Problem 1[30 points, each part has 10 points]
(a) Determine if the following system is linear: y(n) = Ax(n) + B
Solution: The system is nonlinear, assuming that B ¹ 0 (this is because zero input does not result in zero
output).
(b) Determine if the following system is causal: y(n) = ax(n)+ x(n2 )
Solution: The system is non-causal. For example, for n=2, we have y(2) = ax(2)+ x(4). In other words,
the value of the output depends on the future value of the inputs.
(c) Determine if the following system is time-invariant: y(n) = sign[x(n)]
The Sign function can be written as follows:
ìï 1 x(n) ³ 0
y(n) = Sign[x(n)] = í
ïî -1 x(n) < 0
This system is time-invariant. To show this, let’s shift the input first and calculate the output:
ìï 1 x(n - n ) ³ 0
0
y1 (n) = Sign[x(n - n0 )] = í
îï -1 x(n - n0 ) < 0
Now, let’s shift the original output by n0 samples. We get
ìï 1 x(n - n ) ³ 0
0
y2 (n) = í
ïî -1 x(n - n0 ) < 0
Because y1 and y2 are the same, the system is therefore time-invariant.
Problem 2[20 points]
We know that for a system to be stable in BIBO sense, we shall have the following property:
If the system’s x(n) is bounded, then the output y(n) shall be bounded as well. In other words, there
exist constants M x and M y such that
x(n) £ M x < ¥ Þ y(n) £ M y < ¥
1
Show that for an LTI system, the above condition is simplified as an “absolute summability” condition on
the impulse response of the system, h(n) , i.e.,
¥
å h(k) < ¥
k=-¥
Solution: First, let’s use the convolution formula:
¥
y(n) =
å h(k)x(n - k)
k=-¥
If we take the absolute values of both sides, we get
y(n) =
¥
å h(k)x(n - k)
k=-¥
We know that the absolute values of some of terms is always less than or equal to the absolute value of
those terms. Hence,
y(n) £
¥
å h(k) x(n - k)
(1)
k=-¥
If the input is bounded, then x(n) £ M x < ¥ (2)
Using (1) and (2), we can conclude that
y(n) £
¥
å h(k) M
k=-¥
¥
x
= M x å h(k) (3)
k=-¥
From (3), we can conclude that y(n) is bounded is the system satisfies the following condition:
¥
å h(k) < ¥
k=-¥
2
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