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 + ∑ (an cos (nw0t) + bn sin (nw0t)) (1) n=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 t0+T ∞ t0+T ∞ t0+T x(t)dt = a0dt + ∑ an cos (nw0t)dt + ∑ bn sin (nw0t)dt n=1 n=1 tZ0 tZ0 tZ0 tZ0 Since t0+T cos (nw dt) = 0 t0 0

R t0+T

sin (nw0dt) = 0

tZ0

t0+T 1 a = x(t)dt (2) 0 T tZ0

To evaluate an and bn, we use the following result:

t0+T 0 m = n cos (nw0t)cos (mw0t)dt = 6 T/2 m = n = 0 tZ0  6

94 96 Basic System Analysis •

Multiply eqn. (1) by sin (mw0t) and integrate over one period

t0+T t0+T ∞ t0+T x(t)sin (mw0t)dt = a0 sin (mw0t)dt + ∑ an cos (nw0t)sin (mw0t)dt + n=1 tZ0 tZ0 tZ0

∞ t0+T ∑ bn sin (mw0t)sin (nw0t)dt n=1 tZ0

t0+T 2 b = x(t)sin (nw t)dt (4) n T 0 tZ0 Example 1:

1.0 −

−3 −2 −1 0 1 2 3 − −1.0

Fig. 2.1.

T 1to1 T = 2 w0 = π x(t) = t, 1 < t < 1 → − − 1 1 1 a0 = t dt = (1 1) = 0 2 4 − Z1 − an = 0

1 π π 1 π t cos nt cos nt bn = t sin ntdt = − π π n − n 1 Z1  − − 1 π π 1 1 π π π = − [t cos nt + cos nt] 1 = [2cos + cos cos ] nπ − −nπ − 2 2 ( 1)n b = − cos nπ = − − n nπ π n   b1 b2 b3 b4 b5 b6 2 2 2 2 2 2 − − − ··· π 2π 3π 4π 5π 6π ∞ 2 ( 1)n x(t) = ∑ − − sin nπt π n n=1   2 1 1 1 = sin πt sin 2πt + sin 3πt sin 4πt + π − 2 3 − 4 ···   Fourier Series and Fourier Transform 97 •

Example 2:

1.0

−2π 0 2π 4π 6π t

Fig. 2.2.

t 2π x(t) = T = 2π w = = 1 2π 0 T 2π π 1 1 1 2 1 a = x(t)dt = t2 = 0 π2 T 4 2 0 2 Z0  

2π π 2 1 t sin t sin nt 2 a = t cos ntdt = + n π2 π2 4 2 n n 0 Z0   1 2πsin 2nπ sin 2nπ = + = 0 2π2 n n   2π 2 1 t cos nt cos nt 2π bn = t sin ntdt = − + 4π2 2π2 n n 0 Z0 h i 1 2πcos 2nπ cos 2nπ 1 = − + 2π2 n n − n   1 b = − n nπ 1 ∞ 1 1 1 ∞ 1 x(t) = + ∑ − sin nt = + ∑ cos (nt + π/2) 2 nπ 2 π n n=1   n=1 1 1 sin 2t sin 3t = sin t + + + 2 − π 2 3 ···  

Example 3:

A x(t)

−T/2 −T/4 T/4 T/2 t

Fig. 2.3. Rectangular waveform 98 Basic System Analysis •

Figure shows a periodic rectangular waveform which is symmetrical to the vertical axis. Obtain its F.S. representation. ∞ x(t) = a0 + ∑ (an cos nw0t + bn sin nw0t) n=1 ∞ x(t) = a0 + ∑ an cos (nw0t) bn = 0 n=1 T T x(t) = 0 for − < t < − 2 4 T T + A for − < t < 4 4 T T 0 for < t < 4 2 T/4 1 A a = Adt = 0 T 2 TZ/4 − T/4 2 2A T T an = Acos (nw0t)dt = sin nw0 + sin nw0 T Tnw0 4 4 TZ/4   − 4A nπ 2A nπ 2π a = sin = sin w = n 2πn 2 πn 2 0 T 4A 2A    a = = 1 2π π

a2 = 0 2A 3π 2A 2A a3 = sin = ( 1) = − 3π 2 3π − 3π A 2A 1 1 x(t) = + cos w0t cos 3w0t + cos 5w0t + 2 π − 3 5 ···  

Example 4: Find the trigonometric Fourier series for the periodic signal x(t).

x(t) 1.0

−9 −7 −5 −3 −1 0 1 3 5 7 9 11 t

T

Fig. 2.4. Fourier Series and Fourier Transform 99 •

SOLUTION:

1 1 < t < 1 bn = 0 x(t) = − 1 1 < t < 3 ( − 3 1 3 1 1 a0 = x(t)dt = dt + ( 1)dt T = 4 T T  −  Z1 Z1 Zt − − 1   2π 2π π = [2 2] = 0 ∴ w0 = = = T − T 4 2 1 3 2 an = cos (nw0t)dt + cos (nw0t)dt T   Z1 Z1 − 2 πn 3nπ nπ  = 2sin sin sin 2πn 2 − 2 − 2 h i   1 nπ 3nπ 3nπ nπ nπ = 3sin sin sin = sin π + = sin nπ 2 − 2 2 2 − 2     4 nπ a = sin n nπ 2   0 n = even 4 a =  n = 1,5,9,13 n  nπ  4  − n = 3,7,11,15 nπ  4 π 4 3π 4 5π 4 7π x(t) = cos t cos t + cos t cos t + π 2 − 3π 2 5π 2 − 7π 2 ···         4 π 1 3π 1 5π x(t) = cos t cos t + cos t π 2 − 3 2 5 2 ···         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

x(t) 1.5

0 1 23 4 5

−1.5

Fig. 2.5. 100 Basic System Analysis •

SOLUTION:

2π T = = 2, w0 = π w0

a0 = an = 0

1 2

bn = 1.5sinnπtdt 1.5sinnπtdt − Z0 Z1 1.5 = [ cosnπ + 1] + [cos2nπ cosnπ] nπ − − n o 3 bn = [1 cosnπ] nπ − 3 2 2 x(t) = 2sinπt + sin3πt + sin5πt + π 3 5 ···   6 1 1 sinπt + sin3πt + sin5πt + π 3 5 ···   1 2 1 C0 = 1.5dt 1.5 dt = 0 2  −  Z0 Z1 OR   By using complex exponential Fourier series 1 2 1 jnπt jnπt Cn = 1.5e− dt 1.5 e− dt 2  −  Z0 Z1  2  1 3 jnπt jnπt Cn = e− e 4 jnπ  − −  0 − 1     3 jnπ j2nπ jnπ = − e− 1 e− + e− 4 jnπ − −   3 jnπ 3 = 1 e− = [1 cosnπ] 2 jnπ − 2 jnπ − ∞   jnπt x(t) = ∑ Cne− n= ∞ − ∞ 3 jnπ jnπt ∑ π 1 e− e n= ∞ 2 jn − − ∞   3 jnπt jnπt π = ∑ π e e cos n n= ∞ 2 jn − −   102 Basic System Analysis • for n = 1 π π A A = sin t sin tdt = (1 cos2t)dt 2π 2π − Z0 Z0 A A = [π] = 2π 2 When n is even A 2 2 2A = = 2π n + 1 − 1 n π(1 n2)  −  − Example 7:

x(t) 2 b

−3 −2 −1 0 1 2 3 t

a −2 T

Fig. 2.7. SOLUTION: 2π T = 2 w = = π 0 T 2t 1 < t < 1 x(t) = − 0  Point (a) ( 1, 2) − − Point (b) (1,2) 2 ( 2) y ( 2) = − − (x ( 1)) − − 1 ( 1) − − − − 4 y + 2 = (x + 1) 2 y + 2 = 2x + 2 y = 2x x(t) = 2t Since function is an odd function 1 1 1 an = 0, a0 = 2tdt = 0 = 0 T 2 × Z1 − 1 1 1 2 2 t cosnπt 1 b = t sin (nπt)dt = + cosnπt n − π 2π2 T T  n n  Z1 1 1 − − −  

104 Basic System Analysis •

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 x(t) dt < ∞ Z 0

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)

t0+T 1 A B a = [Ax (t) + Bx (t)]e jnw0t = x (t)e jnw0t dt + x (t)e jnw0t dt n T 1 2 − T 1 − T 2 − tZ0 ZT ZT

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 − jnw0 t0 FS[x(t t0)] = e− Cn −

Let t t0 = τ − dt = dτ

1 jnw0t Bn = x(t t0)e− dt T − ZT

1 jnw (t +τ) 1 jnw τ jnw t = x(τ)e− 0 0 dτ = x(τ)e− 0 dτ e− 0 0 T T · ZT ZT

jnw t Bn = e− 0 0 Cn · (3) Time reversal: FS[x( t)] = C n − − 1 1 jnw0t j( n)w0T Bn = x( t)e− dt = x( t)e− − dt T − T − ZT ZT

t = τ − dt = dτ − 1 j( n)w0τ = x(τ)e− − dτ = C n T − ZT − 106 Basic System Analysis •

Example 8: Compute the exponential series of the following signal.

x(t)

2.0 1.0

−5 −4 −3 −2 −1 0 1 2 3 4 5 6 t

T

Fig. 2.8.

SOLUTION:

π T = 4 w = 0 2 T 1 2 1 1 3 C0 = x(t)dt = 2dt + dt = T 4   4 Z0 Z0 Z1 1 πt  2  π 1 jn jn t Cn = 2e− 2 dt + e− 2 dt 4   Z0 Z1  π  π 1 4 jn 2 jnπ jn = − e− 2 1 e− e− 2 4  jnπ  −  − jnπ −   h i   π π π  jnπ  1 jnπ jn 1 jn jn = − 2e− 2 2 + e− e− 2 = − e− 2 + e− 2 2 2 jnπ  − −  2 jnπ  − 

  ∞  1 1 1 π 3 1 π 1 π 1 n jn 2 jn 2 n jn 2 = π 1 ( 1) e− x(t) = + ∑ π e ( 1) e − jn − 2 − − 2 4 n= ∞ jn − 2 − − 2   −  

Example 9:

x(t) 1.0 a b ↓ ↓ −2 −1 0 1 2 3 4 5 6 7 t

Fig. 2.9. Fourier Series and Fourier Transform 107 •

SOLUTION: 2π T = 5 w = 0 5 t + 2 2 < t < 1 − − x(t) = 1.0 1 < t < 1  − 2 t 1 < t < 2  − ( 2,0)( 1,1) (a) − − 1 (y 1) = − (x + 1) − 1 − y = t + 2 (b) (1,1)(2,0) 1 y 0 = (x 2) − 1 − − y = x + 2 = t + 2 − − 1 1 2 1 − C0 = (t + 2)dt + dt + (2 t)dt 5  −  Z2 Z1 Z1 − −  3 C = 0 5

1 1 2 − π π π 1  j 2n j 2n j 2n  Cn = (t + 2)e− 5 dt + e− 5 dt + (2 t)e− 5 dt 5  −  Z2 Z1 Z1  − −     A B C    1 1 | {z } − | π {z − } π| {z } j 2n t j 2n t A = e− 5 dt + 2e− 5 dt Z2 Z2 − − 1 1 1 − − π − 1 jφ 1 jφ 2 j 2n A = te− + e− + e − jφ   φ2 jφ 5 Z2 Z2 − Z2  −  − − π π π π 5 j 2n j 4n 25 j 2n j 4n 10 =  e 5 +2e 5 + e 5 e 5 j 2nπ − 4n2π2 − − 2nπ j     π π π π 5 j 2n j 4n 25 j 2n j 4n A = e 5 + 4e 5 + e 5 e 5 j2nπ − 4n2π2 −     2nπ 2nπ j j π π e 5 e− 5 5 j 2n j 2n B = − = e 5 e− 5 j 2nπ j 2nπ − 5   π π π π π π 10 j 4n j 2n 10 j 4n 5 j 2n 25 j 4n 25 j n2 C = − e− 5 e− 5 + e− 5 e− 5 e− 5 + e 5 j 2nπ − j 2nπ − j 2nπ − 4n2π2 4n2π2   108 Basic System Analysis •

1 25 j2nπ j4nπ 25 j4nπ j2nπ Cn = e 5 e 5 e− 5 e− 5 5 n24π2 − − 4n2π2 −      5 2πn 4πn Cn = cos cos 2n2π2 5 − 5      Example 10: For the continuous-time periodic signal 2π 5π x(t) = 2 + cos t + 4sin t 3 3     Determine the fundamental frequency w0 and the Fourier series coefficients Cn such that ∞ jnw0t x(t) = ∑ Cne n= ∞ −

