233_Sample-Chapter (1).pdf

2
Fourier Series and Fourier Transform
2.1 INTRODUCTION
Fourier series is used to get frequency spectrum of a time-domain signal, when signal is a periodic function of
time. We have seen that the sum of two sinusoids is periodic provided their frequencies are integer multiple
of a fundamental frequency, w0.
2.2 TRIGONOMETRIC FOURIER SERIES
Consider a signal x(t), a sum of sine and cosine function whose frequencies are integral multiple of w0
x(t) = a0 +a1 cos (w0t)+a2 cos (2w0t)+···
b1 sin (w0t)+b2 sin (2w0t)+···
x(t) = a0 +
∞
∑
n=1
(an cos (nw0t)+bn sin (nw0t)) (1)
a0, a1,..., b1, b2,... are constants and w0 is the fundamental frequency.
Evaluation of Fourier Coefficients
To evaluate a0 we shall integrate both sides of eqn. (1) over one period (t0,t0 + T) of x(t) at an arbitrary
time t0
t0+T
Z
t0
x(t)dt =
t0+T
Z
t0
a0dt +
∞
∑
n=1
an
t0+T
Z
t0
cos (nw0t)dt +
∞
∑
n=1
bn
t0+T
Z
t0
sin (nw0t)dt
Since
R t0+T
t0
cos (nw0dt) = 0
t0+T
Z
t0
sin (nw0dt) = 0
a0 =
1
T
t0+T
Z
t0
x(t)dt (2)
To evaluate an and bn, we use the following result:
t0+T
Z
t0
cos (nw0t)cos (mw0t)dt =

0 m 6= n
T/2 m = n 6= 0
94

96 • Basic System Analysis
Multiply eqn. (1) by sin (mw0t) and integrate over one period
t0+T
Z
t0
x(t)sin (mw0t)dt = a0
t0+T
Z
t0
sin (mw0t)dt +
∞
∑
n=1
an
t0+T
Z
t0
cos (nw0t)sin (mw0t)dt +
∞
∑
n=1
bn
t0+T
Z
t0
sin (mw0t)sin (nw0t)dt
bn =
2
T
t0+T
Z
t0
x(t)sin (nw0t)dt (4)
Example 1:
−3 −2 −1
− −1.0
1.0
1 2 3
0
−
Fig. 2.1.
T → −1 to 1 T = 2 w0 = π x(t) = t,−1 t 1
a0 =
1
2
1
Z
−1
t dt =
1
4
(1−1) = 0
an = 0
bn =
1
Z
−1
t sin πntdt =

−t cos πnt
nπ
−
cos πnt
nπ
1
−1
=
−1
nπ
[t cos πnt +cos πnt]1
−1 = −
1
nπ
[2cos π+cos π−cos π]
bn =
−2
nπ
cos nπ =
2
π

−(−1)n
n

b1 b2 b3 b4 b5 b6
2
π
−2
2π
2
3π
−2
4π
2
5π
−2···
6π
x(t) =
∞
∑
n=1
2
π

−(−1)n
n

sin nπt
=
2
π

sin πt −
1
2
sin 2πt +
1
3
sin 3πt −
1
4
sin 4πt +···

Fourier Series and Fourier Transform • 97
Example 2:
−2π
1.0
2π 4π 6π
0 t
Fig. 2.2.
x(t) =
t
2π
T = 2π w0 =
2π
T
= 1
a0 =
1
T
2π
Z
0
x(t)dt =
1
4π2

1
2
t2
2π
0
=
1
2
an =
2
4π2
2π
Z
0
t cos ntdt =
1
2π2

t sin t
n
+
sin nt
n
2π
0
=
1
2π2

2πsin 2nπ
n
+
sin 2nπ
n

= 0
bn =
2
4π2
2π
Z
0
t sin ntdt =
−1
2π2
ht cos nt
n
+
cos nt
n
i2π
0
=
−1
2π2

2πcos 2nπ
n
+
cos 2nπ
n
−
1
n

bn =
−1
nπ
x(t) =
1
2
+
∞
∑
n=1

−1
nπ

sin nt =
1
2
+
1
π
∞
∑
n=1
1
n
cos (nt +π/2)
=
1
2
−
1
π

sin t +
sin 2t
2
+
sin 3t
3
+···

Example 3:
−T/2 −T/4 T/4
A x(t)
t
T/2
Fig. 2.3. Rectangular waveform

Figure shows a periodic rectangular waveform which is symmetrical to the vertical axis. Obtain its F.S.
representation.
x(t) = a0 +
∞
∑
n=1
(an cos nw0t +bn sin nw0t)
x(t) = a0 +
∞
∑
n=1
an cos (nw0t) bn = 0
x(t) = 0 for
−T
2
t
−T
4
+A for
−T
4
t
T
4
0 for
T
4
t
T
2
a0 =
1
T
T/4
Z
−T/4
Adt =
A
2
an =
2
T
T/4
Z
−T/4
Acos (nw0t)dt =
2A
Tnw0

sin nw0
T
4
+sin nw0
T
4

an =
4A
2πn
sin
nπ
2

=
2A
πn
sin
nπ
2

w0 =
2π
T
a1 =
4A
2π
=
2A
π
a2 = 0
a3 =
2A
3π
sin
3π
2
=
2A
3π
(−1) =
−2A
3π
x(t) =
A
2
+
2A
π

cos w0t −
1
3
cos 3w0t +
1
5
cos 5w0t +···

Example 4: Find the trigonometric Fourier series for the periodic signal x(t).
1.0
0 1
−1
−3
−5
−7
−9
x(t)
3 5 7 9 11 t
T
Fig. 2.4.

SOLUTION:
bn = 0 x(t) =
(
1 −1 t 1
−1 1 t 3
a0 =
1
T
3
Z
−1
x(t)dt =
1
T


1
Z
−1
dt +
3
Z
t
(−1)dt

 T = 4
=
1
T
[2−2] = 0 ∴ w0 =
2π
T
=
2π
4
=
π
2
an =
2
T


1
Z
−1
cos (nw0t)dt +
3
Z
1
cos (nw0t)dt


=
2
2πn
h
2sin
πn
2
i
−

sin
3nπ
2
−sin
nπ
2

=
1
nπ

3sin
nπ
2
−sin
3nπ
2

sin
3nπ
2
= sin

π+
nπ
2

= −sin
nπ
2
an =
4
nπ
sin
nπ
2

an =









0 n = even
4
nπ
n = 1,5,9,13
−4
nπ
n = 3,7,11,15
x(t) =
4
π
cos
π
2
t

−
4
3π
cos

3π
2
t

+
4
5π
cos

5π
2
t

−
4
7π
cos

7π
2
t

+···
x(t) =
4
π

cos
π
2
t

−
1
3
cos

3π
2
t

+
1
5
cos

5π
2
t

···

Example 5: Find the F.S.C. for the continuous-time periodic signal
x(t) = 1.5 0 ≤ t 1
= −1.5 1 ≤ t 2
with fundamental freq. w0 = π
−1.5
1.5
0 1
x(t)
3
2 5
4
Fig. 2.5.

