Quiz 3 Review: Circuits and Bode plots.
Circuits and the impedance method.
capacitor (C)
Z(s) ≡
E(s)
I(s)
ZC =
resistor (R)
ZR = R
inductor (L)
ZL = Ls
2 impedances in series, vs. 2 impedances in parallel.
Z
A
ZB
=
Z +Z
A
Z
B
Z Z
A B
=
Z
A
B
Z +Z
A
B
Series
Parallel
"sum"
"product over sum"
3 impedances all in series with one another.
Zeq = Z1 + Z2 + Z3
3 impedances all in parallel with one another.
�
Zeq = (Z1 |Z2 ) | Z3 =
Z1 Z2
| Z3 =
Z1 + Z2
Z1 Z2
Z1 +Z2
�
�
Z1 Z2
Z1 +Z2
Ex. 1
�
Z3
+ Z3
=
Z1 Z2 Z3
Z1 Z2 + Z2 Z3 + Z1 Z3
Ex. 2
R
C
ei
eo
i
ei
eo
i
C
a)
I (s)
Ei (s)
=?
a)
b)
Eo (s)
Ei (s)
=?
b) Bode plot for
c) Bode plot for
Eo (s)
Ei (s)
?
1
Eo (s)
Ei (s)
1
Cs
R
=?
Eo (s)
Ei (s)
?
Ex. 4
Ex. 3
R1
R1
e
ei
e
i
o
i
eo
C
C
R2
Eo (s)
Ei (s)
a)
=?
b) Bode plot for
a)
Eo (s)
Ei (s)
Eo (s)
Ei (s)
L
R2
=?
Eo (s)
can be written
Ei (s)
Ks(τ s+1)
Eo (s)
= (as2 +bs+1) . What
Ei (s)
b)
?
in the form
are K, τ , a,
and b in term of the electrical compo­
nents (R1 , C, R2 , L)?
Ex. 5
Ex. 6
ei
eo
i
L
L
Iin
R2
R2
R1
a)
I (s)
Ei (s)
R1
=?
a)
b) Sketch the Bode plot for EIi((ss))
(L = 1 H, R1 = 100 Ω, R2 = 900 Ω).
Eo (s)
Iin (s)
=?
b) Sketch the Bode plot for EIi((ss))
(L = 1 H, R1 = 100 Ω, R2 = 900 Ω).
2
Mechanically equivalent circuits.
Force­current (velocity­voltage) analogy.
capacitor (C)
F
in
v1
k2
v2
m
k1
e1=v1
s/k2
Fin= Iin
inductor (L)
1
ms
Zb = 1b
Zk =
s
k
e2=v2
1/(ms)
s/k1
b
resistor (R)
Zm =
1/b
E1 (s)
V1
sX1
=
=
= Zk2 + (Zk1 ||Zb ||Zm )
Iin
Fin
Fin
�
� 1 1 ��
s
s
|| 1ms b 1
=
+
k2
k1
+b
ms
s
=
+
k2
Bode plots.
One method to make (or analyze) a Bode plot.
1) Put the transfer function into the form:
⎡�
[Ksp ] ·
s
⎣ � z1
s
p1
+
��
s
z
�� 2
1 ps2
+1
�
s 1 1
k1 ms b
1 1
s 1
+ ks1 1b + ms
k1 ms
b
+
�
s
z
�
� m
1 . . . psn
+ 1 ...
+1
+1
�⎤
�⎦
(1)
2) At low freq (s ≈ 0), the RHS in eqn 1 is unity (1), so we can now draw a low frequency
asymptote based on the LHS alone. (Note p is of course just an integer equal to the number
of zeros at the origin minus the number of poles at the origin.)
Ksp
We can pick any value of ω, calculate Kω p at this frequency, and draw a line through this
point with slope p. s = 1 is often a useful point to choose, because 1p = 1 for any p! The
low frequency phase will be p · 90◦ (if K > 0).
