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UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department
Page 1 of 7
EECS 40/42/100, Spring 2007 Prof. Chang-Hasnain
Homework #7
Due at 6 pm in 240 Cory on Wednesday, 03/14/07
Total Points: 100 • Put (1) your name and (2) discussion section number on your homework. • You need to put down all the derivation steps to obtain full credits of the problems.
Numerical answers alone will at best receive low percentage partial credits. • No late submission will be accepted expect those with prior approval from Prof.
Chang-Hasnain.
1. Hambley, P6.55 [8 Points] This is a first-order high-pass filter analyzed in Section 6.5. The transfer function is
)/(1
)/()(
B
B
in
out
ffj
ffj
V
VfH
+== [2 Points]
where kHzRC
f B 183.32
1==
! [2 Points]
The Bode plots are shown in Figure 6.21 at Page 280 in the textbook. [4 points]
2. Hambley, P6.58 [8 Points] To attenuate the 60-Hz component by 40 dB, the break frequency must be two decades higher than 60 Hz because the roll-off slope is 20dB/decade. [2 Points] Thus the break frequency must be kHzf B 6= . [2 Points] The 600-Hz component is attenuated by 20 dB. [2 Points]
UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department
Page 2 of 7
SinceRC
f B!2
1= , we have
FRf
CB
µ!!
0265.0600010002
1
2
1=
""== . [2 Points]
3. Hambley, P6.64 [12 Points] Resonant frequency:
MHzLC
f 125.1
2
1
0==
! [2 Points]
1002
0 ==R
LfQs
!
Bandwidth:
kHzQ
fB
s
25.110 == [2 Points]
Half Power frequencies
MHzBffH 131.12/0
=+! [1 Point] MHzBff L 119.12/
0=!" [1 Point]
At the resonant frequency:
°!= 01RV [1 Point]
°!= 90100LV [1 Point]
°!"= 90100CV [1 Point] The phasor diagram is shown below: [1 Point for each, 3 Points in total]
°!= 90100LV
°!= 01RV
°!"= 90100CV
UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department
Page 3 of 7
4. Hambley, P6.66 [8 Points]
850
4000 ===
kHz
kHz
B
fQs [2 Points]
Hf
RQL s µ
!66.63
20
== [3 Points]
pFfRQ
Cs
2487)2(
1
0
==!
[3 Points]
5. Hambley, P6.71 [8 Points]
!== kZR p 10|| max [2 Points]
500 ==B
fQp [2 Points]
HQf
RL
p
µ!
183.32
0
== [2 Points]
pFRf
QC
p58.79
20
==!
[2 Points]
6. Hambley, P6.75 [6 Points] An AM radio signal having a carrier frequency of 980kHz has components ranging in frequency from 970kHz to 990kHz. A bandpass filter is needed to pass this signal and reject the signals from other AM radio transmitters. [4 Points] The cutoff frequencies should be 970kHz and 990kHz. [2 Points] 7. Hambley, P6.77 [10 Points] The circuit diagram of a second-order lowpass filter is: [6 Points]
UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department
Page 4 of 7
L
VinC
+
Vout
_
mHf
RQL s
592.12
0
==!
[2 Points]
pFfRQ
Cs
1592)2(
1
0
==!
[2 Points]
8. Hambley, P6.78 [10 Points] The circuit diagram of a second-order highpass filter is: [6 Points]
LVin
C
+
Vout
_
mHf
RQL s
592.12
0
==!
[2 Points]
pFfRQ
Cs
1592)2(
1
0
==!
[2 Points]
R=1k!
R=1k!
UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department
Page 5 of 7
9. Consider the circuit shown below. Sketch the asymptotic Bode magnitude and phase plots to scale for the transfer function H(f)=Vout/Vin. [14 Points]
Apply the voltage-division principle, we have
in
out
V
VfH =)(
)2/(121
1
fCjRR
R
!++= [3 Points]
))(2/(11
)/(
21
211
RRfCj
RRR
++
+=
!
)(21
)(2
21
21
21
1
RRfCj
RRfCj
RR
R
++
+
+=
!
!
)/(1
)/(
21
1
B
B
ffj
ffj
RR
R
++= [1 Points]
UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department
Page 6 of 7
)/(1
)/(1.0
B
B
ffj
ffj
+= [2 Points]
where HzRRCfB 592.1)(2/1 21 =+= ! [2 Points] The asymptotic Bode plots are:
[3 Points for the magnitude, 3 Points for the phase, 6 Points in total] |H(f)|
dB
0.1fB fB 100fB
-20
-40
-60
10fB
f
Phase
0.1fB fB 100fB10fBf
0
90o
45o
0o
10. Consider the first-order highpass filter shown in Figure P.6.59 in your
textbook. The input signal is given by: Vin(t)=10+20cos(200πt)+10cos(2000πt). Find an expression for the output Vout(t) in steady-state conditions. [16 Points]
UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department
Page 7 of 7
This is the first-order high-pass filter analyzed in Section 6.5 in the text. The transfer function is
)/(1
)/()(
B
B
in
out
ffj
ffj
V
VfH
+== [2 Points]
where HzRC
fB 10002
1==
!. [2 Points]
The input signal is given by
)2000cos(10)200cos(2010)( tttVin
!! ++= This signal has components at HzHzHzf 1000,100,0= . [3 Points] The transfer function values at these frequencies are:
001
0)0( =
+=
j
jH [1 Point]
°!=+
= 29.840995.01.01
1.0)100(
j
jH [1 Point]
°!=+
= 457071.011
1)1000(
j
jH [1 Point]
Applying these transfer-function values to the respective components yields:
01=
inv
0100)0( 11 =!=!=inoutvHv [1 Point]
°!== 020),200cos(20 22 inin
Vtv " °!=°!"°!="= 29.8499.102029.840995.0)100( 22 inout
VHV [1 Point] )29.84200cos(99.12 °+= tv
out! [1 Point]
°!== 010),2000cos(10 22 inin
Vtv " °!=°!"°!="= 45071.7010457071.0)1000( 33 inout
VHV [1 Point] )452000cos(071.73 °+= tv
out! [1 Point]
)452000cos(071.7)29.84200cos(99.1 °++°+= ttV
out!! [1 Point]
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