ENGN 1570 Homework 4
(Solution)
Problem 1
A discrete linear system is causal if and only if
(*) For any input x[n] such that x[n] = 0 for n < n0 , the corresponding output y[n] must
also be zero for n < n0 .
(a) Give a nonlinear system that satisfies (*) but is not causal.
y[n] = x[n]x[n + 1]
(b) Give a nonlinear system that does not satisfy (*) but is causal.
y[n] = x[n] + 1
Problem 2
For each signal below, determine if it is periodic and if so give one of its periods. Justify
your answers.
(a) x(t) = cos( 41 t)
Fundamental period is T = 2π/(1/4) = 8π.
(b) x(t) = cos( π4 t)
1
Fundamental period is T = 2π/(π/4) = 8.
(c) x[n] = cos( 41 n)
Not periodic because 8π is not rational.
(d) x[n] = cos( π4 n)
Fundamental period is 8.
(e) x(t) = cos(4t) + cos(5t − 1)
Integer multiples of π/2 are periods of cos(4t). Integer multiples of 2π/5 are
periods of cos(5t − 1). Any common multiple is a period of the sum. Take
5 ∗ (2π/5) = 4 ∗ (π/2) = 2π.
(f ) x(t) = cos(4πt) + cos(12t)
Integer multiples of 1/2 are periods of cos(4πt). Integer multiples of π/6 are
periods of cos(12t). 1/2 is rational and π/6 is irrational, so they don’t have a
common integer multiple. Therefore, the signal is not periodic.
Problem 3
Determine if each system below is invertible. If so, give the inverse system. If not, give two
input signals that have the same output.
(a) y(t) = 2t2 x(2t)
This system is not invertible because the value of x(0) is “lost” due to the fact
that 2t2 x(2t) = 0 when t = 0. Take any two signals x1 and x2 that differ only at
t = 0. Then x1 and x2 map to the same signal under this system.
(b) y[n] = 2n2 x[2n]
Not invertible because odd values of x are “lost.” As in the previous example,
the value of x[0] is also “lost.”
Take x1 [n] = 1 and

 0 if n is odd
x2 [n] =
 1 if n is even
2
Then x1 and x2 map to the same output.
(c)



x[n − 5]
n≥1


y[n] =
0
n=0



 x[n]
n ≤ −1
This system is invertible with inverse:

 y[n + 5] n > −1
z[n] =
 y[n]
n ≤ −1
(d) y[n] = x[n] + x[n + 2]
Not invertible. Take x1 [n] = 0 and

 1 if n = 4k or n = 4k + 1 for some integer k
x2 [n] =
 −1 otherwise
Then x1 and x2 map to the same output.
Problem 4
For the systems below, determine which ones are “Memoryless,” “Time invariant,” “Linear”
and “Causal”. Here x(t) is an input signal and y(t) is the output.
(a) y(t) = x(t) + x(t − 1)
Not memoryless. Time invariant. Linear. Causal.
(b) y(t) = 5
Memoryless. Time invariant. Not linear. Causal.
(c) y(t) = x(t) + t3 + t
Memoryless. Not time invariant. Not linear. Causal.
(d) y(t) = x(t)(t3 + t)
Memoryless. Not time invariant. Linear. Causal.
3
Problem 5
Let



0
t<0


h(t) =
1 0≤t≤1



 0
t>1



0
t<0


x(t) =
t 0≤t≤1



 0
t>1
(a) Sketch h(t)
(b) Sketch x(t)
(c) Sketch h(t) ∗ h(t)
(d) Sketch h(t) ∗ x(t)
4
Problem 6
Consider the causal system with input x and output y defined by
1
y[n] − y[n − 1] = x[n]
2
Determine the output y[n] when the input is δ[n].
Since δ[n] = 0 for all n < 0 and the system is causal, we have that y[n] = 0 for
all n < 0.
Now we can solve for y[0]:
1
y[0] − y[n − 1] = 1 ⇒ y[0] = 1
2
To solve for y[n] for n > 0 we use:
1
1
y[n] − y[n − 1] = 0 ⇒ y[n] = y[n − 1]
2
2
Therefore for n > 0 we have:
y[n] =
5
1
2n