DSP_FOEHU - MATLAB 02 - The Discrete-time Fourier Analysis

MATLAB(02)
The Discrete-time Fourier
Analysis
Assist. Prof. Amr E. Mohamed

Introduction
 A linear and time-invariant system can be represented using its response
to the unit sample sequence. This response, called the unit impulse
response h(n), allows us to compute the system response to any
arbitrary input x(n) using the linear convolution:
 This convolution representation is based on the fact that any signal can
be represented by a linear combination of scaled and delayed unit
samples.
2

THE DISCRETE-TIME FOURIER TRANSFORM (DTFT)
 If x(n) is absolutely summable, that is, −∞
∞
𝑥 𝑛 < ∞, then its discrete-
time Fourier transform is given by
𝑋 𝑒 𝑗𝜔
≜ ℱ 𝑥 𝑛 =
𝑛=−∞
∞
𝑥 𝑛 𝑒−𝑗𝜔𝑛
 The inverse discrete-time Fourier transform (IDTFT) of X(e jω ) is given
by
𝑥 𝑛 ≜ ℱ−1 𝑒 𝑗𝜔 =
1
2𝜋 −𝜋
𝜋
𝑋 𝑒 𝑗𝜔 𝑒 𝑗𝜔𝑛 𝑑𝜔
 The operator ℱ . transforms a discrete signal x(n) into a complex-
valued continuous function 𝑋 𝑒𝑗𝜔 of real variable 𝜔, called a digital
frequency, which is measured in radians/sample.
3

EXAMPLE #1
 Determine the discrete-time Fourier transform of 𝑥 𝑛 = 0.5 𝑛
𝑢(𝑛).
 Solution:
 The sequence x(n) is absolutely summable; therefore its discrete-time
Fourier transform exists.
𝑋 𝑒 𝑗𝜔 =
𝑛=−∞
∞
𝑥 𝑛 𝑒−𝑗𝜔𝑛 =
𝑛=0
∞
0.5 𝑛 𝑒−𝑗𝜔𝑛
𝑛=0
∞
0.5𝑒−𝑗𝜔 𝑛
=
1
1 − 0.5𝑒−𝑗𝜔
=
𝑒 𝑗𝜔
𝑒 𝑗𝜔 − 0.5
4

EXAMPLE #2
 Determine the discrete-time Fourier transform of the following ﬁnite-
duration sequence:
𝑥(𝑛) = { 1, 2, 3, 4, 5 }
 Solution:
𝑛=−∞
∞
𝑥 𝑛 𝑒−𝑗𝜔𝑛 = 𝑒 𝑗𝜔 + 2 + 3𝑒−𝑗𝜔 + 4𝑒−2𝑗𝜔+ . . +5𝑒−5𝑗𝜔
5
↑

DTFT Spectrum Properties
1. Periodicity:
 The discrete-time Fourier transform 𝑋 𝑒 𝑗𝜔
is periodic in ω with period 2π.
𝑋 𝑒 𝑗𝜔 = 𝑋 𝑒 𝑗[𝜔+2𝜋
 Implication: We need only one period of 𝑋 𝑒 𝑗𝜔
(i.e., 𝜔 ∈ [0, 2𝜋], 𝑜𝑟 [− 𝜋, 𝜋],
etc.) for analysis and not the whole domain −∞ < 𝜔 < ∞ .
 2. Symmetry:
 For real-valued x(n), X(e jω ) is conjugate symmetric.
𝑅𝑒 𝑋 𝑒−𝑗𝜔 = 𝑅𝑒 𝑋 𝑒 𝑗𝜔 (𝑒𝑣𝑒𝑛 𝑆𝑦𝑚𝑚𝑒𝑡𝑟𝑦)
𝐼𝑚 𝑋 𝑒−𝑗𝜔
= −𝐼𝑚 𝑋 𝑒 𝑗𝜔
(𝑂𝑑𝑑 𝑆𝑦𝑚𝑚𝑒𝑡𝑟𝑦)
𝑋 𝑒−𝑗𝜔
= 𝑋 𝑒 𝑗𝜔
(𝑒𝑣𝑒𝑛 𝑆𝑦𝑚𝑚𝑒𝑡𝑟𝑦)
∠𝑋 𝑒−𝑗𝜔
= −∠𝑋 𝑒 𝑗𝜔
(𝑂𝑑𝑑 𝑆𝑦𝑚𝑚𝑒𝑡𝑟𝑦)
 Implication: To plot 𝑋 𝑒 𝑗𝜔 , we now need to consider only a half period of
𝑋 𝑒 𝑗𝜔
. Generally, in practice this period is chosen to be 𝜔 ∈ 0, 𝜋 .
6

