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ECE 382 Quiz # 5 Solution
Problem 1. (10 points) For a system with an unknow transfer function G(s), by setting its input
as 3 sin(4t), its steady-state output is measured to be 3
2 sin(4t +
3 π
4
(a) (6 pts) Find the frequency response G(jω) at ω = 4 rad/s. This should be a complex number.
Soution: Compared to the input, the magnitude of the steady-state output is increased by
a factor of
2 and its phase angle is increased by
3 π
4
. Thus, the frequency response at ω = 4
rad/s is:
G(j4) = |G(j4)|e
j
6 G(j4) =
2 e
j
3 π
) = −1 + j.
(b) (4 pts) Suppose the above G(s) is used in the feedback system below:
For the input U (t) = sin(4t), what is the steady-state signal of the output y(t)? Hint: What
is the frequency response of the closed-loop transfer function at ω = 4 based on part (a)?
Solution: The closed-loop transfer function is
H(s) =
Y (s)
U (s)
G(s)
1 + G(s)
Its frequency response at ω = 4 rad/s is
H(j4) =
G(j4)
1 + G(j4)
−1 + j
1 + (−1 + j)
−1 + j
j
= 1 + j =
2 e
j
π
4 ,
where |H(j4)| =
2 and
6 H(j4) =
π
4
. Therefore, for the input signal U (t) = sin(4t), the
steady-state output is
y ss
(t) =
2 sin(4t +
π
Problem 2. (10 points) Given a transfer function
G(s) = 10
s + 0. 1
s(s + 1)(s + 10)
(a) (2 pts) Write G(s) in the real factorization form, namely, as the product of some constant
K and some first order terms of the form 1 + T s or (1 + T s)
− 1 .
Solution:
G(s) = 0. 1
1 + 10s
s(1 + s)(1 + 0. 1 s)
(b) (2 pts) For each first order term 1 + T s or (1 + T s)
− 1 , find the corner frequency of its
frequency response (there should be three of them).
Solution: The three corner frequencies are 0.1 (for 1 + 10s term), 1 (for (1 + s)
− 1 term), and
10 (for (1 + 0. 1 s)
− 1 term), respectively.
(c) (4 pts) These three corner frequencies partition the frequency axis ω > 0 into four subinter-
vals. On each subinterval, determine the approximate slope of the magnitude plot (in dB/dec)
and the angle of the phase plot (in deg), respectively, starting from the leftmost subinterval.
Solution: On the four subintervals, we have
Slope of magnitude plot =
−20 dB/dec 0 < ω < 0. 1
0 dB/dec 0. 1 < ω < 1
−20dB/dec 1 < ω < 10
−40dB/dec ω > 10
Angle of phase plot =
◦ 0 < ω < 0. 1
◦
- 1 < ω < 1
◦ 1 < ω < 10
◦ ω > 10.
(d) (2 pts) Find the magnitude (in dB) of the frequency response G(jω) at the smallest corner
frequency. Use this information and your answer in part (c) to plot the Bode diagram (both
the magnitude plot and the phase plot) of G(jω) approximately.
Solution At the smallest corner frequency ω = 0.1, the magnitude of the frequency response
is approximately |G(jω)| ≃ |
10
jω
| = 100, which corresponds to 40 dB. (If you count the
contribution of the term 1 + 10s, then add an additional 3 dB).
Thus, on the leftmost subinterval 0 < ω < 0 .1, the magnitude plot is approximately a straight
line with slope −20 dB/dec, and takes the value of 40 dB at ω = 0.1. This help to uniquely
determine a line segment as the magnitude plot on 0 < ω < 0 .1. The magnitude plot on
- 1 ω < 1 is then obtained by concatenating to the end of this line segment another line
segment of slope 0 dB/dec. This procedure is repeated for the other two subintervals. The
bode diagram of G(s) is shown in the figure below (generated by Matlab):
−
−
0
50
Magnitude (dB)
10
− 10
− 10
− 10
0 10
1 10
2 10
3
−
−
−
−
0
Phase (deg)
Bode Diagram
Frequency (rad/sec)
