EE C128 / ME C134 Spring 2014 HW2 - Solutions
UC Berkeley
Homework 2 - Solutions
1. Transfer Functions of Electrical Networks with Operational Amplifiers
Find the transfer function, G(s) = Vo (s)/Vi (s), for each operational amplifier circuit shown in the
Figures below.
(a) Solution:
Calculating the feedback and feedforward impedances.
Zf eedback (s) = Zf b (s) = 100 × 103 +
1
2 × 10−6 s
Zf eedf orward (s) = Zf f (s) = 500 × 103 +
1
2 × 10−6 s
Since the system is in the inverted amplifier configuration, we know that the gain is given by
Zf b (s)
Vo (s)
105 (2 × 10−6 s) + 1
=−
=−
Vi (s)
Zf f (s)
5 × 105 (2 × 10−6 s) + 1
Vo (s)
0.2s + 1
=−
Vi (s)
s+1
(b) Solution:
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EE C128 / ME C134 Spring 2014 HW2 - Solutions
UC Berkeley
Calculating the feedback and feedforward impedances.
5
(s + 10)
5
= 105
Zf eedback (s) = Zf b (s) = 10 1 +
s+5
(s + 5)
Zf eedf orward (s) = Zf f (s) = 105
5
+1
s
= 105
(s + 5)
s
Since the system is in the inverted amplifier configuration, we know that the gain is given by
Zf b (s)
s(s + 10)
Vo (s)
=−
=−
Vi (s)
Zf f (s)
(s + 5)2
2. Transfer Functions of Translational Mechanical Systems
For the system shown below, find the transfer function, G(s) = X1 (s)/F (s).
Solution: Equation of motion (in s-domain) for mass M1 is given by
(s2 + 6s + 9)X1 (s) − (3s + 5)X2 (s) = 0
(1)
and equation of motion (in s-domain) for mass M2 is given by
−(3s + 5)X1 (s) + (2s2 + 5s + 5)X2 (s) = F (s)
⇒ X2 (s) =
F (s) + (3s + 5)X1 (s)
2s2 + 5s + 5
(2)
Plugging Equation (2) in Equation (1) and rearranging yields
X1 (s) =
⇒
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2s4
(3s + 5)F (s)
+ 17s3 + 44s2 + 45s + 20
X1 (s)
3s + 5
= 4
3
F (s)
2s + 17s + 44s2 + 45s + 20
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EE C128 / ME C134 Spring 2014 HW2 - Solutions
UC Berkeley
3. State Space Representation of Electrical Networks
(a) Represent the electrical network shown below in state space, where iR (t) is the output.
Solution: Write the differential equations for each energy storage element.
i2
dv1
=
dt
3
di3
vL
=
dt
2
" #
v1
Therefore the state vector is x =
. Now obtain vL and i2 in terms of the state variables,
i3
vL = v1 − v2 = v1 − 3iR = v1 − 3(i3 + 4v1 ) = −11v1 − 3i3
1
1
1
i2 = i1 − i3 = (vi − v1 ) − i3 = − v1 − i3 + vi
3
3
3
Also, the output is
y = iR = 4v1 + i3
Hence,

− 19
ẋ = 
− 11
2
− 13

x +
− 32
h
i
y= 4 1 x
" #
1
9
0
vi
(b) Find the state space representation of the network shown below if the output is vo (t).
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EE C128 / ME C134 Spring 2014 HW2 - Solutions
UC Berkeley
Solution: Let C1 be the grounded capacitor and C2 be the other. Now, writing the equations for
the energy storage components yields,
diL
= vi − vC1
dt
dvC1
= i1 − i2
dt
dvC2
= i2 − i3
dt
(3)
(4)
(5)

iL
 
Thus the state vector is x = vC1 . Now, find the three loop currents in the terms of state variables
v C2
and the input. Writing KVL around Loop 2 yields

vC1 − vC2 = i2
Writing KVL around the outer loop yields
i3 = vi − i2 = vi − vC1 + vC2
Also, i1 − i3 = iL . Hence,
i1 = iL + i3 = iL + vi − vC1 + vC2
Substituting the loop currents in Equations (3), (4) and (5), yields the results in vector-matrix form,


 
0 −1 0
1


 
ẋ = 1 −2 2  x +  1  vi
0 2 −2
−1
Since v0 = i2 = vC1 − vC2 , the output equation is
h
i
y = 0 1 −1 x
4. Transfer Function to Phase Variable Representation
For the system shown below, write the state equations and the output equation for the phase-variable
representation.
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EE C128 / ME C134 Spring 2014 HW2 - Solutions
Solution: Using the standard form

0

0

ẋ = 
0

0
0
h
c= 6
Rev. 1.0, 02/09/2014
UC Berkeley
derived in the textbook,

 
1 0
0
0
0

 
0
0 1
0
0

 


0 0
1
0 x + 
0 r(t)

 
0 0
0
1
0
0 −8 −13 −9
1
i
7 12 2 1 x
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