The bilinear transformation is a conformal mapping that transforms the jΩ-axis
into the unit circle in the z-plane and all the points are linked as mentioned above.

234. In Nth order differential equation, the characteristics of bilinear transformation, let
z=rejw,s=o+jΩ Then for s = 2T(1−z−11+z−1), the values of Ω, ℴ are
a) ℴ = 2T(r2−11+r2+2rcosω), Ω = 2T(2rsinω1+r2+2rcosω)
b) Ω = 2T(r2−11+r2+2rcosω), ℴ = 2T(2rsinω1+r2+2rcosω)
c) Ω=0, ℴ=0
d) None
Answer: a
Explanation: s = 2T(z−1z+1)
= 2T(rejw−1rejw+1)
= 2T(r2−11+r2+2rcosω+j2rsinω1+r2+2rcosω)(s=ℴ+jΩ)
235. In equation ℴ = 2T(r2−11+r2+2rcosω) if r < 1 then ℴ < 0 and then mapping from
s-plane to z-plane occurs in which of the following order?
a) LHP in s-plane maps into the inside of the unit circle in the z-plane
b) RHP in s-plane maps into the outside of the unit circle in the z-plane
Answer: a
Explanation: In the above equation, if we substitute the values of r, ℴ then we get mapping in
the required way
236. In equation ℴ = 2T(r2−11+r2+2rcosω), if r > 1 then ℴ > 0 and then mapping from
s-plane to z-plane occurs in which of the following order?
a) LHP in s-plane maps into the inside of the unit circle in the z-plane
b) RHP in s-plane maps into the outside of the unit circle in the z-plane
Answer: b
Explanation: In the above equation, if we substitute the values of r, ℴ then we get mapping in
the required way

237. If x(n) and X(k) are an N-point DFT pair, then X(k+N)=?
a) X(-k)
b) -X(k)
c) X(k)
Answer: c
x(n)=1N∑N−1k=0x(k)ej2πkn/N
Let X(k)=X(k+N)
=>x1(n)=1N∑N−1k=0X(k+N)ej2πkn/N=x(n)
Therefore, we have X(k)=X(k+N)
238. If X1(k) and X2(k) are the N-point DFTs of X1(n) and x2(n) respectively, then what is
the N-point DFT of x(n)=ax1(n)+bx2(n)?
a) X1(ak)+X2(bk)
b) aX1(k)+bX2(k)
c) eakX1(k)+ebkX2(k)
Answer: b
Explanation: We know that, the DFT of a signal x(n) is given by the expression
Given x(n)=ax1(n)+bx2(n)
=>X(k)= ∑N−1n=0(ax1(n)+bx2(n))e−j2πkn/N
=a∑N−1n=0x1(n)e−j2πkn/N+b∑N−1n=0x2(n)e−j2πkn/N
=>X(k)=aX1(k)+bX2(k).

239. If x(n) is a real sequence and X(k) is its N-point DFT, then which of the following is
true?
a) X(N-k)=X(-k)
b) X(N-k)=X*(k)
c) X(-k)=X*(k)
Answer: d
Now X(N-k)=∑N−1n=0x(n)e−j2π(N−k)n/N=X*(k)=X(-k)
Therefore,
X(N-k)=X*(k)=X(-k)
240. If x(n) is real and even, then what is the DFT of x(n)?
a) ∑N−1n=0x(n)sin2πknN
b) ∑N−1n=0x(n)cos2πknN
c) -j∑N−1n=0x(n)sin2πknN
Answer: b
Explanation: Given x(n) is real and even, that is x(n)=x(N-n)
We know that XI(k)=0. Hence the DFT reduces to
X(k)=∑N−1n=0x(n)cos2πknN ;0 ≤ k ≤ N-1
241. If x(n) is real and odd, then what is the IDFT of the given sequence?
a) j1N∑N−1k=0x(k)sin2πknN
b) 1N∑N−1k=0x(k)cos2πknN
c) −j1N∑N−1k=0x(k)sin2πknN
Answer: a
Explanation: If x(n) is real and odd, that is x(n)=-x(N-n), then XR(k)=0. Hence X(k) is purely
imaginary and odd. Since XR(k) reduces to zero, the IDFT reduces to
x(n)=j1N∑N−1k=0x(k)sin2πknN

242. If X1(n), x2(n) and x3(m) are three sequences each of length N whose DFTs are given as
X1(k), X2(k) and X3(k) respectively and X3(k)=X1(k).X2(k), then what is the expression for
x3(m)?
a) ∑N−1n=0x1(n)x2(m+n)
b) ∑N−1n=0x1(n)x2(m−n)
c) ∑N−1n=0x1(n)x2(m−n)N
d) ∑N−1n=0x1(n)x2(m+n)N
Answer: c
Explanation: If X1(n), x2(n) and x3(m) are three sequences each of length N whose DFTs are
given as X1(k), x2(k) and X3(k) respectively and X3(k)=X1(k).X2(k), then according to the
multiplication property of DFT we have x3(m) is the circular convolution of X1(n) and x2(n).
That is x3(m) = ∑N−1n=0x1(n)x2(m−n)N.
243. What is the circular convolution of the sequences X1(n)={2,1,2,1} and x2(n)={1,2,3,4}?
a) {14,14,16,16}
b) {16,16,14,14}
c) {2,3,6,4}
d) {14,16,14,16}
Answer: d
Explanation: We know that the circular convolution of two sequences is given by the expression
x(m)= ∑N−1n=0x1(n)x2(m−n)N
For m=0, x2((-n))4={1,4,3,2}
For m=1, x2((1-n))4={2,1,4,3}
For m=2, x2((2-n))4={3,2,1,4}
For m=3, x2((3-n))4={4,3,2,1}
Now we get x(m)={14,16,14,16}.