SOLUTION: Given 2π 5π x(t) = 2 + cos t + 4sin t 3 3     2π The time period of the signal cos 3 t is
2π 2π T1 = = π = 3sec w1 2 3 π The time period of the signal sin 5 2 t is
π 2π 6 T2 = 2 = π = sec w2 5 3 5 T 3 5 1 = = ratio of two integers, rational number, hence periodic. T 6 2 2 5

2T1 = 5T2 The fundamental period of the signal x(t) is

T = 2T1 = 5T2 = 6sec and the fundamental frequency is 2π 2π π w = = = 0 T 6 3 2π 5π x(t) = 2 + cos t + 4sin t 3 3     = 2 + cos (2w0t) + 4sin (5w0t)

j2w t j2w t j5w t j5w t e 0 + e− 0 4 e 0 e− 0 = 2 + + − 2 2 j

j2w t j2w t j5w t j5w t = 2 + 0.5 e 0 + e− 0 2 j e 0 e− 0 − − + j( 5)w t + j( 2
)w t  + j2w t  + j5w t x(t) = 2 je − 0 + 0.5e − 0 + 2 + 0.5e 0 2 je 0 − 110 Basic System Analysis •

π 0 2 4 2t a = x(t)cosntdt = + 1 cosnt dt n T T π Zπ Zπ   − − 0 4 2t sinnt 2 = sint + sinnt dt 2π nπ n − π  Zπ  −  0 1 2t 2  = sinnt + sinnt + cosnt π  π n2π  Zπ  −  2  2 2 4  4 = + cosnt = 1 cosnπ = (1 ( 1)n) π 2π 2π 2π2 2π2 (n n ) n − n − − n o 0 n even 2, 4, 6, 8, ··· an = 8 ( 2π2 n odd 1, 3, 5, 7, n ···

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 ∞ jwt X ( jw) = x(t)e− dt (1) Z∞ Fourier transform− ∞ or j2π ft X (if ) = x(t)e− dt (2) Z∞ − Since w = 2π f Similarly, x(t) can be recovered from its Fourier transform X( jw) by using Inverse Fourier transform ∞ 1 x(t) = X( jw)e jwt dw (3) 2π Z∞ ∞ − π x(t) = X(if )e j2 ft dt (4) Z∞ − 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) = X( jw) e j | | | X( jw) Here X( jw) is the amplitude spectrum of x(t) and | 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.) Fourier Series and Fourier Transform 111 •

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. ∞ x(t) dt < ∞ | | Z∞ − 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. • d j X( jw) = FT(tx(t)) dw d FT(tx(t)) = j X( jw) dw at Example 12: Obtain the F.T. of the signal e− u(t) and plot its magnitude and phase spectrum.

SOLUTION: at x(t) = e− u(t) ∞ ∞ j2π ft (a+ j2π f )t X( f ) = x(t)e− dt = e− dt Z∞ Z0 − 1 X( f ) = a + j 2π f To obtain the magnitude and phase spectrum: a j 2π f a 2π f X( f ) = − = A j B | | a2 + (2π f )2 a2 + 4π2 f 2 − a2 + 4π2 f 2     1 1 X( f ) = A2 + B2 = = | | a2 + 4π2 f 2 √a2 + w2 p π 1 2 f p 1 w X( f ) = tan− − = tan− | | a − a     1 X( f ) 1 for a = 1, X( f ) = , | = tan− w | | √1 + w2 −

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◦ | − − − − − − − − Fourier Series and Fourier Transform 113 •

e at t > 0 (ii) x(t) = e a t = − − | | eat t > 0  e−a|t|

t

a t Fig. 2.13. Graphical representation of e− | |

1 1 2a x(w) = + = a + jw a jw a2 + w2 − 2 for a = 1 X(w) = 1 + w2 2 X(w) X(w) = | = 0 | | 1 + w2 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 | |

|X(w)| 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 −10 .... −5 −4 −3 −2 −1 01 2 3 4 5 .... 10 w

Fig. 2.14. Magnitude plot a t (iii) x(t) = e− | | sgn(t) x(t) = e−a | t| sgn(t)

1.0

t

−1.0

a t Fig. 2.15. Graphical representation of e− | |sgn(t) Fourier Series and Fourier Transform 115 •

(ii) x(t) = 1 ∞ jwt X(w) = e− dt = ∞ Z∞ − This means Dirichlet condition is not satisfied. But its F.T. can be calculated with the help of duality property. FT δ(t) 1 ←→ FT Duality property states that: x(t) X(w) then ←→ FT X(t) 2πx( w) ←→ − Here X(t) = 1, then x( w) will be − x(t) = δ(t); X(w) = 1 FT then X(t) = 1; 1 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 − FT 1 2πδ( w) ←→ − Thus, if x(t) = 1, then X(w) = 2πδ(w) 1 t > 0 (iii) x(t) = sgn(t) sgn(t) = 1 t < 0  −  sgn(t)

1

0 t −1

Fig. 2.17. Graphical representation of sgn(t)

x(t) = 2u(t) 1 − Differentiating both the sides d d x(t) = 2 u(t) = 2δ(t) dt dt Taking the F.T. of both sides d F x(t) = 2F[δ(t)] dt   jwX(w) = 2 2 X(w) = jw ∞ 0 jwt jwt X(w) = e− dt e− dt − Z0 Z∞ − 116 Basic System Analysis •

(iv) x(t) = u(t) sgn(t) = 2u(t) 1 − 2u(t) = 1 + sgn(t) Taking F.T. of both sides 2 2F[2u(t)] = F(1) + F[sgn(t)] = 2πδ(w) + jw FT 2 2u(t) 2πδ(w) + ←→ jw FT 1 u(t) πδ(w) + ←→ jw Properties of unit impulse: ∞ (1) x(t)δ(t) = x(0) Z∞ − (2) x(t)δ(t t0) = x(t0)δ(t t0) ∞ − −

(3) x(t)δ(t t0)dt = x(t0) − Z∞ − δ 1 δ (4) (at) = a (t) ∞ | | (5) x(τ)δ(t x)dt = x(t) − Z∞ − δ d (6) (t) = dt u(t)

Example 15: Obtain the F.T. of a rectangular pulse shown in Fig. 2.18.

x(t) 1

−T/2 0 T/2 t

Fig. 2.18. Rectangular pulse

SOLUTION:

T 2 jwt 1 jw T jw T 2 wT X(w) = e− dt = − e− 2 e 2 = sin jw − w 2 ZT   −2 h i wT wT sin π π wT sin π π X(w) = T 2 = sinc = T 2 π wT 2π π wT 2π
  2π
Fourier Series and Fourier Transform 117 •

Sampling function or interpolating function or filtering function denoted by Sa(x) or sinc(x) as shown in figure. sinπx sinc(x) = πx (1) sinc(x) = 0 when x = nπ ± (2) sinc(x) = 1 when x = 0 (using L’Hospital’s rule) π 1 (3) sinc(x) is the product of an oscillating signal sin x of period 2 and a decreasing signal x . Therefore, π 1 sinc(x) is making sinusoidal of oscillations of period 2 with amplified decreasing continuously as x .

sin c(x)

1.0

−4π −3π −2π −π 0 π 2π 3π 4π 5π x

Fig. 2.19. Sine function

sinπx 0 sincx = ; sinc(0) = = 1 L’Hospital rule πx 0 sinπ sinc(1) = = 0; sinc( 1) = 0 π − 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

.9 .8 .7 .6 .5 .4 .3 .2 .1 −3.5 −3 −2.5 −2 −1.5 −1 −.75 −.5 −.25 0.25 .5.75 11.5 2 2.5 3 3.5 t

Fig. 2.20. Sine function Fourier Series and Fourier Transform 119 •

Example 17: Obtain F.T. and spectrums of following signals: (i) x(t) = cosw0t (ii) x(t) = sinw0t

SOLUTION: 1 1 (i) x(t) = cosw t = e jw0t + e jw0t 0 2 2 −

FT 1 FT 1 2πδ(w); πδ(w) ←→ 2 ←→ FT Frequency shifting property states that e jβt x(t) X(w β) ←→ − 1 jw0t FT e πδ(w w0) 2 ←→ − 1 jw0t FT e− πδ(w + w0) 2 ←→ 1 1 F [x(t)] = FT e jw0t + e jw0t 2 2 −  

X(w) = π[δ(w w0) + δ(w + w0)] −

| X(w) |

π

−w0 w0 w

Fig. 2.22. Magnitude plot of cosw0t

(ii) x(t) = sinw0t π X(w) = [δ(w w0) δ(w + w0)] j − −

| X(w) |

π

−ω0

w0 w

π

Fig. 2.23. Magnitude plot of sinw0t 120 Basic System Analysis •

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

at 1 FT e− = a + jw   d d at d 1 (a + jw) dw (1) 1 dw (a + jw) 1 FT(te− ) = j = j − = dw a + jw (a + jw)2 (a + jw)2  

Inverse Fourier Transform: (IFT)

Example 19: Find the IFT of (i) X w 2 jw+1 by partial fraction expansions ( ) = ( jw+2)2 (ii) X w 1 by convolution property ( ) = (a+ jw)2 w (iii) X(w) = e−| | 2w (iv) X(w) = e− u(w)

SOLUTION:

A B (i) X(w) = + ; 2 jw + 1 = A( jw + 2) + B A = 2 2A + B = 1 B = 3 jw + 2 ( jw + 2)2 − 2 3 X(w) = jw + 2 − ( jw + 2)2 2t 2t x(t) = 2e− u(t) 3te− u(t) − 1 1 (ii) X(w) = = = X (w)X (w) (a + jw)2 (a + jw)(a + jw) 1 2 1 1 X (w) = , X (w) = 1 a + jw 2 a + jw at at x1(t) = e− u(t), x2(t) = e− u(t) Using convolution property

x(t) = x1(t)∗x2(t) FT x(t) X(w) ←→ FT x1(t)∗x2(t) X1(w)X2(w) ∞ ←→ u(τ) = 1 τ 0 at a(t τ) x(t) = e− u(t)e− − u(t τ)dτ ≤ − τ τ Z∞ ( u(t ) = 1 t − − ≤ t at at = e− dτ = te− u(t) Z0 122 Basic System Analysis •

Example 20: Find the F.T. of the function

(t t0) x(t t0) = e− − u(t t0) − −

SOLUTION: If F[x(t)] = X(w)

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

SOLUTION:

e j2πt + e j2πt FT(cos2πt) = FT − 2   FT[1] = 2πδ(w)

jw0t FT[e ] = 2πδ(w w0) − F[cos2πt] = πδ(1w 2π) + πδ(w + 2π) (1) − 1 jwt 1 jw jw 2sinw F[u(t + 1) u(t 1)] = e− dt = e− e = (2) − − − jw − w Z1 −
F[x(t)] = F[ u(t + 1) u(t 1) cos2πt] { − − } x(t) is multiplication of (1) and (2), so by using multiplication property ∞ FT 1 1 x(t)y(t) X1(w)∗Y1(w) = X(τ)Y(w τ)dτ ←→ 2π 2π − Z∞ − ∞ 1 2sinτ X(w) = πδ(w 2π τ) + δ(w + 2π τ) dτ 2π  τ − − −  Z∞ − ∞  ∞  sinτ sinτ X(w) = δ(w 2π τ)dτ + δ(w + 2π τ)dτ τ − − τ − Z∞ Z∞ − − ∞

Since x(t)δ(t t0)dt = x(t0) − Z∞ − X(w) = sin(w 2π)/(w 2π) + sin(w + 2π)/(w + 2π) − − Fourier Series and Fourier Transform 123 •

Example 22: Determine the Fourier transform of a triangular function as shown in figure.

x(t) A

−TT t

Fig. 2.24. Triangular pulse

SOLUTION:

x(t)

(0, A) a → → b

(−T,0) (T,0) t

Equation of line (a) is t x(t) = A + 1 T Equation of line (b) is   t x(t) = A 1 − T Mathematically, we can write x(t) as   t t x(t) = A + 1 [u(t + T) u(t)] + A 1 [u(t) u(t T)] T − − T − −     A A x(t) = (t + T)[u(t + T) u(t)] + (T t)[u(t) u(t T)] T − T − − − A A x(t) = (t + T)u(t + T) (t + T)u(t) + [(T t)u(t) (T t)u(t T)] T − T − − − − n o n o A A x(t) = r(t + T) tu(t) Tu(t) + Tu(t) tu(t) + r(t T) T − − T − − n o n o A A = r(t + T) r(t) Tu(t) + Tu(t) r(t) + r(t T) T − − T − − n o n o A = r(t + T) 2r(t) + r(t T) T − − hn oi A e jwT 2 e jwT X( jw) = + − T ( jw)2 − ( jw)2 ( jw)2   Fourier Series and Fourier Transform 125 •

