Electronic Instrumentation Experiment 2 * Part A: Intro to Transfer Functions and AC Sweeps * Part...

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Electronic InstrumentationExperiment 2

* Part A: Intro to Transfer Functions and AC Sweeps* Part B: Phasors, Transfer Functions and Filters* Part C: Using Transfer Functions and RLC Circuits* Part D: Equivalent Impedance and DC Sweeps

Part A

Introduction to Transfer Functions and Phasors

Complex Polar Coordinates Complex Impedance (Z) AC Sweeps

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.

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

We will define Phasors

A phasor is a function of the amplitude and phase of a sinusoidal signal

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

Phasors are based on complex polar coordinates.

Using phasors and complex numbers we will be able to find transfer functions for circuits.

),( AfV

Review of Polar Coordinates

point P is at( rpcosp , rpsinp )

22

1tan

PPP

P

PP

yxr

x

y

Review of 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

Complex Polar Coordinates

z = x+jy where x is A cos and y is A sin t cycles once around the origin once for each

cycle of the sinusoidal wave (=2f)

Now we can define Phasors

.),(

sincos,

)sin()cos(

,)cos()(

droppedisitsotermeachtocommonist

jAAVsimplyor

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.

The “V=IR” of Phasors

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

Once we have Z, we can use phasors to analyze circuits in much the same way that we analyze resistive circuits – except we will be using the complex polar representation.

ZIV

Magnitude and Phase

Vofphasex

yV

VofmagnitudeAyxV

jyxAjAV

1

22

tan

sincos

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

Influence of Resistor on Circuit

Resistor modifies the amplitude of the signal by R

Resistor has no effect on the phase

RIV RR

)sin(*)(

)sin()(

tARtVthen

tAtIif

R

R

Influence of Inductor on Circuit

Inductor modifies the amplitude of the signal by L

Inductor shifts the phase by +/2

dt

dILV L

L

)2

sin(*)(

)cos(*)(

)sin()(

tALtVor

tALtVthen

tAtIif

L

L

L

Note: cos=sin(+/2)

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

)2

sin(*1

)2

sin(*1

)(

)cos(*1

)cos(*1

)(

)sin()(

tAC

tAC

tVor

tAC

tAC

tVthen

tAtIif

C

C

C

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

AC SweepsAC Source sweeps from 1Hz to 10K Hz

Time

200ms 250ms 300ms 350ms 400msV(R1:2) V(C1:1)

-1.0V

0V

1.0V

Time

20ms 25ms 30ms 35ms 40msV(R1:2) V(C1:1)

-1.0V

0V

1.0V

Time

2.0ms 2.5ms 3.0ms 3.5ms 4.0msV(R1:2) V(C1:1)

-1.0V

0V

1.0V

Transient at 10 Hz Transient at 100 Hz Transient at 1k Hz

Notes on Logarithmic Scales

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

Phasors Complex Transfer Functions Filters

Definition of a Phasor

sincos

,)cos()(

jAAV

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

Vofphasex

yV

VofmagnitudeAyxV

jyxAjAV

1

22

tan

sincos

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

zzthen

rejrrjyxzif

jj

j

j

214214

)(2121214

,

2121

andrrrtherefore

errererzzzand jjj

Manipulating Phasors (1)

2132

13

)(

2

1

2

1)(

2

)(1

2

13

)(

,

)sin()cos(

21

2

1

2

1

XandA

AXtherefore

eA

A

e

e

e

e

A

A

eA

eA

V

VX

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

VX

333

222111

jyxV

jyxVjyxV

2

21

1

11213 tantan

x

y

x

yVVX

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

Complex Impedance

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

• Capacitors: ZC=1/jC

• Inductors: ZL=jL

We can use the rules for resistors to analyze circuits with capacitors and inductors if we use phasors and complex impedance.

ZIV

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

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

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

Taking limits -- Example

523

159)(

2

2

jH

At low frequencies, (lowest power)

At high frequencies, (highest power)

35

15)( jH LO

33

9)(

2

2

jH HI

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

Part C

Using Transfer Functions Capacitor Impedance Proof More Filters Transfer Functions of RLC Circuits

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

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:

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).

The equation for the resonant frequency of an RLC circuit is:

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=1mH, C=0.1uF and R=100

krad/secfkHz

LCf

2

10

radiansH2

Frequency

1.0Hz 100Hz 10KHz 1.0MHzV(C1:1)/V(L1:1)

0

0.4

0.8

1.2p(V(C1:1)/V(L1:1))

-200d

-100d

0d

SEL>>

Our example is a low pass filterPhase

= 0 at 0

= - at

Magnitude

= 1 at 0

= 0 at

fkHz

Actual circuit resonance is only at the theoretical resonant frequency, f0, when there is no resistance.

Part D

Equivalent Impedance Transfer Functions of More

Complex Circuits

Equivalent Impedance

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

We can combine complex impedances like resistors to find the equivalent impedance of the components combined.

Equivalent Impedance

LC

Lj

LCj

Lj

CjLj

CjLj

ZZ

ZZZ

CL

CLCL 222 111

1

Determine H

LC

LjZCL 21

LjLCR

LjjH

)1()(

2

LC

LCbymultiply

LCLj

R

LCLj

jH2

2

2

2

1

1

1

1)(

CL

CL

ZR

ZjH

)(

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 Resonant 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