244. What is the circular convolution of the sequences X1(n)={2,1,2,1} and x2(n)={1,2,3,4},
find using the DFT and IDFT concepts?
a) {16,16,14,14}
b) {14,16,14,16}
c) {14,14,16,16}
Answer: b
Explanation: Given X1(n)={2,1,2,1}=>X1(k)=[6,0,2,0]
Given x2(n)={1,2,3,4}=>X2(k)=[10,-2+j2,-2,-2-j2]
when we multiply both DFTs we obtain the product
X(k)=X1(k).X2(k)=[60,0,-4,0]
By applying the IDFT to the above sequence, we get
x(n)={14,16,14,16}.
245.If X(k) is the N-point DFT of a sequence x(n), then what is the DFT of x*(n)?
a) X(N-k)
b) X*(k)
c) X*(N-k)
Answer: c
Explanation: According to the complex conjugate property of DFT, we have if X(k) is the N-
point DFT of a sequence x(n), then what is the DFT .
246. If δ1 represents the ripple in the pass band for a chebyshev filter, then which of the
following conditions is true?
a) 1-δ1 ≤ Hr(ω) ≤ 1+δ1; |ω|≤ωP
b) 1+δ1 ≤ Hr(ω) ≤ 1-δ1; |ω|≥ωP
c) 1+δ1 ≤ Hr(ω) ≤ 1-δ1; |ω|≤ωP
d) 1-δ1 ≤ Hr(ω) ≤ 1+δ1; |ω|≥ωP
Answer: a
Explanation: Let us consider the design of a low pass filter with the pass band edge frequency
ωP and the ripple in the pass band is δ1, then from the general specifications of the chebyshev

filter, in the pass band the filter frequency response should satisfy the condition
1- δ1 ≤ Hr(ω) ≤ 1+δ1; |ω|≤ωP
247. If the filter has symmetric unit sample response with M odd, then what is the value of
Q(ω)?
a) cos(ω/2)
b) sin(ω/2)
c) 1
d) sinω
Answer: c
Explanation: If the filter has a symmetric unit sample response, then we know that
h(n)=h(M-1-n)
and for M odd in this case, Q(ω)=1.
248. If the filter has anti-symmetric unit sample response with M odd, then what is the
value of Q(ω)?
a) cos(ω/2)
b) sin(ω/2)
c) 1
d) sinω
Answer: d
Explanation: If the filter has a anti-symmetric unit sample response, then we know that
h(n)= -h(M-1-n)
and for M odd in this case, Q(ω)=sin(ω).
249. In which of the following way the real valued desired frequency response is defined?
a) Unity in stop band and zero in pass band
b) Unity in both pass and stop bands
c) Unity in pass band and zero in stop band
d) Zero in both stop and pass band

Answer: c
Explanation: The real valued desired frequency response Hdr(ω) is simply defined to be unity in
the pass band and zero in the stop band.
250. The error function E(ω) should exhibit at least how many extremal frequencies in S?
a) L
b) L-1
c) L+1
d) L+2
Answer: d
Explanation: According to Alternation theorem, a necessary and sufficient condition for P(ω) to
be unique, best weighted chebyshev approximation, is that the error function E(ω) must exhibit
at least L+2 extremal frequencies in S.
251. The filter designs that contain maximum number of alternations are called as
a) Extra ripple filters
b) Maximal ripple filters
c) Equi ripple filters
Answer: b
Explanation: In general, the filter designs that contain maximum number of alternations or
ripples are called as maximal ripple filters.
252. In Parks-McClellan program, an array of maximum size 10 that specifies the weight
function in each band is denoted by?
a) WTX
b) FX
c) EDGE

Answer: a
Explanation: FX denotes an array of maximum size 10 that specifies the weight function in each
band.
253 The filter designs which are formulated using chebyshev approximating problem have
ripples in?
a) Pass band
b) Stop band
c) Pass & Stop band
d) Restart band
Answer: c
Explanation: The chebyshev approximation problem is viewed as an optimum design criterion
on the sense that the weighted approximation error between the desired frequency response and
the actual frequency response is spread evenly across the pass band and evenly across the stop
band of the filter minimizing the maximum error. The resulting filter designs have ripples in both
pass band and stop band.
254. If the filter has symmetric unit sample response with M even, then what is the value of
Q(ω)?
a) cos(ω/2)
b) sin(ω/2)
c) 1
d) sinω
Answer: a
Explanation: If the filter has a symmetric unit sample response, then we know that
h(n)=h(M-1-n)
and for M even in this case, Q(ω)=cos(ω/2)

255. Which of the following defines the rectangular window function of length M-1?
a)
w(n)=1, n=0,1,2...M-1
=0, else where
b)
w(n)=1, n=0,1,2...M-1
=-1, else where
c)
w(n)=0, n=0,1,2...M-1
=1, else where
Answer: a
Explanation: We know that the rectangular window of length M-1 is defined as
w(n)=1, n=0,1,2…M-1
=0, else where.
256. What is the Fourier transform of the rectangular window of length M-1?
a) ejω(M−1)/2sin(ωM2)sin(ω2)
b) ejω(M+1)/2sin(ωM2)sin(ω2)
c) e−jω(M+1)/2sin(ωM2)sin(ω2)
d) e−jω(M−1)/2sin(ωM2)sin(ω2)
Answer: d
Explanation: We know that the Fourier transform of a function w(n) is defined as
W(ω)=∑M−1n=0w(n)e−jωn

For a rectangular window, w(n)=1 for n=0,1,2….M-1
Thus we get
W(ω)=∑M−1n=0w(n)e−jωn=e−jω(M−1)/2sin(ωM2)sin(ω2)
257. What is the magnitude response |W(ω)| of a rectangular window function?
a) |sin(ωM/2)||sin(ω/2)|
b) |sin(ω/2)||sin(ωM/2)|
c) |cos(ωM/2)||sin(ω/2)|
Answer: a
Explanation: We know that for a rectangular window
W(ω)=∑M−1n=0w(n)e−jωn=e−jω(M−1)/2sin(ωM2)sin(ω2)
Thus the window function has a magnitude response
|W(ω)|=|sin(ωM/2)||sin(ω/2)|
258. What is the width of the main lobe of the frequency response of a rectangular window
of length M-1?
a) π/M
b) 2π/M
c) 4π/M
d) 8π/M
Answer: c
Explanation: The width of the main lobe width is measured to the first zero of W(ω)) is 4π/M.
259.With an increase in the value of M, the height of each side lobe
a) Do not vary
b) Does not depend on value of M
c) Decreases
d) Increases
Answer: d
Explanation: The height of each side lobes increase with an increase in M such a manner that
the area under each side lobe remains invariant to changes in M.

260.Which of the following windows has a time domain sequence h(n)=1−2|n−M−12|M−1?
a) Bartlett window
b) Blackman window
c) Hanning window
d) Hamming window
Answer: a
Explanation: The Bartlett window which is also called as triangular window has a time domain
sequence as
h(n)=1−2|n−M−12|M−1, 0≤n≤M-1.
261What is the approximate transition width of main lobe of a Hamming window?
a) 4π/M
b) 8π/M
c) 12π/M
d) 2π/M
Answer: b
Explanation: The transition width of the main lobe in the case of Hamming window is equal to
8π/M where M is the length of the window
262. Which of the following is the difference equation of the FIR filter of length M, input
x(n) and output y(n)?
a) y(n)=∑M+1k=0bkx(n+k)
b) y(n)=∑M+1k=0bkx(n−k)
c) y(n)=∑M−1k=0bkx(n−k)
Answer: c
Explanation: An FIR filter of length M with input x(n) and output y(n) is described by the
difference equation
y(n)=∑M−1k=0bkx(n−k)
where {bk} is the set of filter coefficients.

