Electronic InstrumentationExperiment 5

* AC Steady State Theory

* Part A: RC and RL Circuits

* Part B: RLC Circuits

AC Steady State Theory

Transfer Functions Phasors Complex Impedance Complex Transfer Functions

Transfer Functions

The transfer function describes the behavior of a circuit at Vout for all possible Vin.

in

out

V

VH

Simple Example

321

32*

RRR

RRVV inout

KKK

KKVV inout 321

32*

6

5

in

out

V

VH

VKtVtVthen

VKtVtVif

out

in

10)2

2sin(5)(

12)2

2sin(6)(

More Complicated Example

What is H now?

H now depends upon the input frequency ( = 2f) because the capacitor and inductor make the voltages change with the change in current.

Influence of Resistor on Circuit

Resistor modifies the amplitude of the signal by R

Resistor has no effect on the phase

RIV RR

)2

cos(*)sin(*)(

)2

cos()sin()(

tARtARtVthen

tAtAtIif

R

R

Influence of Capacitor on Circuit

Capacitor modifies the amplitude of the signal by 1/C

Capacitor shifts the phase by -/2

dtIC

V CC

1

)cos(*1

)2

sin(*1

)(

)2

sin(*1

)cos(*1

)(

)2

cos()sin()(

tAC

tAC

tVor

tAC

tAC

tVthen

tAtAtIif

C

C

C

Influence of Inductor on Circuit

Inductor modifies the amplitude of the signal by L

Inductor shifts the phase by +/2

dt

dILV L

L

)cos(*)2

sin(*)(

)2

sin(*)cos(*)(

)2

cos()sin()(

tALtALtVor

tALtALtVthen

tAtAtIif

L

L

L

How do we model H? We want a way to combine the effect of

the components in terms of their influence on the amplitude and the phase.

We can only do this because the signals are sinusoids• cycle in time• derivatives and integrals are just phase

shifts and amplitude changes

Phasors

Phasors allow us to manipulate sinusoids in terms of amplitude and phase changes

Phasors are based on complex polar coordinates

The influence of each component is given by Z, its complex impedance

ZIV

Phasor References

http://ccrma-www.stanford.edu/~jos/filters/Phasor_Notation.html

http://www.ligo.caltech.edu/~vsanni/ph3/ExpACCircuits/ACCircuits.pdf

http://robotics.eecs.berkeley.edu/~cssharp/Phasors/Phasors.html

Phasor Applet

Complex Polar Coordinates

tAtV cos)(

Review of Polar Coordinates

point P is at( rpcosp , rpsinp )

22

1tan

PPP

P

PP

yxr

x

y

Introduction to Complex Numbers

zp is a single number represented by two numbers

zp has a “real” part (xp) and an “imaginary” part (yp)

jj

jj

j

1

1

1

Now we can define Phasors

)sin()cos(

,)cos()(

tjAtAV

letthentAtVif

The real part is our signal. The two parts allow us to determine the

influence of the phase and amplitude changes mathematically.

After we manipulate the numbers, we discard the imaginary part.

Magnitude and Phase

Vofphasetx

yV

VofmagnitudeAyxV

jyxtAjtAV

1

22

tan

)sin()cos(

Phasors have a magnitude and a phase derived from polar coordinates rules.

Euler’s Formula

sincos je j

2132

13

)(

2

1

2

1

2

13

,

sincos

21

2

1

andr

rrtherefore

er

r

er

er

z

zz

rejrrjyxz

jj

j

j

Phasor Applet

Manipulating Phasors (1)

2132

13

)(

2

1

2

1)(

2

)(1

2

13

)(

,

)sin()cos(

21

2

1

2

1

VandA

AVtherefore

eA

A

e

e

e

e

A

A

eA

eA

V

VV

AetjtAV

jj

tj

tj

tj

tj

tj

Note t is eliminated by the ratio• This gives the phase change between

signal 1 and signal 2

Manipulating Phasors (2)

22

22

21

2

2

1

3

1

yx

yx

V

VV

333

222111

jyxV

jyxVjyxV

2

21

1

11213 tantan

x

y

x

yVVV

Understanding the influence of Phase

x

yV 1tan

180

)(0

tan

00:

9020

tan

00:

9020

tan

00:

00

tan

00:

1

1

1

1

orx

Vthen

xandyifreal

yVthen

yandxifj

yVthen

yandxifj

xVthen

xandyifreal

Complex Impedance

Z defines the influence of a component on the amplitude and phase of a circuit• Resistors: ZR = R

• change the amplitude by R

• Capacitors: ZC=1/jC • change the amplitude by 1/C

• shift the phase -90 (1/j=-j)

• Inductors: ZL=jL • change the amplitude by L

• shift the phase +90 (j)

ZIV

Capacitor Impedance Proof

)(1

)()()(

)(

)()cos()(

Re)(

)()(

)sin()cos()(

)cos()()(

)(

)()(

)(

tICj

tVtVCjdt

tdVCtI

tVjtAjdt

jVd

dt

tdV

jVjeAjdt

dAe

dt

jVd

AetjAtAjV

tAtVanddt

tdVCtI

CCCC

C

CCC

Ctj

tjC

tjC

CC

C

CjZC

1Prove:

Complex Transfer Functions

If we use phasors, we can define H for all circuits in this way.