SOLUTION:
T =
2π
w0
= 2, w0 = π
a0 = an = 0
bn =
1
Z
0
1.5sinnπtdt −
2
Z
1
1.5sinnπtdt
=
1.5
nπ
n
[−cosnπ+1]+[cos2nπ−cosnπ]
o
bn =
3
nπ
[1−cosnπ]
x(t) =
3
π

2sinπt +
2
3
sin3πt +
2
5
sin5πt +···

6
π

sinπt +
1
3
sin3πt +
1
5
sin5πt +···

C0 =
1
2


1
Z
0
1.5dt −1.5
2
Z
1
dt

 = 0
OR
By using complex exponential Fourier series
Cn =
1
2


1
Z
0
1.5e−jnπt
dt −1.5
2
Z
1
e−jnπt
dt


Cn =
3
−4jnπ


e−jnπt
1
−e−jnπt
0
2
1



=
−3
4jnπ

e−jnπ
−1−e−j2nπ
+e−jnπ

=
3
2jnπ

1−e−jnπ

=
3
2jnπ
[1−cosnπ]
x(t) =
∞
∑
n=−∞
Cne−jnπt
∞
∑
n=−∞
3
2jnπ

1−e−jnπ

ejnπt
=
∞
∑
n=−∞
3
2jnπ

ejnπt
−ejnπt
cosπn

for n = 1
=
A
2π
π
Z
0
sin t sin tdt =
A
2π
π
Z
0
(1−cos2t)dt
=
A
2π
[π] =
A
2
When n is even
=
A
2π

2
n+1
−
2
1−n

=
2A
π(1−n2)
Example 7:
−2
−3 −1
−2
2
1 2
T
3
0
x(t)
a
b
t
Fig. 2.7.
SOLUTION:
T = 2 w0 =
2π
T
= π
x(t) =

2t −1 t 1
0
Point (a) (−1,−2)
Point (b) (1,2)
y−(−2) =
2−(−2)
1−(−1)
(x−(−1))
y+2 =
4
2
(x+1)
y+2 = 2x+2
y = 2x
x(t) = 2t
Since function is an odd function
an = 0, a0 =
1
T
1
Z
−1
2tdt =
1
2
×0 = 0
bn =
2
T
1
Z
−1
t sin (nπt)dt =
2
T

−t cosnπt
nπ
1
−1
+
1
n2π2
cosnπt
1
−1



2.3 CONVERGENCE OF FOURIER SERIES – DIRICHLET CONDITIONS
Existence of Fourier Series: The conditions under which a periodic signal can be represented by an F.S.
are known as Dirichlet conditions. F.P. → Fundamental Period
(1) The function x(t) has only a finite number of maxima and minima, if any within the F.P.
(2) The function x(t) has only a finite number of discontinuities, if any within the F.P.
(3) The function x(t) is absolutely integrable over one period, that is
T
Z
0
x(t) dt ∞
2.4 PROPERTIES OF CONTINUOUS FOURIER SERIES
(1) Linearity: If x1(t) and x2(t) are two periodic signals with period T with F.S.C. Cn and Dn then F.C. of
linear combination of x1(t) and x2(t) are given by
FS[Ax1(t)+Bx2(t)] = ACn +BDn
Proof: If z(t) = Ax1(t)+Bx2(t)
an =
1
T
t0+T
Z
t0
[Ax1(t)+Bx2(t)]e−jnw0t
=
A
T
Z
T
x1(t)e−jnw0t
dt +
B
T
Z
T
x2(t)e−jnw0t
dt
an = ACn +BDn
(2) Time shifting: If the F.S.C. of x(t) are Cn then the F.C. of the shifted signal x(t −t0) are
FS[x(t −t0)] = e−jnw0 t0Cn
Let t −t0 = τ
dt = dτ
Bn =
1
T
Z
T
x(t −t0)e−jnw0t
dt
=
1
T
Z
T
x(τ)e−jnw0(t0+τ)
dτ =
1
T
Z
T
x(τ)e−jnw0τ
dτ·e−jnw0t0
Bn = e−jnw
0
t
0 ·Cn
(3) Time reversal: FS[x(−t)] = C−n
Bn =
1
T
Z
T
x(−t)e−jnw0t
dt =
1
T
Z
T
x(−t)e−j(−n)w0T
dt
−t = τ
dt = −dτ
=
1
T
Z
−T
x(τ)e−j(−n)w0τ
dτ = C−n

Example 8: Compute the exponential series of the following signal.
−5 −4 −3 −2 −1 0 1
T
1.0
2.0
x(t)
2 3 4 5 6 t
Fig. 2.8.
SOLUTION:
T = 4 w0 =
π
2
C0 =
1
T
T
Z
0
x(t)dt =
1
4


1
Z
0
2dt +
2
Z
1
dt

 =
3
4
Cn =
1
4


1
Z
0
2e
−jn
πt
2 dt +
2
Z
1
e−jn πt
2 dt


=
1
4



−4
jnπ

e
−jn
π
2 −1

−
2
jnπ
h
e−jnπ
−e−jn π
2
i



=
−1
2jnπ

2e
−jnπ
2 −2+e−jnπ
−e
−jn
π
2

 =
−1
2jnπ

e
−jn
π
2 +e
−jn
π
2 −2


= −
1
jnπ

1−
1
2
(−1)n
−
1
2
e−jn π
2

x(t) =
3
4
+
∞
∑
n=−∞
1
jnπ

ejn π
2 −
1
2
(−1)n
ejn π
2 −
1
2

Example 9:
−2 −1
a b
0 1
1.0
x(t)
2 3 4 5
↓
↓
6 7 t
Fig. 2.9.

SOLUTION:
T = 5 w0 =
2π
5
x(t) =



t +2 −2 t −1
1.0 −1 t 1
2−t 1 t 2
(a) (−2,0)(−1,1)
(y−1) =
−1
−1
(x+1)
y = t +2
(b) (1,1)(2,0)
y−0 =
1
−1
(x−2)
y = −x+2 = −t +2
C0 =
1
5


−1
Z
−2
(t +2)dt +
1
Z
−1
dt +
2
Z
1
(2−t)dt


C0 =
3
5
Cn =
1
5







−1
Z
−2
(t +2)e−j 2nπ
5 dt
| {z }
A
+
1
Z
−1
e−j 2nπ
5 dt
| {z }
B
+
2
Z
1
(2−t)e−j 2nπ
5 dt
| {z }
C







A =
−1
Z
−2
e−j 2nπ
5 t
dt +
−1
Z
−2
2e−j 2nπ
5 t
dt
A = −
1
jφ



te−jφ
−1
Z
−2



+
1
φ2
e−jφ
−1
Z
−2
+
2
−jφ
ej 2nπ
5
−1
Z
−2
=
5
j2nπ

−ej 2nπ
5 +2ej 4nπ
5

+
25
4n2π2

ej 2nπ
5 −ej 4nπ
5

−
10
2nπ j
A =
5
j2nπ

−ej 2nπ
5 +4ej 4nπ
5

+
25
4n2π2

ej 2nπ
5 −ej 4nπ
5

B =
ej 2nπ
5 −e−j 2nπ
5
j2nπ
5
=
5
j2nπ

ej 2nπ
5 −e−j 2nπ
5

C =
−10
j2nπ

e−j 4nπ
5 −e−j 2nπ
5

+
10
j2nπ
e−j 4nπ
5 −
5
j2nπ
e−j 2nπ
5 −
25
4n2π2
e−j 4nπ
5 +
25
4n2π2
ej n2π
5

Cn =
1
5

25
n24π2

e
j2nπ
5 −e
j4nπ
5

−
25
4n2π2

e− j4nπ
5 −e− j2nπ
5

Cn =
5
2n2π2

cos

2πn
5

−cos

4πn
5

Example 10: For the continuous-time periodic signal
x(t) = 2+cos

2π
3
t

+4sin

5π
3
t

Determine the fundamental frequency w0 and the Fourier series coefficients Cn such that
x(t) =
∞
∑
n=−∞
Cnejnw0t
SOLUTION:
Given
x(t) = 2+cos

2π
3
t

+4sin

5π
3
t

The time period of the signal cos 2π
3 t

is
T1 =
2π
w1
=
2π
2π
3
= 3sec
The time period of the signal sin 5π
2 t

is
T2 = 2
π
w2
=
2π
5π
3
=
6
5
sec
T1
T2
=
3
6
5
=
5
2
ratio of two integers, rational number, hence periodic.
2T1 = 5T2
The fundamental period of the signal x(t) is
T = 2T1 = 5T2 = 6sec
and the fundamental frequency is
w0 =
2π
T
=
2π
6
=
π
3
x(t) = 2+cos