3) Find all break points (poles and zeros) by looking at the RHS and indicate where they
will occur on the plot, so we can get ready for step 4. (Remember to use ωn for any complex
pairs.)
4) Start on the low frequency asymptote at a frequency below any breakpoints. Mov­
ing toward higher frequencies, “break” at each breakpoint. The slope will change by +1
(20dB/dec) for each zero or by −1 (­20dB/dec) for each pole at a particular breakpoint.
(Remember there can be two or more at one particular frequency.) The phase will change
by +90◦ for each zero or by −90◦ for each pole at a particular frequency (if these poles or
zeros are in the left­half plane).
3
Ex. 1 On the axes provided, sketch the Bode plot for the transfer function
H(s) =
(s + 10)
s
Magnitude
Be
sure to label the axes as appropriate!
Phase
Frequency (rad/s)
Frequency (rad/s)
Ex. 2
On the axes provided, sketch the Bode plot for the transfer function
H(s) =
20(s + 0.1)(s + 2)
s(s + 40)
Magnitude
Be sure to label the axes as appropriate!
Phase
Frequency (rad/s)
Frequency (rad/s)
4
Ex. 3
What is the transfer function H(s) represented by the Bode plot below?
Magnitude (dB)
60
40
20
0
−20
−1
10
0
1
10
Frequency (rad/sec)
10
Phase (degrees)
0
−45
−90
−135
−180
−1
0
10
1
10
Frequency (rad/sec)
10
The plot below shows a good starting point. You should recognize this transfer func­
2
ωn
We can find ωn where the phase is −90◦ :
tion is of the form H(s) = Kdc s2 +2ζω
2.
n s+ωn
ωn = 2 rad/s. Kdc is the DC (low frequency) of the magnitude: Kdc = 10(40/20) = 100.
At ωn , the magnitude is at M (ωn ) = 10(60/20) = 1000. M (ωn )/Kdc = 2ζ1 = 1000/100. So
ζ = 1/(2 · 10) = 0.05.
�
�
�
22
H(s) = 100s · 2
s + 2 · .05 · 2 + 22
400
= 2
s + 0.2s + 4
0
Mp = 1/(2*ζ) = ?
60
Magnitude (dB)
�
40
K=?
20
0
−20
−1
10
0
10
Frequency (rad/sec)
1
10
Phase (degrees)
0
−45
−90
ωn = ?
−135
−180
−1
10
0
10
Frequency (rad/sec)
5
1
10
Ex. 4
What is the transfer function H(s) represented by the Bode plot below?
Magnitude (dB)
40
20
0
−20
−40
−3
10
−2
10
−1
10
0
10
Frequency (rad/sec)
1
10
2
10
3
10
Phase (degrees)
90
45
0
−45
−90
−3
−2
10
10
−1
10
0
10
Frequency (rad/sec)
1
10
2
10
3
10
We can assume there is a zero at the origin, because the slope is +20 dB/dec and the
phase is +90◦ at low frequency. Therefore the LHS of the TF will be Ks1 = Ks. To find
K , we can extend the low frequency asymptote, and its value at 1 rad/sec is K. This is
difficult to estimate, so we can look at any other frequency that looks more convenient. At
ω = 0.02 rad/s, the magnitude is 0dB=1, so K · 0.02 = 1.0, K = 1/0.2 = 50. There are
then two breakpoints, both poles, so the phase goes to −90◦ and the slope is ­20 dB/dec at
high frequency. We can use either phase (at ­45 and ­135) or magnitude (estimating where
asymptotes intersect) to find that the poles are at approximately 0.2 and 5.0 rad/s.
�
1
H(s) = [50s] · s
( .2 + 1)( 5s + 1)
50s
=
(s + .2)(s + 5)
�
40
K=?
30
20
10
0
−10
−20
−3
10
−2
10
−1
10
0
10
1
10
2
10
3
10
90
45
0
−45
−90
−3
10
−2
10
−1
10
0
10
6
1
10
2
10
3
10