Solutions - HW #2Physics 426

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1. A steady-state sinusoidal voltage waveform v t V t( ) cos= ω is applied to the input of the RC couplingnetwork presented in the diagram below.

(a.) Using phasors, derive a general expression for the voltage v to ( ) . 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 inthe grading of this problem.

Solution: First, let’s state a general procedure for analyzing circuits using phasors. It is as follows:

Now let’s apply this procedure to Part (a.) of Problem 1.

Step 1. Let $V be the phasor transform of v t( ) . Let $Vo be the phasor transform of v to ( ) .

Step 2. Recognize that the RC coupling network shown in the figure is nothing more than a voltagedivider, really. (It’s true that it is a frequency-dependent voltage divider, since the reactance of Cdepends on frequency. But it’s still basically a voltage divider.) Therefore, applying the voltagedivider equation:

v t V t( ) cos= ω

+

-

+

-

v to ( )

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 voltagesand 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 domainwill be easy.

Step 4. Transform back to the time domain, if asked to get voltage or current as function of time.

Solutions - HW #2Physics 426

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$ $

$

$

V VZ

Z Z

VR

Rj C

VR

R jC

oR

R C

=+

FHG

IKJ

=+

F

H

GGG

I

K

JJJ

=−

F

HGGG

I

KJJJ

1

1

ω

ω

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’spurely real, purely imaginary, or complex). So we make the following aside:

We know that v t V t( ) cos= ω . (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 VR

R jC

o =−

F

HGGG

I

KJJJ1

ω

There are at least a couple of ways you could proceed now to get $Vo in polar form. One way that’s

frequently a quick one is to immediately put into polar form any complex expressions appearing in thephasor expression. So that is what we will do:

Aside: The procedure for finding the phasor transform of a sinusoidal function of time can be statedsimply as follows. If we have a function f t( ) which, when expressed as a cosine, has the form

f t A t( ) cos( )= +ω θ , then the phasor transform of f t( ) (let’s call the phasor transform $F ) will be

$F Ae j= θ .

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 t( ) cos= ω ) then the

phasor transform will be just the amplitude of f t( ) :

$F Ae Aj= =0 .

Solutions - HW #2Physics 426

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$/

V VR

RC

e

o

j

=

+ FHG

IKJ

FHG

IKJ

L

N

MMMMMM

O

Q

PPPPPP−2

2 1 21

ωθ

, θω

= FHG

IKJ

−tan 1 1

RC.

$/

/

V VR

RC

e

V

RC

e

oj

j

=

+ FHG

IKJ

FHG

IKJ

L

N

MMMMMM

O

Q

PPPPPP

=

+ FHG

IKJ

FHG

IKJ

L

N

MMMMMM

O

Q

PPPPPP

22 1 2

2 1 2

1

1

11

ω

ω

θ

θ

Step 4. Now we are ready to convert back to the time domain.

v t V e V

RC

e eo oj t j j t( ) Re $ Re

/= =

+ FHG

IKJ

FHG

IKJ

L

N

MMMMMM

O

Q

PPPPPP

ω θ ω

ω

1

11

2 1 2

v tV

RC

to ( ) cos/

=

+ FHG

IKJ

FHG

IKJ

+

11

2 1 2

ω

ω θb g , θω

= FHG

IKJ

−tan 1 1

RC .

(b.) Show analytically that at sufficiently high frequency the output is an essentially perfect replica ofthe input.

If we let ω → ∞ , then 1

0ωRC

→ and θ → =−tan 1 0 0b g . So

v t V to ( ) cos→ ω

which is exactly the same as the input voltage v t( ) .

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(c.) Show that at sufficiently low frequency the output is a severly attenuated waveform which isapproximately the derivative of the input.

If we let ω → 0 , something interesting happens: 1

2

ωRCFHG

IKJ → ∞ (i.e.,

12

ωRCFHG

IKJ gets much larger than

1) and θπ

→ ∞ =−tan 1

2b g radians, or 90º. So

v t V RC t RC V to ( ) ( ) cos sin→ + = −ω ω ω ω90ob g b g

But the factor in parentheses is the derivative of the input:

v t RCd

dtv to ( ) ( )→

So at low frequencies, the RC coupling network gives you an output that’s proportional to thederivative of the input. For this reason, the RC coupling network is called a differentiator at lowfrequencies.

Now, why does the problem say the output should be a “severly attenuated” waveform? Well,

v t V RC to ( ) sin→ − ω ωb g

and the factor ωRC goes to zero (gets much smaller than 1) as ω → 0 . So at low frequencies, theamplitude 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 t( ) cos= ω

Solutions - HW #2Physics 426

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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 tR ( ) across the resistor R.