2π
3
t

+4sin

5π
3
t

= 2+cos (2w0t)+4sin (5w0t)
= 2+
ej2w0t +e−j2w0t

2
+
4 ej5w0t −e−j5w0t

2j
= 2+0.5 ej2w0t
+e−j2w0t

−2j

ej5w0t
−e−j5w0t

x(t) = 2je+j(−5)w0t
+0.5e+j(−2)w0t
+2+0.5e+j2w0t
−2je+j5w0t

an =
2
T
π
Z
−π
x(t)cosntdt =
4
T
0
Z
−π

2t
π
+1

cosnt dt
=
4
2π



2t
nπ
sint +
sinnt
n
−
0
Z
−π
2
π
sinnt dt



=
1
π



2t
π
sinnt +sinnt +
2
n2π
cosnt
0
Z
−π



=
2
π
(
2
n2π
+
2
n2π
cosnt
)
=
4
n2π2
n
1−cosnπ
o
=
4
n2π2
(1−(−1)n
)
an =
(
0 n even 2, 4, 6, 8,···
8
n2π2 n odd 1, 3, 5, 7,···
2.5 FOURIER TRANSFORM
2.5.1 Definition
Let x(t) be a signal which is a function of time t. The Fourier transform of x(t) is given as
X (jw) =
∞
Z
−∞
x(t)e−jwt
dt (1)
Fourier transform or
X (i f) =
∞
Z
−∞
x(t)e−j2π ft
dt (2)
Since w = 2π f
Similarly, x(t) can be recovered from its Fourier transform X(jw) by using Inverse Fourier transform
x(t) =
1
2π
∞
Z
−∞
X(jw)ejwt
dw (3)
x(t) =
∞
Z
−∞
X(i f)ej2π ft
dt (4)
Fourier transform X(jw) is the complex function of frequency w. Therefore, it can be expressed in the
complex exponential form as follows:
X(jw) = |X(jw)|ej
|X(jw)
Here |X(jw)| is the amplitude spectrum of x(t) and |X( jw)
is phase spectrum.
For a real-valued signal
(1) Amplitude spectrum is symmetric about vertical axis c (even function.)
(2) Phase spectrum is anti-symmetrical about vertical axis c (odd function.)

2.5.2 Existence of Fourier transform (Dirichlet’s condition)
The following conditions should be satisfied by the signal to obtain its F.T.
(1) The function x(t) should be single valued in any finite time interval T.
(2) The function x(t) should have at the most finite number of discontinuities in any finite time interval T.
(3) The function x(t) should have finite number of maxima and minima in any finite time interval T.
(4) The function x(t) should be absolutely integrable, i.e.
∞
Z
−∞
|x(t)|dt ∞
• These conditions are sufficient, but not necessary for the signal to be Fourier transformable.
• A physically realizable signal is always Fourier transformable. Thus, physical realizability is the
sufficient condition for the existence of F.T.
• All energy signals are Fourier transformable.
j
d
dw
X(jw) = FT(tx(t))
FT(tx(t)) = j
d
dw
X(jw)
Example 12: Obtain the F.T. of the signal e−atu(t) and plot its magnitude and phase spectrum.
SOLUTION:
x(t) = e−atu(t)
X(f) =
∞
Z
−∞
x(t)e−j2π ft
dt =
∞
Z
0
e−(a+j2π f)t
dt
X(f) =
1
a+ j2π f
To obtain the magnitude and phase spectrum:
|X(f)| =
a− j2π f
a2 +(2π f)2
=

a
a2 +4π2 f2

A− j

2π f
a2 +4π2 f2

B
|X(f)| =
p
A2 +B2 =
1
p
a2 +4π2 f2
=
1
√
a2 +w2
|X(f)| = tan−1

−2π f
a

= −tan−1
w
a

for a = 1, |X(f)| =
1
√
1+w2
,
|X(f)
= −tan−1
w
w 0 1 2 3 4 5 10 15 25 8
|X(w)| 1 .707 0.447 0.316 0.242 0.196 0.09 0.066 0.03 0
|X(w) 0 45◦ −63.4 −71.5 −75.9 −78.6 −84.2 −86.2 −87.7 −90◦

(ii) x(t) = e−a|t| =

e−at t 0
eat t 0
e−a|t|
t
Fig. 2.13. Graphical representation of e−a|t|
x(w) =
1
a+ jw
+
1
a− jw
=
2a
a2 +w2
for a = 1 X(w) =
2
1+w2
|X(w)| =
2
1+w2
|X(w)
= 0
w (in radians) −∞ −10 −5 −3 −2 −1 0 1 2 3 4 5 10 ∞
|X(w)| 0 0.019 0.0769 0.2 0.4 1 2 1 0.4 0.2 .1176 0.0769 0.019 0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
0 2
−2 1
−1 3 4 5 . . . .
. . . . 10
−10 −3
−4
−5 w
|X(w)|
Fig. 2.14. Magnitude plot
(iii) x(t) = e−a|t| sgn(t)
t
1.0
x(t) = e−a |t| sgn(t)
−1.0
Fig. 2.15. Graphical representation of e−a|t|sgn(t)

(ii) x(t) = 1
X(w) =
∞
Z
−∞
e−jwt
dt = ∞
This means Dirichlet condition is not satisfied. But its F.T. can be calculated with the help of duality
property.
δ(t)
FT
←→ 1
Duality property states that: x(t)
FT
←→ X(w) then
X(t)
FT
←→ 2πx(−w)
Here X(t) = 1, then x(−w) will be
x(t) = δ(t); X(w) = 1
then X(t) = 1; 1
FT
←→ 2πδ(−w)
We know that δ(w) will be an even function of w, since it is impulse function.
Hence, δ(−w) = δ(w). Then above equation becomes
1
FT
←→ 2πδ(−w)
Thus, if x(t) = 1, then X(w) = 2πδ(w)
(iii) x(t) = sgn(t) sgn(t) =

1 t 0
−1 t 0

t
1
sgn(t)
−1
0
Fig. 2.17. Graphical representation of sgn(t)
x(t) = 2u(t)−1
Differentiating both the sides
d
dt
x(t) = 2
d
dt
u(t) = 2δ(t)
Taking the F.T. of both sides
F

d
dt
x(t)

= 2F[δ(t)]
jwX(w) = 2
X(w) =
2
jw
X(w) =
∞
Z
0
e−jwt
dt −
0
Z
−∞
e−jwt
dt

(iv) x(t) = u(t)
sgn(t) = 2u(t)−1
2u(t) = 1+sgn(t)
Taking F.T. of both sides
2F[2u(t)] = F(1)+F[sgn(t)] = 2πδ(w)+
2
jw
2u(t)
FT
←→ 2πδ(w)+
2
jw
u(t)
FT
←→ πδ(w)+
1
jw
Properties of unit impulse:
(1)
∞
Z
−∞
x(t)δ(t) = x(0)
(2) x(t)δ(t −t0) = x(t0)δ(t −t0)
(3)
∞
Z
−∞
x(t)δ(t −t0)dt = x(t0)
(4) δ(at) = 1
|a| δ(t)
(5)
∞
Z
−∞
x(τ)δ(t −x)dt = x(t)
(6) δ(t) = d
dt u(t)
Example 15: Obtain the F.T. of a rectangular pulse shown in Fig. 2.18.
t
0
x(t)
−T/2
1
T/2
Fig. 2.18. Rectangular pulse
SOLUTION:
X(w) =
T
2
Z
−T
2
e−jwt
dt =
−1
jw
h
e−jw T
2 −ejw T
2
i
=
2
w
sin

