ENGN 1570 Homework 1
(Solution)
Problem 1
For each signal below, determine if it is periodic and if so give the fundamental period.
Justify your answers.
We say a continuous signal x(t) is periodic with fundamental period T0 ∈ R+ iff
x(t + T0 ) = x(t)
and T0 is the smallest possible positive real number for which the equation holds.
We say a discrete signal x[n] is periodic with fundamental period N0 ∈ Z+ iff
x[n + N0 ] = x[n]
and N0 is the smallest possible positive integer for which the equation holds.
(a) x(t) = cos(14πt)
(b) x(t) = cos(πt) sin( π8 t)
If f (t) = g(t)h(t) any T which is a period of both g and h is also a period of f .
The fundamental period of cos(πt) is 2, and the fundamental period of sin( π8 t) is
16. The fundamental period of x(t) is the least common multiple, which is 16.
(c) x(t) = cos(4t) sin( π6 t)
(d) x[n] = cos( 3π
n)
2
1
The fundamental period of cos(t) is 2π; therefore, we are looking for N0 ∈ Z+
and m ∈ Z such that:
3
3
3
3
cos
π(n + N0 ) = cos
πn + πN0 = cos
πn + 2mπ
2
2
2
2
(1)
This suggests that 32 πN0 = 2mπ and therefore N0 = 4m/3. Setting m = 3 we
get N0 = 4.
(e) x[n] = cos(3n)
Similar to part (d), we are looking for N0 ∈ Z+ and m ∈ Z such that:
cos (3(n + N0 )) = cos (3n + 2mπ)
(2)
This suggests that N0 = 32 mπ. Since π is irrational @m ∈ Z s.t. 32 mπ ∈ Z+ , and
therefore x[n] is not periodic.
(f) x(t) = e(2+j)t
Not periodic, growing sinusoidal. Alternatively, the period is ( 52 + 54 i)π, which is
not a real number.
Example 1.2
Problem 2
We are given the signal x(t):
x(t)
1
−1
0
1
2
3
4
t
5
Let us transform the given x(t) to x(−2t + 6). We need to use all three types of transformations (a shift, a
scale and a flip), but what in what order shall we do them? How do we do it? The following guide explains:
For the signal x(t) above sketch the graph of the following transformed signals
(a) x(t − 3)
What to do when you need to do multiple transformations:
• We can do the transformations in any order. However, the amount and
direction of the shift depends on whether it is performed before or after
the reversal and the scale.
(b) x(2t)
(c) 2x(t/2 + 2)
• We can think of cascaded transformations as repeated substitutions of
the independent variable t.
(d) −x(t2 )
To demonstrate these principles, let’s do the transformation in three different orders.
(e)x(ttx(t)
+ 6)
• x(t)
Advance by 6
−→
• x(t)
Compress by 2
−→
x(2t)
• x(t)
Reverse
−→
x(−t)
Reverse
−→
Advance by 3
−→
Delay by 6
−→
x(−t + 6)
x(2(t + 3))
x(−(t − 6))
Compress by 2
−→
x(−2t + 6).
Reverse
−→
x(2(−t + 3)).
Compress by 2
−→
x(−(2t − 6)).
2
x(t)
x(2t)
2.5
2
−x(t )
tx(t)
2
1.5
1
0.5
0
−0.5
Quadratic t^2/2 - 2
−1
−1.5
−1
0
1
2
3
4
5
time
Figure 1: Graph of signals in Problem 2.
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 system is invertible if it produces unique outputs for distinct inputs. In other
words, there exists a one-to-one mapping from inputs to the outputs of an invertible system.
If a system is NOT invertible there has to exist two distinct inputs x1 (t) 6= x2 (t)
(or x1 [n] 6= x2 [n] in the discrete case) that get mapped to the same output by
the system.
(a) y(t) = x(t + 2)
Invertible. x(t) = y −1 (t) = y(t − 2).
(b) y(t) = cos(x(t))
Not invertible. Take any x1 (t) that doesn’t have period 2π and let x2 (t) =
x1 (t + 2π).
(c) y[n] = 2nx[n]
Not invertible. Let x1 [n] and x2 [n] be two input signals that are identical every3
where except at n = 0. That is ∀n 6= 0 : x1 [n] = x2 [n] and x1 [0] 6= x2 [0].
Although x1 and x2 are distinct (they differ at n = 0), but the output of the
system is identical for both of them.
(d) y[n] = x[2n]
Not invertible. x1 [n] = δ[n] and x2 [n] = δ[n] + δ[n − 1] give y[n] = δ[n].
(e)



