ECE 382 Quiz # 1 Solution
Problem 1. (8 points) Suppose the Laplace transform of a signal f(t) is given as follows. In
each case, determine the final value of f(t): limt→∞ f(t), if it exists. Otherwise, state the reason
why such a final value does not exist.
(a) (2 points) F(s) = s2+1
s(s2+s+3) .
Solution: f(∞) = lims→0sF (s) = 1
3.
(b) (2 points) F(s) = 2s+1
(s2−4)(s+3) .
Solution: The final value theorem does not apply as F(s) has a pole at s= 2 with positive
real part. The final value of f(t) can be thought of as infinity.
(c) (2 points) F(s) = s+1
s2+4s+4 .
Solution: f(∞) = lims→0sF (s) = 0.
(d) (2 points) F(s) = 1 + s
s2+4 .
Solution: The final value theorem does not apply as F(s) has two poles ±j2 on the purely
imaginary axis which are not the origin. The final value of f(t) does not exist in this case.
Problem 2. (12 points) Consider the following linear time-invariant system:
y′′(t) + 4y′(t) + 4y(t) = u′(t)−u(t), t ≥0,
with the (nonzero!) initial conditions: y′(0) = 1, y(0) = 0, u(0) = 1. Suppose the input signal is
u(t) = e−t,t≥0.
(a) (6 points) Derive the Laplace transform Y(s) of the output signal y(t).
Solution: Note that U(s) = 1
s+1 . By taking the Laplace transform of the ODE, we have
[s2Y(s)−sy(0) −y′(0)] + 4[sY (s)−y(0)] + 4Y(s) = [sU (s)−u(0)] −U(s)
⇒Y(s) = s−1
(s+ 1)(s+ 2)2.
(b) (6 points) Find the output signal y(t).
Solution: The partial fraction expansion of Y(s) is
Y(s) = −2
s+ 1 +2
s+ 2 +3
(s+ 2)2.
Therefore,
y(t) = L−1[Y(s)] = [−2e−t+ 2e−2t+ 3te−2t]·1(t).