wT
2

X(w) = T
sin πwT
2π

πwT
2π
= sinc

wT
2π

= T
sin πwT
2π

πwT
2π

Sampling function or interpolating function or filtering function denoted by Sa(x) or sinc(x) as shown in
figure.
sinc(x) =
sinπx
πx
(1) sinc(x) = 0 when x = ±nπ
(2) sinc(x) = 1 when x = 0 (using L’Hospital’s rule)
(3) sinc(x) is the product of an oscillating signal sin x of period 2π and a decreasing signal 1
x . Therefore,
sinc(x) is making sinusoidal of oscillations of period 2π with amplified decreasing continuously as 1
x .
1.0
sin c(x)
0 2π
−2π
−3π
−4π π
−π 4π
3π 5π x
Fig. 2.19. Sine function
sincx =
sinπx
πx
; sinc(0) =
0
0
= 1 L’Hospital rule
sinc(1) =
sinπ
π
= 0; sinc(−1) = 0
sinc(2) = 0; sinc(−2) = 0
sinc(1/4) = 0.9 sinc(−1/4) = 0.9
sinc(2/4) = .6366 sinc(−0.5) = .6366
sinc(3/4) = 0.3 sinc(−7.5) = .3
sinc(1.5) = −.2122 sinc(−1.5) = −.2122
sinc(2.5) = .1273 sinc(2.5) = .1273
.1
.2
.3
.4
.5
.6
.7
.8
.9
0 .5 1 2 2.5 3 3.5
.25 1.5
.75
−3 −2 −1 −.75 −.5 −.25
−3.5 −1.5
−2.5 t
Fig. 2.20. Sine function

Example 17: Obtain F.T. and spectrums of following signals:
(i) x(t) = cosw0t (ii) x(t) = sinw0t
SOLUTION:
(i) x(t) = cosw0t =
1
2
ejw0t
+
1
2
e−jw0t
1
FT
←→ 2πδ(w);
1
2
FT
←→ πδ(w)
Frequency shifting property states that ejβtx(t)
FT
←→ X(w−β)
1
2
ejw0t FT
←→ πδ(w−w0)
1
2
e−jw0t FT
←→ πδ(w+w0)
F [x(t)] = FT

1
2
ejw0t
+
1
2
e−jw0t

X(w) = π[δ(w−w0)+δ(w+w0)]
| X(w) |
−w0 w0 w
π
Fig. 2.22. Magnitude plot of cosw0t
(ii) x(t) = sinw0t
X(w) =
π
j
[δ(w−w0)−δ(w+w0)]
| X(w) |
−ω0
w0 w
π
π
Fig. 2.23. Magnitude plot of sinw0t

Example 18: Obtain the F.T. of
x(t) = te−at
u(t)
from property of Fourier transform FT[tx(t)] = j d
dw X(w)
FT

e−at

=
1
a+ jw
FT(te−at
) = j
d
dw

1
a+ jw

= j
(a+ jw) d
dw (1)−1 d
dw (a+ jw)
(a+ jw)2
=
1
(a+ jw)2
Inverse Fourier Transform: (IFT)
Example 19: Find the IFT of
2 jω+1
( jω+2)2
by partial fraction expansions
(i) X(ω) =
(ii) X(ω) = 1
(a+jω)2
by convolution property
(iii) X(w) = e−|w|
(iv) X(w) = e−2wu(w)
SOLUTION:
(i) X(w) =
A
jw+2
+
B
(jw+2)2
; 2jw+1 = A(jw+2)+B A = 2 2A+B = 1 B = −3
X(w) =
2
jw+2
−
3
(jw+2)2
x(t) = 2e−2t
u(t)−3te−2t
u(t)
(ii) X(w) =
1
(a+ jw)2
=
1
(a+ jw)(a+ jw)
= X1(w)X2(w)
X1(w) =
1
a+ jw
, X2(w) =
1
a+ jw
x1(t) = e−at
u(t), x2(t) = e−at
u(t)
Using convolution property
x(t) = x1(t)∗
x2(t)
x(t)
FT
←→ X(w)
x1(t)∗
x2(t)
FT
←→ X1(w)X2(w)
x(t) =
∞
Z
−∞
e−at
u(t)e−a(t−τ)
u(t −τ)dτ
(
u(τ) = 1 τ ≤ 0
u(t −τ) = 1 t ≤ τ
=
t
Z
0
e−at
dτ = te−at
u(t)

Example 20: Find the F.T. of the function
x(t −t0) = e−(t−t0)
u(t −t0)
SOLUTION:
If F[x(t)] = X(w)
then FT[x(t −t0)] = e−jwt0 X(w)
F

e−t
u(t)

=
1
1+ jw
F
h
e−(t−t0)
u(t −t0)
i
=
e−jwt0
1+ jw
Example 21: Find the F.T. of the function
x(t) = [u(t +1)−u(t −1)]cos2πt
SOLUTION:
FT(cos2πt) = FT

ej2πt +e−j2πt
2

FT[1] = 2πδ(w)
FT[ejw0t
] = 2πδ(w−w0)
F[cos2πt] = πδ(1w−2π)+πδ(w+2π) (1)
F[u(t +1)−u(t −1)] =
1
Z
−1
e−jwt
dt = −
1
jw
e−jw
−ejw

=
2sinw
w
(2)
F[x(t)] = F[{u(t +1)−u(t −1)}cos2πt]
x(t) is multiplication of (1) and (2), so by using multiplication property
x(t)y(t)
FT
←→
1
2π
X1(w)∗
Y1(w) =
1
2π
∞
Z
−∞
X(τ)Y(w−τ)dτ
X(w) =
1
2π


∞
Z
−∞
2sinτ
τ
πδ(w−2π−τ)+δ(w+2π−τ)

dτ
X(w) =
∞
Z
−∞
sinτ
τ
δ(w−2π−τ)dτ+
∞
Z
−∞
sinτ
τ
δ(w+2π−τ)dτ
Since
∞
Z
−∞
x(t)δ(t −t0)dt = x(t0)
X(w) = sin(w−2π)/(w−2π)+sin(w+2π)/(w+2π)

Example 22: Determine the Fourier transform of a triangular function as shown in figure.
x(t)
T
A
−T t
Fig. 2.24. Triangular pulse
SOLUTION:
x(t)
(T,0)
(0, A)
(−T,0) t
a b
→
→
Equation of line (a) is
x(t) = A
t
T
+1

Equation of line (b) is
x(t) = A

1−
t
T

Mathematically, we can write x(t) as
x(t) = A
t
T
+1

[u(t +T)−u(t)]+A

1−
t
T

[u(t)−u(t −T)]
x(t) =
A
T
(t +T)[u(t +T)−u(t)]+
A
T
(T −t)[u(t)−u(t −T)]
x(t) =
A
T
n
(t +T)u(t +T)−(t +T)u(t)
o
+
A
T
n
[(T −t)u(t)−(T −t)u(t −T)]
o
x(t) =
A
T
n
r(t +T)−tu(t)−Tu(t)
o
+
A
T
n
Tu(t)−tu(t)+r(t −T)
o
=
A
T
n
r(t +T)−r(t)−Tu(t)
o
+
A
T
n
Tu(t)−r(t)+r(t −T)
o
=
A
T
hn
r(t +T)−2r(t)+r(t −T)
oi
X(jw) =
A
T

ejwT
(jw)2
−
2
(jw)2
+
e−jwT
(jw)2

Π(t) = rect(t) =

1 −1
2 t 1
2
0 otherwise
rect(t −5) =

1 −1
2 ≤ t −5 1
2
0 otherwise
rect(t −5) =
(
1 9
2 ≤ t ≤ 11
2
0 otherwise
X(jw) =
∞
Z
−∞
x(t)e−jwt
dt =
∞
Z
−∞
rect(t −5)e−jwt
dt
=
11/2
Z
9/2
e−jwt
dt =
e−jwt
−jw
11/2
9/2
=
e− j11w
2 −e− 9 jw
2
−jw
=
e−9 j w
2 −e−11 j w
2
jw
=
e−5 jwejw/2 −e−5 jwe−jw/2
jw
=
2e−5 jw ejw/2 −e−jw/2

w2j
=
2e−5 jw
w
sin
w
2
= e−5 jw

sin w
2
w
2

X(jw) = e−5 jw
Sa
w
2

2.6 PROPERTIES OF CONTINUOUS-TIME FOURIER TRANSFORM
(1) Linearity
If FT (x1(t)) = X1(jw)
and FT (x2(t)) = X2(jw)
Then linearity property states that
FT(Ax1(t)+Bx2(t)) = AX1(jw)+BX2(jw)
where A and B are constants.
Proof:
Let r(t) = Ax1(t)+Bx2(t)
FT(r(t)) = R(jw) =
∞
Z
−∞
r(t)e−jwt
dt
=
∞
Z
−∞
(Ax1(t)+Bx2(t))e−jwt
dt

