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EECS 40, Spring 2006
Prof. Chang-Hasnain
Midterm
April 6, 2006 Total Time Allotted: 80 minutes Total Points: 100
- This is a closed book exam. However, you are allowed to bring two pages (8.5” x
11”), double-sided notes
- No electronic devices, i.e. calculators, cell phones, computers, etc.
- 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.
- Draw BOXES around your final answers.
- Remember to put down units. Points will be taken off for answers without units. 6. NOTE: μ = - ; k=
3 ; M=
6
Last (Family) Name:_____________________________________________________
First Name: ____________________________________________________________
Student ID: ______________________Lab Session: __________Dis. Session: ______
Signature: _____________________________________________________________
Score:
Problem 1 (20 pts)
Problem 2 (35 pts):
Problem 3 (15 pts):
Problem 4 (30 pts):
Total
1.(20 pts) Match the transfer function to the Bode plot. Each transfer function matches to
exactly one Bode plot. Also, there is no partial credit for this question.
a.
jf
- 5 100
jf
2 ^ +
H f =
b.
2 1 100
jf
H f =
c.
jf
H f =
d.
jf
H f =
e.
jf
^ +
H f =
Mark your answer here
Magnitude Plot (dB)
b
e
c
a
d
100 101 102 103 104 105
0
20
40
100 101 102 103 104 105
0
20
40
100 101 102 103 104 105
0
20
40
100 101 102 103 104 105
0
20
40
100 101 102 103 104 105
0
20
40
Frequency (Hz)
From this, we have that the magnitude plots match as: (b), (e), (c), (a), (d).
For the phase plot, we again split the list into first- and second-order terms. For first-order terms,
the phase plot is -45 degrees at the breakpoint. For second order terms, decreasing values of zeta
gives rise to a sharper phase transition.
2.(35 pts) The circuit schematic for a functional block known as a lead compensator is:
2
R =
2 π
C =100 μ F
C =100 2 μF
1
R =
2 π
C =100 1 μF
C =100 2 μF
V
in V out
a (15 pts) Let R 1 = 10/(2π) ohms, R 2 = 100/(2π) ohms, C 1 = 100 uF, and C 2 = 100 uF. Show that
the transfer function of the circuit shown above is:
jf
jf
(f)
H =
Method 1: all at once
1
in R^^0
j ω C
V I =
1 2
V I I
j ω C j ω C
1 2
I I
j ω C j ω C
2 2
j C C I I I j C C
ω
ω
2
V I R Vout j ω C
2 2
V out I
R
j ω C
1 1 2 2 2
C V out I C R
j ω C
^
Combine with 1): 1 1 2 1 2 2
out in
C V
V R
C j R j C
ω
ω
− ^
− ^ +
^
C
2b (12 pts) In the following table, write the magnitude and phase values for H(f) for f=100Hz,
f=1000 Hz, very low f values ( f → 0 Hz ) and very high f values ( f → ∞ Hz ). These answers
only need to be within 1.5 times the correct answer (but only because of rounding errors or
sketching inaccuracies that you might have. Do not use the “straight line” approximation if it will
cause your answer will be off from the exact value by more than 1.5 times).
Note – terms in red should be f, not ω. Was announced during midterm
f value (Hz) 10 log |H( ω )|
2
∠ H( ω )
Very low f ( f → 0 Hz ) 3dB 39.7 deg
f = 100Hz 17dB 39.7 deg
f = 1000Hz 0dB 0 deg
Very high f ( f → ∞ Hz ) 20dB 0 deg
Given terms:
tan
tan
tan
tan
tan
Magnitude:
4
4 6
6
2
2 2 2 2
log | ( ) | 10 log 10 log 1 10 log 1 10 10 1 10
f
f f H f f
^ ^ ^ ^
= ^ = + −
^
Phase:
1 1 tan tan 100 1000
− ^^ f^ ^ − f −
For f=100Hz, becomes 3dB – 0 dB = 3dB
For f=1000Hz, becomes 20dB – 3dB = 17dB
Low f becomes 0dB – 0dB
High f becomes (^) [ ]
4
6
2
2
log 10 log 100 20
f
dB f
^
^
For f=100Hz, becomes (^) ( ) ( ) 1 1 tan 1 tan .1 45 5.7 39. − − − = ° − ° = °
For f=1000Hz, becomes (^) ( ) ( ) 1 1 tan 10 tan 1 84.3 45 39. − − − = ° − ° = °
For f-> 0, becomes (^) ( ) ( ) ° 1 1 tan 0 tan 0 0 − − − =
For f-> infinity, becomes (^) ( ) ( ) 1 1 tan tan 90 90 0 − − ∞ − ∞ = ° − ° = °
2c (8 pts) Sketch the Bode plot of this transfer function. Sketch BOTH the magnitude and phase
plot. Make sure to label the slopes of segments, the two break points of the transfer function, the
low frequency magnitude, the high frequency magnitude, and the highest value on the phase plot.
