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EE3054 - Sample Midterm Test 2
Total Points: 50
Time: 1 hour
1. (12 points) The impulse response of an LTI system is given to be h(t) =
e−2t u(t) + e−3t u(t).
(a) Find the Laplace transform of h(t) and indicate the region of convergence.
(b) Using convolution, find the output of the system if the input is
x(t) = et u(t). Assume zero initial conditions.
(c) Find the differential equation relating the input x(t) and the output y(t).
Solution:
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(a) The Laplace transform of eαt u(t) is s−α
with the ROC Re(s) > α.
Hence, the Laplace transform of h(t) = e−2t u(t) + e−3t u(t) is
1
2s + 5
1
+
= 2
s+2 s+3
s + 5s + 6
with the ROC Re(s) > max(−2, −3) = −2.
(b) The output of the system under the given input is
(1)
H(s) =
y(t) = h(t) ∗ x(t) = [e−2t u(t) + e−3t u(t)] ∗ et u(t)
= e−2t u(t) ∗ et u(t) + e−3t u(t) ∗ et u(t).
(2)
In general, given any real numbers α and β, we can evaluate
αt
βt
e u(t) ∗ e u(t) =
Z
∞
eατ u(τ )eβ(t−τ ) u(t − τ )dτ
−∞
= eβt
= eβt
Z
t
e(α−β)τ dτ u(t)
0
(α−β)t
e
−1
eαt − eβt
u(t) =
u(t). (3)
α−β
α−β
Hence,
et − e−2t et − e−3t
e−2t − et e−3t − et
+
+
u(t) =
u(t).
y(t) =
−3
−4
3
4
#
"
1
"
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(c) From (1), we have Y (s)[s2 + 5s + 6] = X(s)[2s + 5]. Hence, the
differential equation relating input x(t) and the output y(t) is
ÿ(t) + 5ẏ(t) + 6y(t) = 2ẋ(t) + 5x(t).
(4)
2. (12 points) Consider a circuit formed by connecting a voltage source
v(t), a resistor R = 1Ω, a capacitor C = 0.01F , and an inductor
L = 1H in series. Consider the voltage v(t) provided by the voltage
source to be the input to this system and consider the voltage vC (t)
across the capacitor to be the output of the system. At time t = 0,
when the circuit is powered on, the capacitor has an initial voltage of
1 V and the current through the circuit is initially 0 A.
(a) Show that the differential equation relating the input x(t) = v(t)
and the output y(t) = vC (t) is
0.01ÿ(t) + 0.01ẏ(t) + y(t) = x(t).
(b) Find the output signal vC (t) if the input voltage signal is v(t) =
e−t u(t) taking into account the initial conditions specified above.
Solution:
(a) Denoting the voltages across the resistor, the capacitor, and the
inductor by vR , vC , and vL , respectively, and denoting the current
through the circuit to be i, we obtain the following equations:
vR (t) + vC (t) + vL (t) = v(t)
dvC (t)
i(t) = C
dt
dvC (t)
dt
di(t)
d2 vC (t)
vL (t) = L
= LC
.
dt
dt2
vR (t) = i(t)R = RC
(5)
Hence,
LC
dvC (t)
d2 vC (t)
+ RC
+ vC (t) = v(t).
2
dt
dt
2
(6)
Substituting R = 1Ω, C = 0.01F , L = 1H, v(t) = x(t), vC (t) =
y(t), we get the differential equation
0.01ÿ(t) + 0.01ẏ(t) + y(t) = x(t).
(7)
(b) Initially at time t = 0, the capacitor has a voltage of 1 V, i.e.,
y(0) = 1. Also, the current through the circuit is initially 0 A.
C (t)
Hence, i(t) = C dvdt
= 0 at time t = 0, i.e., ẏ(0) = 0. Therefore,
we need to solve the differential equation (7) with the input signal
x(t) = e−t u(t) and the initial conditions y(0) = 1 and ẏ(0) = 0.
This can be done using either the one-sided Laplace transform or
the method of guessing exponentials.
Using one-sided Laplace transform: Taking one-sided Laplace transforms of both sides of the differential equation (7), we get
0.01[s2 Y (s)−sy(0)− ẏ(0)]+0.01[sY (s)−y(0)]+Y (s)=X(s).(8)
Hence,
X(s) + 0.01sy(0) + 0.01ẏ(0) + 0.01y(0)
0.01s2 + 0.01s + 1
1
+ 0.01s + 0.01
= s+1 2
0.01s + 0.01s + 1
1 + 0.01(s + 1)2
=
(0.01s2 + 0.01s + 1)(s + 1)
1
−0.0501j
0.0501j
=
+
+
s+1 s+0.5000−9.9875j s+0.5000+9.9875j
Y (s) =
(9)
implying that, for all t ≥ 0,
y(t) = e−t − 0.0501je(−0.5+9.9875j)t + 0.0501je(−0.5−9.9875j)t
= e−t + e−0.5t [−0.0501je9.9875jt + 0.0501je−9.9875jt ]
h
= e−t + e−0.5t − 0.0501j cos(9.9875t) + 0.0501 sin(9.9875t)
+0.0501j cos(9.9875t) + 0.0501 sin(9.9875t)
= e−t + 0.1002e−0.5t sin(9.9875t).
i
(10)
Using the method of guessing exponentials: The poles of the system are at −0.5 + 9.9875j and −0.5 − 9.9875j. The pole of the
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input signal is at −1. Hence, for t ≥ 0, the output signal should
be of the form
y(t) = c1 e−t + c2 e(−0.5+9.9875j)t + c3 e(−0.5−9.9875j)t .
