Page 1

Chapter 8

Frequency Response Methods

Review of a simple feedback system:

α( )sR ( )sY+

_

Process

Then ( ) ( )sRsY ⋅+

=α

α1

for α larger ( 1>>α ) ( ) ( ) ( )sRsRsY =⋅≅αα

for α small ( 11

>>α

) ( ) ( ) ( )sRsRsY ⋅=⋅≅ αα1

Now consider that α and R(s) are functions of frequency … ( )jwα and ( )jwsR = .

If α varies in magnitude based on frequency, ( )jwα , the system output would either pass the input (when α is large) or significantly attenuate the input (when α is small) based on frequency.

If we select inputs that are pure tones in frequency, we should be able to observe what ( )jwα is at each frequency. This develops a frequency response curve … the power spectrum plot of a filter or process. The output may have both magnitude and phase components that vary in frequency.

While a spectrum analyzer will output a signals magnitude spectrum, you can use a network analyzer to plot both the magnitude and phase.

Welcome to the world of audio and radio frequencies.

Page 2

Using a sine or cosine waveform as an input.

Laplace Transform Notes:

( ) ( )wtAtr sin⋅= with transform ( )22 ws

wAsR

+⋅=

( ) ( )wtAtr cos⋅= with transform ( )22

2

wss

AsR+

⋅=

Using the a generalized transfer function relationship …

( ) ( ) ( ) ( ) ( )( )wtAthtrthty sin⋅∗=∗=

( ) ( ) ( ) ( )( ) 22 ws

wA

ssN

KsRsTsY+

⋅⋅∆

⋅=⋅=

( ) ( ) ( ) ( )( ) ( ) ( ) 22

21

21

wsw

pspspszszszs

KAsYN

N

+⋅

+⋅⋅+⋅++⋅⋅+⋅+

⋅⋅=LL

( )22

2

2

1

1

wss

psk

psk

psk

sYN

N

++⋅

++

+++

++

=βα

K

( )

++⋅

+⋅++⋅+⋅= −−−−22

121

21

wss

Lekekekty tpN

tptp Nβα

K

For large t ( )

++⋅

= −22

1arg ws

sLty el

βα

( )

+⋅+

+⋅= −

22221

arg wsw

wwss

Lty elβ

α

( ) ( ) ( )wtw

wtty el sincosarg ⋅+⋅=β

α

Page 3

Simplify using trig. Identities as follows:

Given ( ) ( ) ( ) ( ) ( )bababa sincoscossinsin ⋅+⋅=+

( ) ( ) ( )

⋅

+

+⋅

+

⋅

+= wt

w

wwt

w

wty el sincos2

22

2

22

argβα

β

βα

αβα

( ) ( ) ( ) ( ) ( )( )wtwtwty el sincoscossin2

2arg ⋅+⋅⋅

+= θθβα

where

⋅=

=

βα

βα

θw

w

arctanarctan

Finally ( ) ( )θβα +⋅+⋅= wtww

ty el sin1 222

arg

for ( ) ( )wtAtr sin⋅=

As a result, an input sine wave has been modified in gain and phase once the steady state condition has been achieved.

Gain A

wwwgain

2221

)(βα +⋅

=

Phase θ=)(wphase

Page 4

Computing the frequency response of a system

From the previous derivation, the transfer function of the sine wave will result in a phase-shifted sine wave with a new magnitude (i.e. gain and phase changes). If this is true, then when we go back to the transfer function at the oscillation frequency:

( ) ( ) ( ) ( ) ( )22 ws

wewKsRsTsY wj

Gain +⋅⋅=⋅= θ

or ( )

( )( ) ( ) ( )jwj

Gain ejwKjwTjwRsY θ⋅==

Therefore, ( ) ( ) ( ) ( ) ( )wXjwRejwKjwT jwjGain ⋅+=⋅= θ

or ( ) ( ) ( ) ( ) ( ) ( )( )

⋅⋅+=⋅+= wR

wXajwXwRwXjwRjwT tanexp22

This relationship describes the gain and phase changes to an input sine wave due to the system at different frequencies.

For the frequency response of the system, we equate s with jw (s=jw) and evaluate the result!

The frequency response (in jw) is plotted as a gain and a magnitude using a Bode Plot.

Page 5

Network Analyzers

From: Agilent, Network, Spectrum, and Impedance Evaluation of Electronic Circuits and Components, Application Note 1308-1.

Page 6

From: Agilent, Network, Spectrum, and Impedance Evaluation of Electronic Circuits and Components, Application Note 1308-1.

