ECE 301: Signals and Systems

Homework Assignment #6Due on November 30, 2015

Professor: Aly El Gamal

TA: Xianglun Mao

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Aly El Gamal ECE 301: Signals and Systems Homework Assignment #6 Problem 1

Note: Homework 6 will have only 4 problems since there will be a midterm exam and then Thanksgiving

break, each problem is assigned 12.5 points.

Problem 1

Let x1[n] be the discrete-time signal whose Fourier transform X1(ejw) is depicted in Figure 1(a).

(a) Consider the signal x2[n] with Fourier transform X2(ejw), as illustrated in Figure 1(b). Express x2[n]

in terms of x1[n]. (Hint : First express X2(ejw) in terms of X1(ejw), and then use properties of the

Fourier transform.)

(b) Repeat part (a) for x3[n] with Fourier transform X3(ejw), as shown in Figure 1(c).

(c) Let

α =

∑∞n=−∞ nx1[n]∑∞n=−∞ x1[n]

This quantity, which is the center of gravity of the signal x1[n], is usually referred to as the delay time

of x1[n]. Find α. (You can do this without first determining x1[n] explicitly.)

(d) Consider the signal x4[n] = x1[n] ∗ h[n], where

h[n] =sin(πn/6)

πn

Sketch X4(ejw).

Figure 1: The graph of Fourier transforms of signals X1(ejw), X2(ejw), X3(ejw).

Problem 1 continued on next page. . . 2

Aly El Gamal ECE 301: Signals and Systems Homework Assignment #6 Problem 1 (continued)

Solution

(a) We may express X2(ejw) as

X2(ejw) = Re{X1(ejw)}+ Re{X1(ej(w−2π/3))}+ Re{X1(ej(w+2π/3))}.

Therefore,

x2[n] = Ev{x1[n]}[1 + ej2π/3 + e−j2π/3].

(b) We may express X3(ejw) as

X3(ejw) = Im{X1(ej(w−π))}+ Im{X1(ej(w+π))}.

Therefore,

x3[n] = Od{x1[n]}[ejwn + e−jwn] = 2(−1)nOd{x1[n]}.

(c) We may express α as

α =j dX1(e

jw)dw |w=0

X1(ejw)|w=0=j(−6j/π)

1=

6

π

(d) Using the fact that H(ejw) is the frequency response of an ideal lowpass filter with cutoff frequencyπ6 , we may draw X4(ejw) as shown in Figure 2.

Figure 2: The graph of Fourier transforms of signals X4(ejw).

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Aly El Gamal ECE 301: Signals and Systems Homework Assignment #6 Problem 2

Problem 2

The signals x[n] and g[n] are known to have Fourier transforms X(ejw) and G(ejw), respectively. Further-

more, X(ejw) and G(ejw) are related as follows:

1

2π

∫ π

−πX(ejθ)G(ej(w−θ))dθ = 1 + e−jw

(a) If x[n] = (−1)n, determine a sequence g[n] such that its Fourier transform G(ejw) satisfies the above

equation. Are there other possible solutions for g[n]?

(b) Repeat the previous part for x[n] = (12 )nu[n].

Solution

Let1

2π

∫ π

−πX(ejθ)G(ej(w−θ))dθ = 1 + e−jw = Y (ejw)

Taking the inverse Fourier transform of the above equation, we obtain

g[n]x[n] = δ[n] + δ[n− 1] = y[n].

(a) If x[n] = (−1)n,

g[n] = δ[n]− δ[n− 1]

(b) If x[n] = (1/2)nu[n], g[n] has to be chosen such that

g[n] =

1, n = 0

2, n = 1

0, n > 1

any value, otherwise

Therefore, there are many possible choices for g[n].

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Aly El Gamal ECE 301: Signals and Systems Homework Assignment #6 Problem 3

Problem 3

In lecture, we indicated that the continuous-time LTI system with impulse response

h(t) =W

πsinc(

Wt

π) =

sin(Wt)

πt

plays a very important role in LTI system analysis. The same is true of the discrete-time LTI system with

impulse response

h[n] =W

πsinc(

Wn

π) =

sin(Wn)

πn

(a) Determine and sketch the frequency response for the system with impulse response h[n].

(b) Consider the signal

x[n] = sin(πn

8)− 2cos(

πn

4).

Suppose that this signal is the input to LTI systems with the following impulse responses. Determine

and sketch the frequency response of the output in each case.

(i) h[n] = sin(πn/6)πn

(ii) h[n] = sin(πn/6)πn + sin(πn/2)

πn

(iii) h[n] = sin(πn/6)sin(πn/3)π2n2

(iv) h[n] = sin(πn/6)sin(πn/3)πn

(c) Consider an LTI system with unit sample response

h[n] =sin(πn/3)

πn.