263.Which of the following condition should the unit sample response of a FIR filter satisfy
to have a linear phase?
a) h(M-1-n) n=0,1,2…M-1
b) ±h(M-1-n) n=0,1,2…M-1
c) -h(M-1-n) n=0,1,2…M-1
Answer: b
Explanation: An FIR filter has an linear phase if its unit sample response satisfies the condition
h(n)= ±h(M-1-n) n=0,1,2…M-1.
264. If H(z) is the z-transform of the impulse response of an FIR filter, then which of the
following relation is true?
a) zM+1.H(z-1)=±H(z)
b) z-(M+1).H(z-1)=±H(z)
c) z(M-1).H(z-1)=±H(z)
d) z-(M-1).H(z-1)=±H(z)
Answer: d
Explanation: We know that H(z)=∑M−1k=0h(k)z−k and h(n)=±h(M-1-n) n=0,1,2…M-1
When we incorporate the symmetric and anti-symmetric conditions of the second equation into
the first equation and by substituting z-1 for z, and multiply both sides of the resulting equation
by z-(M-1) we get z-(M-1).H(z-1)=±H(z)
265.The roots of the equation H(z) must occur in
a) Identical
b) Zero
c) Reciprocal pairs
d) Conjugate pairs
Answer: c
Explanation: We know that the roots of the polynomial H(z) are identical to the roots of the
polynomial H(z-1). Consequently, the roots of H(z) must occur in reciprocal pairs.

266. What is the value of h(M-1/2) if the unit sample response is anti-symmetric?
a) 0
b) 1
c) -1
Answer: a
Explanation: When h(n)=-h(M-1-n), the unit sample response is anti-symmetric. For M odd, the
center point of the anti-symmetric is n=M-1/2. Consequently, h(M-1/2)=0.
267. What is the number of filter coefficients that specify the frequency response for h(n)
symmetric?
a) (M-1)/2 when M is odd and M/2 when M is even
b) (M-1)/2 when M is even and M/2 when M is odd
c) (M+1)/2 when M is even and M/2 when M is odd
d) (M+1)/2 when M is odd and M/2 when M is even
Answer: d
Explanation: We know that, for a symmetric h(n), the number of filter coefficients that specify
the frequency response is (M+1)/2 when M is odd and M/2 when M is even.
268. What is the number of filter coefficients that specify the frequency response for h(n)
anti-symmetric?
a) (M-1)/2 when M is even and M/2 when M is odd
b) (M-1)/2 when M is odd and M/2 when M is even
c) (M+1)/2 when M is even and M/2 when M is odd
d) (M+1)/2 when M is odd and M/2 when M is even
Answer: b
Explanation: We know that, for a anti-symmetric h(n) h(M-1/2)=0 and thus the number of filter
coefficients that specify the frequency response is (M-1)/2 when M is odd and M/2 when M is
even.

269. Which of the following is not suitable either as low pass or a high pass filter?
a) h(n) symmetric and M odd
b) h(n) symmetric and M even
c) h(n) anti-symmetric and M odd
d) h(n) anti-symmetric and M even
Answer: c
Explanation: If h(n)=-h(M-1-n) and M is odd, we get H(0)=0 and H(π)=0. Consequently, this is
not suitable as either a low pass filter or a high pass filter.
270. The anti-symmetric condition with M evenis not used in the design of which of the
following linear-phase FIR filter?
a) Low pass
b) High pass
c) Band pass
d) Bans stop
Answer: a
Explanation: When h(n)=-h(M-1-n) and M is even, we know that H(0)=0. Thus it is not used in
the design of a low pass linear phase FIR filter
271. Which of the following rule is used in the bilinear transformation?
a) Simpson’s rule
b) Backward difference
c) Forward difference
d) Trapezoidal rule
Answer: d
Explanation: Bilinear transformation uses trapezoidal rule for integrating a continuous time
function.

272. Which of the following substitution is done in Bilinear transformations?
a) s = 2T[1+z−11−z1]
b) s = 2T[1+z−11+]
c) s = 2T[1−z−11+z−1]
Answer: c
Explanation: In bilinear transformation of an analog filter to digital filter, using the trapezoidal
rule, the substitution for ‘s’ is given as
s = 2T[1−z−11+z−1].
273. What is the value of ∫nT(n−1)Tx(t)dt according to trapezoidal rule?
a) [x(nT)−x[(n−1)T]2]T
b) [x(nT)+x[(n−1)T]2]T
c) [x(nT)−x[(n+1)T]2]T
d) [x(nT)+x[(n+1)T]2]T
Answer: b
Explanation: The given integral is approximated by the trapezoidal rule. This rule states that if
T is small, the area (integral) can be approximated by the mean height of x(t) between the two
limits and then multiplying by the width. That is
∫nT(n−1)Tx(t)dt=[x(nT)+x[(n−1)T]2]T
274. What is the value of y(n)-y(n-1) in terms of input x(n)?
a) [x(n)+x(n−1)2]T
b) [x(n)−x(n−1)2]T
c) [x(n)−x(n+1)2]T
d) [x(n)+x(n+1)2]T
Answer: a
Explanation: We know that the derivative equation is
dy(t)/dt=x(t)
On applying integrals both sides, we get
∫nT(n−1)Tdy(t)=∫nT(n−1)Tx(t)dt
=> y(nT)-y[(n-1)T]=∫nT(n−1)Tx(t)dt

On applying trapezoidal rule on the right hand integral, we get
y(nT)-y[(n-1)T]=[x(nT)+x[(n−1)T]2]T
Since x(n) and y(n) are approximately equal to x(nT) and y(nT) respectively, the above equation
can be written as
y(n)-y(n-1)=[x(n)+x(n−1)2]T
275. What is the expression for systemfunction in z-domain?
a) 2T[1+z−11−z1]
b) 2T[1+z−11−z1]
c) T2[1+z−11−z1]
d) T2[1−z−11+z−1]
Answer: c
y(n)-y(n-1)= [x(n)+x(n−1)2]T
Taking z-transform of the above equation gives
=>Y(z)[1-z-1]=([1+z-1]/2).TX(z)
=>H(z)=Y(z)/X(z)=T2[1+z−11−z1].
276. In bilinear transformation, the left-half s-plane is mapped to which of the following in
the z-domain?
a) Entirely outside the unit circle |z|=1
b) Partially outside the unit circle |z|=1
c) Partially inside the unit circle |z|=1
d) Entirely inside the unit circle |z|=1
Answer: d
Explanation: In bilinear transformation, the z to s transformation is given by the expression
z=[1+(T/2)s]/[1-(T/2)s].
Thus unlike the backward difference method, the left-half s-plane is now mapped entirely inside
the unit circle, |z|=1, rather than to a part of it.