=
∞
Z
−∞
x(τ)e−j(−w)τ
dτ
F(x(t)) = X(−jw)
(4) Time shifting
If FT (x(t)) = X(jw)
then FT (x(t −t0)) = e−jwt0 X(jw)
Proof:
Let r(t) = x(t −t0)
R(jw) =
∞
Z
−∞
r(t)e−jwt
dt =
∞
Z
−∞
x(t −t0)e−jwt
dt
R(jw) = FT(x(t −t0)) =
∞
Z
−∞
x(t −t0)e−jwt
dt
Let t −t0 = τ dt = dτ
FT (x(t −t0)) =
∞
Z
−∞
x(τ)e−jw(t0+τ)
dτ
=
∞
Z
−∞
x(τ)e−jwt
e−jwt0 dτ
= e−jwt0
∞
Z
−∞
x(τ)e−jwτ
dτ
FT (x(t −t0)) = e−jwt0 X(jw). Similarly, FT (x(t +t0)) = ejwt0 X(jw)
So FT (x(t ±t0)) = e±jwt0 X(jw)
(5) Frequency shifting
FT (ejw
0
t
x(t)) = X(j(w−w0))
Let r(t) = ejw
0
t
x(t)
FT (r(t)) = FT ejw0t
x(t)

= R(jw) =
∞
Z
−∞
ejw0t
x(t)e−jwt
dt
FT ejw0t
x(t)

=
∞
Z
−∞
x(t)e−j(w−w0)t
dt

Let w−w0 = w0
=
∞
Z
−∞
x(t)e−jw0t
dt
FT ejw0t
x(t)

= X(jw0
) = X(j(w−w0))
Similarly, FT(e−jw0t
x(t)) = X(j(w+w0))
We can write as FT e±jw0t
x(t)

= X(j(w∓w0))
(6) Duality or symmetry property
then FT (x(t)) = 2πx(−jw)
Proof:
We know that x(t) = 1
2π
R ∞
−∞ X(jw)ejwtdw
Replacing t by −t, we get
x(−t) =
1
2π
∞
Z
−∞
X(jw)e−jwt
dw
2π x(−t) =
2π
2π
∞
Z
−∞
X(jw)e−jwt
dw
2π x(−t) =
∞
Z
−∞
X(jw)e−jwt
dw
Interchanging t by jw
2π x(−jw) =
∞
Z
−∞
X(t)e−jwt
dt
2π x(−jw) = FT(X(t))
(7) Convolution in time domain
If FT (x1(t)) = X1(jw) and FT (x2(t)) = X2(jw)
then FT (x1(t)∗x2(t)) = X1(jw)X2(jw)
i.e., convolution in time domain becomes multiplication in frequency domain.

Proof:
r(t) = x1(t)∗
x2(t) =
∞
Z
−∞
x1(τ)x2(t −τ)dτ
FT(r(t)) = R(jw) =
∞
Z
−∞
r(t)e−jwt
dt
=
∞
Z
−∞


∞
Z
−∞
x1(τ)x2(t −τ)dτ

e−jwt
dt
=
∞
Z
−∞
∞
Z
−∞
x1(τ)x2(t −τ)dτ e−jwt
dt
=
∞
Z
−∞
x1(τ)dτ
∞
Z
−∞
x2(t −τ) e−jwt
dt
Let t −τ = ∝ so dt = d ∝
FT[x1(t)∗
x2(t)] =
∞
Z
−∞
x1(t)dτ
∞
Z
−∞
x2(∝) e−jw(∝+τ)
d ∝
=
∞
Z
−∞
x1(τ)dτ
∞
Z
−∞
x2(∝) e−jw∝
e−jwτ
d ∝
=
∞
Z
−∞
x1(τ) e−jwτ
dτ
∞
Z
−∞
x2(∝) e−jw∝
d ∝
FT[x1(t)∗
x2(t)] = X1(jw) X2(jw)
(8a) Integration in time domain
then FT
R t
−∞ x(τ)dτ

= 1
jw ×(jw)
Proof: Let r(t) =
R t
−∞ x(τ)dτ
Differentiating w.r.t. t
dr(t)
dt
= x(t) ⇒ FT(x(t)) = FT

d
dt
r(t)

From differentiation in time domain
X(jw) = jwX(jw)
R(jw) =
1
jw
X(jw)
FT(r(t)) = FT


t
Z
−∞
x(τ)dτ

 =
1
jw
X(jw)

(8b) Differentiation in time domain
then d
dt x(t)

= jw×(jw)
Proof: We know that x(t) =
1
2π
∞
Z
−∞
X(jw)ejwt
dw. Differentiating both sides w.r.t. t
d
dt
x(t) =
1
2π
∞
Z
−∞
X(jw)

d
dt
ejwt

dw
=
1
2π
∞
Z
−∞
jwX(jw)ejwt
dw
= j
1
2π
∞
Z
−∞
(wX(jw))ejwt
dw
d
dt
x(t) = j FT−1
(wX(jw))
yields FT d
dt x(t)

= jwX(jw). On generalizing we get FT

dn
dtn x(t)

= (jw)nX(jw)
(9) Differentiation in frequency domain
then FT (tx(t)) = j d
dw X(jw)
Proof: We know that X(jw) =
R ∞
−∞ x(t)e−jwtdt
On differentiating both sides w.r.t. w
d
dw
X(jw) =
∞
Z
−∞
x(t)

d
dw
e−jwt

dt = −
∞
Z
−∞
j t x(t)e−jwt
dt
Multiplying both sides by j
j
d
dw
X(jw) =
∞
Z
−∞
(tx(t))e−jwt
dt since j2 = −1 or −j2 = 1
j
d
dw
X(jw) = FT[t x(t)]
FT[t x(t)] = j
d
dw
X(jw)
(10) Convolution in frequency domain (multiplication in time domain (multiplication theorem))
If FT(x1(t)) = X1(jw) and FT[x2(t)] = X2(jw)
FT(x1(t)x2(t)) =
1
2π
(X1(jw)∗
X2(jw))

Proof:
E =
∞
Z
−∞
x(t)2
dt =
∞
Z
−∞
x(t)x∗
(t)dt (1)
We know that x(t) =
1
2π
∞
Z
−∞
X(jw)e+jwt
dw
So x∗
(t) =
1
2π
∞
Z
−∞
X(jw)e−jwt
dw (2)
on putting (1)
=
∞
Z
−∞
x(t)

 1
2π
∞
Z
−∞
X∗
(jw)e−jwt
dw

dt
=
1
2π
∞
Z
−∞
X∗
(jw)
∞
Z
−∞
x(t)e−jwt
dt dw
=
1
2π
∞
Z
−∞
X(jw)X∗
(jw)dw
=
∞
Z
−∞
x(t)2
dt =
1
2π
∞
Z
−∞
X(jw)
2
dw
Relation between Laplace Transform and Fourier Transform
Fourier transform X(jw) of a signal x(t) is given as
X(jw) =
∞
Z
−∞
x(t)e−jwt
dt (1)
F.T. can be calculated only if x(t) is absolutely integrable
=
∞
Z
−∞
x(t) dt ∞ (2)
Laplace transform X(s) of a signal x(t) is given as
X(s) =
∞
Z
−∞
x(t)e−st
dt (3)
We know that s = σ+ jw
X(s) =
∞
Z
−∞
x(t)e−(σ+jw)t
dt
X(s) =
∞
Z
−∞

x(t)e−σt

e−jwt
dt (4)

