Solutions - HW #2
Physics 426
1.
A steady-state sinusoidal voltage waveform v (t ) = V cos ωt is applied to the input of the RC coupling
network presented in the diagram below.
+
v (t ) = V cos ωt
-
+
vo (t )
-
(a.) Using phasors, derive a general expression for the voltage vo (t ) . DO NOT base your analysis on
the current which flows in the circuit. If you use the current in your analysis you will get no credit in
the grading of this problem.
Solution:
First, let’s state a general procedure for analyzing circuits using phasors. It is as follows:
General Procedure - Analyzing Circuits Using Phasors
Step 1. Establish notation.
• capital letters and “hats” for phasor voltages and currents
• lowercase script letters and “(t)” for time-domain voltages and currents
Step 2. Apply Ohm’s law, the voltage divider equation, KVL, KCL, etc. using phasor voltages
and currents.
Step 3. Solve for the phasor form of the unknown voltage or current.
• if you know you are going to need to transform back to the time domain, get the
phasor form of the unknown in polar form. Then transforming to the time domain
will be easy.
Step 4. Transform back to the time domain, if asked to get voltage or current as function of time.
Now let’s apply this procedure to Part (a.) of Problem 1.
Step 1. Let V$ be the phasor transform of v (t ) . Let V$o be the phasor transform of vo (t ) .
Step 2. Recognize that the RC coupling network shown in the figure is nothing more than a voltage
divider, really. (It’s true that it is a frequency-dependent voltage divider, since the reactance of C
depends on frequency. But it’s still basically a voltage divider.) Therefore, applying the voltage
divider equation:
C:\TEACHING\PHYS426\SPRING00\HW2SOL.DOC
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Solutions - HW #2
Physics 426
FG Z IJ
HZ +Z K
F
I
G
JJ
R
= V$ G
GH R + jω1C JK
F
I
R
G
J
$
=VG
1 J
GH R − j ωC JK
V$o = V$
R
R
C
Now we need to express this result as a complex number in polar form in preparation for converting
back to the time domain. But in order to do this, we need to know what V$ is (specifically, whether it’s
purely real, purely imaginary, or complex). So we make the following aside:
Aside: The procedure for finding the phasor transform of a sinusoidal function of time can be stated
simply as follows. If we have a function f (t ) which, when expressed as a cosine, has the form
f (t ) = A cos(ωt + θ ) , then the phasor transform of f (t ) (let’s call the phasor transform F$ ) will be
F$ = Ae jθ .
And it will always turn out this way: The phasor form will always look like
(amplitude)e j ( phase) .
in which (amplitude) and (phase) are the amplitude and phase of the time-domain function.
If the time-domain function has a phase of zero (for example, suppose f (t ) = A cos ωt ) then the
phasor transform will be just the amplitude of f (t ) :
F$ = Ae j 0 = A .
We know that v (t ) = V cos ωt . (This was given in the problem statement.) So V$ = V , where V (no
“hat”, notice) is the amplitude of v (t ) . So it is a real number. Therefore:
V$o = V
F
I
GG R JJ
GH R − j ω1C JK
There are at least a couple of ways you could proceed now to get V$o in polar form. One way that’s
frequently a quick one is to immediately put into polar form any complex expressions appearing in the
phasor expression. So that is what we will do:
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Solutions - HW #2
Physics 426
V$o
LM
M
R
=VM
MM F F 1 I I
MN GH R + GH ωC JK JK e
LM
M
R
V$ = V M
MM F F 1 I I
MN GH R + GH ωC JK JK
LM
M
1
=VM
MM F F 1 I I
MN GH1 + GH ωRC JK JK
2 1/ 2
− jθ
2
o
2
OP
PP
PP ,
PQ
θ = tan −1
e jθ
1/ 2
2
2 1/ 2
OP
PP
PPe
PQ
FG 1 IJ .
H ωRC K
OP
PP
PP
PQ
jθ
Step 4. Now we are ready to convert back to the time domain.
LM
M
1
= Re MV
MM F F 1 I I
MN GH1 + GH ωRC JK JK
vo (t ) = Re V$o e jωt
vo (t ) =
2 1/ 2
V
F1 + FG 1 IJ I
GH H ωRC K JK
/
2 12
b
cos ωt + θ
g
e jθ e jωt
, θ = tan −1
OP
PP
PP
PQ
FG 1 IJ .
H ωRC K
(b.) Show analytically that at sufficiently high frequency the output is an essentially perfect replica of
the input.
If we let ω → ∞ , then
bg
1
→ 0 and θ → tan −1 0 = 0 . So
ωRC
vo (t ) → V cos ωt
which is exactly the same as the input voltage v (t ) .
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Solutions - HW #2
Physics 426
(c.) Show that at sufficiently low frequency the output is a severly attenuated waveform which is
approximately the derivative of the input.
If we let ω → 0 , something interesting happens:
bg
1) and θ → tan −1 ∞ =
FG 1 IJ
H ωRC K
2
→ ∞ (i.e.,
FG 1 IJ
H ωRC K
2
gets much larger than
π
radians, or 90º. So
2
b
g
b
vo (t ) → V (ωRC) cos ωt + 90o = RC −ωV sin ωt
g
But the factor in parentheses is the derivative of the input:
vo (t ) → RC
d
v (t )
dt
So at low frequencies, the RC coupling network gives you an output that’s proportional to the
derivative of the input. For this reason, the RC coupling network is called a differentiator at low
frequencies.