DTFT - MATLAB Implementation
 If 𝑥(𝑛) is of inﬁnite duration, then MATLAB cannot be used directly to
compute 𝑋 𝑒𝑗𝜔
from 𝑥(𝑛). However, we can use it to evaluate the
expression 𝑋 𝑒𝑗𝜔 over [0, 𝜋] frequencies and then plot its magnitude
and angle (or real and imaginary parts).
 EXAMPLE #3:
 Determine the discrete-time Fourier transform of 𝑥 𝑛 = 0.5 𝑛
𝑢(𝑛) .
Evaluate 𝑋 𝑒 𝑗𝜔 at 501 equispaced points between [0, π] and plot its
magnitude, angle, real, and imaginary parts.
7

EXAMPLE #3- Solution
 The discrete-time Fourier transform is.
𝑛=−∞
∞
𝑥 𝑛 𝑒−𝑗𝜔𝑛 =
𝑛=0
∞
0.5 𝑛 𝑒−𝑗𝜔𝑛
𝑛=0
∞
0.5𝑒−𝑗𝜔 𝑛
=
1
1 − 0.5𝑒−𝑗𝜔
=
𝑒 𝑗𝜔
𝑒 𝑗𝜔 − 0.5
 MATLAB script:
8

EXAMPLE #3- Solution (Cont.)
9

DTFT – Matrix Form
 If x(n) is of ﬁnite duration, then MATLAB can be used to compute 𝑋 𝑒 𝑗𝜔
numerically at any frequency ω.
 we evaluate X(e jω ) at equispaced frequencies between [0, π],
 then DTFT can be implemented as a matrix-vector multiplication operation.
 To understand this, let us assume that the sequence 𝑥(𝑛) has 𝑁 samples
between 𝑛1 ≤ 𝑛 ≤ 𝑛 𝑁 (i.e., not necessarily between [0, 𝑁 − 1]) and
that we want to evaluate 𝑋 𝑒 𝑗𝜔
at
𝜔 𝐾 ≜
𝜋
𝑀
𝑘, 𝑘 = 0, 1, . . . , 𝑀
 which are (M + 1) equispaced frequencies between [0, π]. Then (3.1)
can be written as
𝑋 𝑒 𝑗𝜔 𝐾 =
ℓ=1
𝑁
𝑥 𝑛ℓ 𝑒
−𝑗
𝜋
𝑀 𝐾𝑛ℓ
, 𝑘 = 0, 1, . . . , 𝑀
10

DTFT – Matrix Form (Cont.)
 When {𝑥 𝑛ℓ } and {𝑋 𝑒𝑗𝜔 𝐾 } are arranged as 𝑐𝑜𝑙𝑢𝑚𝑛 vectors 𝑥 and 𝑋, respectively, we
have
𝑿 = 𝑾𝒙
 where 𝑾 is an (𝑀 + 1) × 𝑁 matrix given by
𝑾 ≜ 𝑒
−𝑗
𝜋
𝑀 𝐾𝑛ℓ
; 𝑛1 ≤ 𝑛 ≤ 𝑛 𝑁, 𝐾 = 0,1, … , 𝑀
 In addition, if we arrange { k } and { n } as row vectors k and n respectively, then
𝑾 = exp(−𝑗
𝜋
𝑀
𝑲 𝑇 𝒏)
 In MATLAB we represent sequences and indices as row vectors; therefore taking the
transpose of 𝑿 = 𝑾𝒙, we obtain
𝑿 𝑇 = 𝒙 𝑇 exp(−𝑗
𝜋
𝑀
𝑲 𝑇 𝒏)
 Note that 𝐧 𝑇
𝐊 is an 𝑁 × (𝑀 + 1) matrix. Now we can be implemented in MATLAB as
follows.
11

Example #4
 Numerically compute the discrete-time Fourier transform of the sequence
𝑥(𝑛) given in Example #2 at 501 equispaced frequencies between [0, 𝜋].
 Solution:
 MATLAB script:
12