Comparing (1) and (4), we find that L.T. of x(t) is basically the F.T. of [x(t)e−σt].
If s = jw, i.e. σ = 0, then eqn. (4) becomes X(s) =
R ∞
−∞ x(t)e−jwtdt = X(jw)
Thus, X(s) = X(jw) when σ = 0 or s = jw
This means L.T. is same as F.T. when s = jw. The above equation shows that F.T. is special case of L.T.
Thus, L.T. provides broader characterization compared to F.T., s = jw indicates imaginary axis in complex
s-plane.
2.7 APPLICATIONS OF FOURIER TRANSFORM OF NETWORK ANALYSIS
Example 24: Determine the voltage Vout(t) to a current source excitation i(t) = e−tu(t) for the circuit shown
in figure.
1Ω F
i(t) ↑
+
Vout(t)
1
2
−
Fig. 2.26.
SOLUTION:
1Ω F
i(t)
↓ i1(t) ↓ i2(t)
↑
+
Vout(t)
1
2
−
i(t) = i1(t)+i2(t)
i(t) =
Vout(t)
1
+
1
2
dVout(t)
dt
(
since i = V
R
and i = cdv
dt or v = 1
c
R
idt
e−t
u(t) = Vout(t)+
1
2
dVout(t)
dt
(1)
On taking the z-transform on both sides
1
1+ jw
= Vout(jw)

1+
jw
2

=
(2+ jw)
2
Vout(jw)
Vout(jw) =
2
(1+ jw)(2+ jw)
=
A
1+ jw
+
B
2+ jw
Vout(jw) =
2
1+ jw
−
2
2+ jw









A(2+ jw)+B(1+ jw) = 2
2A+B = 2
A+B = 0 s0 A = −B
2A−A = 2; A = 2,B = −2

V0(jw) =
2
6(jw)2 +7(jw)+1
=
2
(6jw+1)(jw+1)
V0(jw) =
1/3
(jw+1/6)(jw+1)
=
A
1
6 + jw
+
B
1+ jw
V0(jw) =
2
5 1
6 + jw
−
2
5(1+ jw)
(5)
Taking inverse Fourier transform, we get
V0(t) =
2
5

e−t/6
−e−t

u(t) (6)
Example 26: Determine the response of current in the network shown in Fig. 2.28(a) when a voltage having
the waveform shown in Fig. 2.28(b) is applied to it by using the Fourier transform.
1Ω
∼
v(t) 1F
0
v(t)
π wt
(a) (b)
Fig. 2.28.
SOLUTION:
Waveform V(t) is defined as
V(t) = sint(u(t)−u(t −π)) (1)
1Ω
∼
u(t) 1F
i(t)
a
Let i(t) be the current in the loop. Applying KVL in loop
V(t) = 1·i(t)+
1
1
t
Z
0
i(t)dt = i(t)+
t
Z
0
i(t)dt (2)
On taking Fourier transform of
V(jw) =
1
(jw)2 +1
+
e−jπw
(jw)2 +1
Since FT

sintu(t)

=
1
(jw)2 +1
FT

sintu(t −π)

=
e−jπw
(jw)2 +1

Solve using F.T. formula
V(jw) =
1+e−jπw
(jw)2 +1
(3)
V(jw) = I(jw)+
1
jw
I(jw)
V(jw) =

1+
1
jw

I(jw) =
jw+1
jw
I(jw)
I(jw) =
jw
jw+1
V(jw) (4)
I(jw) =
jw
jw+1
·
(1+e−jπw)
((jw)2 +1)
From (3)
I(jw) =
jw
jw+1
·

1
(jw)2 +1
+
e−jπw
(jw)2 +1

=
jw
(jw+1)
·
1
((jw)2 +1)
+
jw
(jw+1)
·
1
((jw)2 +1)
·e−jπw
| {z }
I2( jw)
I1(jw)
I1(jw) =
A
jw+1
+
Bjw+c
((jw)2 +1)
=
−1/2
(jw+1)
+
1
2 (jw+1)
((jw)2 +1)
i1(t) = −
1
2
e−t
u(t)+
1
2
costut +
1
2
sintδt +
1
2
sintu(t)
Since IFT
n
1
( jw)2+1
o
= sintu(t)
so IFT

jw
( jw)2+1

= d
dt sintu(t)
Using differential in time domain property
IFT

jw
(jw)2 +1

= costu(t)+sintδ(t)
I2(jw) =
jw
(jw+1)
·
1
((jw)2 +1)
·e−jπw
I2(jw) = I3(jw)·e−jπw
Since I3 = I1(jw)
so i3(t) = −
1
2
e−t
u(t)+
1
2
costu(t)+
1
2
sintδ(t)+
1
2
sintu(t)

From time shifting property FT(x(t ±t0)) = e±jwt0 ×(jw)
i2(t) = i3(t −π)
= −
1
2
e−(t−π)
u(t −π)+
1
2
cos(t −π)u(t −π)+
1
2
sin(t −π)δ(t −π)+
1
2
sin(t −π)u(t −π)
so i(t) =
1
2
−

−e−t
+cost +sint

u(t)+
1
2
sintδ(t)+
1
2
h
−e−(t−π)
+cos(t −π)+sin(t −π)
i
u(t −π)+
1
2
sin(t −π)δ(t −π)
Example 27: For the RC circuit shown in figure.
R
C
i(t)
x(t) 1 y(t)
Fig. 2.29.
(a) Determine frequency response of the circuit.
(b) Find impulse response.
(c) Plot the magnitude and phase response for RC = 1.
SOLUTION:
Applying KVL in loop (1)
x(t)−Ri(t)−
1
C
t
Z
−∞
i(t)dt = 0
x(t) = Ri(t)+
1
C
t
Z
−∞
i(t)dt (1)





Since
VR = iR
Vc = 1
C
R
i(t)dt
and y(t) =
1
C
t
Z
−∞
i(t)dt (2)

A = B−C
H(jw) =
1
√
1+w2
(8)
H(jw) = 1−(1+ jw)
= tan−1 0
1
−tan−1
w = −tan−1
w (9)
For different values of w, we find H(jw) and H(jw)
S. No w |H(jw)| H(jw)
1− −∞ 0 90◦
2− −50 0.0199 88.9◦
3− −20 0.0499 87.1◦
4− −10 0.099 84.3◦
5− −5 0.196 78.7◦
6− −2 0.447 63.4◦
7− −1 0.707 45◦
8− 0 1 0
9− 1 0.707 −45◦
10− 2 0.447 −63.4◦
11− 5 0.196 −78.7◦
12− 10 0.099 −84.3◦
13− 20 0.0499 −87.1◦
14− 50 0.0199 −88.9◦
15− ∞ 0 −90◦
0 20
−20 10
1
−10 30 40 50
−30
−40
−50
| H(jw) |
w
Fig. 2.30. Magnitude plot frequency response of the circuit

90°
−90°
−45°
45°
0 20
−20 10
−10 30 40 50
−30
−40
−50
∠H(jw)
w
Fig. 2.31. Phase plot
Example 28: For the circuit shown in figure, determine the output voltageV0(t) to a voltage source excitation
Vi(t) = e−tu(t) using Fourier transform
2Ω
+
−
+
−
V0(t)
Vin(t) 1H
1
Fig. 2.32.
SOLUTION:
Since Vin(t) = e−t
u(t) (1)
Vin( jw) =
1
1+ jw
(2)
Applying KVL in loop (1)
Vin(t) = 2i(t)+1·
di(t)
dt
Vin(t) = 2i(t)+
di(t)
dt
(3)
V0(t) = 1·
di(t)
dt
V0(t) =
di(t)
dt
(4)

