ECE 314
Solution to Test #1
Georghiades
Given: September 26, 2008.
Instructions: Exam is open-book, open notes. Please give your solution in the allocated space. If
more space is needed, use the back of the page. All problems carry equal weight.
1. Consider a linear, time-invariant system. It is known that when the input is a unit-step, the
output is given by 12 (1 − e−2t )u(t). Find the output, y(t), when the input is
x(t) = e−t u(t)
and compute its energy.
Solution
The impulse response, h(t), of the system is the derivative of the unit-step response:
h(t) =
d 1
1
(1 − e−2t )u(t) = (1 − e−2t )δ(t) + e−2t u(t) = e−2t u(t).
dt 2
2
Then the out of the system when the input is x(t) is y(t) = x(t) ∗ h(t), i.e.,
y(t) =
Z
∞
e−2τ u(τ )e−(t−τ ) u(t − τ )dτ
−∞
= e
−t
Z
∞
e−τ u(τ )u(t − τ )dτ
−∞
= e−t u(t)
=
Z
t
e−τ dτ
0
−2t
e−t − e
u(t).
The energy in y(t) is:
E=
Z
∞
−∞
y 2 (t)dt =
Z
0
∞
1
e−t − e−2t
2
(t)dt =
1
.
12
2. The impulse response of a LTI system is given by h(t) = e−t u(t). Find the output of the
system when the input is x(t) as given in the figure below.
x(t)
1
1
t
Solution
Let the output of the system when the input is x(t) be y(t). We have:
x(t) = u(t) − u(t − 1),
and thus
y(t) = h(t) ∗ [u(t) − u(t − 1)] .
If we let yu (t) = h(t) ∗ u(t), then we have:
y(t) = yu (t) − yu (t − 1).
Now,
yu (t) = h(t) ∗ u(t) = u(t)
Z
t
e−τ dτ = 1 − e−t u(t),
0
and, thus,
y(t) = 1 − e−t u(t) − 1 − e−(t−1) u(t − 1).
2
3. In a financial system modeled as LTI, f (t) represents the amount of money in dollars accumulated in an account as a function of time using a particular investment scheme and starting
with an initial investment at t = 0. After thorough analysis of the system, the Laplace
transform of the output response f (t) was found to be:
F (s) =
6 · 103 (s + 1)2
.
s(s + 2)(s + 3)
How much will the investment scheme loose or gain if allowed to (theoretically) work for an
infinite amount of time?
Solution
The loss or gain is the difference between the initial invested dollar amount and the final
(theoretically at infinite time) value of the investment. The initial and final values of the
investment are obtained by using the initial and final value theorems:
6 · 103 (s + 1)2
= 103 .
s→0 (s + 2)(s + 3)
f (∞) = lim sF (s) = lim
s→0
Similarly, using the initial value theorem:
6 · 103 (s + 1)2
= 6 · 103 .
s→∞ (s + 2)(s + 3)
f (0) = lim sF (s) = lim
s→∞
Thus, the investment scheme actually looses money to the tune of 6 · 103 − 103 = 5 · 103 .
3
4. Find the Laplace transform of the function f (t) given by:
f (t) =
e−2t , 0 ≤ t ≤ 1
0
otherwise.
Solution
We can express f (t) as:
f (t) = e−2t [u(t) − u(t − 1)] = e−2t u(t) − e−2 e−2(t−1) u(t − 1).
Thus, using the time shift property of the Laplace transform:
F (s) =
1
e−s
1 − e−(s+2)
− e−2
=
.
s+2
s+2
s+2
4
5. Find the Laplace transform of the function f (t) given by:
f (t) =
∞
X
δ(t − kT ),
k=0
where δ(t) is the impulse function. Simplify as much as possible.
Solution
The Laplace transform of δ(t − kT ) (using the time shift property, for example) is e−sT k .
Thus, the Laplace transform of f (t) is
F (s) =
∞
X
e−sT k =
k=0
5
1
.
1 − e−sT
Scratch
Hint: For 0 ≤ r < 1,
∞
X
rk =
k=0
6
1
.
1−r