Example #5
 Let 𝑥(𝑛) = (0.9 exp(𝑗𝜋/3)) 𝑛 , 0 ≤ n ≤ 10. Determine 𝑋 𝑒 𝑗𝜔 and
investigate its periodicity.
 Solution:
 MATLAB script:
13

Example #5 – Solution (Cont.)
 However, we will evaluate and plot it at 401 frequencies over two periods
between [ − 2π, 2π] to observe its periodicity.
14

THE PROPERTIES OF THE DTFT
 DSP Lecture Notes No. 5
16

The Frequency Domain Representation Of LTI Systems
 the Fourier transform representation is the most useful signal
representation for LTI systems.
 Response To A Complex Exponential 𝑒 𝑗𝜔0 𝑛
 RESPONSE TO SINUSOIDAL SEQUENCES
17

RESPONSE TO ARBITRARY SEQUENCES
 Finally, we can be generalized to arbitrary absolutely summable
sequences. Let 𝑋 𝑒 𝑗𝜔 = ℱ[𝑥(𝑛)] and Y 𝑒 𝑗𝜔 = ℱ[𝑦(𝑛)]; then using the
convolution property, we have
Y 𝑒 𝑗𝜔
= H 𝑒 𝑗𝜔
X 𝑒 𝑗𝜔
 Therefore an LTI system can be represented in the frequency domain by
18

Example #13
 Determine the frequency response H 𝑒 𝑗𝜔 of a system characterized by
h(𝑛) = (0.9) 𝑛 𝑢(𝑛). Plot the magnitude and the phase responses.
 Solution:
𝐻 𝑒 𝑗𝜔 =
𝑛=−∞
∞
ℎ 𝑛 𝑒−𝑗𝜔𝑛 =
𝑛=0
∞
0.9 𝑛 𝑒−𝑗𝜔𝑛 =
1
1 − 0.9𝑒−𝑗𝜔
 Hence, the magnitude and the phase are
19

Example #13 – Solution (Cont.)
 MATLAB Script:
20

Example #14
 Let an input to the system in Example #13 be x n = 0.1𝑢(𝑛). Determine
the steady-state response y ss (n).
 Solution:
 Since the input is not absolutely summable, the discrete-time Fourier
transform is not particularly useful in computing the complete response.
However, it can be used to compute the steady-state response. In the steady
state (i.e., 𝑛 → ∞ ), the input is a constant sequence (or a sinusoid with
𝜔0 = 𝜃0 = 0). Then the output is
𝑦𝑠𝑠(𝑛) = 0.1 × H 𝑒 𝑗0
= 0.1 × 10 = 1
 where the gain of the system at ω = 0 (also called the DC gain) is H 𝑒𝑗0 = 10.
21

Frequency Response Function From Difference Equations
 When an LTI system is represented by the diﬀerence equation
𝑦 𝑛 +
ℓ=1
𝑁
𝑎ℓ 𝑦 𝑛 − ℓ =
𝑚=0
𝑀
𝑏 𝑚 𝑥 𝑛 − 𝑚
 then to evaluate its frequency response, we would need the impulse
response ℎ(𝑛). However, we can easily obtain H 𝑒 𝑗𝜔
. We know that
when x(n) = 𝑒 𝑗𝜔𝑛, then y(n) must be H 𝑒 𝑗𝜔 𝑒 𝑗𝜔n. Then we have
22

EXAMPLE #15
 An LTI system is speciﬁed by the diﬀerence equation
𝑦(𝑛) = 0.8𝑦(𝑛 − 1) + 𝑥(𝑛)
a. Determine H 𝑒 𝑗𝜔 .
b. Calculate and plot the steady-state response 𝑦𝑠𝑠(𝑛) to 𝑥(𝑛) = cos(0.05𝜋𝑛)𝑢(𝑛)
23

EXAMPLE #15 - Solution
 Rewrite the diﬀerence equation as 𝑦(𝑛) − 0.8𝑦(𝑛 − 1) = 𝑥(𝑛).
 a. The H 𝑒 𝑗𝜔
is
𝐻 𝑒 𝑗𝜔 =
1
1 − 0.8𝑒−𝑗𝜔
 b. In the steady state the input is 𝑥(𝑛) = cos(0.05𝜋𝑛) with frequency 𝜔0
= 0.05𝜋 and 𝜃0 = 0◦. The response of the system is
𝐻 𝑒 𝑗0.05𝜋 =
1
1 − 0.8𝑒−𝑗0.05𝜋
= 4.0928𝑒− 𝑗0.5377
 Therefore
𝑦𝑠𝑠(𝑛) = 4.0928 cos(0.05𝜋𝑛 − 0.5377) = 4.0928 cos[0.05𝜋(𝑛 − 3.42)]
 This means that at the output the sinusoid is scaled by 4.0928 and shifted by
3.42 samples. This can be veriﬁed using MATLAB.
24

