Olin College of Engineering
ENGR2410 – Signals and Systems
Assignment 5
Problem 1 (6 points) In lecture, we showed that if the input current i(t) = Q0 δ(t) in the
circuit below, then the output voltage v(t) = QC0 e−t/τ , τ = RC. Since i(t) is an impulse
input, v(t) is the impulse response of the system.
A. Find the impedance of the circuit Z(jω) = V (jω)/I(jω).
Solution:
Z(jω) =
1
1
(
C jω +
1
RC
)
B. Find the frequency content of the impulse response v(t), V (jω), by multiplying the
Fourier transform of the input i(t) with the impedance Z(jω).
Solution:
I(jω) = Q0
V (jω) =
Q0
1
(
C jω +
1
RC
)
C. Take the Fourier transform of the impulse response v(t) using the integral definition
of the Fourier transform directly. Note that you have just found another important
Fourier transform pair!
Solution:
∫
∫
Q0 ∞ −t/τ −jωt
Q0 ∞ −(1/τ +jω)t
Q0
−t/τ
F {e
}=
e
e
dt =
e
dt
V (jω) = F {v(t)} =
C
C t=0
C t=0
]∞
Q0
Q0
Q0
−1
−1
1
=
e−(1/τ +jω)t t=0 =
[0−1] =
, τ = RC
C (1/τ + jω)
C (1/τ + jω)
C (1/τ + jω)
Note that we have just found that:
F {e−t/τ } =
This implies that:
e
−t/τ
=F
1
1/τ + jω
{
−1
1
1/τ + jω
}
More generally, we write:
e−t/τ
F
⇐⇒
1
1/τ + jω
This transform will be useful in part A of problem 2.
D. Compare the answers to the previous two parts. Explain the relationship between
them clearly.
Solution:
We have arrived at the Fourier transform of the impulse response by taking the Fourier
transform directly and indirectly by multiplying the transfer function of the system
times the transform of the impulse. In system notation,
Q0 δ(t) −−−→


Fy
Q0
−−−→
−−−→
RC ckt
1
1
(
C jω +
1
RC
) −−−→
Q0 −t/τ
e
C


yF
Q0
1
C (jω+ 1 )
RC
The remaining parts should convince you that since the impulse is the derivative of the
step, the impulse response must be the derivative of the step response.
E. Find v(t) if i(t) = I0 u(t). Use either a differential equation, or circuit analysis and
your knowledge of first-order systems.
Solution:
v(t) = I0 R(1 − e−t/τ )
τ = RC
F. Find the operator that transforms the input from part E into the input from part A.
Be careful with any constants needed.
Solution:
[
]
Q0 d
Q0 δ(t) =
I0 u(t)
I0 dt
| {z }
operator
2
G. Apply the operator you found in the previous part to the step response you found in
part E and compare it to the impulse response of the circuit. Explain your result.
Solution:
[
]
Q0 d
Q0 R −t/τ
Q0 R −t/τ
I0 R(1 − e−t/τ ) =
e
=
e
=
{z
}
I0 dt |
τ
RC
| {z } step response
operator
Q0 −t/τ
e
|C {z }
impulse response
The derivative can be taken before or after finding the response of the system. In
either case, we find the impulse response. In system notation,
I0 u(t) −−−→
[
]
Q0 d 
I0 dt y
RC ckt
−−−→ I0 R(1 − e−t/τ )
[
 Q0 d ]
y I0 dt
Q0 δ(t) −−−→
RC ckt
−−−→
Q0 −t/τ
e
C
3
Problem 2 (4 points) This problem emphasizes that distributions like the impulse are
mathematical idealizations. Thus, the details of the distribution are not important as long
as the effect is the same. In other words, if it behaves like an impulse, it is an impulse.
A. Find an algebraic expression for the impulse response vC (t) if vS (t) = Λ0 δ(t) using
frequency analysis. Λ0 is the product of voltage and time, V·s, and has those units
(which turn out to be the units of magnetic flux for a really good reason if you’re
curious!)
Solution:
vS (t) = Λ0 δ(t)
vs (jω) = F {vS (t)} = Λ0
H(jω) =
1
τ
1
τ
+ jω
,
τ = RC
Λ0
τ
vC (jω) = H(jω)vs (jω) = 1
+ jω
}τ
{
Λ0
Λ0 − t
=
e τ u(t)
vC (t) = F −1 1 τ
τ
+ jω
τ
4
B. For the same circuit as in part A, assume that vS (t) is as shown below and find an
algebraic expression for vC (t). Let ∆ → 0. Show that vC (t) approaches the impulse
response you found in the previous part. Note that vS (0) = 0 for all values of ∆. Think
about how to define the function it approaches. Hint: Note that in the limit only one
part of the solution matters. Also, you will find it useful to remember that ex ≈ 1 + x
if x ≪ 1.
Solution:


