33-228
Electronics
Spring 2016
Version of February 15, 2016
Carnegie Mellon University
Department of Physics
Homework 5
SOLUTIONS
Problems 3.23, 3.29 and 3.D45
3.23 (a) In discussing low and high frequencies, we mean relative to the characteristic fre1
1
, and fRC = 2πRC
. For our circuit, we have that
quency, ωRC = RC
1
1
2π 4700 Ω 339 × 10−12 F
≈ 100 kHz .
fRC =
fRC
Thus, 20 kHz is a low frequency for this circuit. We know that Zout = ZR || ZC , so if
we define ωRC = 1/RC, then we have that
ZR ZC
ZR + ZC
R/jωC
=
R + 1/jωC
1/jωRC
= R
1 + 1/jωRC
−jωRC /ω
= R
1 − jωRC /ω
Zout =
Zout
Zout
Zout
and the magnitude of Zout is
| Zout | = R q
ωRC /ω
2
1 + ωRC
/ω 2
So at ω = ωRC , we have that
−jR
1−j
R
| = √
2
Zout =
| Zout
(c) At high frequencies, the capacitor impedance goes to zero, or we the capacitor
behaves like a wire. Thus, at high frequency, all the voltage will be across the resistor,
and the gain will be one. For low frequency, the capacitor impedance is very large,
and all the voltage will be across C. Thus the gain will be quite small. So, we have a
high-pass filter.
1
3.29 (a) From about 10 kHz to about 100 kHz. (b) There are two −3 dB points. One at
about 1 kHz and the other at about 103 kHz. (c) We have that
20 dB log (Vo /Vi ) = K − 10 dB log f
1
log (Vo /Vi ) = K 0 − log f
2
1
0
log (Vo /Vi ) = K + log f − 2
thus, we find that
K 00
Vo = Vi · √
f
Vi
Vo ∼ √
f
3.D45 This problem needs to be solved using the Kirchhoff rules, and to facilitate that, it is
useful to redraw the circuit as shown in Figure 1. In doing so, we have indicated the
currents I1 , I2 and I3 flowing through the loops, and we can imediately write down
from the node rules that
i1 = i2 + i3
We also note that the output voltage will be given as
€€
vout = vin − 2R i2
so we need to solve for i2 to get the gain of the circuit. In order to proceed, we will
i1
vin
€ …ƒ‚ÿ€ ‚ú€ ‚ € …‚ÿ€ ‚ò
„ƒ i i 2R „„
ûƒ …ù€
ƒ ‚ € ‚…ÿ€ ‚ € „ƒ‚
„
v
C ƒ C
R/2
/
„„
i ù
„ƒ€
ÿƒ‚
€
‚ò
3
2
out
1
Figure 1: A bridged T-filter for solution to problem D3.45 which shows the currents flowing
around the relevant loops.
need to write the three loop equations for this circuit. These are
0 = vin − 2R i2 − ZC i2 −
R
i1
2
R
i1
2
0 = −2R i2 − ZC i2 + ZC i3
0 = vin − ZC i3 −
2
The third equation can be written to express i3 in terms of i2 as
ZC + 2R
i2 ,
ZC
i3 =
and combining with our current rule, we have
ZC + 2R
ZC
ZC + R
i2 .
= 2
ZC
i1 =
i1
1+
i2
Putting these currents into the second loop equation, we get that
ZC + 2R
R ZC + R
i2 + 2
i2
ZC
2
ZC
ZC + R
= (ZC + 2R) + R
i2
ZC
ZC2 + R2 + 3ZC R
=
i2
ZC
vin = ZC
vin
vin
which yields that
i2 =
ZC2
ZC
vin .
+ R2 + 3ZC R
From this, we get that
ZC
vin
+ R2 + 3ZC R
(ZC2 + R + 3ZC R) − (2ZC R)
vin
=
ZC2 + R2 + 3ZC R
ZC2 + R2 + ZC R
=
vin
ZC2 + R2 + 3ZC R
vout = vin − 2R
vout
vout
ZC2
2
and from this, we can express the gain as
ZC2 + R2 + ZC R
.
ZC2 + R2 + 3ZC R
G =
Now using the fact that ZC = (jωC)−1 , we can and defining ωRC = (RC)−1 , we can
write the gain as
1−
ω
ωRC
2 ω
+ j ωRC
G = 2 ω
ω
1 − ωRC
+ 3j ωRC
3
(a) From this, we can write that the magnitude of the gain is given as
|G| =
v
u 2 2
u
ω
u 1−
u
ωRC
u
u
2 2
t
ω
[1 −
ωRC
+
h
i2
ω
ωRC
+9
h
ω
i2
ωRC
(b) We can examine some limits of this gain function. If we look at ω << ωRC , we
have that | G |→ 1. For the case of ω >> ωRC , we also have that | G |→ 1. Finally,
for the case of ω = ωRC , we get that the gain is | G |= 13 . (c) In order to get the
phase, we need to rewrite the complex gain as a proper complex number. If we define
x = ω/ωRC , then we can write our complex gain as
[(1 − x2 ) + jx] [(1 − x2 ) − 3jx]
(1 − x2 )2 + 9x2
h
i
1
2 2
2
2
(1
−
x
)
+
3x
−
j
2x(1
−
x
)
G =
(1 − x2 )2 + 9x2
G =
From this, we can write that
−2x(1 − x2 )
.
(1 − x2 )2 + 3x2
20db log(|G(ω)|)
tan φ =
10
8
6
4
2
0
-2
-4
-6
-8
-10 -2
10
10
-1
1
10
10
2
10
3
Phase φ [ Degrees ]
Frequency (ω/ωRC)
40
30
20
10
0
-10
-20
-30
-40
10
-2
10
-1
1
10
10
2
10
3
Frequency (ω/ωRC)
Figure 2: The Bode and phase plots for the bridged T-filter in problem D3.45.
4