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RC2017 Signals and Systems Assign: Mon 17 April 2017

Homework 8 Due for RC2: Practical Problems

Instruction: Hand in your work with name and code to my desk by 10.00 am. of the due date. DO NOT copy homework from your classmates or lend it to others. Anyone who violates this regulation will be given -10 for the homework.

1. Compute the Fourier transform of each of the following siganals: (a) [e−αt cos ω0 t]u(t), α > 0 Solution: The given signal is 1 1 e−αt cos(ω0 t)u(t) = e−αt ejω0 t u(t) + e−αt e−jω0 t u(t) 2 2 Therefore, X(jω) =

1 1 1 1 − 2 jω + α − jω0 2 jω + α − jω0 

(b) e−3|t| sin 2t Solution: The given signal is x(t) = e−3t sin(2t)u(t) + e3t sin(2t)u(−t) We have x1 (t) = e−3t sin(2t)u(t)

FT ←→

X1 (jω) =

1 1 1 1 − 2j jω + 3 − j2 2j jω + 3 + j2

Also, x2 (t) = e3t sin(2t)u(−t) = −x1 (−t) X2 (jω) = −X1 (−jω) =

FT ←→

1 1 1 1 − 2j −jω + 3 − j2 2j −jω + 3 + j2

Therefore, X(jω) = X1 (jω) + X2 (jω) =

3j 3j − 2 9 + (ω + 2) 9 + (ω − 2)2 

1

{ 1 + cos πt, |t| ≤ 1 (c) x(t) = 0, |t| > 1 Solution: Using the equation ∫

∞

X(jω) =

x(t)e−jωt dt

−∞

we have X(jω) =

(d)

∞ ∑

2 sin ω sin ω sin ω + − ω π−ω π+ω 

αk δ(t − kT ), |α| < 1

k=0

Solution: Using the equation ∫

∞

X(jω) =

x(t)e−jωt dt

−∞

we have X(jω) =

1 1 − αe−jωT 

−2t

(e) [te sin 4t]u(t) Solution: We have x(t) =

1 1 −2t j4t te e u(t) − te−2t e−j4t u(t) 2j 2j

Therefore, X(jω) = [ (f)

][

sin πt sin 2π(t − 1) πt π(t − 1) Solution: We have x1 (t) =

1 1 1 1 − 2 2j (jω + 2 − j4) 2j (−jω + 2 + j4)2 

]

sin πt πt

FT ←→

2

{ 1, |ω| < π X1 (jω) = 0, otherwise

Also

{ e−2ω , |ω| < 2π FT X (jω) = 2 ←→ 0, otherwise 1 X(jω) = {X1 (jω) ∗ X2 (jω)} . 2π

sin 2π(t − 1) x2 (t) = π(t − 1) FT ←→

x(t)x1 (t)x2 (t) Therefore,

 −jω e , |ω| < π    (1/2π)(2π + ω)e−jω , −3π < ω < −π X(jω) =  (1/2π)(3π − ω)e−jω , π < ω < 3π    0, otherwise  (g) x(t) as shown in Figure 1 . x(t) 1

t

−2

2

0 −1

Figure 1: Question 1g Solution: Using Fourier Transform formula we obtain [ ] 2j sin ω X(jω) = cos 2ω − ω ω  (h) x(t) as shown in Figure 2 . Solution: If x1 (t) =

∞ ∑

δ(t − 2k),

k=−∞

then x(t) = 2x1 (t) + x1 (t − 1). 3

x(t) 2

1

−6

−5

−4

−3

−2

−1

t 1

0

2

3

4

5

6

Figure 2: Question 1h Therefore, X(jω) = X1 (jω)[2 + e

−ω

]=π

∞ ∑

δ(ω − kπ)[2 + (−1)k ]

k=−∞

{ 1 − t2 , 0 < t < 1 (i) x(t) = 0, otherwise Solution: Using the Fourier transform formula, we obtain X(jω) =

(j)

∞ ∑



2e−jω 2e−jω − 2 1 + − jω −ω 2 jω 2 

e−|t−2n|

n=−∞

Solution: x(t) is periodic with period 2. Therefore, X(jω) = π

∞ ∑

ˆ X(jkπ)δ(ω − kπ),

k=−∞

ˆ where X(jω) is the Fourier transform of one period of x(t). That is, ] [ −2 −2(1+jω) −2(1+jω) e [1 − e ] 1 1 − e ˆ X(jω) = − 1 − e−2 1 + jω 1 − jω  2. Determine the continuous-time signal corresponding to each of the following transforms 2 sin[3(ω − 2π)] (ω − 2π) Solution: From table, we have