0
vC (t) =
Λ0
(1
∆

 Λ0
(1
∆
t<0
∆ < t < 2∆ ,
t > 2∆
− e−(t−∆)/τ )
− e−∆/τ )e−(t−2∆)/τ
τ = RC
Letting ∆ → 0 we can show that vC (t) approaches the impulse response we found in
the previous part.
Λ0
(1 − e−∆/τ )e−(t−2∆)/τ ,
∆→0 ∆
lim vC (t) = lim
∆→0
t > 2∆
Λ0 2∆/τ
(e
− e∆/τ )e−t/τ , t > 2∆
∆→0
∆→0 ∆
(
)
Λ0
2∆
∆ −t/τ
lim vC (t) = lim
1+
−1−
e
, t > 2∆
∆→0
∆→0 ∆
τ
τ
( )
Λ0 ∆ −t/τ
lim vC (t) = lim
e
, t > 2∆
∆→0
∆→0 ∆
τ
lim vC (t) = lim
lim vC (t) =
∆→0
Λ0 −t/τ
e
,
τ
t>0
5
We can graphically illustrate the effect as ∆ → 0. The rectangle pulse becomes taller
and thinner. Eventually, it will have a height of ∞ and a width of 0, like an impulse.
Also, similar to an impulse, the area of the rectangle is constant, even when ∆ → 0:
Area = Λ∆0 · ∆ = Λ0 . Thus, as ∆ → 0, vS (t) becomes an impulse with area Λ0 .
6
Problem 3 (0 points) This optional problem illustrates the inverse frequency-time relationship of the Fourier transform. This relationship is reflected in quantum physics as
the Heisenberg Uncertainty Principle because the momentum and position of a particle are
Fourier transforms of each other.
A. Using the fact that
∫ ∞
√
2
e−t dt = π
−∞
show that the Fourier transform of the Gaussian e−t is the Gaussian
Complete the square of the exponential and make a substitution.
2
Solution:
∫
{ 2}
2
−t
F e
=
e−t e−jωt dt
∫t
2
e−t −jωt dt
=
∫t
ω2
ω2
2
=
e−t −jωt+ 4 e− 4 dt
∫t
jω 2
ω2
=
e−(t− 2 ) e− 4 dt
t
jω
Let u = t −
, du = dt
2
2
− ω4
∫
e−u du
2
= e
t
√ − ω2
=
πe 4
7
√ − ω2
πe 4 . Hint:
B. Show the scaling property of the Fourier Transform, F {x(at)} =
1
X( jω
).
|a|
a
Solution:
{
F −1
if a > 0,
}
∫
1
jω
1
jω
X( )
=
X( )ejωt dω
|a|
a
|a| ω
a
ω
, dω = adu
a
{
}
∫
1
jω
a
−1
F
X( )
=
X(ju)eju(at) du
|a|
a
|a| u
= x(at)
if a < 0,
{
}
∫
1
jω
−|a| ∞
−1
F
X( )
X(ju)eju(at) du
=
|a|
a
|a| −∞
∫
|a| −∞
=
X(ju)eju(at) du
|a| ∞
= x(at)
Let u =
t2
C. Use the scaling property to find the Fourier Transform of e− σ2 .
Solution:
1
jω
F {x(at)} =
X( )
|a|
a
{ 2}
√
ω2
F e−t
=
πe− 4
{ t 2}
√
ω2 σ2
= |σ| πe− 4
F e− σ
D. Define the width of the Gaussian as the time or frequency where the function reaches
1
of its peak value. Show that the width of the frequency Gaussian is inversely proe
portional to the width of the time Gaussian.
Solution:
t2
The peak value of the function, e− σ2 , occurs at t = 0.
t2
w
e− σ 2
−
=
1
e
t2w
= −1
σ2
t2w = σ 2
tw = ±σ
8
√
ω2 σ2
The peak value of the function, |σ| πe− 4 , occurs at ω = 0.
2 σ2
ωw
√
|σ| πe− 4
e−
2 σ2
ωw
4
√ 1
= |σ| π ·
e
1
=
e
ωw2 σ 2
=1
4
4
σ2
2
= ±
σ
ωw2 =
ωw
Therefore, |tw | · |ωw | = 2. Thus, the width of the frequency Gaussian is inversely
proportional to the width of the time Gaussian.
Cultural note: In quantum physics, quantities such as position and momentum are
described by random variables. The quantity σ 2 in this problem is equivalent to the
variance of one these random variables, say position. Since the momentum is the
Fourier transform of the position, the variance of the momentum is proportional to
1/σ 2 . Using any other probability distribution instead of Gaussians increases the product of the variances. That is, the uncertainty product is minimized when the particle
position and momentum are described by Gaussians.
9