ECE2111 Signals and systems exam
This is the same exam as is on the mock eAssessment platform. For questions where you are expected to answer in a text box, both typeset and plain text solutions are given here. This is intended to model how we expect you to answer these questions in plain text. For questions where you are expected to answer on paper, only typeset solutions are provided.
Question 1 (2 marks)
A discrete-time linear time-invariant system with input x and output y is described by the difference equation
y[n]=x[n]−0.5x[n−1]−0.5x[n−2]+x[n−3] foralln.
Write down an expression for the impulse response of the system in terms of the discrete-time unit
impulse signal. (2 marks)
Solution: The impulse response is the output h of the system if the input is δ, the unit impulse. In other words
h[n]=δ[n]−0.5δ[n−1]−0.5δ[n−2]+δ[n−3] foralln. Plain text solution:
The impulse response is the output h of the system if the input is x=delta,
the unit impulse. This means that
h[n] = delta[n] – 0.5delta[n-1] – 0.5delta[n-2] + delta[n-3] for all n.
Question 2 (4 marks)
Drag and drop the system labels onto the diagram so that the output y and input x are related by the difference equation
y[n]=x[n]−0.5x[n−1]−0.5x[n−2]+x[n−3] foralln.
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Question 3 (7 marks)
Let x(t) = u(t + 1) − u(t − 1) be a pulse of length two. The signal xm(t) = cos(aπt)x(t) is shown below for −1 ≤ t ≤ 2.
(a) What is the value of a? Briefly explain your answer. (2 marks) Solution: There are 10 periods of cos(aπt) between t = −1 and t = 1. Therefore, the
fundamental period of cos(aπt) is 0.2. It follows that 2π = 2/a = 0.2 and so a = 10. aπ
Plain text solution:
There are 10 periods of cos(a pi t) between t=-1 and t=1.
Therefore, the fundamental period of cos(a pi t) is 0.2.
It follows that (2 pi)/(a pi) = 2/a = 0.2 and so a=10.
(b) Write down an expression for the continuous-time Fourier transform X of x. Solution: From the Fourier transform table we have that
X(ω)=2sin(ω) forallω. ω
Plain text solution:
From the Fourier transform table we have that
X(w) = 2 sin(w)/w for all w.
(c) Find an expression for the Fourier transform Xm of xm, in terms of X. Briefly explain how you obtained your answer. (3 marks)
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Solution: First we note that
xm(t) = cos(10πt)x(t) = 0.5ej10πtx(t) + 0.5e−j10πtx(t) for all t.
By the modulation property of the Fourier transform and the linearity of the Fourier transform we see that
Xm(ω) = 0.5X(ω − 10π) + 0.5X(ω + 10π) for all ω.
Plain text solution
First we note that
x_m(t) = cos(10 pi t)x(t) = 0.5e^(j 10 pi t) x(t) + 0.5e^(-j 10 pi t)x(t)
for all t. By the modulation property of the Fourier transform
and the linearity of the Fourier transform we see that
X_m(w) = 0.5 X(w-10 pi) + 0.5 X(w + 10 pi)
for all omega.
Question 4 (7 marks)
A demodulator system involves multiplication by a cosine at frequency ωc and then low-pass filtering the result using an ideal low-pass filter. The filter has frequency response H(ω) = A(u(ω + B) − u(ω − B)) where u is the unit step and A and B are positive constants. A schematic of this system is shown in the diagram below.
In the diagram, the multiplication system labelled × takes in signals x1 and x2 and outputs the signal defined by y(t) = x1(t)x2(t) for all t.
(a) Is the demodulator system a linear system? Explain your answer. (3 marks)
Solution: The demodulator is a linear system. This is because the system given by mul- tiplication by cos(ωct) is a linear system, and an ideal low-pass filter is a linear system, and the series interconnection of linear systems is a linear system. (Note that the system is not time-invariant.)
Plain text solution:
The demodulator is a linear system. This is because the
system given by multiplication by cos(w_c t) is a
linear system, and an ideal low-pass filter is a linear
system, and the series interconnection of linear
systems is a linear system. (Note that the system is
not time-invariant.)
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(b) The input x to the demodulator is periodic with fundamental frequency ωc, and has Fourier series coefficients that satisfy X1 = X−1 = 0.5. The corresponding output is y(t) = 2 for all t. Write down a possible value for the passband gain A and the cutoff frequency B (in rad/sec) of the low-pass filter. Justify your answers.