EXAMPLE #15 – Solution (Cont.)
 The MATLAB Script
25

 we note that the
amplitude of y ss (n) is
approximately 4. To
determine the shift in the
output sinusoid, we can
compare zero crossings of
the input and the output.
26

EXAMPLE #16
 A 3rd -order lowpass filter is described by the difference equation
𝑦(𝑛) = 0.0181𝑥(𝑛) + 0.0543𝑥(𝑛 − 1) + 0.0543𝑥(𝑛 − 2) + 0.0181𝑥(𝑛 − 3)
+1.76𝑦(𝑛 − 1) − 1.1829𝑦(𝑛 − 2) + 0.2781𝑦(𝑛 − 3)
 Plot the magnitude and the phase response of this filter, and verify that it is a
lowpass filter.
27

 We will implement this procedure in MATLAB and then plot the ﬁlter
responses.
28

 From the plots in the Figure we see that the ﬁlter is indeed a lowpass
ﬁlter.
29

SAMPLING AND RECONSTRUCTION
OF ANALOG SIGNALS
30

Sampling And Reconstruction Of Analog Signals
 In many applications—for example, in digital communications—real-
world analog signals are converted into discrete signals using sampling
and quantization operations (collectively called analog-to-digital
conversion, or ADC). These discrete signals are processed by digital
signal processors, and the processed signals are converted into analog
signals using a reconstruction operation (called digital-to-analog
conversion or DAC).
 Using Fourier analysis, we can describe the sampling operation from the
frequency-domain viewpoint, analyze its effects, and then address the
reconstruction operation. We will also assume that the number of
quantization levels is sufficiently large that the effect of quantization on
discrete signals is negligible.
 We will study the effects of quantization later.
31

Sampling And Reconstruction Of Analog Signals
 DSP Lecture Notes No. 2
32

Review
 Band-limited Signal:
 A signal is band-limited if there exists a ﬁnite radian frequency Ω0such that
𝑋 𝑎(𝑗Ω) is zero for |Ω| > Ω0. The frequency 𝐹0 = Ω0/2𝜋 is called the signal
bandwidth in Hz.
 Sampling Principle:
 A band-limited signal 𝑥 𝑎(𝑡) with bandwidth 𝐹0 can be reconstructed from its
sample values 𝑥(𝑛) = 𝑥 𝑎(𝑛𝑇𝑠) if the sampling frequency 𝐹𝑠 = 1/𝑇𝑠 is greater
than twice the bandwidth 𝐹0 of 𝑥 𝑎(𝑡).
𝐹𝑠 > 2𝐹0
 Otherwise aliasing would result in x(n). The sampling rate of 2𝐹0for an
analog band-limited signal is called the Nyquist rate.
33

EXAMPLE #19
 Let 𝑥 𝑎 𝑡 = 𝑒−1000 𝑡
.
a. Determine and plot its Fourier transform.
b. Sample 𝑥 𝑎(𝑡) at 𝐹𝑠 = 5000 𝑠𝑎𝑚𝑝𝑙𝑒/sec to obtain 𝑥1(𝑛). Determine and
plot 𝑋1 𝑒 𝑗𝜔 .
c. Sample 𝑥 𝑎(𝑡) at 𝐹𝑠 = 1000 𝑠𝑎𝑚𝑝𝑙𝑒/sec to obtain 𝑥2(𝑛). Determine and
plot 𝑋2 𝑒 𝑗𝜔 .
35

a. The Fourier transform is
 which is a real-valued function since 𝑥 𝑎 𝑡 is a real and even signal. To evaluate
𝑋 𝑎(𝑗Ω) numerically, we have to first approximate 𝑥 𝑎 𝑡 by a finite-duration grid
sequence 𝑥 𝐺 𝑚 .
 Using the approximation 𝑒−5 ≈ 0, we note that 𝑥 𝑎 𝑡 can be approximated by a
finite-duration signal over − 0.005 ≤ 𝑡 ≤ 0.005 (or equivalently, over [− 5, 5]
msec).
 Then, 𝑋 𝑎(𝑗Ω) ≈ 0 for Ω ≥ 2𝜋(2000). Hence choosing
36