1 1 < t < 1 Π(t) = rect(t) = − 2 2 0 otherwise  1 1 t 5 < 1 rect(t 5) = − 2 ≤ − 2 − 0 otherwise  9 11 1 2 t 2 rect(t 5) = ≤ ≤ − ( 0 otherwise ∞ ∞ jwt jwt X( jw) = x(t)e− dt = rect(t 5)e− dt − Z∞ Z∞ − − 11/2 e jwt 11/2 = e jwt dt = − − jw Z 9/2 9/2 −

w j11w 9 jw 9 j w e 11 j 2 e 2 e 2 e 2 − = − − − = − − jw jw − 5 jw jw/2 5 jw jw/2 5 jw jw/2 jw/2 e e e e 2e− e e− = − − − − = − jw w2 j
2e 5 jw w sin w = − sin = e 5 jw 2 w 2 − w  2  w X( jw) = e 5 jwS − a 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) ∞ jwt FT(r(t)) = R( jw) = r(t)e− dt Z∞ − ∞ jwt = (Ax1(t) + Bx2(t))e− dt Z∞ − Fourier Series and Fourier Transform 127 • ∞ j( w)τ = x(τ)e− − dτ Z∞ − F(x(t)) = X( jw) − (4) Time shifting If FT (x(t)) = X( jw) jwt then FT (x(t t0)) = e 0 X( jw) − − Proof:

Let r(t) = x(t t0) − ∞ ∞ jwt jwt R( jw) = r(t)e− dt = x(t t0)e− dt − Z∞ Z∞ − − ∞ jwt R( jw) = FT(x(t t0)) = x(t t0)e− dt − − Z∞ − Let t t0 = τ dt = dτ − ∞

jw(t0+τ) FT (x(t t0)) = x(τ)e− dτ − Z∞ − ∞ jwt jwt = x(τ)e− e− 0 dτ Z∞ − ∞ jwt jwτ = e− 0 x(τ)e− dτ Z∞ −

jwt0 jwt FT (x(t t0)) = e− X( jw). Similarly, FT (x(t +t0)) = e 0 X( jw) − jwt0 So FT (x(t t0)) = e± X( jw) ± (5) Frequency shifting If FT (x(t)) = X( jw)

jw t FT (e 0 x(t)) = X( j(w w0)) − jw t Let r(t) = e 0 x(t) ∞ jw t jw t jwt FT (r(t)) = FT e 0 x(t) = R( jw) = e 0 x(t)e− dt Z∞
− ∞ jw t j(w w )t FT e 0 x(t) = x(t)e− − 0 dt Z∞
− 128 Basic System Analysis •

Let w w0 = w0 − ∞ jw t = x(t)e− 0 dt Z∞ −

jw0t FT e x(t) = X( jw0) = X( j(w w0)) − jw0t
Similarly, FT(e− x(t)) = X( j(w + w0))

jw0t We can write as FT e± x(t) = X( j(w w0)) ∓

(6) Duality or symmetry property If FT (x(t)) = X( jw) then FT (x(t)) = 2πx( jw) − Proof: 1 ∞ jwt We know that x(t) = 2π ∞ X( jw)e dw − Replacing t by t, we getR − ∞ 1 jwt x( t) = X( jw)e− dw − 2π Z∞ − ∞ π 2 jwt 2π x( t) = X( jw)e− dw − 2π Z∞ − ∞ jwt 2π x( t) = X( jw)e− dw − Z∞ − Interchanging t by jw

∞ jwt 2π x( jw) = X(t)e− dt − Z∞ − 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. Fourier Series and Fourier Transform 129 •

Proof: ∞

r(t) = x1(t)∗x2(t) = x1(τ)x2(t τ)dτ − Z∞ ∞ − jwt FT(r(t)) = R( jw) = r(t)e− dt Z∞ − ∞ ∞ jwt = x1(τ)x2(t τ)dτ e− dt  −  Z∞ Z∞ − − ∞ ∞  jwt = x1(τ)x2(t τ)dτ e− dt − Z∞ Z∞ − ∞− ∞ jwt = x1(τ)dτ x2(t τ) e− dt − Z∞ Z∞ − − Let t τ = ∝ so dt = d ∝ − ∞ ∞ jw(∝+τ) FT[x1(t)∗x2(t)] = x1(t)dτ x2(∝) e− d ∝ Z∞ Z∞ − ∞ − ∞ jw∝ jwτ = x1(τ)dτ x2(∝) e− e− d ∝ Z∞ Z∞ − ∞ − ∞ jwτ jw∝ = x1(τ) e− dτ x2(∝) e− d ∝ Z∞ Z∞ − − FT[x1(t)∗x2(t)] = X1( jw) X2( jw)

(8a) Integration in time domain If FT (x(t)) = X( jw) t τ τ 1 then FT ∞ x( )d = jw ( jw) − × R t Proof: Let r(t) = ∞ x
(τ)dτ − DifferentiatingR w.r.t. t dr(t) d = x(t) FT(x(t)) = FT r(t) dt ⇒ dt   From differentiation in time domain X( jw) = jwX( jw) 1 R( jw) = X( jw) jw t 1 FT(r(t)) = FT x(τ)dτ = X( jw)   jw Z∞ −   130 Basic System Analysis •

(8b) Differentiation in time domain If FT (x(t)) = X( jw) then d x(t) = jw ( jw) dt × ∞
1 Proof: We know that x(t) = X( jw)e jwt dw. Differentiating both sides w.r.t. t 2π Z∞ − ∞ d 1 d x(t) = X( jw) e jwt dw dt 2π dt Z∞   − ∞ 1 = jwX( jw)e jwt dw 2π Z∞ − ∞ 1 = j (wX( jw))e jwt dw 2π Z∞ − d x(t) = j FT 1(wX( jw)) dt −

d dn n yields FT dt x(t) = jwX( jw). On generalizing we get FT dtn x(t) = ( jw) X( jw)
  (9) Differentiation in frequency domain If FT (x(t)) = X( jw) d then FT (tx(t)) = j dw X( jw) ∞ jwt Proof: We know that X( jw) = ∞ x(t)e− dt − On differentiating both sidesR w.r.t. w ∞ ∞ d d jwt jwt X( jw) = x(t) e− dt = j t x(t)e− dt dw dw − Z∞   Z∞ − − Multiplying both sides by j ∞ d jwt 2 2 j X( jw) = (tx(t))e− dt since j = 1 or j = 1 dw − − Z∞ − d j X( jw) = FT[t x(t)] dw d FT[t x(t)] = j X( jw) dw