277. If s=σ+jΩ and z=rejω, then what is the condition on σ if r<1?
a) σ > 0
b) σ < 0
c) σ > 1
d) σ < 1
Answer: b
Explanation: We know that if = σ+jΩ and z=rejω, then by substituting the values in the below
expression
s = 2T[1−z−11+z−1]
=>σ = 2T[r2−1r2+1+2rcosω]
When r<1 => σ < 0.
278. If s=σ+jΩ and z=rejω and r=1, then which of the following inference is correct?
a) LHS of the s-plane is mapped inside the circle, |z|=1
b) RHS of the s-plane is mapped outside the circle, |z|=1
c) Imaginary axis in the s-plane is mapped to the circle, |z|=1
Answer: c
Explanation: We know that if =σ+jΩ and z=rejω, then by substituting the values in the below
expression
s = 2T[1−z−11+z−1]
=>σ = 2T[r2−1r2+1+2rcosω]
When r=1 => σ = 0.
This shows that the imaginary axis in the s-domain is mapped to the circle of unit radius centered
at z=0 in the z-domain.
279. If s=σ+jΩ and z=rejω, then what is the condition on σ if r>1?
a) σ > 0
b) σ < 0
c) σ > 1
d) σ < 1

Answer: a
Explanation: We know that if = σ+jΩ and z=rejω, then by substituting the values in the below
expression
s = 2T[1−z−11+z−1]
=>σ = 2T[r2−1r2+1+2rcosω]
When r>1 => σ > 0.
280. What is the expression for the digital frequency when r=1?
a) 1Ttan(ΩT2)
b) 2Ttan(ΩT2)
c) 1Ttan−1(ΩT2)
d) 2Ttan−1(ΩT2)
Answer: d
Explanation: When r=1, we get σ=0 and
Ω = 2T[2sinω1+1+2cosω]
=>ω=2Ttan−1(ΩT2).
281. What is the kind of relationship between Ω and ω?
a) Many-to-one
b) One-to-many
c) One-to-one
d) Many-to-many
Answer: c
Explanation: The analog frequencies Ω=±∞ are mapped to digital frequencies ω=±π. The
frequency mapping is not aliased; that is, the relationship between Ω and ω is one-to-one. As a
consequence of this, there are no major restrictions on the use of bilinear transformation
282. Which of the following defines a chebyshev polynomial of order N, TN(x)?
a) cos(Ncos-1x) for all x
b) cosh(Ncosh-1x) for all x
c)
cos(Ncos-1x), |x|<1
cosh(Ncosh-1x), |x|>1

Answer: c
Explanation: In order to understand the frequency-domain behavior of chebyshev filters, it is
utmost important to define a chebyshev polynomial and then its properties. A chebyshev
polynomial of degree N is defined as
TN(x) = cos(Ncos-1x), |x|≤1
cosh(Ncosh-1x), |x|>1.
283. What is the formula for chebyshev polynomial TN(x) in recursive form?
a) 2TN-1(x) – TN-2(x)
b) 2TN-1(x) + TN-2(x)
c) 2xTN-1(x) + TN-2(x)
d) 2xTN-1(x) – TN-2(x)
Answer: d
Explanation: We know that a chebyshev polynomial of degree N is defined as
From the above formula, it is possible to generate chebyshev polynomial using the following
recursive formula
TN(x)= 2xTN-1(x)-TN-2(x), N ≥ 2.
284. What is the value of chebyshev polynomial of degree 0?
a) 1
b) 0
c) -1
d) 2
Answer: a
For a degree 0 chebyshev filter, the polynomial is obtained as
T0(x)=cos(0)=1.

a) 1
b) x
c) -1
d) -x
Answer: b
For a degree 1 chebyshev filter, the polynomial is obtained as
T0(x)=cos(cos-1x)=x.
a) 3x3+4x
b) 3x3-4x
c) 4x3+3x
d) 4x3-3x
Answer: d
TN(x) = cos(Ncos-1x), |x|≤1; TN(x) = cosh(Ncosh-1x), |x|>1
And the recursive formula for the chebyshev polynomial of order N is given as
TN(x)=2xTN-1(x)-TN-2(x)
Thus for a chebyshev filter of order 3, we obtain
T3(x)=2xT2(x)-T1(x)=2x(2x2-1)-x=4x3-3x.
a) 16x5+20x3-5x
b) 16x5+20x3+5x
c) 16x5-20x3+5x
d) 16x5-20x3-5x

Answer: c
= cosh(Ncosh-1x), |x|>1
And the recursive formula for the chebyshev polynomial of order N is given as
TN(x)= 2xTN-1(x)-TN-2(x)
Thus for a chebyshev filter of order 5, we obtain
T5(x)=2xT4(x)-T3(x)=2x(8x4-8x2+1)-(4x3-3x)=16x5-20x3+5x.
288. Chebyshev polynomials of odd orders are
a) Even functions
b) Odd functions
c) Exponential functions
d) Logarithmic functions
Answer: b
Explanation: Chebyshev polynomials of odd orders are odd functions because they contain only
odd powers of x.
289. What is the value of TN(0) for evendegree N?
a) -1
b) +1
c) 0
d) ±1
Answer: d
For x=0, we have TN(0)=cos(Ncos-10)=cos(N.π/2)=±1 for N even

