bruce mayer, pe licensed electrical & mechanical engineer bmayer@chabotcollege

[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx1

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 43

FourierTransfer

Fcn



Fourier Transform A Fourier Transform

is an integral transform that re-expresses a function in terms of different Sine/Cosine waves of varying amplitudes, wavelengths, and phases.

A Conceptual Example• This Irregular Signal

• Is the SAME as the Sum of these Sinusoids



Fourier Transform John Baptiste

Joseph Fourier investigated Time Varying Heat-Flow in a Metal Bar

His great Insight: ANY Periodic Function Could be Expressed as the sum of Sinusoidal Functions

For a Given, arbitrary Periodic Function, f(t), The Fourier Equivalents

1

00 cosm

mm tmBbtf

1

00 sinn

nn tnDdtf



Example: Square Wave

The SquareWave Shown at Bottom-Lt can be described by a sum-of-sines

tA

tA

tA

tA

tA

vsq 00000 9sin9

47sin

7

45sin

5

43sin

3

4sin

4



Transfer Fuction, H(f)

Consider a “Black Box” that takes Input Power, vin & iin Transforms this Power into an Output, vout & iout

• A typical transformation would be to “Filter-Out” certain electrical frequencies.

For Phasor Voltages Vin & Vout Define the voltage Transfer Function as

inv outvini outi

in

outfHV

V



Transfer Function

Note that the Transfer Function• Is a Function of FREQENCY ONLY• Can Change (and usually does change)

the Magnitude and Phase-Angle of many of the incoming, frequency-dependent, electrical signals

Measuring an Unknown “Black Box”

in

outfHV

V

Apply Sinusoidal Vin (Vin0°), Measure Vout (Voutφ°) and Plot: Vout / Vin and φ



Example Transfer Function

Hz f

Hz f

f

H

fH



Example Transfer Function

Hz f

f

H

fH

Find vout for vin = 1.35Vcos(40∙2πt+65°)

−25



Example Transfer Function Then at 40 Hz

(40∙2π rads/sec)

Recall vin

In Phasor for

Thus

Using the Values Taken from the H(f) Mag & Phase Graphs

Or in the Time Domain

in

outHV

V 15025Hz40

65240cos35.1 tVvin

6535.1 VinV

inout H VV Hz40

8575.33

6535.115025

V

V

out

out

V

V

85240cos75.33 tVtvout



MultiFrequency Example 6.2

Note the THREE Frequencies• 0 Hz• 1000 Hz

– 1000∙2π rad/sec

• 2000 Hz– 2000∙2π

rad/sec



Ex6.2 Transfer Function

Apply to vin the Transfer Function

From the Transfer Function find

• Apply To components of vin

60220003031000040 HHH



Example 6.2

Using This H(f) Set find

Note that the above Phasors CanNOT be added as they have DIFFERENTFrequencies.

60220003031000040 HHH

1201203040 11 inout H VV

306023031000 22 inout H VV

1027016022000 33 inout H VV



Example 6.2

Because of Differing Frequencies MUST add TIME-DOMAIN Voltages

Then vout(t) is simply the SUM of the above

10230612 321 outoutout VVV

1022000cos23021000cos612 321 tvtvtv outoutout



1st Order Lo-Pass Filter Consider the RC Ckt

Shown below

In the Frequency Domain the Cap Impedance, Zc

Notice the Limits of Behavior

A cap is• OPEN to Low-Freq• SHORT to Hi-Freq

inV outV

CfjCjC 2

11Z

Cfjf

Cf 2

1limlim

00Z

02

1limlim

CfjfC

f Z



1st Order Lo-Pass Filter Thus the Behavior

of a Cap-Based Impedance• At LO-Frequencies a

Cap acts as an OPEN Circuit

• At HI-Frequencies a Cap Acts as a SHORT Circuit

Now use Phasor V-Divider on RC ckt

Multiplying Top&Bot by j2πfC

inV outV

CfjC 2

1Z

RR Z

inin

tot

Cout CfjR

CfjVV

Z

ZV

21

21

inout RCfjVV

21

1



1st Order Lo-Pass Filter Then the Transfer

Function

ReWriting

Where

fB is the “Break point” Frequency at which H(f) falls to 70.7% of its Original Magnitude Value.

