Final Exam
Signals & Systems
February 9th, 2011
(151-0575-01)
Prof. R. D’Andrea
Solutions
Exam Duration:
150 minutes
Number of Problems:
10
Permitted aids:
One double-sided A4 sheet.
Questions can be answered in English or German.
Use only the prepared sheets for your solutions. Additional
paper is available from the supervisors.
Final Exam – Signals & Systems
Problem 1
5 points
A linear, time invariant system with input x[n] and output y[n] has the
transfer function
10 − 2z −1
Y (z)
=
H(z) =
X(z) a2 + 2az −1 + z −2
a) For what real values of a is the system causal and stable?
(3 points)
b) Write down a causal difference equation of the above system (2 points)
of the following form:
!
M
N
X
X
1
y[n] =
bk x[n − k] −
ak y[n − k]
a0
k=0
k=1
Solution 1
a) The poles of the system are at
0 = a2 + 2az −1 + z −2
= (a + z −1 )2
z −1 = −a
1
z=−
a
For the system to be causal and stable, the poles must lie within the unit
circle. It follows that |a| > 1 must hold.
b)
Y (z)
10 − 2z −1
=
X(z) a2 + 2az −1 + z −2
Y (z)(a2 + 2az −1 + z −2 ) = X(z)(10 − 2z −1 )
a2 y[n] + 2ay[n − 1] + y[n − 2] = 10x[n] − 2x[n − 1]
1
y[n] = 2 (10x[n] − 2x[n − 1] − 2ay[n − 1] − y[n − 2])
a
Final Exam – Signals & Systems
Problem 2
5 points
A system is governed by the difference equation
π
,
y[n] = x[n]x[n − 1] + cos 3πn −
3
where x[n] is the input and y[n] is the output. Is the system
a)
b)
c)
d)
e)
linear?
time invariant?
bounded input bounded output (BIBO) stable?
memoryless?
causal?
Yes
Yes
Yes
Yes
Yes
No
No
No
No
No
Please circle either ‘Yes’ or ‘No’. If you change your mind, please
cross out both ‘Yes’ and ‘No’ and write either ‘Yes’ or ‘No’ alongside, or leave it blank. It is not necessary to justify choices.
You can get a maximum of 5 points and a minimum of 0 points
for this problem. For each subproblem you get: +1 for a correct
answer, -1 for an incorrect answer and 0 for no answer.
Solution 2
a)
b)
c)
d)
e)
No
No
Yes
No
Yes
Final Exam – Signals & Systems
Problem 3
5 points
The output of a system T is y[n] = T{x[n]}. T is a causal, linear, time
invariant system. Given the input and output below, calculate the impulse
response h[n] of the system T for all n.
n
<0 0 1
2
3
4
5 >5
x[n] 0 1 −1 −2 0
0
0
0
y[n] 0 3 −2 −2 −5 −12 −4 0
Solution 3
Because the system is given to be causal, h[n] = 0 for all n < 0. One can
see from the input output behavior that the system has a finite impulse
response (the output goes to zero a finite number of steps after the input
goes to zero). The input is zero for n > 2, and the output is zero for n > 5;
therefore the impulse response must be zero for n > 3.
n = 0:
y[0] = h[0]x[0]
y[0]
h[0] =
x[0]
=3
n = 1:
y[1] = h[0]x[1] + h[1]x[0]
y[1] − h[0]x[1]
h[1] =
x[0]
=1
n = 2:
y[2] = h[0]x[2] + h[1]x[1] + h[2]x[0]
y[2] − h[0]x[2] − h[1]x[1]
h[2] =
x[0]
=5
Final Exam – Signals & Systems
n = 3:
y[3] = h[0]x[3] + h[1]x[2] + h[2]x[1] + h[3]x[0]
y[3] − h[0]x[3] − h[1]x[2] − h[2]x[1]
h[3] =
x[0]
=2
Final Exam – Signals & Systems
Problem 4
5 points
The transfer function of a filter is
b0
Y (z)
=
.