(11)
Differentiating twice,
ẏ(t) = −c1 e−t + c2 (−0.5 + 9.9875j)e(−0.5+9.9875j)t
+c3 (−0.5 − 9.9875j)e(−0.5−9.9875j)t
ÿ(t) = c1 e−t + c2 (−0.5 + 9.9875j)2 e(−0.5+9.9875j)t
+c3 (−0.5 − 9.9875j)2 e(−0.5−9.9875j)t
= c1 e−t + c2 (−99.5000 − 9.9875j)e(−0.5+9.9875j)t
+c3 (−99.5000 + 9.9875j)e(−0.5−9.9875j)t .
(12)
Using the given initial conditions, we get the equations
c1 + c2 + c3 = 1
−c1 + (−0.5 + 9.9875j)c2 + (−0.5 − 9.9875j)c3 = 0. (13)
Using the differential equation 0.01ÿ(t) + 0.01ẏ(t) + y(t) = e−t and
equating the coefficients of e−t , e(−0.5+9.9875j)t , and e(−0.5−9.9875j)t ,
we also get the equation
c1 = 1.
(14)
Solving, we obtain c1 = 1, c2 = −0.0501j, and c3 = 0.0501j
implying that for all t ≥ 0,
y(t) = e−t − 0.0501je(−0.5+9.9875j)t + 0.0501je(−0.5−9.9875j)t
= e−t + 0.1002e−0.5t sin(9.9875t).
(15)
3. (10 points) You are given a system with the differential equation
ÿ(t) + 2ẏ(t) + 2y(t) = x(t).
(a) Find the frequency response H(jω) of the system. Find the magnitude response |H(jω)| and the phase response 6 H(jω).
(b) If the input signal is x(t) = 2 sin( π3 t + π3 )u(t), find the steady-state
output signal.
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Solution:
(a) The transfer function of the system is
1
.
+ 2s + 2
Hence, the frequency response of the system is
H(s) =
H(jω) =
−ω 2
s2
1
1
=
.
2
+ 2jω + 2
(2 − ω ) + 2jω
(16)
(17)
The magnitude response and the phase response are
|H(jω)| = q
6
1
(2 − ω 2 )2 + 4ω 2
2ω
−1
H(jω) = − tan
.
2 − ω2
(18)
(19)
(b) The frequency of the given input signal is ω = π3 . Hence, we expect
the steady-state output signal to be the same as the input signal
except for a magnitude gain of |H(j π3 )| = 0.4384 and a phase
shift of 6 H(j π3 ) = −1.1636, i.e., the steady-state output signal is
2(0.4384) sin( π3 t + π3 − 1.1636)u(t) = 0.8768 ∗ sin( π3 t − 0.1164).
4. (6 points) You are given a causal LTI system with transfer function
1
.
H(s) = s2 +s+1
(a) Find the impulse response of the system.
(b) Find the step response of the system.
Solution:
(a) The given transfer function can be decomposed into partial fractions as
0.5774j
−0.5774j
+
.
(20)
H(s) =
s + 0.5 − 0.866j s + 0.5 + 0.866j
Since the system is given to be causal, h(t) is the causal inverse
Laplace transform of H(s):
h(t) = −0.5774je(−0.5+0.866j)t u(t) + 0.5774je(−0.5−0.866j)t u(t)
= 1.1548e−0.5t sin(0.866t)u(t).
(21)
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(b) The step response is obtained by integrating the impulse response:
s(t) =
Z
t
−∞
h(τ )dτ
e(−0.5−0.866j)t − 1
e(−0.5+0.866j)t − 1
u(t) + 0.5774j
u(t)
−0.5 + 0.866j
−0.5 − 0.866j
= [−e−0.5t cos(0.866t) − 0.5774e−0.5t sin(0.866t) + 1]u(t). (22)
= −0.5774j
Alternatively, the above answer can also be obtained by integrating 1.1548e−0.5t sin(0.866t)u(t) using integration by parts.
5. (10 points) You are given an unknown LTI system. You are told that
the transfer function of the system is of the form a + bs with a and b
being unknown constants. It is observed that the output of the system
to the input tu(t) is (1 + 3t)u(t).
(a) Find the constants a and b.
(b) Is this system BIBO stable?
(c) If your answer to part (b) was ”No”, construct a bounded input
signal which results in an unbounded output signal. If your answer
to part (b) was ”Yes”, prove that the system is BIBO stable.
Solution: The transfer function of the system is given to be H(s) =
a+bs. Hence, the relation between the input and output can be written
as y(t) = ax(t) + bẋ(t).
(a) If x(t) = tu(t), then y(t) = atu(t) + b[u(t) + tδ(t)] = (at + b)u(t)
since tδ(t) = 0. Therefore, a = 3 and b = 1.
(b) The system is not BIBO stable since the output to the bounded
input signal u(t) is 3u(t)+δ(t) which is not a bounded signal. The
fact that the system is not BIBO stable can also be inferred by
noticing that |H(s)| → ∞ as |s| → ∞, i.e., the system has a right
half plane pole at ∞.
(c) When the input signal is the bounded signal u(t), the output signal
is the unbounded signal 3u(t) + δ(t).
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