x[n − 1]
n≥1


y[n] =
0
n=0



 x[n]
n ≤ −1
Invertible. The system only inserts a zero-valued sample at n = 0 shifting the
input signal to the right. So no information about the input gets lost in the
output of the system. The inverse of the system is as follows:

 y[n + 1] n ≥ 0
−1
x[n] = y [n] =
 y[n]
n<0
(f) y[n] = x[n] − x[n − 1]
Not invertible. Take two distinct constant functions.
Problem 4
Sketch the even and odd parts of the signal in Problem 2.
Problem 5
We can characterize systems in terms of properties such as “Memoryless”, “Time invariant”,
“Linear”, “Causal”. For the systems below determine which of these properties hold. Justify
your answers. Here x(t) is an input signal and y(t) is the output.
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Let y(t) be the response of a system to arbitrary input signal x(t):
• The system is memoryless if the output of the system at any given time
depends only on the current input (i.e. neither past nor future). That is
y(t) = f (x(t)) where f is some function.
• The system is time invariant if any time-shift of the input shifts the output by the exact same amount without changing the shape of it. That is
response of the system to x[t − τ ] is y[t − τ ].
• The system is linear if response to a linear combination of primary inputs
is the same linear combination of the outputs of the primary signals. That
is if x(t) = αx1 (t) + βx2 (t) then y(t) = αy1 (t) + βy2 (t) where y1 and y2 are
responses of the system to input signals x1 and x2 respectively.
• The system is causal if the output of the system at any given time depends
only on the current or past time steps. That is if x1 and x2 are the same
up to time t then their responses (i.e. y1 and y2 ) are the same up to time
t too.
The above properties are defined similarly for a discrete system.
(a) y(t) = cos(t)x(t)
• Memoryless: output only depends on the current input
• NOT time invariant: y(t + τ ) = cos(t + τ )x(t + τ ) 6= x(t + τ ).
• Linear: output to x(t) = αx1 (t) + βx2 (t) is cos(t) [αx1 (t) + βx2 (t)] =
α cos(t)x1 (t) + β cos(t)x2 (t) = αy1 (t) + βy2 (t).
• Causal: output does not depend on the future inputs.
(b) y(t) = x(t) + x(t − 2)
• NOT Memoryless: output depends on past input.
5
• Time invariant: a time shift of τ unit in the input results in the response
x(t − τ ) + x(t − 2 − τ ) which is y(t − τ ).
• Linear: output to x(t) = αx1 (t)+βx2 (t) is [αx1 (t) + βx2 (t)]+[αx1 (t − 2) + βx2 (t − 2)] =
α [x1 (t) + x1 (t − 2)] + β [x2 (t) + x2 (t − 2)] = αy1 (t) + βy2 (t).
• Causal: output does not depend on the future inputs.
(c) y(t) = x(t/5)
NOT memoryless, NOT time invariant (see pg. 52), linear, NOT causal.
(d) y(t) =
dx(t)
dt
NOT memoryless, time invariant, linear, causal (use the definition of the derivative as a limit approaching from t in the past).
Problem 6
Calculate the value of E∞ and P∞ for the signals
t)
(a) x(t) = 5 cos( 2π
3
(b) x(t) = 3et/2
6