X(z) 1 + a1 z −1
H(z) =
Calculate the coefficients b0 and a1 such that the filter is stable and causal,
and such that the frequency response H(Ω) of the filter fulfills the two
criteria
H(Ω = 0) = 1,
and
π
1
H Ω=
=√ .
2
2
Solution 4
The first criterion yields
1=
b0
b0
,
=
−j0
1 + a1 e
1 + a1
1 + a1 = b0 .
From the second criterion, we obtain
b0
1
√ =
1 − ja1
2
|b0 |
=p
.
1 + a21
Combining the two criteria, we obtain
It follows that
|1 + a1 |
1
√ =p
.
2
1 + a21
(1 + a1 )2 1
− .
0=
1 + a21
2
= a21 + 4a1 + 1.
Final Exam – Signals & Systems
Solving the quadratic equation for a1 , we obtain the two solutions
√
a1 = −2 ± 3.
Since the filter needs to be stable, causal, and has a pole at z = −a1 , the
only possible solution is
√
a1 = −2 + 3,
from which follows
b0 = −1 +
√
3.
Final Exam – Signals & Systems
Problem 5
5 points
Associate the following impulse responses ((a) to (e)) and frequency responses ((1) to (5)) with the corresponding difference equation by filling
out the following table. It is not necessary to justify choices.
Difference equation
Impulse Response Frequency Response
(a) to (e)
(1) to (5)
y[n] = 31 (x[n] + x[n − 1] + x[n − 2])
y[n] = 21 (x[n] − x[n − 1])
y[n] = 0.8x[n] + 0.2y[n − 1]
y[n] = 31 (x[n + 1] + x[n] + x[n − 1])
0.5
0
−0.5
−2
2
4
0
Time index n
6
Impulse response h[n]
Impulse response h[n]
y[n] = 0.2x[n] + 0.8y[n − 1]
0.5
0
−0.5
−2
0.5
0
−0.5
−2
0
2
4
Time index n
6
Impulse response h[n]
(c)
0
−0.5
0
2
4
Time index n
(e)
0.5
0
−0.5
−2
0
2
4
Time index n
(d)
0.5
−2
6
(b)
Impulse response h[n]
Impulse response h[n]
(a)
2
4
0
Time index n
6
6
1
Amplitude
Amplitude
Final Exam – Signals & Systems
0.5
0
1
2
90
0
−90
0
1
2
Frequency Ω
0.5
0
3
Phase (◦ )
Phase (◦ )
0
1
0
1
0
1
2
Frequency Ω
3
Amplitude
Amplitude
0.5
0
1
2
Frequency Ω
3
Phase (◦ )
Amplitude
(3)
0.5
0
0
1
2
0
1
2
Frequency Ω
0.5
0
0
1
2
0
1
2
Frequency Ω
3
90
0
−90
3
3
90
0
−90
(4)
1
(5)
1
3
Phase (◦ )
Phase (◦ )
2
90
0
−90
0
3
(2)
1
1
3
90
0
−90
(1)
0
2
3
Final Exam – Signals & Systems
Solution 5
Difference equation
Impulse Response Frequency Response
(a) to (e)
(1) to (5)
y[n] = 31 (x[n] + x[n − 1] + x[n − 2])
d
1
y[n] = 21 (x[n] − x[n − 1])
c
5
y[n] = 0.8x[n] + 0.2y[n − 1]
e
4
y[n] = 31 (x[n + 1] + x[n] + x[n − 1])
b
3
a
2
y[n] = 0.2x[n] + 0.8y[n − 1]
Final Exam – Signals & Systems
Problem 6
5 points
For a zero mean signal x[n],
Rxx [0] = σx2 ,
where Rxx is the autocorrelation function and σx2 is the variance of the
signal x[n]. The autocorrelation function is defined as
Rxx [k] := E (x[n]x[n − k]) ,
where E(·) is the expected value operator. Let x[n] be white noise with
variance σx2 = 2. The signal x[n] is applied to a linear time invariant system
T to generate the output
y[n] = T {x[n]} .