Determine and sketch the frequency response of the output in each case.

(i) x[n] = the square wave depicted in Figure 3.

(ii) x[n] =∑∞k=−∞ δ[n− 8k]

(iii) x[n] = (−1)n times the square wave depicted in Figure 3.

(iv) x[n] = δ[n+ 1] + δ[n− 1]

Figure 3: The square wave that construct x[n].

Solution

(a) The frequency response of the system is as shown in Figure 4.

(b) The Fourier transform X(ejw) of x[n] is as shown in Figure 4.

(i) The frequency response H(ejw) is as shown in Figure 5. Therefore, y[n] = sin(πn/8).

Problem 3 continued on next page. . . 5

Aly El Gamal ECE 301: Signals and Systems Homework Assignment #6 Problem 3 (continued)

(ii) The frequency response H(ejw) is as shown in Figure 5. Therefore, y[n] = 2sin(πn/8) −2cos(πn/4).

(iii) The frequency response H(ejw) is as shown in Figure 5. Therefore, y[n] = 16sin(πn/8) −

14cos(πn/4).

(iv) The frequency response H(ejw) is as shown in Figure 5. Therefore, y[n] = −sin(πn/4).

(c) (i) The signal x[n] is periodic with period 8. The Fourier series coefficients of the signal are

ak =1

8

7∑n=0

x[n]e−j(2π/8)kn.

The Fourier transform of this signal is

X(ejw) =

∞∑k=−∞

2πakδ(w − 2πk/8).

The Fourier transform Y (ejw) of the output is Y (ejw) = X(ejw)H(ejw). Therefore,

Y (ejw) = 2π[a0δ(w) + a1δ(w − π/4) + a−1δ(w + π/4)]

in the range 0 ≤ |w| ≤ π. Therefore,

y[n] = a0 + a1ejπn/4 + a−1e

−jπn/4 =5

8+ [(1/4) + (1/2)(1/

√2)]cos(πn/4).

(ii) The signal x[n] is periodic with period 8. The Fourier series coefficients of the signal are

ak =1

8

7∑n=0

x[n]e−j(2π/8)kn.

The Fourier transform of this signal is

X(ejw) =

∞∑k=−∞

2πakδ(w − 2πk/8).

The Fourier transform Y (ejw) of the output is Y (ejw) = X(ejw)H(ejw). Therefore,

Y (ejw) = 2π[a0δ(w) + a1δ(w − π/4) + a−1δ(w + π/4)]

in the range 0 ≤ |w| ≤ π. Therefore,

y[n] = a0 + a1ejπn/4 + a1e

−jπn/4 =1

8+

1

4cos(πn/4).

(iii) The Fourier transform X(ejw) of the signal x[n] is of the form shown in part (i). Therefore,

y[n] = a0 + a1ejπn/4 + a1e

−jπn/4 =1

8+ [(1/4)− (1/2)(1/

√2)]cos(πn/4).

(iv) The output is

y[n] = h[n] ∗ x[n] =sin[π/3(n− 1)]

π(n− 1)+sin[π/3(n+ 1)]

π(n+ 1).

Problem 3 continued on next page. . . 6

Aly El Gamal ECE 301: Signals and Systems Homework Assignment #6 Problem 3 (continued)

Figure 4: The resulting frequency responses.

Problem 3 continued on next page. . . 7

Aly El Gamal ECE 301: Signals and Systems Homework Assignment #6 Problem 3 (continued)

Figure 5: The resulting frequency responses.

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Aly El Gamal ECE 301: Signals and Systems Homework Assignment #6 Problem 4

Problem 4

An LTI system X with impulse response h[n] and frequency response H(ejw) is known to have the property

that, when −π ≤ w0 ≤ π,

cos(w0n)→ w0cos(w0n)

(a) Determine H(ejw).

(b) Determine h[n].

Solution

(a) From the given information, it is clear that when the input to the system is a complex exponential

frequency w0, the output is a complex exponential of the same frequency but scaled by the |w0|.Therefore, the frequency response of the system is

H(ejw) = |w|, for 0 ≤ |w0| ≤ π.

Note that H(ejw) here should be a well-defined frequency response regardless of w0.

(b) Taking the inverse Fourier transform of the frequency response, we obtain

h[n] =1

2π

∫ π

−πH(ejw)ejwndw

=1

2π

∫ 0

−π−wejwndw +

1

2π

∫ π

0

wejwndw

=1

π

∫ π

0

wcos(wn)dw

=1

π[cos(nπ)− 1

n2]

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