Be as accurate as you can, i.e., do not use the “straight line” approximation except as a starting
guide if you wish for plotting the actual transfer function.
10
0 10
1 10
2 10
3 10
4 10
5 10
-60 6
0
20
40
60
Bode Plot
Magnitude (dB)
10 0 10 1 10 2 10 3 10 4 10 5 10
-180 6
0
45
90
135
180
Frequency - log scale
Phase (degrees)
3.(15 pts) Find the unknown values in the circuits below. For the diodes, use the “0.8V ON-
OFF” model:
If I_d < 0, then the diode is open or OFF If I_d = 0, then the diode is open or OFF If I_d >= 0, then the diode is a 0.8V source or ON If I_d > 0, then the diode is a 0.8V source or ON
a. (5 pts) Find I_a in the circuit below:
0V diode is off, 5V diode and I_a diode are on.
I_a =
V V
mA
b. (5 pts) Find I_b in the circuit below:
- 10V – I_1(100ohm) = 0; I_1=100mA (I_1 is current in left branch)
- 10V – I_2(100ohm)-0.8V = 0 I_2=92mA (I_2) is current in right branch with diode)
So I_b=I_1+I_2=192mA
c. (5 pts) Let R_1 = 10 ohms and R_2 = 100 ohms. Find V_c in the circuit below, in terms of V_1 and V_2:
1
R =10 1 Ω R =100 2 Ω
V
V C
R =10 1 Ω R =100 2 Ω
V 2
1) V2 – I2(R1) – I2(R2) = 0
1) V2 – I2(R1+R2)=
4.(30 pts) Consider the circuit shown below, in which the RC time constant is very long
compared to the period T of the input VIN(t). Use the Ideal Diode model:
If VD < 0, then the diode is OFF and does not pass current (ID=0) If ID >= 0, then the diode is ON and VD= VD is the voltage drop across the diode and ID is current through the diode. VD =VOUT in this
problem. Analyze the following circuit. Given VIN(t)= Vmsin( 2π t/T) for t>0, and VC(t=
I (^) D
VIN
-
+ VC -
R (^) V OUT
-
I (^) D
VIN
-
VIN
VIN
-
+ VC -
R (^) V OUT
-
VOUT
VOUT
-
V IN
Vm
0
-Vm
T 2T 3T t
V IN
Vm
0
-Vm
t
V IN
t
V IN
Vm
0
-Vm
Vm
0
-Vm
TT 2T2T 3T3T
(a) (8 pts) Sketch VC(t)? Label all key values. The capacitor is initially able to charge up, since V_out starts at 0V and so the diode is a short. However, the capacitor is not able to discharge through the diode since the diode is an open when reverse biased. Thus, the capacitor discharges through the resistor. Since the RC constant is large, we have either:
Or:
(b) (8 pts) Sketch VOUT(t)? Label all key values. Simple application of KVL gives that V_out = V_in – V_c. The respective sketches of V_out are:
V
OUT
t
0 T 2T 3T
V
OUT
t
00 TT 2T2T 3T3T
Or:
V
OUT
t
0 T 2T 3T
V
OUT
t
00 TT 2T2T 3T3T
Note the concavity of the curves above.
(c) (8 pts) Explain what is happening for different time duration.
The capacitor is initially able to charge up, since V_out starts at 0V and so the diode allows
current flow in the positive direction. However, the capacitor is not able to discharge
through the diode since the diode is an open when reverse biased. Thus, the capacitor
discharges through the resistor. Since the RC time constant is large, the capacitor will
discharge very slowly (in the limit it will not discharge at all). When V_C matches V_in,
then V_out is 0V and so the diode will again allow the capacitor to charge up. We repeat this
process.