The frequency response H (Ω) of T can be approximated by
1 −jΩ
e , 0 ≤ |Ω| ≤ Ωc
H (Ω) = 2
.
0,
Ωc < |Ω| ≤ π
Calculate the frequency Ωc such that the variance of the output signal y[n]
is σy2 = 0.4.
Solution 6
The power spectral density function of y[n] is
1 2 1
σ = , 0 ≤ |Ω| ≤ Ωc
.
Syy (Ω) = |H (Ω)|2 Sxx (Ω) = 4 x 2
0,
Ωc < |Ω| ≤ π
The system T is linear, therefore the zero mean input x[n] results in a
zero mean output y[n]. We calculate the variance of y[n] using the inverse
Fourier transform
Z π
1
σy2 = Ryy [k = 0] =
Syy (Ω) ejΩ·0 dΩ
2π −π
Z Ωc
1
1
Ωc
=
dΩ =
.
π 0 2
2π
Solving for Ωc with σy2 = 0.4, we obtain
Ωc =
4π
.
5
Final Exam – Signals & Systems
Problem 7
5 points
Consider the continuous time signal
π
x(t) = 4 cos 6πt +
+ 2 cos (8πt + π) − 2.
3
a) Determine the largest possible sampling time Ts in seconds (1 point)
to sample the signal without aliasing effects.
b) Sample the given continuous time signal with the sampling (1 point)
time you obtained in a), starting at t = 0:
x[0] = x(0),
x[1] = x(Ts ),
...,
x[n] = x(nTs ).
Determine the fundamental period N0 of the discrete time
signal x[n].
c) State the Fourier series coefficients ck given by
(3 points)
j2πk
N0 −1
n
−
1 X
N
0
ck =
,
x[n]e
N0 n=0
k = 0, 1, ..., N0 − 1.
Solution 7
a) The largest possible sampling time is Ts = 1/8 s.
b) The sampled signal is
3
π
x[n] = 4 cos
+ 2 cos (πn + π) − 2.
πn +
4
3
The fundamental period of the signal is N0 = 8.
c) The coefficients are
k
ck
0
1 2
3
π
j
-2 0 0 2e 3
4
5
π
−j
-2 2e 3
6 7
.
0 0
For a detailed solution, please refer to the solution of problem 4 of the 2009
final exam or the recitation notes posted on Nov. 5 on the class website.
Final Exam – Signals & Systems
Problem 8
5 points
A continuous time, linear, time invariant system with input x(t) and output
y(t) is given by the state space description
q̇(t) = Aq(t) + Bx(t)
y(t) = Cq(t) + Dx(t),
where the state vector has two
matrices are
0
A =
0
C =
1
elements q(t) = {q1 (t), q2 (t)}T and the
1
0
B =
0
1
0
D = [0.5]
Assume the input x(t) is piece-wise constant
x(t) = x[k]
kTs ≤ t < (k + 1)Ts
with sampling time Ts . Compute Ad , Bd , Cd , Dd of a discrete time state
space description of the system
q[k + 1] = Ad q[k] + Bd x[k]
y[k] = Cd q[k] + Dd x[k],
when the output y[k] is defined as
y[k] := y(t = kTs ).
Solution 8
We solve this problem using the matrix exponential (see lecture notes for
derivation). We define:


0 1 0
A B
M :=
=  0 0 1 .
0 0
0 0 0
The discrete time matrices Ad , Bd are then obtained by calculating
Ad Bd
= e M Ts .
0 1
Final Exam – Signals & Systems
Since the matrix M is nilpotent, i.e. M 3 = 0, it is straightforward to
calculate the matrix exponential:
1
Ad Bd
= I + M Ts + M 2 Ts2 ,
0 1
2
where I is the identity matrix. The matrices are therefore
1 2
1 Ts
T
Bd = 2 s
Ad =
0 1
.