Page 7

Example:

( )( )

( ) ( )( ) ( ) sRC

sCRsC

sZsZsZ

sTsVsV

+=

+=

+==

11

1

1

21

2

1

2

( )( ) ( )

( )( )( )wRCaj

wRCjwRCjwT

jwVjwV

tanexp1

11

12

1

2 ⋅−⋅+

=+

==

num=1;den=[1 1];freqs(num,den);

Notice: (1) The magnitude is flat until it gets near the critical “frequency” of the pole and then drops off linearly (at 10 dB per decade). (2) The phase transitions from 0 to –90 degrees with -45 degrees at the critical frequency..

Page 8

Generalized Frequency Response and the Bode Plot

( ) ( ) ( )( )jwTjwT

ws

ws

ps

s

zs

KsTN

n nn

nM

m m

R

Q

i i ∠⋅=

+

⋅⋅+⋅

+⋅

+

⋅=

∏∏

∏

==

= exp

211

1

1

2

1

1

ζ

( )( )

( ) ( )( )jwTjwT

wjw

wjw

pjw

jw

zjw

KjwTN

n nn

nM

m m

R

Q

i i ∠⋅=

+

⋅⋅+⋅

+⋅

+

⋅=

∏∏

∏

==

= exp

211

1

1

2

1

1

ζ

( )

∏∏

∏

==

=

⋅⋅+

−⋅

+⋅

+

⋅=N

n n

n

n

M

m m

R

Q

i i

ws

ww

pw

w

zw

KjwT

1

222

1

2

1

2

211

1

ζ

( )

+

⋅⋅+∠−

+∠−∠⋅−

+∠+∠=∠ ∑∑∑

===

2

111

2111nn

nN

nm

M

mi

Q

i ws

ws

ps

jwRzs

KjwTζ

Signal level description in decibels (dB):

A measure of signal power gain defined as

⋅=

in

out

PP

dBinGain 10log10

Equivalently for voltages

⋅=

in

out

V

VdBinGain 10log20

For power defined as ( )

out

outout R

VP

2

= and ( )

in

inin R

VP

2

=

The power and voltage equations have the same result when inout RR = In most RF and audio system, this is always case and part of component designs.

Page 9

Revisiting the magnitude

Since ( ) ( ) ( ) ( )dcbadcba

1010101010 log20log20log20log20log20 ⋅−⋅−⋅+⋅+=

⋅⋅

⋅

( )

∏∏

∏

==

=

⋅⋅+

−⋅

+⋅

+

⋅=N

n n

n

n

M

m m

R

Q

i i

ws

ww

pw

w

zw

KjwT

1

222

1

2

1

2

211

1

ζ

( )( ) ( )

( )

∑

∑

∑

∑

=

=

=

=

⋅⋅+

−⋅−

+⋅−

⋅−

+⋅+

⋅+=⋅=

N

n n

n

n

M

m m

M

m

R

Q

i i

ws

ww

pw

w

zw

KjwTdBinLogMag

1

222

10

1

2

10

110

1

2

10

1010

21log20

1log20

log20

1log20

log20log20

ζ

Simplifying leads to

( )( ) ( )

( )

∑

∑

∑

∑

=

=

=

=

⋅⋅+

−⋅−

+⋅−

⋅⋅−

+⋅+

⋅+=⋅=

N

n n

n

n

M

m m

M

m

Q

i i

ww

ww

pw

wR

zw

KjwTdBinLogMag

1

222

10

1

2

10

110

1

2

10

1010

21log10

1log10

log20

1log10

log20log20

ζ

Page 10

For the phase angle relationship

( ) ( ) ( )( )jwTjwT

ws

ws

ps

s

zs

KsTN

n nn

nM

m m

R

Q

i i ∠⋅=

+

⋅⋅+⋅

+⋅

+

⋅=

∏∏

∏

==

= exp

211

1

1

2

1

1

ζ

( )

+

⋅⋅+∠−

+∠−∠⋅−

+∠+∠=∠ ∑∑∑

===

2

111

2111nn

nN

nm

M

mi

Q

i ws

ws

ps

jwRzs

KjwTζ

using ( )

=⋅+∠

ab

abja tan

Simplifies to

( ) ∑∑∑===

−

⋅⋅

−

−⋅−

+∠=∠

N

n

n

n

nM

m m

Q

i i

ww

ww

apw

aRzw

aKjwT1

211

1

2tantan

2tan

ζπ

Note 9th ed., fix Eq. 8.28, p. 416 in the textbook It is wrong!!!!

Note 10th ed., fix Eq. 8.28, p. 442 in the textbook It is wrong!!!!

