University of California, Berkeley
EE 42/100
Spring 2012
Prof. A. Niknejad
Problem Set 6
Solutions
Please note that these are merely suggested solutions. Many of these problems can be
approached in different ways.
1. (a) First we note that we simply have three impedances in series, where ZC =
and ZR = R. Then we derive Vo using a voltage divider:
H(ω) =
1
jωC
Vo
1
1
ZC
1/jωC
=
=
=
=
Vi
2ZC + ZR
2/jωC + R
2 + jωRC
2 + jω(0.02)
(b) The magnitude is found by considering the magnitude of the numerator divided
by that of the denominator. Recall that the magnitude of a complex number in
rectangular form can be found using the Pythagorean theorem.
|H(ω)| =
1
|1|
=p
|2 + jω(0.02)|
4 + (0.02ω)2
For the phase, we look at the phase of the numerator minus the phase of the
denominator. Note that the phase is just the angle of the complex number
above the real axis. In general, this is given by the inverse tangent of the
imaginary part divided by the real part.
0.02ω
−1
6 H(ω) = 6 1 − 6 (2 + jω(0.02)) = 0 − tan
= − tan−1 (0.01ω)
2
(c) The phasor representation compacts the signal’s amplitude and phase as follows:
Vi = 56 π4 V. Note that the frequency is not expressed in phasor form.
(d) For any LTI system in general (and thus the circuits that we analyze in this
class), the output in response to an input of the form vi (t) = A cos(ωt + φ) is
vo (t) = |H(ω)|A cos(ωt + φ + 6 H(ω)). In our case, ω = 100. So
π
π π
vo (t) = |H(100)|5 cos(100t+ +6 H(100)) = (0.354)5 cos(100t+ − ) = 1.77 cos(100t)
4
4 4
In phasor notation, this is simply Vo = 1.776 0 V.
Note that you could also have found the solution by first considering
Vo = H(ω)Vi = H(100)56
π
5ejπ/4
=
= 1.77e0 = 1.776 0 V
4
2 + j2
Then you can convert this back to the time domain by writing this as a cosine
function with the same frequency as the input, namely ω = 100.
2. Turning off the current source will make it an open circuit. Then V1a can be found
using a voltage divider:
100 − j10
−j10 + 100
(206 90◦ ) =
(j20) = 0.164 + j18.2 V
V1a =
10 − j10 + 100
110 − j10
Turning off the voltage source makes it a short. So we can write a single KCL
equation at V1b only taking into account the current source:
V1b
V1b
j20 − 200
+
+ 2 = 0 ⇒ V1b =
= −18.2 + j0.164 V
10
100 − j10
11 − j
Notice that we kept the two solutions in rectangular form for easier addition. The
final solution is
V1 = V1a + V1b = −18 + j18.4 = 25.76 134◦ V
3. Starting with Voc , notice that Ix must remain 0 for KCL to hold at the bottom right
node. Since no current flows across the capacitor, there is no voltage drop across it,
and Voc = 206 45◦ V. For Isc , adding a short between terminals a and b no longer
automatically forces all the current to be 0. By KCL, we infer that Isc = 2Ix .
− j 10 Ω
+
20 45 V
−
Ix
a
Ix
2I x
b
100 Ω
The most straightforward way to solve for Ix is to write a KVL loop. We start from
the reference node and work our way clockwise (note that there is no voltage drop
across the current source as it is shorted).
206 45◦
−206 45◦ − j10Ix + 100Ix = 0 ⇒ Ix =
= 0.26 50.7◦ A
100 − j10
Thus, the short-circuit current is Isc = 2Ix = 0.46 50.7◦ A.
We thus have VTh = Voc = 206 45◦ V and IN = Isc = 0.46 50.7◦ A. The last step is to
find the Thévenin/Norton impedance, which is given by Ohm’s law:
ZT h
VTh
206 45◦ V
= ZN =
=
= 50 − j5Ω = 50.26 − 5.71◦ Ω
◦
6
IN
0.4 50.7 A
4. If we view the components in terms of their complex impedances, we simply have an
inverting amplifier. Recall that the transfer function is given by
Vo
Z2
R
jωRC
H(ω) =
=−
=−
=−
Vi
Z1
R + 1/jωC
1 + jωRC
This is a high-pass filter. As ω → 0 at low frequencies, the magnitude of the transfer
function goes to 0; as ω → ∞ at higher frequencies, the transfer function approaches
1 and faithfully passes the input signal.
2