290. What is the value of |TN(±1)|?
a) 0
b) -1
c) 1
Answer: c
Thus |TN(±1)|=1.
291.If NB and NC are the orders of the Butterworth and Chebyshev filters respectively to
meet the same frequency specifications, then which of the following relation is true?
a) NC=NB
b) NC<NB
c) NC>NB
d) Cannot be determined
Answer: b
Explanation: The equi-ripple property of the chebyshev filter yields a narrower transition band
compared with that obtained when the magnitude response is monotone. As a consequence of
this, the order of a chebyshev filter needed to achieve the given frequency domain specifications
is usually lower than that of a Butterworth filter.
292. What is the equation for magnitude frequency response |H(jΩ)| of a low pass
chebyshev-I filter?
a) 11−ϵT2N(ΩΩP)√
b) 11+ϵT2N(ΩΩP)√
c) 11−ϵ2T2N(ΩΩP)√
d) 11+ϵ2T2N(ΩΩP)√
Answer: d
Explanation: The magnitude frequency response of a low pass chebyshev-I filter is given by
|H(jΩ)|=11+ϵ2T2N(ΩΩP)√

where ϵ is a parameter of the filter related to the ripple in the pass band and TN(x) is the Nth order
chebyshev polynomial.
293. What is the number of minima’s present in the pass band of magnitude frequency
response of a low pass chebyshev-I filter of order 4?
a) 1
b) 2
c) 3
d) 4
Answer: b
Explanation: In the magnitude frequency response of a low pass chebyshev-I filter, the pass
band has 2 maxima and 2 minima(order 4=2 maxima+2 minima).
294. What is the number of maxima present in the pass band of magnitude frequency
response of a low pass chebyshev-I filter of order 5?
a) 1
b) 2
c) 3
d) 4
Answer: c
Explanation: In the magnitude frequency response of a low pass chebyshev-I filter, the pass
band has 3 maxima and 2 minima(order 5=3 maxima+2 minima).
295.Which of the following is the characteristic equation of a Chebyshev filter?
a) 1+ϵ2TN
2(s/j)=0
b) 1-ϵ2TN
2(s/j)=0
c) 1+ϵ TN
2(s/j)=0
Answer: a
Explanation: We know that for a chebyshev filter, we have
|H(jΩ)|=11+ϵ2T2N(ΩΩP)√
=>|H(jΩ)|2=11+ϵ2T2N(ΩΩP)√
By replacing jΩ by s and consequently Ω by s/j in the above equation, we get

N
N
=>|HN(s)|2=11+ϵ2T2N(s/j)
The poles of the above equation is given by the equation 1+ϵ2T 2(s/j)=0 which is called as the
characteristic equation.
296. The poles of HN(s).HN(-s) are found to lie on
a) Circle
b) Parabola
c) Hyperbola
d) Ellipse
Answer: d
Explanation: The poles of HN(s).HN(-s) is given by the characteristic equation 1+ϵ2T 2(s/j)=0.
The roots of the above characteristic equation lies on ellipse, thus the poles of HN(s).HN(-s) are
found to lie on ellipse.
297. If the discrimination factor ‘d’ and the selectivity factor ‘k’ of a chebyshev I filter are
0.077 and 0.769 respectively, then what is the order of the filter?
a) 2
b) 5
c) 4
d) 3
Answer: b
Explanation: We know that the order of a chebyshev-I filter is given by the equation,
N=cosh-1(1/d)/cosh-1(1/k)=4.3
Rounding off to the next large integer, we get N=5
298. What is the cutoff frequency of the Butterworth filter with a pass band gain KP=-1 dB
at ΩP=4 rad/sec and stop band attenuation greater than or equal to 20dB at ΩS=8 rad/sec?
a) 3.5787 rad/sec
b) 1.069 rad/sec
c) 6 rad/sec
d) 4.5787 rad/sec

Answer: d
Explanation: We know that the equation for the cutoff frequency of a Butterworth filter is given
as
ΩC = ΩP(10−KP/10−1)1/2N
We know that KP=-1 dB, ΩP=4 rad/sec and N=5
Upon substituting the values in the above equation, we get
ΩC=4.5787 rad/sec.
299. What is the system function of the Butterworth filter with specifications as pass band
gain KP=-1 dB at ΩP=4 rad/sec and stop band attenuation greater than or equal to 20dB at
ΩS=8 rad/sec?
a) 1s5+14.82s4+109.8s3+502.6s2+1422.3s+2012.4
b) 1s5+14.82s4+109.8s3+502.6s2+1422.3s+1
c) 2012.4s5+14.82s4+109.8s3+502.6s2+1422.3s+2012.4
Answer: c
Explanation: From the given question,
KP=-1 dB, ΩP=4 rad/sec, KS=-20 dB and ΩS=8 rad/sec
We find out order as N=5 and ΩC=4.5787 rad/sec
We know that for a 5th order normalized low pass Butterworth filter, system equation is given as
H5(s)=1(s+1)(s2+0.618s+1)(s2+1.618s+1)
The specified low pass filter is obtained by applying low pass-to-low pass transformation on the
normalized low pass filter.
That is, Ha(s)=H5(s)|s→s/Ωc
=H5(s)|s→s/4.5787
upon calculating, we get
Ha(s)=2012.4s5+14.82s4+109.8s3+502.6s2+1422.3s+2012.4

300. If H(s)=1s2+s+1 represent the transfer function of a low pass filter (not Butterworth)
with a pass band of 1 rad/sec, then what is the systemfunction of a low pass filter with a
pass band 10 rad/sec?
a) 100s2+10s+100
b) s2s2+s+1
c) s2s2+10s+100
Answer: a
Explanation: The low pass-to-low pass transformation is
s→s/Ωu
Hence the required low pass filter is
Ha(s)=H(s)|s→s/10
=100s2+10s+100.
with a pass band of 1 rad/sec, then what is the systemfunction of a high pass filter with a
cutoff frequency of 1rad/sec?
a) 100s2+10s+100
b) s2s2+s+1
c) s2s2+10s+100
Answer: b
Explanation: The low pass-to-high pass transformation is
s→Ωu/s
Hence the required high pass filter is
Ha(s)= H(s)|s→1/s
=s2s2+s+1

with a pass band of 1 rad/sec, then what is the systemfunction of a high pass filter with a
cutoff frequency of 10 rad/sec?
a) 100s2+10s+100
b) s2s2+s+1
c) s2s2+10s+100
Answer: c
Explanation: The low pass-to-high pass transformation is
s→Ωu/s
Hence the required low pass filter is
Ha(s)=H(s)|s→10/s
=s2s2+10s+100
with a pass band of 1 rad/sec, then what is the systemfunction of a band pass filter with a
pass band of 10 rad/sec and a center frequency of 100 rad/sec?
a) s2s4+10s3+20100s2+105s+1
b) 100s2s4+10s3+20100s2+105s+1
c) s2s4+10s3+20100s2+105s+108
d) 100s2s4+10s3+20100s2+105s+108
Answer: d
Explanation: The low pass-to-band pass transformation is
s→s2+ΩuΩls(Ωu−Ωl)
Thus the required band pass filter has a transform function as
Ha(s)=100s2s4+10s3+20100s2+105s+108.

