1

Solution of Assignment 3 Q1: We know that

  t rect ←→ T sinc(πf T ) T

Therefore, S(f ) = T 2 sinc2 (πf T ). Option (c) is correct. 1 Q2: S(f0 ) = 0 ⇒ sin(πf0 T ) = 0 ⇒ f0 = T Option (a) is correct. Q3: Fourier transform can be calculated as: Z X(f ) =

∞

exp(−σt) exp(−2πjf t)dt =

0

1 σ + 2πjf

1 Therfore, S(f ) = |X(f )|2 = 2 σ + 4π 2 f 2 Option (c) is correct. Q4: Energy contained within the band [−f, f ] is calculated as:   Z f df 0 1 2πf −1 E(f ) = = tan 2 2 02 πσ σ −f σ + 4π f Therefore, total energy of the signal is: lim E(f ) =

f →∞

1 2σ

From definition, σ 0.9 ⇒ f90 = tan E(f90 ) = 2σ 2π



0.9π 2



Option (b) is correct. Q5: Using differentiation property, x(t) = sinc(tT ) ←→ Therefore, S(f ) =

4π 2 f 2 rect2 T2

1 rect T

    t t dx(t) 2πjf ⇒ ←→ rect T dt T T

  t T

Option (d) is correct. Q6: Note that, Z E(f ) =

f

S(f 0 )df 0

0

dE(f ) Then, S(f ) = = f 2 exp(−σf ). df Option (d) is correct.   σf S(f ) = f exp − , f > 0. The Fourier transform therefore, can 2   σf exp (−2πjf T ) , f > 0 X(f ) = f exp − 2

Q7: The magnitude response is given by: |X(f )| = be calculated as:

p

2

The signal is given by Inverse Fourier transform: Z ∞ Z 0 X(f ) exp(2πjf t)df X(f ) exp(2πjf t)df + x(t) = 0 Z Z−∞ ∞ ∞ X(f ) exp(2πjf t)df X(−f ) exp(−2πjf t)df + = 0 0 Z ∞ Z ∞ X(f ) exp(2πjf t)df X ∗ (f ) exp(−2πjf t)df + = 0

0

Now, ∞

Z

∞

Z X(f ) exp(2πjf t)df =

0

0

  σf f exp − exp(2πjf (t − T ))df 2

and Z

∞

X ∗ (f ) exp(2πjf t)df =

0

Z 0

Therefore,

Z x(t) = 0

∞

∞

  σf exp(2πjf (t + T ))df f exp − 2

  σf 2f exp − cos(2πf T ) exp(2πjf t)df 2

Option (a) is correct. Q8: Autocorrelation can be calculated as follows: Z ∞ Rx (τ ) = x(t)x(t + τ )dt −∞   Z ∞  1 1 exp − t2 + (t + τ )2 dt = 2 −∞ 2π    Z ∞ 1 τ2 1 exp − 2(t + τ /2)2 + dt = 2 2 −∞ 2π  2Z ∞  2 τ 1 1 y √ exp − = √ exp − dy 4 2 2 π 2π  2  −∞ 1 τ = √ exp − 4 2 π

Replacing y =

√

2(t + τ /2)

Option (b) is correct. Q9: Fourier transform of Autocorrelation function, Rx (τ ) (which equals to the ESD of x(t)) can be calculated as: Z ∞ S(f ) = Rx (τ ) exp(−2πjf τ )dτ −∞   Z ∞ 1 1 = √ exp − (τ + 4πjf )2 − 4π 2 f 2 dτ 4 2 π −∞  2 Z ∞ (τ + 4πjf ) 1 y √ = √ exp(−4π 2 f 2 ) exp − dy Replacing y = 2 2π 2 −∞ 2 2 = exp(−4π f ) Using the identity given in Q1 Option (b) is correct. Q10: Note that x(t) can be rewritten as follows: ( exp(−σt) [exp(jωt) + exp(−jωt)] t ≥ 0 x(t) = 0 elsewhere

3

Fourier transform, X(f ) of the signal can be found as follows: Z ∞ x(t) exp(−2πjf t)dt X(f ) = −∞ Z ∞ Z = exp(−(σ − jω + 2πjf )t)dt + 0

∞

exp(−(σ + jω + 2πjf )t)dt

0

1 1 + σ − jω + 2πjf σ + jω + 2πjf 2(σ + 2πjf ) 2(σ + 2πjf ) = 2 = 2 2 (σ + 2πjf ) + ω (σ − 4π 2 f 2 + ω 2 ) + 4πσjf =

Therefore, S(f ) = |X(f )|2 = Option (c) is correct.

4(σ 2 + 4π 2 f 2 ) . (σ 2 − 4π 2 f 2 + ω 2 )2 + 16π 2 σ 2 f 2