ECE 301: Signals and Systems
Homework Assignment #6
Due on November 30, 2015
Professor: Aly El Gamal
TA: Xianglun Mao
1
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
P∞
nx1 [n]
α = Pn=−∞
∞
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)n Od{x1 [n]}.
(c) We may express α as
jw
(e )
|w=0
j dX1dw
6
j(−6j/π)
=
=
α=
jw
X1 (e )|w=0
1
π
(d) Using the fact that H(ejw ) is the frequency response of an ideal lowpass filter with cutoff frequency
π
jw
6 , we may draw X4 (e ) as shown in Figure 2.
Figure 2: The graph of Fourier transforms of signals X4 (ejw ).
3
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. Furthermore, X(ejw ) and G(ejw ) are related as follows:
Z π
1
X(ejθ )G(ej(w−θ) )dθ = 1 + e−jw
2π −π
(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] = ( 21 )n u[n].
Solution
Let
1
2π
Z
π
X(ejθ )G(ej(w−θ) )dθ = 1 + e−jw = Y (ejw )
−π
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)n u[n], g[n] has to be chosen such that

1,
n=0



2,
n=1
g[n] =

0,
n>1


any value, otherwise
Therefore, there are many possible choices for g[n].
4
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
Wt
sin(W t)
sinc(
)=
π
π
πt
plays a very important role in LTI system analysis. The same is true of the discrete-time LTI system with
impulse response
W
Wn
sin(W n)
h[n] =
sinc(
)=
π
π
πn
(a) Determine and sketch the frequency response for the system with impulse response h[n].
(b) Consider the signal
πn
πn
) − 2cos( ).
8
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.
x[n] = sin(
(i) h[n] =
(ii) h[n] =
(iii) h[n] =
(iv) h[n] =
sin(πn/6)
πn
sin(πn/6)
+ sin(πn/2)
πn
πn
sin(πn/6)sin(πn/3)
π 2 n2
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.
P∞
(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.
2cos(πn/4).
Therefore, y[n] = 2sin(πn/8) −
(iii) The frequency response H(ejw ) is as shown in Figure 5.
1
4 cos(πn/4).
Therefore, y[n] =
1
6 sin(πn/8)
(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 =
7
1X
x[n]e−j(2π/8)kn .
8 n=0
The Fourier transform of this signal is
X(ejw ) =
∞
X
2πak δ(w − 2πk/8).
k=−∞
The Fourier transform Y (ejw ) of the output is Y (ejw ) = X(ejw )H(ejw ). Therefore,
Y (ejw ) = 2π[a0 δ(w) + a1 δ(w − π/4) + a−1 δ(w + π/4)]
in the range 0 ≤ |w| ≤ π. Therefore,
y[n] = a0 + a1 ejπn/4 + a−1 e−jπn/4 =
√
5
+ [(1/4) + (1/2)(1/ 2)]cos(πn/4).
8
(ii) The signal x[n] is periodic with period 8. The Fourier series coefficients of the signal are
ak =
7
1X
x[n]e−j(2π/8)kn .
8 n=0
The Fourier transform of this signal is
X(ejw ) =
∞
X
2πak δ(w − 2πk/8).
k=−∞
The Fourier transform Y (ejw ) of the output is Y (ejw ) = X(ejw )H(ejw ). Therefore,
Y (ejw ) = 2π[a0 δ(w) + a1 δ(w − π/4) + a−1 δ(w + π/4)]
in the range 0 ≤ |w| ≤ π. Therefore,
y[n] = a0 + a1 ejπn/4 + a1 e−jπn/4 =
1 1
+ cos(πn/4).
8 4
(iii) The Fourier transform X(ejw ) of the signal x[n] is of the form shown in part (i). Therefore,
y[n] = a0 + a1 ejπn/4 + a1 e−jπn/4 =
√
1
+ [(1/4) − (1/2)(1/ 2)]cos(πn/4).
8
(iv) The output is
y[n] = h[n] ∗ x[n] =
Problem 3 continued on next page. . .
sin[π/3(n − 1)] sin[π/3(n + 1)]
+
.
π(n − 1)
π(n + 1)
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.
8
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(w0 n) → w0 cos(w0 n)
(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
Z π
1
h[n] =
H(ejw )ejwn dw
2π −π
Z 0
Z π
1
1
=
−wejwn dw +
wejwn dw
2π −π
2π 0
Z
1 π
wcos(wn)dw
=
π 0
1 cos(nπ) − 1
]
= [
π
n2
9