with a pass band of 1 rad/sec, then what is the systemfunction of a stop band filter with a
stop band of 2 rad/sec and a center frequency of 10 rad/sec?
a) (s2+100)2s4+2s3+204s2+200s+104
b) (s2+10)2s4+2s3+204s2+200s+104
c) (s2+10)2s4+2s3+400s2+200s+104
Answer: a
Explanation: The low pass-to- band stop transformation is
s→s(Ωu−Ωl)s2+ΩuΩl
Hence the required band stop filter is
Ha(s)=(s2+100)2s4+2s3+204s2+200s+104
305. What is the stop band frequency of the normalized low pass Butterworth filter used to
design a analog band pass filter with -3.0103dB upper and lower cutoff frequency of 50Hz
and 20KHz and a stop band attenuation 20dB at 20Hz and 45KHz?
a) 2 rad/sec
b) 2.25 Hz
c) 2.25 rad/sec
d) 2 Hz
Answer: c
Explanation: Given information is
Ω1=2π*20=125.663 rad/sec
Ω2=2π*45*103=2.827*105rad/sec
Ωu=2π*20*103=1.257*105 rad/sec
Ωl=2π*50=314.159 rad/sec
We know that
A=−Ω21+ΩuΩlΩ1(Ωu−Ωl) and B=Ω22−ΩuΩlΩ2(Ωu−Ωl)
=> A=2.51 and B=2.25
Hence ΩS=Min{|A|,|B|}=>ΩS=2.25 rad/sec.

306. What is the order of the normalized low pass Butterworth filter used to design a
analog band pass filter with -3.0103dB upper and lower cutoff frequency of 50Hz and
20KHz and a stop band attenuation 20dB at 20Hz and 45KHz?
a) 2
b) 3
c) 4
d) 5
Answer: b
Explanation: Given information is
Ω1=2π*20=125.663 rad/sec
Ω2=2π*45*103=2.827*105 rad/sec
Ωu=2π*20*103=1.257*105 rad/sec
Ωl=2π*50=314.159 rad/sec
We know that
A=−Ω21+ΩuΩlΩ1(Ωu−Ωl) and B=Ω22−ΩuΩlΩ2(Ωu−Ωl)
=> A=2.51 and B=2.25
Hence ΩS=Min{|A|,|B|}=> ΩS=2.25 rad/sec.
The order N of the normalized low pass Butterworth filter is computed as follows
N=log[(10−KP/10−1)(10−Ks/10−1)]2log(1ΩS)=2.83
Rounding off to the next large integer, we get, N=3.
307. Which of the following condition is true?
a) N ≤ log(1k)log(1d)
b) N ≤ log(k)log(d)
c) N ≤ log(d)log(k)
d) N ≤log(1d)log(1k)
View Answer
Answer: d
Explanation: If ‘d’ is the discrimination factor and ‘K’ is the selectivity factor, then
N ≤ log(1d)log(1k)

308. Which of the following is true in the case of Butterworth filters?
a) Smooth pass band
b) Wide transition band
c) Not so smooth stop band
Answer: d
Explanation: Butterworth filters have a very smooth pass band, which we pay for with a
relatively wide transmission region.
309. What is the magnitude frequency response of a Butterworth filter of order N and
cutoff frequency ΩC?
a) 11+(ΩΩC)2N√
b) 1+(ΩΩC)2N
c) 1+(ΩΩC)2N−−−−−−−−−√
Answer: a
Explanation: A Butterworth is characterized by the magnitude frequency response
|H(jΩ)|=11+(ΩΩC)2N√
where N is the order of the filter and ΩC is defined as the cutoff frequency.
310. What is the factor to be multiplied to the dc gain of the filter to obtain filter magnitude
at cutoff frequency?
a) 1
b) √2
c) 1/√2
d) 1/2
Answer: c
Explanation: The dc gain of the filter is the filter magnitude at Ω=0.
We know that the filter magnitude is given by the equation
|H(jΩ)|=11+(ΩΩC)2N√
At Ω=ΩC, |H(jΩC)|=1/√2=1/√2(|H(jΩ)|)
Thus the filter magnitude at the cutoff frequency is 1/√2 times the dc gain.

311. What is the value of magnitude frequency response of a Butterworth low pass filter at
Ω=0?
a) 0
b) 1
c) 1/√2
Answer: b
Explanation: The magnitude frequency response of a Butterworth low pass filter is given as
|H(jΩ)|=11+(ΩΩC)2N√
At Ω=0 => |H(jΩ)|=1 for all N.
312. As the value of the frequency Ω tends to ∞, then |H(jΩ)| tends to _
a) 0
b) 1
c) ∞
Answer: a
Explanation: We know that the magnitude frequency response of a Butterworth filter of order N
is given by the expression
|H(jΩ)|=11+(ΩΩC)2N√
In the above equation, if Ω→∞ then |H(jΩ)|→0.
313.What is the magnitude squared response of the normalized low pass Butterworth
filter?
a) 11+Ω−2N
b) 1+Ω-2N
c) 1+Ω2N
d) 11+Ω2N
Answer: d
Explanation: We know that the magnitude response of a low pass Butterworth filter of order N
is given as
|H(jΩ)|=11+(ΩΩC)2N√

For a normalized filter, ΩC =1
=> |H(jΩ)|=11+(Ω)2N√ => |H(jΩ)|2=11+Ω2N
Thus the magnitude squared response of the normalized low pass Butterworth filter of order N is
given by the equation,
|H(jΩ)|2=11+Ω2N.
314. What is the transfer function of magnitude squared frequency response of the
normalized low pass Butterworth filter?
a) 11+(s/j)2N
b) 1+(sj)−2N
c) 1+(sj)2N
d) 11+(s/j)−2N
Answer: a
Explanation: We know that the magnitude squared frequency response of a normalized low pass
Butterworth filter is given as
H(jΩ)|2=11+Ω2N => HN(jΩ).HN(-jΩ)=11+Ω2N
Replacing jΩ by ‘s’ and hence Ω by s/j in the above equation, we get
HN(s).HN(-s)=11+(sj)2N which is called the transfer function.
315. Where does the poles of the transfer function of normalized low pass Butterworth
filter exists?
a) Inside unit circle
b) Outside unit circle
c) On unit circle
Answer: c
Explanation: The transfer function of normalized low pass Butterworth filter is given as
HN(s).HN(-s)=11+(sj)2N
The poles of the above equation is obtained by equating the denominator to zero.
=> 1+(sj)2N=0
=> s=(-1)1/2N.j

