113_1_Quiz1
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Transcript of 113_1_Quiz1
EE113: Digital Signal Processing Fall 2011Prof. Abeer Alwan October 13, 2011TA: Jianshu Chen and Sen Li
Solution to Quiz 1
1. Solution:
(a) From sampling theory, we have x(n) = x(t)|t=nTs , where Ts is the sampling interval. Weknow that sampling rate is 180 samples/sec, so Ts = 1/180 sec. Thus, we get
x(n) = cos(120πt) + sin(60πt) + 10|t=n/180
= cos(2π
3n) + sin(
π
3n) + 10
(b) Let x1(n) = cos(2π3 n) and x2(n) = sin(π3n), therefore, x(n) = x1(n) + x2(n) + 10.
ω1 =2π
3ω1N1 = 2πk =⇒ N1 = 3k =⇒ N1 = 3
ω2 =π
3ω2N2 = 2πk =⇒ N2 = 6k =⇒ N2 = 6
the period N should be the least common multiple of N1 and N2, which is 6.
2. Solution:
(a) The system is linear. Assume x1(n) is an arbitrary input with corresponding outputy1(n), and x2(n) is another arbitrary input with output y2(n). Then, y1(n) and y2(n)can be written as
y1(n) = sin(π
4n)x1(n− 1) + x1(n)
y2(n) = sin(π
4n)x2(n− 1) + x2(n)
Then, the output corresponding to input x(n) = ax1(n) + bx2(n) is
y(n) = sin(π
4n)
[ax1(n− 1) + bx2(n− 1)] + [ax1(n) + bx2(n)]
= a[sin(π
4n)x1(n− 1) + x1(n)
]+ b
[sin(π
4n)x2(n− 1) + x2(n)
]= ay1(n) + by2(n)
(b) The system is causal, because the output y(n) at time n is only determined by thecurrent and previous inputs: {x(n), x(n− 1)}.
(c) The system is not time-invariant. Let us denote y0(n) as the response to input x0(n).Then, we have
y0(n) = sin(π
4n)x0(n− 1) + x0(n)
EE113 - Fall 2011 1 of 3
Then, the response to x(n) = x0(n− n0) can be written as
y(n) = sin(π
4n)x(n− 1) + x(n)
= sin(π
4n)x0(n− n0 − 1) + x0(n− n0)
6= y0(n− n0) = sin(π
4(n− n0)
)x0(n− n0 − 1) + x0(n− n0)
(d) The system is BIBO stable. Assume x(n) is an arbitrary bounded input, i.e.,
|x(n)| ≤ Bx, ∀n
Then, the corresponding output satisfies
|y(n)| =∣∣∣sin(π
4n)x(n− 1) + x(n)
∣∣∣≤∣∣∣sin(π
4n)x(n− 1)
∣∣∣+ |x(n)|
≤∣∣∣sin(π
4n)∣∣∣ · |x(n− 1)|+ |x(n)|
≤ 1×Bx +Bx ≤ 2Bx
3. Solution:
(a) According to the figure, we can write the system difference equation as
y(n) = −3y(n− 1) + x(n) + 2x(n− 1)
(b) With initial condition y(−1) = 0 and x(n) = δ(n), we need to solve the impulse responsefrom the following difference equation:
h(n) = −3h(n− 1) + δ(n) + 2δ(n− 1), h(−1) = y(−1) = 0
from which we can write:
h(0) = −3h(−1) + δ(0) + 2δ(−1) = 1
h(1) = −3h(0) + δ(1) + 2δ(0) = −3 + 2 = −1
h(2) = −3h(1) + δ(2) + 2δ(1) = −3× (−1)
h(3) = −3h(2) + δ(3) + 2δ(2) = (−3)2 × (−1)
...
h(n) = −(−3)n−1, n ≥ 1
In summary, the closed form expression for h(n) should be
h(n) = δ(n) +1
3(−3)nu(n− 1)
EE113 - Fall 2011 2 of 3
(c) Given that the input x(n) = u(n), the output of the system is
y(n) = h(n) ? u(n)
=
[δ(n) +
1
3(−3)nu(n− 1)
]? u(n)
= δ(n) ? u(n) +1
3(−3)nu(n− 1) ? u(n)
= u(n) +1
3
n∑k=1
(−3)ku(n− 1)
= u(n) +1
3
(−3)− (−3)n+1
1− (−3)u(n− 1)
= u(n)− 1
12
[3 + (−3)n+1
]u(n− 1)
EE113 - Fall 2011 3 of 3