Now, why does the problem say the output should be a “severly attenuated” waveform? Well,
b g
vo (t ) → −V ωRC sin ωt
and the factor ωRC goes to zero (gets much smaller than 1) as ω → 0 . So at low frequencies, the
amplitude of the output is much less than V .
2.
Consider the series LRC circuit shown below with a steady-state sinusoidal voltage source:
v (t ) = V cos ωt
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Solutions - HW #2
Physics 426
Once again, do NOT base your analysis required below on knowledge of the current in the circuit.
(a.) Using phasors, derive a general expression for the voltage v R (t ) across the resistor R.
Solution:
Step 1. Let V$ be the phasor transform of v (t ) and V$R be the phasor transform of v R (t ) .
Step 2. Recognizing the circuit as a voltage divider, we apply the voltage divider equation:
V$R
F
= V$ G
HZ
R
IJ
K
ZR
= V$
+ Z L + ZC
F
I
GG
JJ
R
GG R + jFGHωL − 1 IJK JJ
ωC K
H
Step 3. Get V$R in polar form. To do this, we need to know whether V$ is purely real, purely imaginary, or
complex. Since v (t ) = V cos ωt (given), V$ will be just the amplitude of v (t ) :
V$ = V
So now we make this substitution in the expression for V$R and write the denominator in polar form to get:
V$R
F
GG
R
=VG
GG LMR + FGωL − 1 IJ OP
GH NM H ωC K PQ
/
2 12
2
e jθ
I
JJ
JJ ,
JJ
K
θ = tan −1
LM FGωL − 1 IJ OP
MM H RωC K PP .
MN
PQ
Now take the complex exponential factor to the numerator:
V$R
F
I
GG
JJ
R
=VG
e
GG LMR + FGωL − 1 IJ OP JJJ
GH NM H ωC K PQ JK
− jθ
2 1/ 2
2
Step 4: We now have V$R in polar form and are ready to transform back to the time domain to get v R (t ) :
v R (t ) = Re V$R e jω t
F
GG
R
=VG
GG LMR + FGωL − 1 IJ
GH MN H ωC K
2
C:\TEACHING\PHYS426\SPRING00\HW2SOL.DOC
OP
PQ
2 1/ 2
I
JJ
JJ Re e b
JJ
K
j ω t −θ
g
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Solutions - HW #2
Physics 426
vR
F
GG
R
(t ) = G
GG LM R + FGωL − 1 IJ
GH MN H ωC K
2
OP
PQ
2 1/ 2
I
JJ
JJV cosbω t − θ g
JJ
K
θ = tan −1
,
LM FGωL − 1 IJ OP
MM H RωC K PP
MN
PQ
(b.) Derive a general expression for the voltage across the capacitor.
Step 1. Let V$ be phasor transform of v (t ) , V$C be phasor transform of vC (t ) .
Step 2. Recognize circuit as a voltage divider. Then apply voltage divider equation:
V$C = V$
FG
HZ
R
ZC
+ Z L + ZC
F
I
1
I
G
IJ = V$ G jωC JJ = V$ FG
1
K GG R + jFGω L − 1 IJ JJ GH e1 − ω LCj + jωRC JJK
H H ωC K K
2
Step 3. Get V$C in polar form. Once again, V$ = V , so putting the denominator in polar form and taking the
exponential factor to the numerator, we have:
V$C
F
G
1
=VG
GG Le1 − ω LCj + bωRCg O
H MN
QP
2
2
2
1/ 2
I
JJ e
JJ
K
− jθ
,
θ = tan −1
ωRC
e1 − ω LCj
2
Step 4. Now V$C is in polar form, so we are ready to transform back to get vC (t ) :
vC (t ) = Re V$C e jω t
F
G
1
=VG
GG Le1 − ω LCj + bωRCg O
PQ
H MN
2
2
2
1/ 2
I
JJ cos ω t − θ
JJ b g
K
, θ = tan −1
ωRC
e1 − ω LCj
2
(c.) Derive a general expression for the voltage across the inductor.
Step 1. Let V$ be phasor transform of v (t ) , V$L be phasor transform of v L (t ) .
Step 2. Recognize circuit as a voltage divider. Then apply voltage divider equation:
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Solutions - HW #2
Physics 426
V$L = V$
FG
HZ
R
ZL
+ Z L + ZC
I
F
I FG
JJ
IJ = V$ GG jω L JJ = V$ G
1
K GG R + jFGω L − 1 IJ JJ GG F1 − 1 I − j R JJ .
H H ωC K K GH GH ω LC JK ω L JK
2
Step 3. Get V$L in polar form. Once again, V$ = V , so putting the denominator in polar form and taking the
exponential factor to the numerator, we have:
V$L
F
GG
1
=VG
GG LF 1 I F R I
GG MMGH1 − ω LC JK + GH ω L JK
HN
2
2
2
OP
PQ
1/ 2
I
JJ
JJ e
JJ
JK
jθ
,
θ = tan −1
LM R OP
MM ω L PP
MM FG1 − 1 IJ PP
N H ω LC K Q
2
Step 4. Now V$L is in polar form, so we are ready to transform back to get v L (t ) :
v L (t ) = Re V$L e jω t
F
GG
1
=VG
GG LF 1 I F R I
GG MMGH1 − ω LC JK + GH ω L JK
HN
2
2
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OP
PQ
1/ 2
I
JJ
JJ cosbω t + θ g
JJ
JK
. θ = tan −1
LM R OP
MM ω L PP .
MM FG1 − 1 IJ PP
N H ω LC K Q
2
7