37

b. Since the bandwidth of 𝑥 𝑎(𝑡) is 2𝐾𝐻𝑧, the Nyquist rate is 4000 𝑠𝑎𝑚𝑝𝑙𝑒/sec,
which is less than the given 𝐹 𝑠. Therefore aliasing will be (almost) nonexistent.
 MATLAB script:
38

39

c. Here 𝐹𝑠 = 1000 < 4000.
Hence there will be a
considerable amount of
aliasing. This is evident
from the shown Figure, in
which the shape of
𝑋2 𝑒 𝑗𝜔
is diﬀerent from
that of 𝑋 𝑎(𝑗Ω) and can be
seen to be a result of
adding overlapping
replicas of 𝑋 𝑎(𝑗Ω).
40

Reconstruction
 From the sampling theorem and the preceding examples, it is clear that
if we sample band-limited 𝑥 𝑎(𝑡) above its Nyquist rate, then we can
reconstruct 𝑥 𝑎(𝑡) from its samples 𝑥(𝑛). This reconstruction can be
thought of as a 2-step process:
1) First the samples are converted into a weighted impulse train.
2) Then the impulse train is ﬁltered through an ideal analog lowpass ﬁlter
band-limited to the [ −𝐹𝑠/2, 𝐹𝑠/2] band.
41

Reconstruction
 This two-step procedure can be described mathematically using an
interpolating formula
 where 𝑠𝑖𝑛𝑐(𝑥) =
sin 𝜋𝑥
𝜋𝑥
is an interpolating function. The physical
interpretation of the above reconstruction formula is given in the shown
Figure, from which we observe that this ideal interpolation is not practically
feasible because the entire system is noncausal and hence not realizable.
42

Practical D/A converters
 In practice we need a different approach than reconstruction formula.
The two-step procedure is still feasible, but now we replace the ideal
lowpass filter by a practical analog lowpass filter.
 Another interpretation of reconstruction formula is that it is an infinite-
order interpolation. We want finite-order (and in fact low-order)
interpolations. There are several approaches to do this.
1) Zero-order-hold (ZOH) interpolation
2) 1st-order-hold (FOH) interpolation
3) Cubic spline interpolation
44

Zero-order-hold (ZOH) interpolation
 In this interpolation a given sample value is held for the sample interval until
the next sample is received.
 which can be obtained by filtering the impulse train through an interpolating
filter of the form
 which is a rectangular pulse. The resulting signal is a piecewise-constant
(staircase) waveform which requires an appropriately designed analog
postfilter for accurate waveform reconstruction.
45

1st -order-hold (FOH) interpolation
 In this case the adjacent samples are joined by straight lines. This can
be obtained by ﬁltering the impulse train through
 Once again, an appropriately designed analog postﬁlter is required for
accurate reconstruction. These interpolations can be extended to
higher orders.
46

Cubic spline interpolation
 This approach uses spline interpolants for a smoother, but not
necessarily more accurate, estimate of the analog signals between
samples. Hence this interpolation does not require an analog postﬁlter.
The smoother reconstruction is obtained by using a set of piecewise
continuous third-order polynomials called 𝑐𝑢𝑏𝑖𝑐 𝑠𝑝𝑙𝑖𝑛𝑒𝑠, given by
 where {𝛼𝑖(𝑛), 0 ≤ 𝑖 ≤ 3 } are the polynomial coeﬃcients, which are
determined by using least-squares analysis on the sample values.
(Strictly speaking, this is not a causal operation but is a convenient one
in MATLAB.)
47

EXAMPLE #23
 Plot the reconstructed signal from the samples 𝑥 𝑎(𝑡) in Example #19
using the ZOH and the FOH interpolations. Comment on the plots.
 Solution:
48

 The plots are shown in the
Figure, from which we observe
that the ZOH reconstruction is a
crude one and that the further
processing of analog signal is
necessary.
 The FOH reconstruction appears
to be a good one, but a careful
observation near t = 0 reveals
that the peak of the signal is not
correctly reproduced.
 In general, if the sampling
frequency is much higher than
the Nyquist rate, then the FOH
interpolation provides an
acceptable reconstruction.
49

DSP_FOEHU - MATLAB 02 - The Discrete-time Fourier Analysis

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