(10) Convolution in frequency domain (multiplication in time domain (multiplication theorem))

If FT(x1(t)) = X1( jw) and FT[x2(t)] = X2( jw) 1 FT(x (t)x (t)) = (X ( jw) X ( jw)) 1 2 2π 1 ∗ 2 132 Basic System Analysis •

Proof: ∞ ∞ 2 E = x(t) dt = x(t)x∗(t)dt (1) Z∞ Z∞ − ∞ − 1 We know that x(t) = X( jw)e+ jwt dw 2π Z∞ − ∞ 1 So x (t) = X( jw)e jwt dw (2) ∗ 2π − Z∞ − on putting (1)

∞ ∞ 1 jwt = x(t) X∗( jw)e− dw dt 2π  Z∞ Z∞ − − ∞  ∞  1 = X ( jw) x(t)e jwt dt dw 2π ∗ − Z∞ Z∞ − ∞ − 1 = X( jw)X ( jw)dw 2π ∗ Z∞ ∞ − ∞ 1 2 = x(t)2 dt = X( jw) dw 2π Z∞ Z∞ − − Relation between Laplace Transform and Fourier Transform Fourier transform X( jw) of a signal x(t) is given as ∞ jwt X( jw) = x(t)e− dt (1) Z∞ − F.T. can be calculated only if x(t) is absolutely integrable ∞ = x(t) dt < ∞ (2) Z∞ − Laplace transform X(s) of a signal x(t) is given as ∞ st X(s) = x(t)e− dt (3) Z∞ − We know that s = σ + jw ∞ (σ+ jw)t X(s) = x(t)e− dt Z∞ − ∞ σt jwt X(s) = x(t)e− e− dt (4) Z∞ −   Fourier Series and Fourier Transform 133 •

σt Comparing (1) and (4), we find that L.T. of x(t) is basically the F.T. of [x(t)e− ]. ∞ jwt If s = jw, i.e. σ = 0, then eqn. (4) becomes X(s) = ∞ x(t)e− dt = X( jw) σ − Thus, X(s) = X( jw) when = 0 or s = jw R 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 t Example 24: Determine the voltage Vout (t) to a current source excitation i(t) = e− u(t) for the circuit shown in figure. + 1 1Ω F i(t) ↑ 2 Vout(t) −

Fig. 2.26.

SOLUTION: + ↓ i 1(t) ↓ i2(t) 1 i(t) ↑ 1Ω 2 F Vout(t) −

i(t) = i1(t) + i2(t) V (t) 1 dV (t) i(t) = out + out 1 2 dt V since i = R dv 1 ( and i = c dt or v = c idt 1 dV (t) R e t u(t) = V (t) + out (1) − out 2 dt On taking the z-transform on both sides 1 jw (2 + jw) = V ( jw) 1 + = V ( jw) 1 + jw out 2 2 out   2 A B V ( jw) = = + out (1 + jw)(2 + jw) 1 + jw 2 + jw 2 2 Vout ( jw) = 1 + jw − 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 − −   Fourier Series and Fourier Transform 135 • 2 2 V ( jw) = = 0 6( jw)2 + 7( jw) + 1 (6 jw + 1)( jw + 1) 1/3 A B V ( jw) = = + 0 jw 1 6 jw 1 1 1 jw ( + / )( + ) 6 + jw + 2 2 V ( jw) = (5) 0 1 5 1 jw 5 6 + jw − ( + ) Taking inverse Fourier transform, we get
2 t/6 t V0(t) = e− e− u(t) (6) 5 −   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)

v(t)∼ 1F 0 π wt

(a) (b)

Fig. 2.28.

SOLUTION: Waveform V(t) is defined as V(t) = sint(u(t) u(t π)) (1) − −

1Ω

a u(t) 1F ∼ i(t)

Let i(t) be the current in the loop. Applying KVL in loop t t 1 V(t) = 1 i(t) + i(t)dt = i(t) + i(t)dt (2) · 1 Z0 Z0 On taking Fourier transform of 1 e jπw V( jw) = + − ( jw)2 + 1 ( jw)2 + 1 1 Since FT sintu(t) = ( jw)2 + 1   e jπw FT sintu(t π) = − − ( jw)2 + 1   136 Basic System Analysis •

Solve using F.T. formula

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

I1( jw) I2( jw) A B jw + c | {z } I ( jw) = + 1 jw + 1 (( jw)2 + 1)

1/2 1 ( jw + 1) = − + 2 ( jw + 1) (( jw)2 + 1)

1 t 1 1 1 i1(t) = e− u(t) + costut + sintδt + sintu(t) −2 2 2 2

Since IFT 1 sintu t ( jw)2+1 = ( ) n o so IFT jw d sintu t ( jw)2+1 = dt ( ) Using differential in time domain property

jw IFT = costu(t) + sintδ(t) ( jw)2 + 1   jw 1 jπw I2( jw) = e− ( jw + 1) · (( jw)2 + 1) · jπw I2( jw) = I3( jw) e− · Since I3 = I1( jw)

1 t 1 1 1 so i3(t) = e− u(t) + costu(t) + sintδ(t) + sintu(t) −2 2 2 2 Fourier Series and Fourier Transform 137 •

jwt From time shifting property FT(x(t t0)) = e 0 ( jw) ± ± × i2(t) = i3(t π) −

1 (t π) 1 1 1 = e− − u(t π) + cos(t π)u(t π) + sin(t π)δ(t π) + sin(t π)u(t π) −2 − 2 − − 2 − − 2 − −

1 t 1 1 (t π) so i(t) = e− + cost + sint u(t) + sintδ(t) + e− − + cos(t π) + sin(t π) u(t π)+ 2 − − 2 2 − − − −   h i 1 sin(t π)δ(t π) 2 − −

Example 27: For the RC circuit shown in figure.

