EE40 Lec 12 Transfer Function Bode Plots Filters Transfer Function

EE40 Lec 12Transfer Function Bode Plots FiltersTransfer Function, Bode Plots, Filters

Prof. Nathan Cheung

10/08/2009

Reading: Hambley Chapter 6.1-6.5

Slide 1EE40 Fall 2009 Prof. Cheung

OUTLINE

– Fourier Analysis Concept– Transfer Function – dB scale– First Order Low-Pass and High-Pass Filters– Bode PlotsBode PlotsLog magnitude vs log frequency plotLinear Phase vs log frequency plotLinear Phase vs log frequency plot– Asymptotic Frequency Behavior


Fourier Synthesis of Waveforms

Any waveform can be constructed by adding sinusoids that have the proper amplitudes, f i d h

Example: Fourier Components of A Square Wave

frequencies, and phase.

p p q


Transfer Function

• Transfer function is a function of frequency– A complex quantity– Both magnitude and phase are function of

frequency

Two Port filter networkVin Vout

( )VV ( )( )

( )

outout in

in

VfV

H f

θ θ

θ

= = ∠ −

∠

out

in

VHV

H(f)


( )H f θ= ∠H(f)

Example: Response of a Two-Frequency Signal

H(f1) ≠ H(f2)i lin general


Cascade of Two-Port Networks

( ) ( ) ( )fHfHfH 21 ×=


Origin of Decibel (dB) unitThe bel (abbreviated B) is named after Alexander Graham Bell, who did much pioneering work with sound and the way o r ears respond to so nd O r earsour ears respond to sound. Our ears respond to sounds ranging from an intensity less than 10 -16 W/cm2 to intensities larger than 10 -4 W/cm2intensities larger than 10 4 W/cm(where we begin to experience pain). This is a range of more than 1x1012 times from the softest to the loudest sounds. Logarithm provide a convenient way to represent these values, because they compress this scale into a range of 12, rather than a range of a billion. A bel is defined as the logarithm of a power ratio. It gives us a way to compare power l l i h h h d i h


levels with each other and with some reference power.

The Decibel (dB) Unit

• A bel (symbol B) is a unit of measure of ratios of power levels

B = log (P /P ) where P and P are– B = log10(P1/P2) where P1 and P2 are power levels.

– one bel corresponds to a ratio of 10:1 10dB• The bel is too large for everyday

use, so the decibel (dB), equal to 0.1B, is more commonly used sc

ale

, y1dB = 10 log10(P1/P2)

Log 1

0s

1B• dB are used to measure

– Electric power, Gain or loss of amplifiers, transmission loss of optical

1B


amplifiers, transmission loss of optical fibers…..

Decibel Exercise

• Exercise: – Express a power of 50 mW in decibels relative to 1 watt.

10 l (50 10 3) 13 dB10 log10 (50 x 10-3) = - 13 dB

• Exercise:• Exercise: – Express a power of 100 mW in decibels relative to 2 mW.

10 log10 (50) = 17 dB. g10 ( )

• dBm to express absolute values of power relative to a milliwatt. – dBm = 10 log10 (power in milliwatts / 1 milliwatt)

100 mW = 20 dBm


– 100 mW = 20 dBm– 10 mW = 10 dBm

Decibel (dB) as voltage or current ratios

22 IVP ∝∝1

2

1 Vlog20Vlog10 =

ale)20dB

22 Vlog20

Vlog10 =

g 10

sca

1

2

1 Ilog20Ilog10 =

or I

( lo

1B22 I

log20I

log10

V

o1B


Note: 1 Bel of voltage yields 2 Bels of Power

Some useful dB values to memorize

|H(f)| |H(f)|dB

100 4010 201 01 0

0.1 -200 01 400.01 -40

2 6√2 3√2 3

1/√2 -35 (=10/2) 14 (=20-6)


( ) ( )

Bode PlotsA Bode plot is a straight line approximation of H(ω)

• Plot of transfer function magnitude vs. frequency

i i th 20 l f th it d f th t f– y-axis is the 20•log of the magnitude of the transfer function in dB and x-axis is ω

– x -axis is ω or f in log scalex axis is ω or f in log scale

dB

log(f) or log(ω)


Bode Plots• Plot of transfer function phase vs. frequency

– y-axis is the phase of transfer function in degrees (li l )(linear scale)

– x -axis is ω or f in log scale

degrees

log(f) or log(ω)


Decade , Octave and Log Scale

Log10 scale

How about the Log ω axis ?

Shifted by Log (2 ) or 0 798Shifted by Log10(2π) or 0.798

Note: Log x•y = Log x + Log y !!ω =2πf (rad/sec)


g y g g y

Frequency Response

• The shape of the frequency response of the complex ratio of phasors VOUT/VIN is a convenient means of classifying a circuit behavior and identifying keyclassifying a circuit behavior and identifying key parameters.

OUT

VV

GainBreak point OUT

VV

G i

Break point

Low Pass

INV Gain INV

High Pass

Gain

Frequency

Low Pass

Frequency

High Pass


*These are log ratio vs log frequency plots

A Quick Quiz: High Pass or Low Pass?


