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EECS 40, Fall 2006
Prof. Chang-Hasnain
Midterm
October 25, 2006 Total Time Allotted: 50 minutes Total Points: 100 / Bonus: 10 pts
- This is a closed book exam. However, you are allowed to bring one page (8.5” x 11”), single-sided notes PLUS your 1-page notes from midterm 1.
- No electronic devices, i.e. calculators, cell phones, computers, etc.
- Slide rules are allowed.
- SHOW all the steps on the exam. Answers without steps will be given only a small percentage of credits. Partial credits will be given if you have proper steps but no final answers.
- Remember to put down units. Points will be taken off for answers without units.
Last (Family) Name:__Perfect______________________________________________
First Name: __Peter______________________________________________________
Student ID: __314159265_________________Discussion Session: __2718__________
Signature: __PP_________________________________________________________
Score: 110
Problem 1 (16 pts)
Complex Impedances
16
Problem 2 (54 pts):
Bode Plots
54
Bonus (10 pts): 10
Problem 3 (30 pts):
Second-order Circuits
30
Total 110
1. [16 points] Parallel and Series Complex Impedance a) [8 pts] What is the complex impedance Z 1?
LC j RC
RLC j L
j RC j LC
j L j RC j L j RC
j RC j L j C j RC
j C j L
j C
j RC R j L j C
j L
Z ZL ZC R
−^ −
2
2
2 2
1 1
1
(subscripts omitted for clarity)
b) [8 pts] What is the complex impedance Z 2?
LC j RC
R RLC j L
j RC j LC
R j L j RLC j C R j L
R j L j C j RL R j L
j RL j C
j C j L R
j L R j C
Z ZC ZL R
ω ω
ω ω
ω ω
ω ω ω ω
ω ω ω ω
ω ω
ω ω
ω ω
−
2
2
2 2
2 2
1
1
(c) [22 pt] Bode Magnitude Plot. You must put down all the steps leading to your results.
Hint: You may consider f (^) 1 << f 2
[4 points] Write down the expression for
2 y = 10 log H ( f )
2 1
2
2
2 2
10 log 1 10 log 1
10 log 1
10 log 1 2
f
f f
f
j j
y f
f f
f
Units are: dB Note: The other acceptable expression can be found by multiplying the terms in H(f), and finding the magnitude of the resulting product.
[4 points] As frequency goes to a very small value, what is the slope of y as a function of log f?
Constant 1 dominates in left term, f −^1 dominates in the right term :
( )
1
1
2 1
20 log 20 log
10 log 1 10 log 0 20 log
f f
f
f f
f y
= −
=^ +
Slope: 20 dB / decade
[4 points] As frequency goes to a very large value, what is the slope of y as a function of log f?
f dominates in left term, constant 1 dominates in the right term :
( )
20 log ( 20 log )
10 log 10 log 1 20 log 0
2
2
2
2 f f
f
f f
f y
=− − −
− =^ −
Slope: -20 dB / decade
[4 points] What is y, f (^) 1 << f << f 2?
In both terms, constant 1 dominates:
( ) ( ) dB
y 0
10 log 1 10 log 1 0 0
[2 points] What is y at f 1?
2
1
1
2
2
10 log 1 1 10 log 1 f
f f
f y
Since f 1 (^) << f 2 , ( ) ( ) dB
y 10 log 2 3
10 log 1 10 log 1 1 =− = −
[2 points] What is y at f 2?
2
2
1
2
2
10 log 1 2 10 log 1 f
f f
f y
Again, f 1 (^) << f 2 ,
[2 points] What filter is this?
Bandpass filter
Bonus [5 points] If the input Vin = 1 V and the frequency is 1 MHz, what is the output Vout?
f (^) 2 = 1 MHz , so y = − 3 dB = 10 log(^12 ) 2 2 ( ) 2
in
out V
V
= H f = Vout Vin 0. 707 V 2
Bonus [5 points] If the input Vin = 1 V and the frequency is 10 MHz, what is the output Vout?
10 MHz = 10 f 2 : one decade past break frequency
At large f , slope is -20 dB/decade. Thus, y = -20 dB, since f is 1 decade higher than the break frequency, where the Bode approximation is 0 dB.
y = − 20 dB = 10 log( 1001 ) 2 2 ( ) 100
in
out V
V
= H f = Vout Vin 0. 1 V 100
(d) [16 pt total] Bode Phase Plot. You must put down all the steps leading to your results. Hint: You may consider f 1 (^) << f 2
[4 points] Write down the expression for ∠ H ( f )
^ −
− −
− − − −
2
1 1 1
(^111) 2
1 1
tan tan
( ) tan 0 tan tan 0 tan
f
f f
f
f
f f
f H f
[4 points] What does the value of ∠ H ( f )approaches to as f → 0?
H f ( ) ( ) radians 2
( ) tan^1 tan^10
( ) ( ) dB
y 10 log 2 3
10 log 1 1 10 log 1 =− = −
3. [30 points] Second-order Circuits:
Assume the switch has been to the left for a long time before switching to the right at t = 0.
(a) Find the following values: [18 points] (Hint: What is vo(t) in terms of vC(t) ?)
mA k
V
iL 2 5
Inductor is short circuit right before switch is thrown.
iL ( ∞)= 0 A
All power is dissipated through resistors.
vC ( 0 +)= 0 V
No stored charge.
vC ( ∞)= 0 V
Discharges through resistors.
vo ( 0 +)= 0 V
All current initially flows through capacitor (short).
vo ( ∞)= 0 V
Power dissipated, no current.
i A s dt
d L (^0 +)=^0 /
i V V dt
d v (^) L ( 0 +)= L L ( 0 +)= C ( 0 +)= 0
v MV s dt
d C (^0 +)=−^2 /
V s nF
mA C
i v dt
d v dt
d i (^) C C C C L 2 10 / 1
( 0 ) ( 0 ) ( 0 ) =− ×^6
v MV s dt
d o (^0 +)=−^1.^5 /
Voltage divider: () 4
v t dt
d v t dt
d v k k
k v (^) o C ⇒ o = C Ω+ Ω
(b) [6 points] Write the differential equation in terms of vc.
KCL @ node above capacitor, (the 2 resistors are combined into R):
v t R
v t dt
d v tdt C L
v t R
v t dt
d i t C
C C C
L C C
Differentiating,
2 v t dt
d R
v t dt
d v t C L C^ C C
2 v t LC
v t dt
d RC
v t dt
d = C + C + C
(c) [6 points] What are the values of the natural frequency (ω 0 ) and the damping ratio (ζ)?
Damped harmonic oscillation:
2 v t v t dt
d v t dt
d = C + α (^) C + ω C
krad s
LC mH nF s 250 /
(^0122)
=
×
krad s
RC k nF s 125 /
6
×
α= = −
0
krad s
krad s
ζ (underdamped behavior)
