EE3054 - Homework 10 - Solutions
1.
(a) From the properties of an op-amp, we can find the current through the series RC combination
to be x(t)
R . If vR (t) and vC (t) are the voltages across the resistor and capacitor in the series RC
combination with the polarities being such that y(t) = vR (t) + vC (t), then
x(t)
R = −x(t)
R
x(t)
= −
.
R
vR (t)
C
dvC (t)
dt
C
dy(t)
dt
= −
(1)
(2)
Therefore,
dvR (t)
dvC (t)
+C
dt
dt
dx(t) x(t)
= −C
−
.
dt
R
= C
(3)
Hence, the differential equation relating the input signal x(t) and the output signal y(t) is
ẏ(t) = −ẋ(t) −
x(t)
.
RC
(b) Let the step response of the system be denoted by s(t). Noting that δ(t) =
= −δ(t) −
ṡ(t)
(4)
du(t)
dt ,
we have
u(t)
RC
(5)
implying that
s(t) − s(0)
Z t
u(τ )
−δ(τ ) −
dτ
RC
0
tu(t)
= −u(t) −
.
RC
=
(6)
Since the capacitor is not given to have an initial charge, we can take s(0) = vR (0) + vC (0) = 0.
Hence, the step response of the system is
tu(t)
.
RC
= −u(t) −
s(t)
(7)
(c) Since the unit impulse is the derivative of the unit step, the impulse response h(t) is the derivative
of the step response s(t). Therefore,
= −δ(t) −
h(t)
u(t)
.
RC
(8)
(d) Taking the Laplace transform of h(t), we find the transfer function H(s) of the system to be
H(s)
= −1 −
1
.
sRC
(9)
Alternatively, the transfer function H(s) can be found directly from the given electrical circuit
1
by using the fact that the impedance of a capacitor is sC
to get
H(s)
=
1
R + sC
1
Y (s)
= −
= −1 −
.
X(s)
R
sRC
1
(10)
(e) We can check BIBO stability using either of the following methods:
Method 1 (From the impulse response): For the impulse response given in (8), we have
Z ∞
Z ∞
1
1dt = ∞.
(11)
|h(t)|dt = 1 +
RC 0
−∞
R∞
Hence, −∞ |h(t)|dt is not finite implying that the system is not BIBO stable.
Method 2 (From the transfer function): The transfer function H(s) has one pole at 0. We
know that a system is BIBO stable if and only if all poles have real parts less than zero (i.e., if
all poles lie in the left half plane). Since the pole 0 does not lie in the left half plane, the given
system is not BIBO stable.
(f) From the relation between the input signal x(t) and the output signal y(t) given in (4), we have
Z t
1
x(τ )dτ
y(t) = −x(t) −
RC −∞
Z t
cos(2t) − 1
1
u(t). (12)
sin(2τ )u(τ )dτ = − sin(2t)u(t) +
= − sin(2t)u(t) −
RC −∞
2RC
Alternatively, given that the input signal is the sinusoidal signal x(t) = sin(2t)u(t), we can find
the output signal using convolution as follows:
Z ∞
x(τ )h(t − τ )dτ
y(t) = x(t) ∗ h(t) =
−∞
Z ∞
u(t − τ )
sin(2τ )u(τ )[−δ(t − τ ) −
=
]dτ
RC
−∞
Z ∞
u(t − τ )
sin(2τ )[−δ(t − τ ) −
]dτ
=
RC
0
Z ∞
u(t − τ )
[− sin(2t)δ(t − τ ) − sin(2τ )
=
]dτ
RC
0
Z ∞
u(t − τ )
dτ
sin(2τ )
= − sin(2t)u(t) −
RC
0
Z t
1
sin(2τ )
= − sin(2t)u(t) −
dτ u(t)
RC
0
τ =t
cos(2τ ) = − sin(2t)u(t) +
u(t)
2RC τ =0
cos(2t) − 1
u(t).
(13)
= − sin(2t)u(t) +
2RC
2.
(a) The impulse response of System 1 is given to be h1 (t) = e−2t u(t). We know that the Laplace
1
transform of e−αt u(t) is s+α
. Therefore, the Laplace transform of h1 (t), i.e., the transfer function
1
1
. Therefore,
of System 1, is s+2 . The transfer function of System 2 is given to be H2 (s) = s+1
the transfer function of the overall system is
H(s)
= H1 (s)H2 (s) =
1
1
= 2
.
(s + 1)(s + 2)
s + 3s + 2
(14)
−1
1
+
.
s+1 s+2
(15)
(b) Using partial fractions,
H(s)
=
Taking the (causal) inverse Laplace transform of H(s), we get the impulse response of the overall
system to be
h(t)
= e−t u(t) − e−2t u(t).
2
(16)
To find the step response s(t), we can integrate the impulse response:
s(t) =
Z
=
Z
t
h(τ )dτ =
Z
t
[e−τ − e−2τ ]u(τ )dτ
−∞
−∞
t
[e−τ − e−2τ ]dτ u(t)
0
t
1 −2τ −τ
=
−e + e
u(t)
2
0
1
=
1 − e−t + (e−2t − 1) u(t).
2
(17)
(c) The impulse response was found above to be as in (16).
(d) The overall system being the cascade interconnection of two causal systems is also causal.
(e) The poles of H(s) are -1 and -2, both of which are in the left half plane (i.e., have real parts
smaller than zero). Hence, the system is BIBO stable. Alternatively, the system is BIBO stable
since
Z ∞
Z ∞
Z ∞
3
−t
< ∞.
(18)
e−2t dt =
e dt +
|h(t)|dt ≤
2
0
0
−∞
3