Example: Low Pass CircuitR2

++A = 100

R1 = 100,000 Ω

−

+AVT

R1+

VTVOUT

CVIN R2 = 1000 Ω

C = 10 uF

OUT

VV)(HFunctionTransfer =ω

C = 10 uF

INV

cOUT AZ=

V/100C/1i

]/[j1A)(H

Bωω+=ω

AA)jwC/1(AV

cRIN ZZ +V s/rad100CR/1with 2B ==ω


)]CR/1/(j1[A

)CRj1(A

)Cj/1R)jwC/1(A

VV

222IN

OUT

ω+=

ω+=

ω+=

First Order Bode Magnitude Plot

• The break frequency ωΒ describes the frequency where the trends on the Bode plot are changed

12/12

B ])/(1[1log20Alog20)(Hlog20ωω+

+=ω

dB dB

40 dB

0 dBωB ωB

0 dB

-20dB/dec


Bode Magnitude Plot – First-order Low-Pass Filter1

dB2/12

B ])/(1[1log20Alog20)(Hlog20ωω+

+=ω

10010040 ωB = 100A

2100

|1|100

=+ j

0

20Actual value =

0

-2010 100 10001 Radian

Frequency

logω10000


Slope = -20dB/dec or -6dB/octave for ω>> ωB

Bode Phase Plot – First-order Low-pass Filter

180)/(tan

]/[j1A)(H B

1

B

ωω=ωω+

∠=ω∠ −

90

Phas

e s/rad100CR/1with 2B ==ω

-90

010 100 10001 Radian

Frequency (log scale)

-180

4545001000100 ==∠

=∠ PhasePhase

Actual value at ωB is -45o


45450452

|1|

−=−=∠

=+

Phasej

Phase

Bode Plot: Actual versus Straight Line Approximation

First Order Low-Pass Filter

Magnitude Phase


Bode Plot of a High-Pass Filter

( )( )

cout

/j1/j

RLjLj

VV)(H ωω

=ω

==ω LR/=ω( )cs /j1RLjV)(

ωω++ω LRc /ωPole: will cause decrease by

20db/dec after ω


20db/dec after ωc=> so magnitude is flat after ωc

Bode Magnitude Plot of a High-Pass Filter

( ) ( ) |/j1

1|log20|/j|log20|)(H|log20c

c ωω++ωω=ω


Bode Phase Plot of a High-Pass Filter

( ) 1/j

)(H

∠+ωω∠=

ω∠

( ) ( )

( )o

cc

190

/j1/j

∠+=

ωω+∠+ωω∠=

( )c/j190

ωω+∠+=


Poles and Zeros

• The transfer function H(ω) = M(ω) ejφ(ω)

can be written as:H(ω) = A1(ω) A2(ω) … An(ω)H(ω) A1(ω) A2(ω) … An(ω)

Zero

( ) ( )

⋅ωω+⋅=ωω+

•=ω z 1/j110)/(j110)(H

ZeroExample

( ) ( )

ωω+

⋅ωω+⋅=ωω+

•=ωp

zp /j1

/j110)/(j1

10)(H


Pole

Poles and Zeros

• With H(ω) = A1(ω) A2(ω) … An(ω)

MagnitudeM [dB] = 20 log|A1| + 20 log|A2| + … + 20 log|An|

= A [dB] + A [dB] + + A [dB]= A1 [dB] + A2 [dB] + … + An [dB]

Phaseφ(ω) = φA1(ω) + φA2(ω) + … + φAn(ω)

Thi t B d l t ( it d d


• This means we can generate Bode plot (magnitude and phase) by adding the poles and zeros!

Poles And Zero Exercise

• Generate the Magnitude Bode Plots of H(ω).

( )( )ω+ω+ω

j1000j10010)(H( )( )( )( )ω+ω+

=ωj10j10

)(H 4


( )( )Poles And Zero Exercise

( )( )( )( )

( )( )4

1000/j1100/j1100010010j10j10

j1000j10010)(H

ω+ω+••ω+ω+

ω+ω+=ω

( )( )( )( )

( )( )44

1000/j1100/j11010/j110/j11010

1000/j1100/j1100010010

ω+ω+ω+ω+•

ω+ω+••=

( )( )( )( )410/j110/j1

jjω+ω+

=


A More Challenging Bode Plot Example

( ) ( )( )50/j14000j

5/j1400H2

ω+•ω•ω+•

=ω

Zeros: Two zeros at 5 rad/s ( a double zero)

( )50/j14000j ω+•ω•

( )Poles: one pole at 0 rad/s and one at 50 rad/s;

1 |400/j4000| l t1. |400/j4000| plotRatio = 1/10 => -20dB constant for all ω

2. |1/ω| plot| | plog 0 does not show up in log scale ( -∞ !!! )However we know magnitude is decreasing at -20 dB/dec and equals -20 dB at 1 rad/s. We can now


dB/dec and equals 20 dB at 1 rad/s. We can now draw the 1/ω plot around ω =1

A More Challenging Bode Plot Example

( ) ( )( )50/j14000j

5/j1400H2ω+•

=ω( ) ( )50/j14000j ω+•ω•

3. |(1+jω/5)2| plot0dB below 5 rad/sslope= +40 dB/dec beyond 5 rad/sslope= +40 dB/dec beyond 5 rad/s

4. |1/(1+jω/50)| plot0dB b l 50 d/0dB below 50 rad/sslope= -20 dB/dec beyond 50 rad/s


Magnitude

3

11 42


Phase

( ) ( ) ( )50/j115/j1

j1

4000400H 2

ω+∠+ω+∠+

ω∠+∠=ω∠

( ) ( )50/j115/j12)90(0 o

ω+∠+ω+∠+−+=

5 10 20 50050 5000


Bode Plot: Quick Summary

In addition, we need to checkneed to check H(ω) as ω -> 0 and H(ω) as ω -( )> ∞.


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EE40 Lec 12 Transfer Function Bode Plots Filters Transfer Function

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