Solution: Let w denote the output of the multiplication system. Then, by the modulation property of Fourier series, we know that the kth Fourier series coefficient of w is
Wk = 0.5Xk−1 + 0.5Xk+1. The output of the system has Fourier coefficients given by
Yk = H(kωc)Wk = 0.5H(kωc)Xk−1 + 0.5H(kωc)Xk+1 foreachk. WeknowthatYk =0unlessk=0andthatY0 =2. Thistellsusthat
2 = Y0 = 0.5H(0)X−1 + 0.5H(0)X1 = 0.25A + 0.25A = 0.5A. ItfollowsthatA=4. ToensurethatYk =0wheneverk̸=0itisenoughtochoose0
(b) If the Laplace transform of the output is
Yˆ ( s ) = 1
with region of convergence RoC(y) = {s ∈ C : Re(s) > 1} find an expression for y(t), the
output at time t. (4 marks)
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Solution: First we perform a partial fraction decomposition to obtain Yˆ(s)= 1 −1.
Since the region of convergence is {s ∈ C : Re(s) > 1} we take the inverse Laplace transform
of each term to see that
(c) Find the transfer function Hˆ (s) of the system.
Hˆ(s)= 1 ·s+2=s+2. s(s−1) 1 s(s−1)
Question 8 (9 marks)
A continuous-time signal x with bandwidth 20π/3 rad/sec is sampled with sampling period Ts = 0.1 seconds to obtain a discrete-time signal
xs[n] = x(nTs) for all n.
The discrete-time Fourier transform Xs of the sampled signal is shown below:
y(t) = etu(t) − u(t) for all t.
Solution: The transfer function is Hˆ (s) = Yˆ (s)/Xˆ (s). Evaluating this we obtain
ω −3π −2π −π−2π 0 2π π 2π 3π
(a) Let X be the continuous-time Fourier transform of the input signal x. Sketch X(ω) vs ω for −10π ≤ ω ≤ 10π. Label all features of your plot. (4 marks)
Solution: We know that
1∞ ω−2πk Xs(ω) = T X T
shown below. Note the scaling of the horizontal and vertical axes.
and that the bandwidth of X is 20π/3. From this we can deduce that X(ω) has the form
−10π −20π 0 20π 10π 33
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(b) Find the energy of the signal xs, i.e., compute
E = |xs[n]|2. n=−∞
By Parseval’s relation we have that 1π21π2
E=2π −π|Xs(ω)|dω=22π 0 |Xs(ω)| dω where the second equality holds because |Xs(ω)|2 is even.
Now for ω > 0 we can see that
Xs(ω)= 0 if2π/3<ω<π.
and so that
(20 − 30ω/π)2 |Xs(ω)|2 = 0
Computing the integral for E gives E=π
if 0 ≤ ω ≤ 2π/3 if2π/3<ω<π.
20−30ω/π if0≤ω≤2π/3
1 2π/3 0
(20−30ω/π) dω = 1 −π(20 − 30ω/π)32π/3
1 8000π =π 0+ 90
(c) Given that the bandwidth of x is 20π/3 rad/sec, what is the longest sampling period we could
use so that no aliasing occurs when sampling x? (2 marks) Solution: To avoid aliasing we need that TsB ≤ π where B = 20π/3 is the bandwidth in
rad/sec and Ts is the sampling period. Thus we need
Ts(20π/3) ≤ π which is equivalent to Ts ≤ 3/20 = 0.15 seconds.
Question 9 (6 marks)
The input of a discrete-time LTI system is x[n] = (0.5)nu[n] and the output is y[n] = (0.25)nu[n].
(a) Find the transfer function Hˆ of the system. Show your working/explain your answer. (4 marks)
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Solution: The transfer function is Hˆ (z) = Yˆ (z) where Yˆ (z) and Xˆ (z) are the Z-transforms Xˆ ( z )
of y and x respectively.
The Z-transform of x, from the tables, is
Xˆ ( z ) = 1
1 − 0.5z−1
The Z-transform of y, from the tables, is
Yˆ ( z ) = 1
H(z)= 1−0.25z−1 = z−0.25.
(b) Is the system an infinite impulse response (IIR) system or a finite impulse response (FIR)
system? Explain your answer. (2 marks) Solution: The system is an IIR system. We know this because the transfer function of a
FIR system either has no poles, or has a pole at z = 0. This system has a pole at z = 0.25. An alternative solution would be to find the impulse response explicitly. To do this we rewrite
The transfer function is, therefore,
1 − 0.25z−1
ˆ 1 − 0.5z−1 z − 0.5
the transfer function as
Hˆ ( z ) = 2 − 1 . 1 − 0.25z−1
Taking the inverse Z-transform then gives an impulse response of h[n] = 2δ[n] − (0.25)nu[n].
This is non-zero for infinitely many value of n, so this is an IIR system.
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