=> sk=ejπ(2k+12N)ejπ/2, k=0,1,2…2N-1
The poles are therefore on a circle with radius unity.
316. What is the general formula that represent the phase of the poles of transfer function
of normalized low pass Butterworth filter of order N?
a) πNk+π2N k=0,1,2…N-1
b) πNk+π2N+π2 k=0,1,2…2N-1
c) πNk+π2N+π2 k=0,1,2…N-1
d) πNk+π2N k=0,1,2…2N-1
Answer: d
Explanation: The transfer function of normalized low pass Butterworth filter is given as
HN(s).HN(-s)=11+(sj)2N
The poles of the above equation is obtained by equating the denominator to zero.
=> 1+(sj)2N=0
=> s=(-1)1/2N.j
=> sk=ejπ(2k+12N)ejπ/2, k=0,1,2…2N-1
The poles are therefore on a circle with radius unity and are placed at angles,
θk=πNk+π2N k=0,1,2…2N-1
317. What is the Butterworth polynomial of order 3?
a) (s2+s+1)(s-1)
b) (s2-s+1)(s-1)
c) (s2-s+1)(s+1)
d) (s2+s+1)(s+1)
Answer: d
Explanation: Given that the order of the Butterworth low pass filter is 3.
Therefore, for N=3 Butterworth polynomial is given as B3(s)=(s-s0) (s-s1) (s-s2)
We know that, sk=ejπ(2k+12N)ejπ/2
=> s0=(-1/2)+j(√3/2), s1= -1, s2=(-1/2)-j(√3/2)
=> B3(s)= (s2+s+1)(s+1).

318. What is the Butterworth polynomial of order 1?
a) s-1
b) s+1
c) s
Answer: b
Explanation: Given that the order of the Butterworth low pass filter is 1.
Therefore, for N=1 Butterworth polynomial is given as B3(s)=(s-s0).
We know that, sk=ejπ(2k+12N)ejπ/2
=> s0=-1
=> B1(s)=s-(-1)=s+1.
319. What is the transfer function of Butterworth low pass filter of order 2?
a) 1s2+2√s+1
b) 1s2−2√s+1
c) s2−2–√s+1
d) s2+2–√s+1
Answer: a
Explanation: We know that the Butterworth polynomial of a 2nd order low pass filter is
B2(s)=s2+√2 s+1
Thus the transfer function is given as 1s2+2√s+1.
320. What is the duration of the unit sample response of a digital filter?
a) Finite
b) Infinite
c) Impulse(very small)
d) Zero
Answer: b
Explanation: Digital filters are the filters which can be designed from analog filters which have
infinite duration unit sample response.

321. Which of the following methods are used to convert analog filter into digital filter?
a) Approximation of Derivatives
b) Bilinear transformation
c) Impulse invariance
Answer: d
Explanation: There are many techniques which are used to convert analog filter into digital filter
of which some of them are Approximation of derivatives, bilinear transformation, impulse
invariance and many other methods.
322. Which of the following is the difference equation of the FIR filter of length M, input
x(n) and output y(n)?
a) y(n)=∑M+1k=0bkx(n+k)
b) y(n)=∑M+1k=0bkx(n−k)
c) y(n)=∑M−1k=0bkx(n−k)
Answer: c
Explanation: An FIR filter of length M with input x(n) and output y(n) is described by the
difference equation
y(n)=∑M−1k=0bkx(n−k)
where {bk} is the set of filter coefficients.
323. What is the relation between h(t) and Ha(s)?
a) Ha(s)=∫∞−∞h(t)e−stdt
b) Ha(s)=∫∞0h(t)estdt
c) Ha(s)=∫∞−∞h(t)estdt
Answer: a
Explanation: We know that the impulse response h(t) and the Laplace transform Ha(s) are
related by the equation.
Ha(s)=∫∞−∞h(t)e−stdt

324. Which of the following is a representation of systemfunction?
a) Normal system function
b) Laplace transform
c) Rational system function
Answer: d
Explanation: There are many ways how we represent a system function of which one is normal
representation i.e., output/input and other ways like Laplace transform and rational system
function.
325. For an analog LTI systemto be stable, where should the poles of systemfunction H(s)
lie?
a) Right half of s-plane
b) Left half of s-plane
c) On the imaginary axis
d) At origin
Answer: b
Explanation: An analog linear time invariant system with system function H(s) is stable if all its
poles lie on the left half of the s-plane.
326. If the conversion technique is to be effective, then the LHP of s-plane should be
mapped into
a) Outside of unit circle
b) Unit circle
c) Inside unit circle
d) Does not matter
Answer: c
Explanation: If the conversion technique is to be effective, then the LHP of s-plane should be
mapped into the inside of the unit circle in the z-plane. Thus a stable analog filter will be
converted to a stable digital filter.

327.What is the condition on the systemfunction of a linear phase filter?
a) H(z)=z−NH(z−1)
b) H(z)=zNH(z−1)
c) H(z)=±zNH(z−1)
d) H(z)=±z−NH(z−1)
Answer: d
Explanation: A linear phase filter must have a system function that satisfies the condition
H(z)=±z−NH(z−1)
where z(-N) represents a delay of N units of time.
328. What is the order of operations to be performed in order to realize linear phase IIR
filter?
(i) Passing x(-n) through a digital filter H(z)
(ii) Time reversing the output of H(z)
(iii) Time reversal of the input signal x(n)
(iv) Passing the result through H(z)
a) (i),(ii),(iii),(iv)
b) (iii),(i),(ii),(iv)
c) (ii),(iii),(iv),(i)
d) (i),(iii),(iv),(ii)
Answer: b
Explanation: If the restriction on physical reliability is removed, it is possible to obtain a linear
phase IIR filter, at least in principle. This approach involves performing a time reversal of the
input signal x(n), passing x(-n) through a digital filter H(z), time reversing the output of H(z),
and finally, passing the result through H(z) again

