MAE143A In Class Midterm 2 - Tuesday February 23, 2016
Instructions
This midterm is open book. You may use whatever written materials you choose including your
class notes and textbooks etc. You may NOT use electronic devices; computers, calculators, phones,
music players, tablet computers, etc. These permissions should be taken to indicate the limited help
that either written material or computational assistance is likely to provide — please do not spend
significant time looking up books. That is not what is being tested here. Marks are awarded for
concepts, methods and the expression of ideas.
There are three equally valued questions. You should attempt to answer all.
Question 2 follow from Question 1. Question 3 is independent from the other two.
You have 70 minutes. Please mark your papers with your name and student number. If your answer
to any specific question occurs on non-consecutive pages, please indicate this on the departing and
arriving pages.
HINTS:
- Read the questions very carefully.
- Marks are awarded for methods, concepts and expression.
- The symbol θc means that angle θ is measured in radians.
- A gain G expressed in decibels is 20 log10 G dB.
Question 1 – Continuous-time system response
A continuous-time linear system and its initial conditions are described by the ordinary differential
equation
du(t)
dy(t) d2 y(t)
dy(t)
+ 10y(t) = −
+ u(t),
+2
= −1, y(0) = 2,
(1)
dt2
dt
dt
dt t=0
where u(t) is the input signal and y(t) is the output signal.
(i) Using Laplace transforms on (1), find the Laplace transform of: the zero-input response, and
the zero-state response.
(ii) Determine the poles of the system and state whether the system is stable in the sense of Lyapunov. Is this system also BIBO stable?
(iii) Using the residue formula, find the partial fraction expansion of the zero-input response and,
thereby show that the zero-input response is given by
√
37 −t
e cos[3t + atan2(−1/6, 1)]1(t).
yz.i. (t) =
3
(iv) [Bonus] M ATLAB gives the following
>> dsolve(’D2y+2*Dy+10*y=0’,’Dy(0)=-1’,’y(0)=2’)
ans =
2*cos(3*t)*exp(-t) + (sin(3*t)*exp(-t))/3
Show that the answer in Part (iii) concurs with M ATLAB. Or, if you like, show that M ATLAB is
correct.
(i) Take Laplace transforms
s2 Y (2) − sy(0) − ẏ(0) + 2sY (s) − 2y(0) + 10Y (s) = −sU (s) + U (s),
(s2 + 2s + 10)Y (s) − 2s + 1 − 4 = (−s + 1)U (s),
2s + 3
−s + 1
Y (s) = 2
+ 2
U (s).
s + 2s + 10 s + 2s + 10
2s+3
The first term, s2 +2s+10
, is the zero-input response transform. The second term,
is the zero-state response.
−s+1
U (s),
s2 +2s+10
(ii) The poles of the system are at the roots of s2 + 2s + 10, i.e. s = −1 ± 3j. Since these roots have
negative real part, they lie in the open left half-plane and the system is stable in the sense of
Lyapunov. That is, the zero-input response decays to zero as t → ∞ for any initial conditions.
Since the poles lie in the open left half-plane it is also BIBO stable. That is, the total response
(zero-state plus zero-input) to a bounded input (possessing a Laplace transform with poles only
in the open left half-plane with the possible exception of simple poles on the imaginary axis)
yields a bounded output (same condition on its transform’s poles).
(iii) The partial fraction expansion of the zero-input response has the form
a∗
a
+
.
s + 1 − 3j s + 1 − 3j
Yzi (s) =
To compute a, use the residue formula
(s + 1 − 3j)(2s + 3)
,
s→−1+3j (s + 1 − 3j)(s + 1 + 3j)
−2 + 6j + 3
,
=
6j
1 + 6j
=
,
6j
1
= 1 − j.
6
1
a∗ = 1 + j.
6
a=
lim
So the zero-input response is given by
yz.i. (t) = [ae−t ej3t + a∗ e−t e−3jt ]1(t),
= 2|a| cos(3t + ∠a)1(t).
Now,
r
|a| =
1
1+
=
36
√
37
, ∠a = atan2(−1/6, 1).
