ECE 301: Signals and Systems Homework Assignment...
Transcript of ECE 301: Signals and Systems Homework Assignment...
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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