329. What is the Fourier series representation of a signal x(n) whose period is N?
a) ∑N+1k=0ckej2πkn/N
b) ∑N−1k=0ckej2πkn/N
c) ∑Nk=0ckej2πkn/N
d) ∑N−1k=0cke−j2πkn/N
Answer: b
Explanation: Here, the frequency F0 of a continuous time signal is divided into 2π/N intervals.
So, the Fourier series representation of a discrete time signal with period N is given as
x(n)=∑N−1k=0ckej2πkn/N
where ck is the Fourier series coefficient
330. What is the expression for Fourier series coefficient ck in terms of the discrete signal
x(n)?
a) 1N∑N−1n=0x(n)ej2πkn/N
b) N∑N−1n=0x(n)e−j2πkn/N
c) 1N∑N+1n=0x(n)e−j2πkn/N
d) 1N∑N−1n=0x(n)e−j2πkn/N
Answer: d
Explanation: We know that, the Fourier series representation of a discrete signal x(n) is given as
x(n)=∑N−1n=0ckej2πkn/N
Now multiply both sides by the exponential e-j2πln/N and summing the product from n=0 to n=N-
1. Thus,
∑N−1n=0x(n)e−j2πln/N=∑N−1n=0∑N−1k=0ckej2π(k−l)n/N
If we perform summation over n first in the right hand side of above equation, we get
∑N−1n=0e−j2πkn/N = N, for k-l=0,±N,±2N…
= 0, otherwise
Therefore, the right hand side reduces to Nck
So, we obtain ck=1N∑N−1n=0x(n)e−j2πkn/N

331. Which of the following represents the phase associatedwith the frequency component
of discrete-time Fourier series(DTFS)?
a) ej2πkn/N
b) e-j2πkn/N
c) ej2πknN
Answer: a
x(n)=∑N−1k=0ckej2πkn/N
In the above equation, ck represents the amplitude and ej2πkn/N represents the phase associated
with the frequency component of DTFS.
332. What are the Fourier series coefficients for the signal x(n)=cosπn/3?
a) c1=c2=c3=c4=0,c1=c5=1/2
b) c0=c1=c2=c3=c4=c5=0
c) c0=c1=c2=c3=c4=c5=1/2
Answer: a
Explanation: In this case, f0=1/6 and hence x(n) is periodic with fundamental period N=6.
Given signal is x(n)=cosπn/3=cos2πn/6=12ej2πn/6+12e−j2πn/6
We know that -2π/6=2π-2π/6=10π/6=5(2π/6)
Therefore, x(n)=12ej2πn/6+12ej2π(5)n/6
Compare the above equation with x(n)=∑N−1k=0ckej2πkn/N
So, we get c1=c2=c3=c4=0 and c1=c5=1/2.
333. What is the Fourier series representation of a signal x(n) whose period is N?
a) ∑∞k=0|ck|2
b) ∑∞k=−∞|ck|
c) ∑0k=−∞|ck|2
d) ∑∞k=−∞|ck|2

Answer: b
Explanation: The average power of a periodic signal x(t) is given as 1Tp∫t0+Tpt0|x(t)|2dt
=1Tp∫t0+Tpt0x(t).x∗ (t)dt
=1Tp∫t0+Tpt0x(t).∑∞k=−∞c∗ ke−j2πkF0tdt
By interchanging the positions of integral and summation and by applying the integration, we get
=∑∞k=−∞|ck|2
334. What is the average power of the discrete time periodic signal x(n) with period N?
a) 1N∑Nn=0|x(n)|
b) 1N∑N−1n=0|x(n)|
c) 1N∑Nn=0|x(n)|2
d) 1N∑N−1n=0|x(n)|2
Answer: d
Explanation: Let us consider a discrete time periodic signal x(n) with period N.
The average power of that signal is given as
Px=1N∑N−1n=0|x(n)|2
335. What is the equation for average power of discrete time periodic signal x(n) with
period N in terms of Fourier series coefficient ck?
a) ∑N−1k=0|ck|
b) ∑N−1k=0|ck|2
c) ∑Nk=0|ck|2
d) ∑Nk=0|ck|
Answer: b
Explanation: We know that Px=1N∑N−1n=0|x(n)|2
=1N∑N−1n=0x(n).x∗ (n)
=1N∑N−1n=0x(n)∑N−1k=0ck∗ e−j2πkn/N
=∑N−1k=0ck∗ 1N∑N−1n=0x(n)e−j2πkn/N
=∑N−1k=0|ck|2

336. What is the Fourier transform X(ω) of a finite energy discrete time signal x(n)?
a) ∑∞n=−∞x(n)e−jωn
b) ∑∞n=0x(n)e−jωn
c) ∑N−1n=0x(n)e−jωn
Answer: a
Explanation: If we consider a signal x(n) which is discrete in nature and has finite energy, then
the Fourier transform of that signal is given as
X(ω)=∑∞n=−∞x(n)e−jωn
337. What is the period of the Fourier transform X(ω) of the signal x(n)?
a) π
b) 1
c) Non-periodic
d) 2π
Answer: d
Explanation: Let X(ω) be the Fourier transform of a discrete time signal x(n) which is given as
Now X(ω+2πk)=∑∞n=−∞x(n)e−j(ω+2πk)n
=∑∞n=−∞x(n)e−jωne−j2πkn
=∑∞n=−∞x(n)e−jωn=X(ω)
So, the Fourier transform of a discrete time finite energy signal is periodic with period 2π.
338. What is the synthesis equation of the discrete time signal x(n), whose Fourier
transform is X(ω)?
a) 2π∫20πX(ω)ejωndω
b) 1π∫2π0X(ω)ejωndω
c) 12π∫2π0X(ω)ejωndω
Answer: c
Explanation: We know that the Fourier transform of the discrete time signal x(n) is

By calculating the inverse Fourier transform of the above equation, we get
x(n)=12π∫2π0X(ω)ejωndω
The above equation is known as synthesis equation or inverse transform equation.
339. What is the value of discrete time signal x(n) at n=0 whose Fourier transform is
represented as below?
a) ωc.π
b) -ωc/π
c) ωc/π
Answer: c
Explanation: We know that, x(n)=12π∫π−πX(ω)ejωndω
=12π∫ωc−ωc1.ejωndω
At n=0,
x(n)=x(0)=∫ωc−ωc1dω=12π(2ωc)=ωcπω
Therefore, the value of the signal x(n) at n=0 is ωc/π.
340. What is the value of discrete time signal x(n) at n≠0 whose Fourier transform is
represented as below?
a) ωcπ.sinωc.nωc.n
b) −ωcπ.sinωc.nωc.n
c) ωc.πsinωc.nωc.n
Answer: a
Explanation: We know that, x(n)=12π∫π−πX(ω)ejωndω
=12π∫ωc−ωc1.ejωndω=sinωc.nωc.n =ωcπ.sinωc.nωc.n

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