Solution:

Step 1. Let $V be the phasor transform of v t( ) and $VR be the phasor transform of v tR ( ) .

Step 2. Recognizing the circuit as a voltage divider, we apply the voltage divider equation:

$ $ $V VZ

Z Z ZV

R

R j LC

RR

R L C

=+ +

FHG

IKJ =

+ −FHG

IKJ

F

H

GGGG

I

K

JJJJωω1

Step 3. Get $VR in polar form. To do this, we need to know whether $V is purely real, purely imaginary, or

complex. Since v t V t( ) cos= ω (given), $V will be just the amplitude of v t( ) :

$V V=

So now we make this substitution in the expression for $VR and write the denominator in polar form to get:

$/

V VR

R LC

e

R

j

=

+ −FHG

IKJ

LNMM

OQPP

F

H

GGGGGG

I

K

JJJJJJ22 1 2

1ω

ωθ

, θω

ω=

−FHG

IKJ

L

N

MMMM

O

Q

PPPP−tan 1

1L

C

R.

Now take the complex exponential factor to the numerator:

$/

V VR

R LC

eRj=

+ −FHG

IKJ

LNMM

OQPP

F

H

GGGGGG

I

K

JJJJJJ

−

22 1 2

1ω

ω

θ

Step 4: We now have $VR in polar form and are ready to transform back to the time domain to get v tR ( ) :

v t V e VR

R LC

eR Rj t j t( ) Re $ Re

/= =

+ −FHG

IKJ

LNMM

OQPP

F

H

GGGGGG

I

K

JJJJJJ

−ω ω θ

ωω

22 1 2

1

b g

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v tR

R LC

V tR ( ) cos/

=

+ −FHG

IKJ

LNMM

OQPP

F

H

GGGGGG

I

K

JJJJJJ−

22 1 2

1ω

ω

ω θb g , θω

ω=

−FHG

IKJ

L

N

MMMM

O

Q

PPPP−tan 1

1L

C

R

(b.) Derive a general expression for the voltage across the capacitor.

Step 1. Let $V be phasor transform of v t( ) , $VC be phasor transform of v tC ( ) .

Step 2. Recognize circuit as a voltage divider. Then apply voltage divider equation:

$ $ $ $V VZ

Z Z ZV

j C

R j LC

VLC j RC

CC

R L C

=+ +

FHG

IKJ =

+ −FHG

IKJ

F

H

GGGG

I

K

JJJJ=

− +

FHGG

IKJJ

1

1

1

1 2

ω

ωω

ω ωe j

Step 3. Get $VC 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 V

LC RC

eCj=

− +LNM

OQP

F

H

GGGG

I

K

JJJJ−1

1 2 2 21 2

ω ω

θ

e j b g , θ

ω

ω=

−−tan 1

21

RC

LCe j

Step 4. Now $VC is in polar form, so we are ready to transform back to get v tC ( ) :

v t V e V

LC RC

tC Cj t( ) Re $ cos

/= =

− +LNM

OQP

F

H

GGGG

I

K

JJJJ−ω

ω ω

ω θ1

1 2 2 21 2

e j b gb g , θ

ω

ω=

−−tan 1

21

RC

LCe j

(c.) Derive a general expression for the voltage across the inductor.

Step 1. Let $V be phasor transform of v t( ) , $VL be phasor transform of v tL ( ) .

Step 2. Recognize circuit as a voltage divider. Then apply voltage divider equation:

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$ $ $ $V VZ

Z Z ZV

j L

R j LC

V

LCj

R

L

LL

R L C

=+ +

FHG

IKJ =

+ −FHG

IKJ

F

H

GGGG

I

K

JJJJ=

−FHG

IKJ −

F

H

GGGGG

I

K

JJJJJ

ω

ωω ω ω1

1

11

2

.

Step 3. Get $VL 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 V

LC

R

L

eLj=

−FHG

IKJ +

FHG

IKJ

LNMM

OQPP

F

H

GGGGGGG

I

K

JJJJJJJ

1

11

2

2 21 2

ω ω

θ , θω

ω

=

−FHG

IKJ

L

N

MMMMM

O

Q

PPPPP

−tan 1

21

1

R

L

LC

Step 4. Now $VL is in polar form, so we are ready to transform back to get v tL ( ) :

v t V e V

LC

R

L

tL Lj t( ) Re $ cos

/= =

−FHG

IKJ +

FHG

IKJ

LNMM

OQPP

F

H

GGGGGGG

I

K

JJJJJJJ

+ω

ω ω

ω θ1

11

2

2 21 2 b g . θ

ω

ω

=

−FHG

IKJ

L

N

MMMMM

O

Q

PPPPP

−tan 1

21

1

R

L

LC

.