R

i(t) x(t) 1 C 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)

t 1 x(t) Ri(t) i(t)dt = 0 − − C Z∞ − t 1 x(t) = Ri(t) + i(t)dt (1) C Z∞ − Since

 VR = iR  1  Vc = C i(t)dt  t R 1 and y(t) = i(t)dt (2) C Z∞ − Fourier Series and Fourier Transform 139 •

A = B C − 1 H( jw) = (8) √1 + w2

H( jw) = 1 (1 + jw) −

1 0 1 1 = tan− tan− w = tan− w (9) 1 − − 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 − − ◦

| H(jw) | 1

−50 −40 −30 −20−100 10 20 30 40 50 w

Fig. 2.30. Magnitude plot frequency response of the circuit 140 Basic System Analysis •

∠H(jw) 90°

45°

−50 −40 −30 −20−100 10 20 30 40 50 w

−45°

−90°

Fig. 2.31. Phase plot

Example 28: For the circuit shown in figure, determine the output voltage V0(t) to a voltage source excitation t Vi(t) = e− u(t) using Fourier transform 2Ω

+ + Vin (t)− 1 1H V0(t) −

Fig. 2.32. SOLUTION:

t Since Vin(t) = e− u(t) (1) 1 V = (2) in( jw) 1 + jw Applying KVL in loop (1) di(t) V = 2i(t) + 1 in(t) · dt di(t) V = 2i(t) + (3) in(t) dt di(t) V0(t) = 1 · dt di(t) V (t) = (4) 0 dt 142 Basic System Analysis •

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;

x(t)

1 t −1 −0.5 0 0.5 t −1

Fig. 2.3 P.

Q5: Obtain the Fourier series component of the periodic square wave signals.

x(t) 1

−T/2 −T/4 0 T/4 T/2 t

−1

Fig. 2.4 P.

Q6: Determine the Fourier transform of the Gate function

x(t) A

−T/2 T/2 t

Fig. 2.5 P.

Q7: Determine the Fourier series representation of the signal t t2 for π t π x(t) = − − ≤ ≤ 0 elsewhere  Fourier Series and Fourier Transform 143 •

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 ∞ jnw0t x(t) = ∑ Cne n= ∞ − 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.

∠X(w) π/2 | X(w) | 1 w0

−w0 w −π/2 − w0 w0 w (a) (b)

Fig. 2.6 P. Q12: Find the Fourier transform of the following waveform.

x(t)

1

−b −a 0 a b t

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) 144 Basic System Analysis •

Q16: Find the Fourier transform of the signal x(t) shown in figure below.

x(t) A

0 T 2T 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

d2y(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.

x(t) K

−T 0 T 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

Asin (w0t), 0 t T0/2 x(t) = ≤ ≤ 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 1 1 1 x(t) = δ(t + 1) + δ(t 1) + δ t + δ + t 2 − 2 − 2      146 Basic System Analysis •

1 ( 1)n a = − − n n2π2 1 b = n nπ

Q3: x(t) 1

−1 −0.5 0 0.5 1 t −1

T = 1

w0 = 2π rad/sec

y2 y1 y y1 = − (x x1) − x2 x1 − − x(t) = 2t + 1 − t0+T 2 a = x(t)cos nw tdt n T 0 tZ0

an = 0

Q4: x(t) 1.0

−T/2 −T/4 T/4 T/2 t

−1.0

T T 3T 2π 8π = ; w0 = = 2 − − 4 4 3T 3T   4 1 T t T x t − 4 ≤ ≤ 4 ( ) = T T ( 1 t − 4 ≤ ≤ 2
T T 4 2
1 4 T 1 a0 = dt + ( 1)dt = = 3T  −  3T 4 3 4  ZT ZT  − 4 4   Fourier Series and Fourier Transform 147 •

T T 4 2 8 8nπ 8nπ an = cos dt cos tdt 3T  3T − 3T   ZT ZT  − 4 4 π π 1  2n 4n  an = 3sin sin nπ 3 − 3   bn = 0, since even function 1 1 2π 4π 3 4π 1 8π x(t) = + 3sin sin + sin sin + 3 π 3 − 3 2 3 − 2 3 ···   Q5: x(t) A

−T/2 T/2 t

A T t T x(t) = − 2 ≤ ≤ 2 ( 0 elsewhere

T 2 2A wT AT wT X( jw) = A e jwt dt = sin = sin − w 2 wT 2 ZT 2 − 2 X(if ) = AT sinc f T

Q6: T0 = 2π;

w0 = 1;

π π2 1 2 a0 = t t dt = − 2π − 3 Zπ −
π n 1 2 4( 1) an = t t cos ntdt = − − π − n2 Zπ −
π n 1 2 2( 1) bn = t t sin ntdt = − − π − n Zπ −
Fourier Series and Fourier Transform 149 •

Taking inverse Fourier transform w0 jw0t 1 jwt 1 e x1(t) = j e dw = − 2π − 2πt Z0 0 1 1 e jw0t x (t) = j e jwt dw = − − 2 2π 2πt Zw − 0 1 jw0t jw0t x(t) = x1(t) + x2(t) = (1 e + 1 e− ) 2πt − − 2 w0t 1 2sin 2 = (2 2cos w0t) = 2πt − πt

Q11:

x(t) 1.0

−b−a0 a b t

t+b b a for b < t < a − − − x(t) = 1 for a < t < a  −  t b  a−b for a < t < b −  2 X( jw) =  (cos wa cos wb) w2(b a) − −

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 d FT(tx(t)) = j X( jw) dw 1 X ( jw) = 1 ( jw)2 1 jwt 1 jw X2( jw) = 1.e− dt = (1 e− ) jw − Z0 1 jw X( jw) = X1( jw)X2( jw) = (1 e− ) ( jw)3 − 150 Basic System Analysis •

Q13: Prove convolution in time domain property.