6
(iv) Denote θ = atan2(−1/6, 1). Note from the following plot of a = 1 − j 61 in the complex
.
s-plane, cos θ = √637 and sin θ = √−1
37
Im s
1
θ
1
6
p
37
6
1
x
j
Re s
1
6
Then
cos(3t + θ) = cos(3t) cos θ − sin(3t) sin θ,
6
1
= √ cos(3t) + √ sin(3t).
37
37
M ATLAB’s answer follows directly.
Question 2 — Transfer function, sinusoidal steady-state
(i) Define the impulse response, h(t), of a linear system. Define the transfer function of a linear
system.
(ii) From your Laplace transform in Question 1, Part (i), show that the transfer function of this
system is given by
H(s) =
−s + 1
.
s2 + 2s + 10
(iii) If the input to this system is chosen as a step, i.e. u(t) = 1(t), does the step response tend to
a limit as t → ∞? If so, then use the Final Value Theorem to determine the limit. If not, then
explain what happens to the step response as t → ∞.
(iv) If the input signal is chosen as sinusoid u(t) = A cos(ωt + φ)1(t), then describe the output
response. What happens to the zero-input response, yz.i. (t), and to the zero-state response,
yz.s. (t) as t → ∞?
(v) Show that the sinusoidal steady-state response to a sinusoid of zero frequency has gain of -20 dB
and a phase shift of 0◦ .
(vi) Show that the approximate response to a sinusoid of frequency 3 rad/s has a gain less than 0 dB
and has a phase shift in the third quadrant?
(vii) What is the approximate response to a sinusoid of frequency greater than 100 rad/s?
(viii) [Bonus] Plot the poles and zero of this system in the complex s-plane and argue for the connection to the numerical results in Parts (v)-(vii) above.
(i) The impulse response, h(t), is the zero-state response when the input signal is u(t) = δ(t). The
transfer function, H(s), is the Laplace transform of the impulse response.
(ii) The Laplace transform of the zero-state response is, from Question 1 Part (i),
s2
−s + 1
U (s).
+ 2s + 10
When u(t) = δ(t), i.e. U (s) = 1, the above is the Laplace transform of the impulse response
−s+1
and, hence, the transfer function. Thus, H(s) = s2 +2s+10
.
(iii) Since the system has all of its poles in the open left half-plane, Re s < 0, the system is stable
and the Laplace transform of the step response has all the poles in the left half-plane except for
single pole at s = 0 due to the step function. Thus the output signal (the step response) has a
limit as t → ∞. Denote the step response as ys.r. (t). Then the Final Value Theorem says
lim ys.r. (t) = lim sYs.r. (s),
t→∞
s→0
−s + 1
U (s),
s→0
+ 2s + 10
−s + 1
1
= lim s × 2
× ,
s→0
s + 2s + 10 s
−s + 1
,
= lim 2
s→0 s + 2s + 10
1
= .
10
= lim s ×
s2
In decibels, 10−1 is -20 dB. Since 1/10 is a positive number, its phase is 0◦ . This is the answer
to Part (v) below as well.
(iv) Since the system is stable the output will tend to the sinusoidal steady state signal. The zeroinput response tends to zeros as t → ∞. The zero-state response tends to a sinusoid of the same
frequency, ω but with the amplitude changed by gain |H(jω)| and phase changed by ∠H(jω).
(v) This is already done in Part (iii) of this question above.
(vi) For radian frequency 3 rad/sec, the gain is
1 − 3j
1
−s + 1
=
= (−17 − j19).
H(j3) =
(s + 1 − 3j)(s + 1 + 3j) s=j3 1 + 6j
37
This has magnitude less than one, since 172 + 192 = 249 + 361 = 610 < 1369 = 372 . A gain
less than one is less than 0 dB. Because both the real part and the imaginary part of H(j3) are
negative, its angle is in the third quadrant.
−jω
j
(vii) For a sinusoid with frequency above 100 rad/sec, H(jω) ≈ (jω)
2 = ω . This is a phase of -270
degrees (90 degrees is fine too) and amplitude 1/ω. The amplitude goes down by -20 dB per
decade of frequency.
(viii) [Bonus] See the figure nearby for the pole-zero plot. The zero at s = 1 is close to the imaginary
axis for low frequencies. So the transfer function has low magnitude here. The poles at s =
−1 ± j3 is where the transfer function H(s) blows up and is close to the imaginary axis at
s = j3. So the magnitude of the transfer function is much larger as s gets close to a pole.
