ENGR2410S10 Problem Set 8
Spring 2010
Problem 1 Solution
Problem 1 A
Find an algebraic expression for the impulse response of vC (t) and graph it for = .1 and = .01.
Problem 1 A Solution
Note that the impulse response h(t) = F −1 {H(jω)}. Now, let’s break up the circuit and find the
transfer function.
ZX
vmid
vS
vC
vmid
= ZC ||(R2 + ZC )
ZX
=
R1 + ZX
ZC
=
R2 + ZC
Cascaded transfer functions:
vmid vC
·
vs vmid
vC
vS
vC
vS
=
=
=
ZX
ZC
·
R1 + ZX R2 + ZC
ZC ||(R2 + ZC )
ZC
·
R1 + ZC ||(R2 + ZC ) R2 + ZC
ZC2
R1 R2 + 2R1 ZC + R2 ZC + ZC2
1
Substitute in ZC =
1
jωC ,
R1 = 1, R2 = , C = 1.
1
vC
(jωC)2
=
1
1
1 2
vS
R1 R2 + 2R1 Cjω + R2 Cjω
+ ( Cjω
)
1
2
(jω) + 2jω + jω + 1
H(jω) =
=
1
=
(jω)2 + ( 2 + 1)jω =
1
The roots of the denominator are
jω = − 21 −
1
±
q
1
4
+
1
2
Since 1, we can approximate
jω ≈ − 12 −
jω ≈
H(jω) ≈
≈
1
− 21 , − 2
±
1
1
(jω + 12 )(jω + 2 )
1
jω +
1
2
·
1
jω +
2
Partial fraction expansion gives us the following
H(jω) ≈
1
2
jω +
1
2
+
− 12
jω +
2
Now we can take the inverse Fourier transform to obtain the impulse response.
(
)
(
)
1
1
−
2
2
h(t) = F −1
+ F −1
jω + 21
jω + 2
=
t
−2
1
2 u(t)e
2t
− 12 u(t)e− Problem 1 B
Show that the circuit can be decomposed into two cascaded first-order systems as shown.
Problem 1 B Solution
From part A we know
H(jω) ≈
≈
1
(jω + 12 )(jω + 2 )
1
jω +
2
1
2
·
1
jω +
2
Thus we have two subsystems.
1
jω +
H1 (jω) =
1
2
1
H2 (jω) =
jω +
2
Problem 1 C
Find the impulse response of each subsystem. For the second subsystem, find the total area under
the impulse response. Does it depend on ? Let → 0. What is the transfer function of the second
subsystem in this limit? What is its impulse response?
Problem 1 C Solution
1
jω +
H1 (jω) =
1
2
t
vC1 = D0 u(t)e− 2
1
H2 (jω) =
jω + 2
2t
D0
u(t)e− vC2 =
Z
Area =
=
=
∞
D0 − 2t
e dt
0
D0 − 2t ∞
[− e ]0
2
D0
2
The area under the impulse response of the second subsystem does not depend on .
lim
→0
1
jω +
=
2
vC2 =
1
2
D0
δ(t)
2
Problem 1 D
Collapse the two subsystems under the limit of the previous part. Find the impulse response. What
is the order of the system? Explain this result using the original circuit.
3
Problem 1 D Solution
1
2
H(jω) =
jω + 21
t
D0
vC (t) =
u(t)e− 2
2
The above system is a first order system. When → 0, R2 goes away leaving two capacitors in
parallel. These capacitors can be combined into one with twice the capacitance.
1
τ
H(jω) =
jω +
1
τ
1
2
=
jω +
1
2
X
Figure 1: As gets smaller the impulse response looks more like a single eponential times the step
function rather than a sum of exponentials.
Problem 2 Solution
The impulse response is h(t) = 34 e−2t u(t) + 41 e2t u(t) and the input is x(t) = 23 e−t u(t) − 21 et u(t).
Problem 2 A
Graph the input x(t) and the impulse response of the system h(t). Do they have a Fourier transform? If so, find them.
4
Problem 2 A Solution
3 −2t
1
e u(t) + e2t u(t)
4
Z4 ∞
3 −2t −jωt 1 2t −jωt
F {h(t)} =
e e
+ e e
dt
4
4
0
Z ∞
3 −(2+jω)t 1 −(−2+jω)t
=
e
+ e
dt
4
4
0
3
−1
−1
= 0+
+
e∞ e−∞jω −
4(2 + jω) 4(−2 + jω)
4(−2 + jω)
h(t) =
The integral does not converge and thus, the fourier transform does not exist.
3 −t
1
e u(t) − et u(t)
2
Z2 ∞
3 −(1+jω)t 1 −(−1+jω)t
F {x(t)} =
e
− e
dt
2
2
0
x(t) =
The input function x(t) has the same problem with the Fourier transform as does h(t) and does
not have a fourier transform.
Problem 2 B
Find the Laplace transform of the input and the system. Show the region of convergence in a
pole-zero diagram for each one.
5
Problem 2 B Solution
3 −2t
1
e u(t) + e2t u(t)
4
4
Z ∞
3 −2t −st 1 2t −st
L {h(t)} =
e e
+ e e dt
4
4
0
Z ∞
3 −(2+s)t 1 −(−2+s)t
=
e
+ e
dt
4
4
0
<{s} > −2,
<{s} > 2
h(t) =
=
=
x(t) =
L {x(t)} =
=
<{s} > −1,
=
=
3
4
+
1
4
s+2 s−2
s−1
, <{s} > 2
s2 − 4
3 −t
1
e u(t) − et u(t)
2
2
Z ∞
3 −t −st 1 t −st
e e
− e e dt
2
2
0
Z ∞
3 −(s+1)t 1 −(s−1)
e
− e
dt
2
2
0
<{s} > 1
3
2
−
1
2
s+1 s−1
s−2
, <{s} > 1
s2 − 1
Show the region of convergence in a pole-zero diagram for each one.
Problem 2 C
Find the Laplace transform of the output y(t). Show the region of convergence in a pole-zero
diagram.
6
Problem 2 C Solution
Find the Laplace transform of the ouput y(t).
Y (s) = H(s)X(s)
(s − 1)(s − 2)
=
(s2 − 4)(s2 − 1)
1
=
(s + 2)(s + 1)
Show the region of convergence in a pole-zero diagram.
Problem 2 D
Find the output y(t) and graph it.
Problem 2 D Solution
1
(s + 2)(s + 1)
1
1
=
−
s+1 s+2
y(t) = L −1 {Y (s)} = u(t)e−t − u(t)e−2t
Y (s) =
7
Does it have a Fourier transform? Find it.
Yes, the jω axis is included in the region of convergence.
s = jω
Y (jω) =
1
1
−
jω + 1 jω + 2
8