If we use complex impedances, we can combine all components the way we combine resistors.

H and V are now functions of j and

)(

)()(

jV

jVjH

in

out

Simple Example

1

1)(

1

1

)(

)(

)()(

RCjjH

Cj

Cj

CjR

CjjH

ZZ

Z

IZZ

IZ

jV

jVjH

CR

C

CR

C

in

out

CjZ

RZ

C

R

1

Simple Example (continued)

1

1)(

RCjjH

222

22

)(1

1

)(1

01

1

01)(

RCRCRCj

jjH

)(tan1

tan1

0tan)(

)1()01()(

111 RCRC

jH

RCjjjH

2)(1

1)(

RCjH

)(tan)( 1 RCjH

Using H to find Vout

in

out

in

out

jin

jout

tjjin

tjjout

in

out

eA

eA

eeA

eeA

V

VjH

)(

inout

inout

jin

jHjjout

jin

jout

eAejHeA

eAjHeA

)()(

)(

inout AjHA )( inout jH )(

Simple Example (with numbers)

157.0)2(1

1)(

2

2

jH 41.11

2tan0)( 1

jH

)4

2cos(2)(

11

tKVtV

KRFC

in

)41.1785.02cos(2*157.0)( tKVtVout

12

1

1112

1

1

1)(

jKKjRCjjH

)625.02cos(314.0)( tKVtVout

Adding Phasors & Other Applets

Part A -- RC and RL Circuits

High and Low Pass Filters H and Filters

High and Low Pass FiltersHigh Pass Filter

H = 0 at 0

H = 1 at

at c

Frequency

1.0Hz 100Hz 10KHz 1.0MHz 100MHzV(C1:2) / V(R1:2)

0

0.5

1.0

Frequency

1.0Hz 100Hz 10KHz 1.0MHz 100MHzV1(R1) / V(C1:2)

0

0.5

1.0

Low Pass Filter

H = 1 at 0

H = 0 at

at c

c=2fc

fc

fc

c=2fc

Corner Frequency The corner frequency of an RC or RL circuit

tells us where it transitions from low to high or visa versa.

We define it as the place where

For RC circuits:

For RL circuits:

2

1)( cjH

L

Rc

RCc

1

Corner Frequency of our example

212 RC 2

2

1 RC

RCjjH

1

1)(

2

1)( jH

2

1

)(1

1

2

1

)(1

1)(

22

RCRCjH

RCc

1

H(j), c, and filters We can use the transfer function, H(j), and the

corner frequency, c, to easily determine the characteristics of a filter.

If we consider the behavior of the transfer function as approaches 0 and infinity and look for when H nears 0 and 1, we can identify high and low pass filters.

The corner frequency gives us the point where the filter changes:

2

ccf

Our example at low frequencies

)(01

0tan)(

110)(

1 axisxonjH

asjH

LOW

LOW

RCjjH

1

1)(

101

1)(

jH LOW

Our example at high frequencies

220

0tan

1

0tan)(

011

)(

11

RCjH

RCjasjH

HIGH

HIGH

RCjjH

1

1)(

RCjjH HIGH 1

)(

Our example is a low pass filter

Frequency

1.0Hz 100Hz 10KHz 1.0MHz 100MHzV(C1:2) / V(R1:2)

0

0.5

1.0

RCf c

c

2

1

2

01 HIGHLOW HH

What about the phase?

Our example has a phase shift

0

1

HIGH

LOW

H

H

Frequency

1.0Hz 10KHz 100MHzVP(C1:2)

-100d

-50d

0dV(R1:2) / V(V1:+)

0

0.5

1.0

SEL>>

90)(

0)(

jH

jH

HIGH

LOW

Taking limits

012

2

012

2)(bbb

aaajH

At low frequencies, (ie. =10-3), lowest power of dominates

At high frequencies (ie. =10+3), highest power of dominates

0

00

03

16

2

00

31

62

101010

101010)(

b

a

bbb

aaajH

2

20

03

16

2

00

31

62

101010

101010)(

b

a

bbb

aaajH

Capture/PSpice Notes Showing the real and imaginary part of the signal

• in Capture: PSpice->Markers->Advanced• ->Real Part of Voltage

• ->Imaginary Part of Voltage

• in PSpice: Add Trace• real part: R( )

• imaginary part: IMG( )

Showing the phase of the signal• in Capture:

• PSpice->Markers->Advanced->Phase of Voltage

• in PSPice: Add Trace• phase: P( )

Part B -- RLC Circuits

Band Filters Another example A more complex example

Band Filters

Frequency

1.0Hz 100Hz 10KHz 1.0MHz 100MHzV(L1:1) / V(V1:+)

0

0.5

1.0

Frequency

1.0Hz 100Hz 10KHz 1.0MHz 100MHzV(R1:1)/ V(R1:2)

0

0.5

1.0

f0

f0

Band Pass Filter

H = 0 at 0

H = 0 at

at 0=2f0

Band Reject Filter

H = 1 at 0

H = 1 at

at 0 =2f0

Resonant Frequency The resonant frequency of an RLC circuit tells

us where it reaches a maximum or minimum. This can define the center of the band (on a band

filter) or the location of the transition (on a high or low pass filter).