(a) X(ω) =

W sin(W t) π 4

F ←→

rect

( ω ) 2W

Using delay and duality properties, then 3 sin[3(t + 2π)] π 3(t + 2π) ( ) t rect ej2πt 6

F ←→

(ω )

ej2πω 6 3 sin[3(−ω + 2π) sin(3(ω − 2π) 2π =2 π 3(−ω + 2π) (ω − 2π) rect

F ←→

Note: sin(−A) = − sin(A). Therefore, ( ) t ej2πt x(t) = rect 6 . (b) X(ω) = cos(4ω + π/3) Solution: From cos(4t +

π ) 3

F ←→

π[δ(ω + 4)ej 3 + δ(ω − 4)e−j 3 π

π

By duality, we obtain π[δ(t + 4)ej 3 + δ(t − 4)e−j 3 ] π

π

F ←→

2π cos(−4ω +

π π ) = 2π cos(4ω + ) 3 3

Therefore cos(4ω +

π ) 3

π π 1 [δ(t + 4)ej 3 + δ(t − 4)e−j 3 ] 2

F ←→

 (c) X(ω) as given by the magnitude and phase plots of Figure 3. ̸

1

|X(ω)| 1

−1

−1

ω

1

(a)

Figure 3: Question 2c

5

1

0 −1

ω 0

X(ω)

−3ω

Solution: From

∫ ∞ 1 F (ω)ejωt dω x(t) = 2π −∞ ∫ ∞ 1 = |F (ω)|ej∠X(ω) ejωt dω 2π −∞ ] [∫ 0 ∫ 1 1 jω(t−3) jω(t−3) = −ωe dω + ωe dω 2π −1 0

1 Using by-part integral, u = ω, du = dω and dv = ejω(t−3) dω, v = j(t−3) ejω(t−3) , we have [ ( ) 0 ∫ 0 1 ω 1 − x(t) = ejω(t−3) − ejω(t−3) dω 2π j(t − 3) −1 j(t − 3) −1 ( ] 1 ) ∫ 1 ω 1 + ejω(t−3) − ejω(t−3) dω j(t − 3) j(t − 3) 0 0 [( ) 1 1 1 1 −j(t−3) −j(t−3) e − + e = − 2π j(t − 3) (t − 3)2 (t − 3)2 ( )] 1 1 1 j(t−3) j(t−3) + e + e − j(t − 3) (t − 3)2 (t − 3)2 [ ] 1 sin(t − 3) cos(t − 3) − 1 + = 2π t−3 (t − 3)2

 (d) X(ω) = 2[δ(ω − 1) − δ(ω + 1)] + 3[δ(ω − 2π) + δ(ω + 2π)] Solution: From table, x(t) =

2j 3 sin(t) + cos(2πt) π π 

6

(e) X(ω) as in Figure 4. |X(ω)| 1

−4

−3

−2

−1

ω 0

1

2

3

4

−1

Figure 4: Question 2e Solution: We have ∫ ∞ 1 F (ω)ejωt dω x(t) = 2π −∞ [∫ −2 ∫ −1 1 jωt = −1e dω + (ω + 1)ejωt dω 2π −3 −2 ] ∫ 3 ∫ 2 jωt jωt 1e dω (ω − 1)e dω + + 2 1 [ −2 −1 ∫ −1 −1 1 −1 jωt 1 jωt ω jωt 1 jωt = e + e − e dω + e 2π jt jt jt −2 jt −3 −2 −2 2 ∫ 2 2 3 ] ω 1 jωt 1 1 + ejωt − e dω − ejωt + ejωt jt jt jt 1 jt 1 1 2 [ 1 1 2 1 1 1 1 1 −1 −j2t e + e−j3t − e−jt + e−j2t + 2 e−jt − 2 e−j2t + e−jt − e−j2t = 2π jt jt jt jt t t jt jt ] 2 1 1 1 1 1 1 1 + ej2t − ejt + 2 ej2t − 2 ejt − ej2t + ejt + ej3t − ej2t jt jt t t jt jt jt jt [ ] 1 −2 cos(2t) 2 cos(3t) 4 cos(2t) 2j sin(t) 2j sin(2t) 2 cos(2t) + + − = + − 2π jt jt jt t2 t2 jt [ ] 1 2 cos(3t) 2j sin(t) 2j sin(2t) = − + 2π jt t2 t2 cos(3t) sin(t) sin(2t) + = − jπt jπt2 jπt2 

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