Q3: (i) State and prove the following properties of Fourier series:
(a) Time shifting property (b) Frequency shifting property
(ii) What are Dirichlet’s conditions?
Q4: Find the fundamental period T, the fundamental frequency w0 and the Fourier series coefficients an of
the following periodic signal;
0.5
−0.5
−1 t
x(t)
0
1
t
−1
Fig. 2.3 P.
Q5: Obtain the Fourier series component of the periodic square wave signals.
T/2
T/4
−T/4
−T/2
x(t)
1
0 t
−1
Fig. 2.4 P.
Q6: Determine the Fourier transform of the Gate function
T/2
−T/2
x(t)
A
t
Fig. 2.5 P.
Q7: Determine the Fourier series representation of the signal
x(t) =

t −t2 for −π ≤ t ≤ π
0 elsewhere

Q8: For the continuous-time periodic signal
x(t) = 2+cos[2πt/3]+4sin[5πt/3]
determine the fundamental frequency w0 and the Fourier series coefficients Cn such that
x(t) =
∞
∑
n=−∞
Cnejnw0t
Q9: Find the Fourier transform of the following signals:
(a) x(t) = δ(t) (b) x(t) = 1 (c) x(t) = sgn (t) (d) x(t) = u(t)
(e) x(t) = exp(−at)u(t) (f) x(t) = cos [w0t]sin [w0t]
Q10: Show that the Fourier transform of rect (t − 5) is Sa(w/2)exp(j5w). Sketch the resulting amplitude
and phase spectrum.
Q11: Find the inverse Fourier transform of spectrum shown in figure.
−w0 w0
| X(w) |
1
w
(a)
−w0
w0
∠X(w)
w
π/2
−π/2
(b)
Fig. 2.6 P.
Q12: Find the Fourier transform of the following waveform.
x(t)
0
1
a b t
−a
−b
Fig. 2.7 P.
Q13: State and prove duality property of CTFT.
Q14: Determine the Fourier transform of the signal
x(t) = {tu(t)∗[u(t)−u(t −1)]}, where u(t) is unit step function and ∗ denotes the convolution operation.
Q15: Show that the frequency response of a CTLTIS is Y(w) = H(w)X(w)
where X(w) = Fourier transform of the signal x(t)
H(w) = Fourier transform of LTIS response h(t)

Q16: Find the Fourier transform of the signal x(t) shown in figure below.
A
0 T 2T t
x(t)
Fig. 2.8 P.
Q17: Determine the frequency response H(jw) and impulse response h(t) for a stable CTLTIS characterized
by the linear constant coefficient differential equation given as
d2
y(t)/dt2
+4dy(t)/dt +3y(t) = dx(t)/dt +2x(t)
Q18: Find the Fourier transform of the signal x(t) shown in figure below.
K
0 T
−T t
x(t)
Fig. 2.9 P.
Q19: If g(t) is a complex signal given by g(t) = gr(t) + jgi(t) where gr(t) and gi(t) are the real and
imaginary parts of g(t) respectively. If G(f) is the Fourier transform of g(t), express the Fourier transform
of gr(t) and gi(t) in terms of G(f).
Q20: Find the coefficients of the complex exponential Fourier series for a half wave rectified sine wave
defined by
x(t) =

Asin (w0t), 0 ≤ t ≤ T0/2
0, T0/2 ≤ t ≤ T0
with x(t) = x(t +T0)
Q21: (a) Show that the Fourier transform of the convolution of two signals in the time domain can be given
by the product of the Fourier transform of the individual signals in the frequency domain.
(b) Determine the Fourier transform of the signal
x(t) =
1
2

δ(t +1)+δ(t −1)+δ

t +
1
2

δ+

t −
1
2

an =
1−(−1)n
n2π2
bn =
1
nπ
Q3:
0.5
−0.5 1
−1 t
x(t)
0
1
−1
T = 1
w0 = 2π rad/sec
y−y1 =
y2 −y1
x2 −x1
(x−x1)
x(t) = −2t +1
an =
2
T
t0+T
Z
t0
x(t)cos nw0t dt
an = 0
Q4:
T/2
T/4
−T/4
−T/2
x(t)
1.0
t
−1.0
T
2
−

−
T
4

=
3T
4
; w0 =
2π
3T
4
=
8π
3T
x(t) =
(
1 −T
4 ≤ t ≤ T
4

−1 T
4 ≤ t ≤ T
2

a0 =
1
3T
4





T
4
Z
− T
4
dt +
T
2
Z
T
4
(−1)dt





=
4
3T
T
4
=
1
3

an =
8
3T





T
4
Z
− T
4
cos
8nπ
3T
dt −
T
2
Z
T
4
cos
8nπ
3T
tdt





an =
1
nπ

3sin
2nπ
3
−sin
4nπ
3

bn = 0, since even function
x(t) =
1
3
+
1
π

3sin
2π
3
−sin
4π
3
+
3
2
sin
4π
3
−
1
2
sin
8π
3
+···

Q5:
T/2
−T/2
x(t)
A
t
x(t) =
(
A− T
2 ≤ t ≤ T
2
0 elsewhere
X(jw) = A
T
2
Z
− T
2
e−jwt
dt =
2A
w
sin
wT
2
=
AT
wT
2
sin
wT
2
X(i f) = AT sincfT
Q6:
T0 = 2π;
w0 = 1;
a0 =
1
2π
π
Z
−π
t −t2

dt =
−π2
3
an =
1
π
π
Z
−π
t −t2

cos nt dt =
−4(−1)n
n2
bn =
1
π
π
Z
−π
t −t2

sin nt dt =
−2(−1)n
n

Taking inverse Fourier transform
x1(t) =
1
2π
w0
Z
0
−j ejwt
dw =
1−ejw0t
2πt
x2(t) =
1
2π
0
Z
−w0
j ejwt
dw =
1−e−jw0t
2πt
x(t) = x1(t)+x2(t) =
1
2πt
(1−ejw0t
+1−e−jw0t
)
=
1
2πt
(2−2cos w0t) =
2sin2 w0t
2
πt
Q11:
1.0
0 a
−a b
−b t
x(t)
x(t) =





t+b
b−a for−b t −a
1 for−a t a
t−b
a−b for a t b
X(jw) =
2
w2(b−a)
(cos wa−cos wb)
Q12:
x(t) = tu(t)∗
[u(t)−u(t −1)]
x1(t) = tu(t) x2(t) = u(t)−u(t −1)
Differentiating in frequency domain property
FT(tx(t)) = j
d
dw
X(jw)
X1(jw) =
1
(jw)2
X2(jw) =
1
Z
0
1.e−jwt
dt =
1
jw
(1−e−jw
)
X(jw) = X1(jw)X2(jw) =
1
(jw)3
(1−e−jw
)

Q13: Prove convolution in time domain property.
Q14:
0
(0,0)
T
x(t)
A
(T,A)
2T t
x(t) =
A
T t 0 t T
A T t 2T
X(jw) =
A
T
T
Z
0
te−jwt
dt +A
2T
Z
T
e−jwt
dt
X(jw) =
A
T

te−jwt
−jw
T
Z
0
−
T
Z
0
e−jwt
jw
dt

+A

e−jwt
−jw
2T
T

=
A
T

Tejwt
−jw
+
1
w2
e−jwT
−1

+A

e−j2wT −e−jwT
−jw

= A

e−jwt
−jw
+
1
w2T
e−jwT
−1

−
A
jw
e−jwT
e−jwT
−1

=
Ae−jwT
jw
+
A
w2T
e−jwT
−
A
w2T
−
A
jw
e−j2wT
+
A
jw
e−jwT
=
A
wT

1
w
e−jwT
−
1
w
+ jTe−2 jwT

Q15:
d2y(t)
dt2
+4
dy(t)
dt
+3y(t) =
dx(t)
dt
+2x(t) (1)
Taking Fourier transform on both sides
(jw)2
Y(jw)+4(jw)Y(jw)+3Y(jw) = (jw)X(jw)+2X(jw)
(jw)2
+4(jw)+3

