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9/21/
ECE 382 Midterm 1 Solution
Problem 1 (20 points) Find the transfer function V I^ ((ss)) of the above circuit. Here, i(t) is the current
generated by the current source, and v(t) is the voltage across the inductor. Solution: The current iL(t) through the inductor has Laplace transform:
IL(s) =
R
R + (^) Cs^1 + Ls
I(s).
Hence,
V (s) I(s)
Ls · IL(s) I(s)
RLs R + (^) Cs^1 + Ls
RLCs^2 LCs^2 + RCs + 1
Problem 2 (20 points) Find the transfer function Y U^ ((ss)) of the above system. Note that the two diagonal lines at the center are not connected. Solution: We use the Mason’s formula. There are three loops:
L 1 = −G 1 H 1 , L 2 = −G 2 H 2 , L 3 = G 1 · 1 · G 2 · 1 = G 1 G 2 ,
of which L 1 and L 2 are nontouching. Therefore
∆ = 1 − (L 1 + L 2 + L 3 ) + L 1 L 2 = 1 + G 1 H 1 + G 2 H 2 − G 1 G 2 + G 1 G 2 H 1 H 2.
There are four forward paths:
P 1 = G 1 , ∆ 1 = 1 + G 2 H 2 P 2 = G 2 , ∆ 2 = 1 + G 1 H 1 P 3 = G 1 · 1 · G 2 , ∆ 3 = 1 P 4 = G 2 · 1 · G 1 , ∆ 4 = 1.
Therefore, the transfer function is
Y (s) U (s)
i=1 Pi∆i ∆
G 1 (1 + G 2 H 2 ) + G 2 (1 + G 1 H 1 ) + 2G 1 G 2
1 + G 1 H 1 + G 2 H 2 − G 1 G 2 + G 1 G 2 H 1 H 2
G 1 + G 2 + 2G 1 G 2 + G 1 G 2 H 1 + G 1 G 2 H 2
1 + G 1 H 1 + G 2 H 2 − G 1 G 2 + G 1 G 2 H 1 H 2
Problem 3 (20 points) Consider the system given by the differential equation:
y′′(t) + 0. 5 y′(t) + 0. 25 y(t) = u(t),
with nonzero initial condition y′(0) = 0 and y(0) = 1. Find the output y(t) under the unit step input u(t) = 1(t). Solution: Taking the Laplace transform:
s^2 Y (s) − s + 0.5[sY (s) − 1] + 0. 25 Y (s) =
s
Y (s) =
s + 0.5 + (^1) s s^2 + 0. 5 s + 0. 25
s^2 + 0. 5 s + 1 s(s^2 + 0. 5 s + 0.25)
s
3 s + 1. 5 (s + 0.25)^2 + 0. 4332
Y (s) =
s
3(s + 0.25) (s + 0.25)^2 + 0. 4332
(s + 0.25)^2 + 0. 4332 y(t) = 4 − 3 e−^0.^25 t^ cos(0. 433 t) − 1. 7321 e−^0.^25 t^ sin(0. 433 t).
Problem 4 (20 points) For the system given above.
(a) (10 pts) Find the transfer function Y U^ ((ss)). Solution: The transfer function is
Y (s) U (s)
K(s−1) (s+1)^2 1 + K (s(+1)s−1) 2
K(s − 1) (s + 1)^2 + K(s − 1)
K(s − 1) s^2 + (K + 2)s + (1 − K)
(b) (5 pts) Suppose K = 0.5, and the input is the unit step function: u(t) = 1(t). Determine if the final value y(∞) of the output exists, and find it if it does. Solution: In this case, U (s) = (^1) s , and
Y (s) =
0 .5(s − 1) s(s^2 + 2. 5 s + 0.5)
Since sY (s) has two negative real poles, we can use the final value theorem to find y(∞) = lim s→ 0 sY (s) = − 1.