We are already familiar with the equation for the resonant frequency of an RLC circuit:

LC

10

Another Example

1

11

1

)(22

LCjRCj

CjLjR

CjjH

CjZ

LjZRZ

C

LR

1

RCjLCjH

)1(

1)(

2

At Very Low Frequencies

0)(

10)(

11

1)(

jH

jH

jH

LOW

LOW

LOW

At Very High Frequencies

orjH

jH

LCjH

HIGH

HIGH

HIGH

)(

01

)(

1)(

2

At the Resonant Frequency

LC

10

RCjLCjH

)1(

1)(

2

LC

RCjRC

LCjLC

LC

jH

)11(

1

1)

11(

1)( 20

2)(

)(

)(

0

0

0

jH

RC

LCjH

RC

LCjjH

if L=10uH, C=1nF and R=1K

rad/secfHz

LCf

21

0

radiansH2

Our example is a low pass filter

Frequency

1.0Hz 10KHz 100MHzV(R1:2) / V(V1:+)

0

0.5

1.0P( V(R1:2))

-200d

-100d

0d

SEL>>

Phase

= 0 at 0

= - at Magnitude

= 1 at 0

= 0 at

fHz

A more complex example

Even though this filter has parallel components, we can still handle it.

We can combine complex impedances like resistors

Determine H

LC

Lj

LCj

Lj

CjLj

CjLj

ZCL 222 111

1

LjLCR

LjjH

)1()(

2

LC

LCbymultiply

LCLj

R

LCLj

ZR

ZjH

CL

CL2

2

2

2

1

1

1

1)(

At Very Low Frequencies

2)(

00)(

)(

jH

jHR

LjjH

LOW

LOW

LOW

At Very High Frequencies

2)(

01

)(

)(2

jH

jH

RC

j

LRC

LjjH

HIGH

HIGH

HIGH

At the Resonance Frequency

11

)1

1(

1

)( 20

LLC

jLCLC

R

LLC

j

jH

0)(1)( 00 jHjH

LC

10

Frequency

1.0Hz 10KHz 100MHzVP(R1:1)

-100d

0d

100d

SEL>>

V1(R1) / V(V1:+)0

0.5

1.0

Our example is a band pass filter

Phase

= 90 at 0

= 0 at

= - at

Magnitude

= 0 at 0

H=1 at

= 0 at

f

Simple Filter Design Step One: Pick a High Pass or Low Pass

Filter Configuration • This is done by looking at the high or low

frequency limit of various simple circuits. • For example, for an RC low pass filter, we

choose the following configuration:R1

0

V1

FREQ = VAMPL = VOFF = C1

Simple Filter Design

At very high frequencies, the capacitor becomes a short circuit. (Impedance goes to zero.)

At very low frequencies, the capacitor becomes an open circuit. (Impedance goes to infinity.) The output appears fully across the capacitor.

Thus, if the output voltage is taken across the capacitor, this is a low pass filter.

R1

0

V1

FREQ = VAMPL = VOFF = C1

V o u tV in

Simple Filter Design

Step Two: Pick values for R1 and C1.• The impedance of C1 decreases from infinity to

zero as frequency increases from zero to infinity.

• At the corner frequency of the circuit, the magnitude of the impedance of C1 equals the impedance of R1

R1

0

V1

FREQ = VAMPL = VOFF = C1

V o u tV in

Simple Filter Design

At the corner frequency,

R1

0

V1

FREQ = VAMPL = VOFF = C1

HV

Vj C

Rj C

j R C jo u t

in

1

11

1

1

1

V o u tV in

Simple Filter Design

The magnitude of H at the corner frequency is 0.707.

R1

0

V1

FREQ = VAMPL = VOFF = C1

HV

V jo u t

in

1

1

1

20 7 0 7.

V o u tV in

Simple Filter Design

Thus, for frequencies less than the corner frequency, Vout is comparable to Vin.

For frequencies greater than the corner frequency, the output is much smaller than the input.

R1

0

V1

FREQ = VAMPL = VOFF = C1

V o u tV in

Example

For this example, we select components from our boxes of parts to pass most frequencies below 100Hz.

For R=50 and C=33uF, the corner frequency is about 96Hz.

Frequency

1.0Hz 3.0Hz 10Hz 30Hz 100Hz 300Hz 1.0KHz 3.0KHz 10KHzV(R2:2)/ V(R1:2)

0

0.2

0.4

0.6

0.8

1.0

VV1

FREQ = 1kVAMPL = 1VOFF = 0

R1

50

R2

50

C1

33uF

V

0

V in V o u t