Q14:

x(t) (T,A) A

0 T 2T t (0,0)

A t 0 < t < T x(t) = T A T < t < 2T  T 2T A X( jw) = te jwt dt + A e jwt dt T − − Z0 ZT

T T A te jwt e jwt e jwt 2T X( jw) = − − dt + A − T  jw − jw  jw T Z Z   − 0 0 −   jwt j 2wT jwT A Te 1 jwT e− e− = + e− 1 + A − T jw w2 − jw  −   − 
jwt e− 1 jwT A jwT jwT = A + e− 1 e− e− 1 jw w2T − − jw −  − 

jw Ae− T A jwT A A j 2wT A jwT = + e− e− + e− jw w2T − w2T − jw jw

A 1 jwT 1 2 jwT = e− + jTe− wT w − w  

Q15: d2y(t) dy(t) dx(t) + 4 + 3y(t) = + 2x(t) (1) dt2 dt dt

Taking Fourier transform on both sides ( jw)2Y( jw) + 4( jw)Y( jw) + 3Y( jw) = ( jw)X( jw) + 2X( jw)

( jw)2 + 4( jw) + 3 Y( jw) = (( jw) + 2)X( jw) (2) Y( jw) 2 + jw Frequency response H( jw) = =
(3) X( jw) ( jw)2 + 4 jw + 3 2 + jw A B H( jw) = = + (3 + jw)(1 + jw) 3 + jw 1 + jw 152 Basic System Analysis •

T0 x(t) = Asinw0t for 0 t ≤ ≤ 2

T0 = 0 for t T0 2 ≤ ≤

T0 2 T 1 A cosw0t 0 C = Asinw tdt = − 2 0 T 0 T w 0 0 Z 0  0  0

A T A A cosw 0 1 cosπ 1 = 2π 0 = π[ ] = −T0 · 2 − −2 − 2 · T0  

T0 2 1 jnw0t Cn = Asinw0te− dt T0 Z0

T0 2 A jw t jnw t jnw t = (e 0 e− 0 )e− 0 dt 2 jT0 − Z0

T0 2 A jw t(1 n) jw t(n+1) = e 0 − e− 0 dt 2 jT0 − Z0  

A e jw0t(1 n) e jw0t(n+1) T0 = − − 2 2 jT jw (1 n) − jw (n + 1) 0 0 0 0 ! − − T T jw 1 n 0 jw n 1 0 A e 0( ) 2 e 0( + ) 2 1 1 = − + − 2 jT0w0 1 n (n + 1) − 1 n − n + 1 − − ! A e jπ(1 n) e jπ(n+1) 1 1 = − + − −4π 1 n n + 1 − 1 n − n + 1 " − − # A e jπe jnπ e jnπ e jπ 1 1 = − + − · − −4π 1 n n + 1 − 1 n − n + 1  − −  jnπ jnπ A e e 1 1 π = − − − Since e j = 1 −4π 1 n − n + 1 − 1 n − n + 1 −  − −  A 2e jnπ 2 = − + 4π 1 n2 1 n2  − − 

A jnπ = (e− + 1) 2π(1 n2) − Fourier Series and Fourier Transform 153 •

Q19: 1 1 1 x(t) = δ(t + 1) + δ(t 1) + δ t + + δ t 2 − 2 − 2      Taking Fourier transform on both sides ∞ jwt X( jw) = x(t)e− dt (1) Z∞ − ∞ 1 1 1 jwt X( jw) = δ(t + 1) + δ(t 1) + δ t + + δ t e− dt 2 − 2 − 2 Z∞      − ∞ ∞ ∞ 1 jwt jwt 1 jwt X( jw) = δ(t + 1)e− dt + δ(t 1)e− dt + δ t + e− dt 2  − 2 Z∞ Z∞ Z∞   − − − ∞ 1 jwt + δ t e− dt − 2  Z∞   −  Since FT(δ(t)) = 1

jwt0 So FT(δ(t t0)) = e± dt using time shifting property ± { } 1 jw jw j w j w X( jw) = e + e + e 2 + e 2 2 − −   j w j w e jw + e jw e 2 + e 2 X( jw) = − + − 2 2 w X( jw) = cos w + cos 2

OBJECTIVE TYPE QUESTIONS

Q1: If the Fourier transform of a function x(t) is X( jw), then X( jw) is defined as ∞ jwt ∞ dx(t) jwt (a) ∞ x(t)e dt (b) ∞ dt e− dt −∞ −∞ jwt (c) R ∞ x(t)dt (d) R ∞ x(t)e− dt − − R R Q2: If X( jw) be the Fourier transform of x(t), then 1 ∞ jwt 1 ∞ jwt (a) x(t) = 2π ∞ X( jw)e dw (b) x(t) = 2π ∞ X( jw)e− dw 1 −∞ jwt 1 −∞ jwt (c) x(t) = 2π R0 X( jw)e dw (d) x(t) = 2π R ∞ X( jw)e− dw − R R Q3: Fourier transform of x(t) = 1 is (a) 2πδ(w) (b) πδ(w) (c) 3πδ(w) (d) 4πδ(w)

Q4: Fourier transform of x(t t0) is jwt −jwt 1 jwt (a) e 0X( jw) (b) e 0X( jw) (c) X( jw) (d) t0e 0X( jw) − t0 − Fourier Series and Fourier Transform 155 •

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 + ∑ (an cosnw0t + bn sinw0t) ◦ n=1 (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 . Also find the Fourier −| | −| | = 1+w2 transform of 4t using duality property. (1+t2)2  

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.

x(t) 1

−T/2 0 T/2 t

Fig. 2.10 P.

T 2 wt Ans. X( jw) = 2 sinc ( 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)] − π Q8: Show that FT [sinw tu(t)] = w0 + j [δ(w + w ) δ(w w )]. 0 w2 w2 2 0 0 0− − − Q9: Find inverse Fourier transform of X jw jw ( ) = (1+ jw)2 d t Ans. x(t) = dt [te− u(t)] 156 Basic System Analysis •

Q10: Sketch and then find the Fourier transform of following signals

x1(t)

3 3 −1 w w (a) x1(t) = π t + +π t Ans. (a) X1( jw) = 2sinc cos3 2 − 2 2 2

−2 −1 1 2 t

Fig. 2.11 P.

x2(t) 2

π t π t −1 (b) x2(t) = 4 + 2 Ans. (b) X2( jw) = 4sinc2w+2sincw

−2 −1 1 2 t

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 y( jw) 1 x( jw) = = x( jw) 1 + jwRC

↑ ↑

R y(l) ↑ x(t) −C ↑

Fig. 2.13 P.

1 Ans. x( jw) = | | √1 + w2 1 x( jw) = tan− w − Q12: Find the Fourier series of the waveform shown in figure. 2A x(t) = for n = 1,3,5,7 jnπ 158 Basic System Analysis • 1 1 2 x(tπ)t = + cos5πt cos10 π 2 − 3π 2 cos15πt − 8π

Q15: The output of a system is given by

Asinw0t for 0 t π x(t) = ≤ ≤ 0 for π t 2π  ≤ ≤ Determine trigonometric form of Fourier series of x(t) A A π ∞ 2A Ans. x(t) = + cos(nt ) + ∑ cosnt π 2 − 2 π(1 n2) " n=2 − #