θ
1
6
p
37
6
1
x
j
1
6
Im s
1 + j3 x
j3
1
1
Re s
1
j3 x
j3
Figure 1: Question 2 Bonus: Poles and zero of H(s). Poles indicated by x and zero by o.
Question 3 — Discrete-time system
[Hint 1: when we say “show that” here, we do not mean “prove”. A few computations is fine. If we
had said “prove” then proof by induction would probably be the simplest approach.
Hint 2: For each problem, since the input signal begins at time n = 0, the initial condition is at time
n = −1.]
A discrete-time system is described by the ordinary difference equation
y1 [n] − 0.8y1 [n − 1] = u1 [n],
(2)
(i) Show that the impulse response of this system is h1 [n] = (0.8)n 1[n]. Then show that the ztransform of this signal is
H1 (z) = Z {h1 [n]} =
1
z
=
.
z − 0.8
1 − 0.8z −1
[Since this is the z-transform of the impulse response, this is the transfer function of this discrete
system.]
(ii) Show that the step response, s1 [n], of the system in (2) is given by
s1 [n] =
n
X
h1 [k].
k=0
(iii) Show that the zero-state response output signal of the system described by difference equation
y2 [n] − y2 [n − 1] = u2 [n],
to input signal u2 [n] is given by
y2 [n] =
n
X
k=0
u2 [k].
(iv) [Bonus] Use the above results to show that the transfer function of the summer in Part (iii) has
a pole at z = 1.
(i) For the impulse response, we take the zero-state response with u1 [n] = δ[n] the discrete impulse, u1 [0] = 1 and u1 [n] = 0 otherwise. The zero state means that h1 [−1] = 0.
h1 [0] = 0.8h1 [−1] + u1 [0] = 0 + 1 = 1,
h1 [1] = 0.8h1 [0] + u1 [1] = 0.8 ∗ 1 + 0 = 0.8,
h1 [2] = 0.8h1 [1] + u1 [2] = 0.8 ∗ 0.8 + 0 = (0.8)2 ,
h1 [3] = 0.8h1 [2] + u1 [3] = 0.8 ∗ (0.8)2 + 0 = (0.8)3 ,
..
.
The z-transform of this sequence is
H1 (z) =
∞
X
n=0
z
−n
h1 [n] =
∞
X
(z −1 0.8)n =
n=0
1
z
=
.
−1
1 − 0.8z
z − 0.8
(ii) Run the recursion (2) with u1 [n] = 1[n], i.e. u1 [n] = 1 if n ≥ 0 else u[n] = 0. The step
response is a zero-state response. So s1 [−1] = 0.
s1 [0] = 0.8s1 [−1] + u1 [0] = 0 + 1 = 1,
s1 [1] = 0.8s1 [0] + u1 [1] = 0.8 + 1,
s1 [2] = 0.8s1 [1] + u1 [2] = (0.8)2 + 0.8 + 1,
s1 [3] = 0.8s1 [2] + u1 [3] = (0.8)3 + (0.8)2 + 0.8 + 1,
..
.
This shows the result.
(iii) Start by substitution form y2 [−1] = 0 since we want the zero-state response.
y2 [0] = y2 [−1] + u2 [0] = u2 [0],
y2 [1] = y2 [0] + u2 [1] = u2 [0] + u2 [1],
y2 [2] = y2 [1] + u2 [2] = u2 [0] + u2 [1] + u2 [2],
..
.
This establishes the result.
(iv) [Bonus] We see that the step response is the summed version of the impulse response. The
transfer function of a discrete-time system is the z-transform of its impulse response. So the
transfer function of (2) is
H1 (z) =
1
.
1 − 0.8z −1
Therefore, the step response has z-transform S1 (z) = H1 (z) × 1−z1 −1 , since 1−z1 −1 is the ztransform of a step sequence. Equally, treating the signal with z-transform H1 (z) as the input
signal, we see that s1 [n] is also the sum of the impulse response or
S1 (z) =
So, the summer has transfer function
1
× H1 (z).
1 − z −1
1
,
1−z −1
which has a pole at z = 1.