Ts
Cd = C
Dd = D
Final Exam – Signals & Systems
Problem 9
5 points
You are given the state space description of a system:
2
2 − 41
q[n] +
x[n]
q[n + 1] =
25 −3
−1
y[n] = −1 1 q[n] + 4 x[n]
a) Is the system bounded input bounded output stable?
b) Is the system controllable?
c) Calculate the step response of the system for
n = 0, 1, 2.
d) Consider the second system
1
−1
−4 2
q[n] +
x[n]
q[n + 1] =
−3 25
2
y[n] = 1 −1 q[n] + 3 x[n]
(2 points)
(1 point)
(1 point)
(1 point)
is the input output behavior of this system identical to the
first one?
Solution 9
a) The eigenvalues of the matrix A are λ1 = λ2 = − 21 . The system is
therefore stable.
2 17
4 . The matrix has full
b) The controllability matrix is [B AB] =
−1 53
row rank and the system is therefore controllable.
c) n = 0:
2 − 14 0
2
q[1] =
+
1
25 −3 0
−1
2
=
−1
0
+ 4 1
y[0] = −1 1
0
=4
Final Exam – Signals & Systems
n = 1:
2
2
2 − 14
+
1
q[2] =
25 −3 −1
−1
25 = 4
52
2
y[1] = −1 1
+ 4 1
−1
=1
n = 2:
y[2] = −1 1
=
199
4
25 4
52
+ 4 1
d) No. An easy way to see this isthe
first value of the step response:
0
The state of both systems is q[0] =
. The output of the first system is
0
therefore y[0] = 4, while the output of the second system is y[0] = 3.
Final Exam – Signals & Systems
Problem 10
5 points
Assume that you are identifying a linear time invariant system T with
input x[n] and output y[n]:
y[n] = T{x[n]}
Input signal x[n]
You applied the input x[n] and measured the output y[n] for 101 samples,
i.e. 0 ≤ n ≤ 100, as shown below:
1
0.5
0
Output signal y[n]
−0.5
−1
0
10
20
30
40
60
70
80
90
100
0
10
20
30
40
60
50
Time Index n
70
80
90
100
50
1
0.5
0
−0.5
−1
You want to use this data to identify a model of the form
b0 + b1 z −1 + b2 z −2 + · · · + bM z −M
b
H(z) =
1 + a1 z −1 + a2 z −2 + · · · + aN z −N
in the frequency domain.
a) Let Y (Ω) be the Discrete Fourier Transform of the mea- (1 point)
sured output y[n]. How many frequency points will the
Transform have in the range 0 ≤ Ω ≤ π?
Now assume that you have calculated the system transfer function
H(Ωk ) =
Y (Ωk )
X(Ωk )
at each frequency point Ωk .
b) When fitting a model to the system transfer function, what (2 points)
are the highest possible model orders N and M that you
can identify using only the given measurement data?
Final Exam – Signals & Systems
c) A colleague has identified a model from the given measure- (2 points)
ment data. A plot of the frequency response of his estib
mated model H(z)
is shown below. Is this result plausible,
and why?
Amplitude
3
2
1
Phase (◦ )
0
0
0.5
1
0
0.5
1
1.5
2
2.5
3
2
2.5
3
90
0
−90
1.5
Frequency Ω
Solution 10
a) There will be 51 points. The Discrete Fourier Transform will result in
101 frequency domain points in a 2π interval. This means that the spacing
between frequency points will be
2π
.
∆Ω =
101
There will be one point at Ω = 0. There will be 50 more points in the
interval, because 50∆Ω < π and 51∆Ω > π.
b) There are 101 pieces of information (the point at Ω = 0 is real, and all
points at Ω 6= 0 are complex.). Therefore,
N + M + 1 ≤ 101
N + M ≤ 100.
c) No. The system has a phase of 90◦ at Ω = 0, which means that the
system has a complex gain. This is clearly not the case, as the inputoutput data is real.