Note The Mag & Ph of H(f) in terms of fB :

RCfjfH

in

out

21

1

V

V

21

1

11

1

21

1

B

B

B

ff

ffjfH

fjfRCjffH

RCfB 2

1

B

B

fffH

fffH

arctan

1

12



Lo-Pass Filter

The LoPass Filter Transfer Function

fB : is also call the Half-Power-Frequency• Recall Full Power to a Resistor:• Then HALF Power:

RVorRI 22

RVorRI22

22

inV outV



LR (LowPass) Filter Find the Transfer

Function for LR Ckt

Use Ohm Find The Single Loop Current

Then also by Ohm

ReWriting

Arrive at Xfer Fcn very similar to RC Ckt

fLjL 2Z

outVinV

I

RfLjRin

L

in

2

V

Z

VI

1211

2

RfLj

RRfLj

R ininout

VVIV

B

inininout

ff

j

fLR

fjRfLj

11

2

12

VVVV

LRf

fjffH

B

Bin

out

2 :where

1

1

V

V



The deciBel (dB) Named after

Alexander Graham Bell, the deciBel (dB) relates two Power Levels

SomeTimes The Power Level is Referenced to a Standard Value, P0

In this case

ReCall a Current or Voltage delivering Power to a Resistor

Then the dB in Current or Voltage Ratios

1

2log10P

PLdB

0

log10P

PLdB

RIPRVP iv22



The deciBel (dB) dB In Terms of

Voltage Ratios

Or dB for Currents

Now we Defined

Since |H(f)| is a Voltage Ratio, define

1

2

2

1

22

1

22

21

22

1

2

log20log10log10

log10log10

V

V

V

V

V

V

RV

RV

P

PLdB

inout

inout

VVfH

VVfH

2

1

2

2

1

22

1

22

22

21

1

2

log20log10log10

log10log10

I

I

I

I

I

I

RI

RI

P

PLdB

fHfHdB

log20



dB Plots (SemiLog) Plot

Plotting H(f) on the logarithmic dB Scale makes it easier to distinguish Very Large (104 vs 105) or Very Small (10−4 vs 10−5) Points on the Plots

0000562.01010log2085 25.42085 fHfHdb



Cascaded NetWork Gain Consider the

Transfer Function of the “BlackBox” at Right

Looking inside the BlackBox find

Note that with Vout1 = Vin2

Or in dB form

in

outfHV

V

1

1

1

2

1

2

1

2 1out

out

in

out

in

out

in

out

in

out

V

V

V

V

V

V

V

V

V

V

fHfHfH

fH

in

out

in

out

in

out

in

out

out

out

in

out

in

out

212

2

1

1

2

1

1

2

1

1

1

2

:so

or

V

V

V

V

V

V

V

V

V

V

V

V

V

V

dBdBdB

fHfHfH

fHfHfH

fHfHfH

fHfHfH

21

21

21

21

log20log20log20

log20log20



Reading Logarithmic Scales

Tools Needed• Ruler• Scientific Calculator

To Find a Value of a Pt Between Decades m & n• Use Ruler to Measure

– Decade Distance, dd

– Distance from Pt to Lower Decade (Decade m), dp

• Then The Value at the Pt

mdd dpV 1010



400 405 410 415 420 425 430 435 440 445 45010

-33

10-32

10-31

10-30

x

y

mm 121.dd

mm 415.pd

3232730032121415 1037510101010 ....V



Octave

An octave is the interval between two points where the frequency at the second point is twice the frequency of the first.

Given Frequencies f1 & f2

2log

log

log

21

2

12

ffN

ORf

fN

oct

oct

Octave 1 2 3 4 5 6 7 8

Frequency (Hz) 63 125 250 500 1k 2k 4k 8k

Wavelength in air (70oF, 21oC) (ft)

17.92 9.03 4.52 2.26 1.129 0.56 0.28 0.14

Wavelength in air (70oF, 21oC) (m)

5.46 2.75 1.38 0.69 0.34 0.17 0.085 0.043

MUSICAL Octaves



WhiteBoard Work

Let’s This Nice Problem

Find the OutPut Voltage for For this Input

tvin

tvout

tVtVVtvin 212000cos3121000cos2317



All Done for Today

79.5 MHzNotchFilter

http://www.soontai.com/curve/TRFch5_fg1.gif



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 43

Appendix



Lo

gar

ith

m C

han

ge

of

Bas

e P

roo

f



White Board RL Filter Problem



LR Filter Transfer Function

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 104

0.4

0.5

0.6

0.7

0.8

0.9

1

f (Hz)

|H(f

)

f = 0:10:20e3HfB = 1./sqrt(1+(f/fB).^2);plot(f,HfB,'LineWidth',3), grid, xlabel('f (Hz)'), ylabel('|H(f)')disp('showing fB plot - hit ANY KEY to continue')pausefB = 2700/(2*pi*68e-3)Hf = abs(2700./(2700 + j*2*pi*f*68e-3));plot(f,Hf,'LineWidth',3), grid, xlabel('f (Hz)'), ylabel('|H(f)')



P5.57 Graphics



P5.81 Graphics

bruce mayer, pe licensed electrical & mechanical engineer bmayer@chabotcollege

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pe engineering

vout vin

voltage transfer function

vin iin

components of vin

transfer functionis

transfer functionapply

measure vout vout