Y(jw) = ((jw)+2)X(jw) (2)
Frequency response H(jw) =
Y(jw)
X(jw)
=
2+ jw
(jw)2 +4jw+3
(3)
H(jw) =
2+ jw
(3+ jw)(1+ jw)
=
A
3+ jw
+
B
1+ jw

x(t) = Asinw0t for 0 ≤ t ≤
T0
2
= 0 for
T0
2
≤ t ≤ T0
C0 =
1
T0
T0
2
Z
0
Asinw0tdt =
A
T0

−cosw0t
w0
T0
2
0

= −
A
T0 · 2π
T0

cosw0 ·
T0
2
−1

= −
A
2π
[cosπ−1] =
A
2
Cn =
1
T0
T0
2
Z
0
Asinw0te−jnw0t
dt
=
A
2jT0
T0
2
Z
0
(ejw0t
−e−jnw0t
)e−jnw0t
dt
=
A
2jT0
T0
2
Z
0

ejw0t(1−n)
−e−jw0t(n+1)

dt
=
A
2jT0
ejw0t(1−n)
jw0(1−n)
−
e−jw0t(n+1)
−jw0(n+1)
T0
2
0
!
=
A
2jT0w0
ejw0(1−n)
T0
2
1−n
+
e−jw0(n+1)
T0
2
(n+1)
−
1
1−n
−
1
n+1
!
= −
A
4π

ejπ(1−n)
1−n
+
e−jπ(n+1)
n+1
−
1
1−n
−
1
n+1
#
= −
A
4π

ejπe−jnπ
1−n
+
e−jnπ ·e−jπ
n+1
−
1
1−n
−
1
n+1

= −
A
4π

−e−jnπ
1−n
−
e−jnπ
n+1
−
1
1−n
−
1
n+1

Since ejπ
= −1
=
A
4π

2e−jnπ
1−n2
+
2
1−n2

=
A
2π(1−n2)
(e−jnπ
+1)

Q19:
x(t) =
1
2

δ(t +1)+δ(t −1)+δ

t +
1
2

+δ

t −
1
2

Taking Fourier transform on both sides
X(jw) =
∞
Z
−∞
x(t)e−jwt
dt (1)
X(jw) =
∞
Z
−∞
1
2

δ(t +1)+δ(t −1)+δ

t +
1
2

+δ

t −
1
2

e−jwt
dt
X(jw) =
1
2


∞
Z
−∞
δ(t +1)e−jwt
dt +
∞
Z
−∞
δ(t −1)e−jwt
dt +
∞
Z
−∞
δ

t +
1
2

e−jwt
dt
+
∞
Z
−∞
δ

t −
1
2

e−jwt
dt


Since FT(δ(t)) = 1
So FT(δ(t ±t0)) = e±jwt0 dt {using time shifting property}
X(jw) =
1
2

ejw
+e−jw
+ej w
2 +e−j w
2

X(jw) =
ejw +e−jw
2
+
ej w
2 +e−j w
2
2
X(jw) = cos w+cos
w
2
OBJECTIVE TYPE QUESTIONS
Q1: If the Fourier transform of a function x(t) is X(jw), then X(jw) is defined as
(a)
R ∞
−∞ x(t)ejwtdt (b)
R ∞
−∞
dx(t)
dt e−jwtdt
(c)
R ∞
−∞ x(t)dt (d)
R ∞
−∞ x(t)e−jwtdt
Q2: If X(jw) be the Fourier transform of x(t), then
(a) x(t) = 1
2π
R ∞
−∞ X(jw)ejwtdw (b) x(t) = 1
2π
R ∞
−∞ X(jw)e−jwtdw
(c) x(t) = 1
2π
R ∞
0 X(jw)ejwtdw (d) x(t) = 1
2π
R ∞
−∞ X(jw)e−jwtdw
Q3: Fourier transform of x(t) = 1 is
(a) 2π δ(w) (b) π δ(w) (c) 3π δ(w) (d) 4π δ(w)
Q4: Fourier transform of x(t −t0) is
(a) e−jwt
0X(jw) (b) ejwt
0X(jw) (c) 1
t0
X(jw) (d) t0e−jwt
0X(jw)

Q19: The trigonometric Fourier series of a periodic time function have
(a) sine terms (b) cosine term
(c) both (a) and (b) (d) DC term
Q20: Fourier series is defined as
x(t) = a◦ +
∞
∑
n=1
(an cosnw0t +bn sinw0t)
(a) True (b) False
Answers: (1) d (2) a (3) a (4) a (5) b
(6) c (7) a (8) a (9) a (10) d
(11) c (12) b (13) b (14) a (15) a
(16) a (17) e (18) c (19) c (20) a
UNSOLVED PROBLEMS
Q1: Show that the Fourier transform of x(t) = δ(t +2)+δ(t)+δ(t −2) is (1+2cos2w).
Q2: Show that the inverse Fourier transform of X(jw) = 2πδ(w) + πδ(w − 4π) + πδ(w + 4π) is x(t) =
1+cos 4πt.
Q3: Calculate the Fourier transform of te−|t|, using the F.T. pair, FT

e−|t|

= 2
1+w2 . Also find the Fourier
transform of 4t
(1+t2)2 using duality property.
Q4: X(jw) = δ(w)+δ(w−π)+δ(w−5); find IFT x(t) and show that x(t) is non-periodic.
Q5: Find the Fourier transform of the triangular pulse as shown in figure.
0
1
x(t)
t
T/2
−T/2
Fig. 2.10 P.
Ans. X(jw) = T
2 sinc2(wt
4 )
Q6: Find the Fourier transform of x(t) = rect(t/2). Ans. X(jw) = 2sincw
Q7: Find the Fourier transform of the signal x(t) = cosw0t by using the frequency shifting property.
Ans: X(jw) = π[δ(w−w0)+δ(w+w0)]
Q8: Show that FT [sinw0tu(t)] = w0
w2
0−w2 + π j
2 [δ(w+w0)−δ(w−w0)].
Q9: Find inverse Fourier transform of X(jw) = jw
(1+jw)2
Ans. x(t) = d
dt [te−tu(t)]

Q10: Sketch and then find the Fourier transform of following signals
(a) x1(t) = π t + 3
2

+π t − 3
2

Ans. (a)
x1(t)
−1 1
1
2
−2
−
t
X1(jw) = 2sincw
2 cos3w
2
Fig. 2.11 P.
(b) x2(t) = π t
4

+π t
2

Ans. (b)
x2(t)
−1 1
1
2
2
−2
−
t
X2(jw) = 4sinc2w+2sincw
Fig. 2.12 P.
Q11: Find the frequency response x(jw) of the RC circuit shown in figure. Plot the magnitude and phase
response for RC = 1
x(jw) =
y(jw)
x(jw)
=
1
1+ jwRC
↑ ↑
R
C y(l)
x(t) −
− ↑
↑
Fig. 2.13 P.
Ans.|x(jw)| =
1
√
1+w2
x(jw) = −tan−1
w
Q12: Find the Fourier series of the waveform shown in figure.
x(t) =
2A
jnπ
for n = 1,3,5,7

x(tπ)t =
1
π
+
1
2
cos5πt −
2
3π
cos10
−
2
8π
cos15πt
Q15: The output of a system is given by
x(t) =

Asinw0t for 0 ≤ t ≤ π
0 for π ≤ t ≤ 2π
Determine trigonometric form of Fourier series of x(t)
Ans.

x(t) =
A
π
+
A
2
cos(nt −
π
2
)+
∞
∑
n=2
2A
π(1−n2)
cosnt
#

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233_Sample-Chapter (1).pdf