Page 11

Summary:

Contributions based on terms Magnitude and Phase angle

K constant: ( )KLogMag 10log20 ⋅= and 0=∠

Zeros

+

izs1 :

+⋅=

2

10 1log20iz

wLogMag and

=∠

izw

a tan

Poles

+

ips1

1 :

+⋅−=

2

10 1log20ip

wLogMag and

−=∠

ipw

a tan

2nd Order

+

⋅⋅+

2

21

1

nn

n

ws

wsζ

:

⋅⋅+

+⋅−=

222

10 21log20n

n

n ws

ww

LogMagζ

and

−

⋅⋅

−=∠ 2

1

2tan

n

n

n

ww

ww

a

ζ

Page 12

Asymptotic Performance (Back-of-the-envelope Bode Plots)

Magnitude approximations

+⋅−=

2

10 1log20ip

wLogMag

Let ipw *1.0= ( ) ( ) 043.001.1log101.01log20 102

10 −=⋅−=

+⋅−=LogMag

Let ipw = ( ) ( ) 010.32log1011log20 102

10 −=⋅−=

+⋅−=LogMag

Let ipw *10= ( ) ( ) 043.20101log10101log20 102

10 −=⋅−=

+⋅−=LogMag

In general a factor of 10 change in w causes a 20 dB change in the magnitude (20 dB/decade).

Phase approximations

−=∠

ipw

a tan

Let ipw *1.0= ( ) π⋅=−=−=−=∠ -0.032rad 0.100deg71.51.0tana

Let ipw = ( ) rad 4

deg0.451tanπ

−=−=−=∠ a

Let ipw *10= ( ) π⋅−=−=−=−=∠ 468.0rad47.1deg0.29.8410tana

In general, a 45 deg. change in the previous decade followed by another in the decade following.

Page 13

Asymptotic Curves

Page 14

General knowledge for a frequency response

From music, what is an octave? An octave is a doubling in frequency.

Using w

w⋅2 02.6

2log20 10 −=

⋅

⋅−=w

wLogMag

Therefore we have a factor of 6 dB per octave, which is the same as 20 dB per decade!

Making a Bode Plot (by hand)

(1) Format the transfer function, ( )sT

(2) Identify the components: constant, zeros, terms in Rs , poles, and/or 2nd order poles.

(3) Construct the asymptotic gain. Start at a reasonable frequency such as w=1 (constant gain plus any resonance), draw the asymptote to the first resonance, determine the asymptote change required, draw the asymptote to the next resonance, determine the asymptote change required, and continue

(4) Construct the asymptotic phase. Start at w=1 (constant gain plus any resonance), draw the asymptote to 1/10th the first resonance, determine the asymptote change required, draw the asymptote to the 10x the first resonance, determine the asymptote change required, and repeat. If there is not a factor of 100 between resonances, sketch in the transition regions and add them as you go from w=1 to infinity.

For examples of 15 typical transfer functions, see Table 8.5, (9th ed. p. 448-450 or 10th ed. P. 474-476).

Page 15

Make a Bode Plot Text book example: (Section 8.3)

( )

⋅+⋅⋅⋅+⋅

⋅+⋅

⋅+⋅

=2

50503.021

21

1015

wj

wj

wjjw

wj

jwT

Constant: 5, Zeros: 10=w and Poles: 50,50,2,0=w

Determine the individual elements asymptotes

Sketch in the composite results remembering each of the zeros and poles as you pass the resonant frequency. (Start somewhere, such as at w=1, for gain location, follow dB/decade)

Page 16

For the phase sketch in the asymptotes and add them as you go

( )

⋅+⋅⋅⋅+⋅

⋅+⋅

⋅+⋅

=2

50503.021

21

1015

wj

wj

wjjw

wj

jwT

Zeros: 10=w and Poles: 50,50,2,0=w

Sketch in the composite results remembering each of the zeros and poles as you pass the resonant frequency. (Start at w=1 for gain location, follow dB/decade)

The actual Bode plot is:

Page 17

Matlab: num=5*[1/10 1]; den1=conv([1 0],[1/2 1]); den2=([(1/50)^2 (0.6/50) 1]); den=conv(den1,den2); sys=tf(num,den) figure(1) bode(sys) grid

Transfer function: 0.5 s + 5 --------------------------------------- 0.0002 s^4 + 0.0064 s^3 + 0.512 s^2 + s

-150

-100

-50

0

50

Mag

nitu

de (

dB)

10-1

100

101

102

103

-270

-225

-180

-135

-90

Pha

se (

deg)

Bode Diagram

Frequency (rad/sec)

Page 18

wref=logspace(-1,2,1000); [magw,phasew]=bode(sys,wref); magw=squeeze(magw)'; phasew=squeeze(phasew)'; magwdB=dBv(magw); figure(2) semilogx(wref,magwdB) title('Bode Plot Power') xlabel('Frequency (rad/sec)') ylabel('Magnitude (dB)'); grid figure(3) semilogx(wref,phasew) title('Bode Plot Phase') xlabel('Frequency (rad/sec)') ylabel('Phase (degrees)'); grid

10-1

100

101

102

-60

-50

-40

-30

-20

-10

0

10

20

30

40Bode Plot Power

Frequency (rad/sec)

Mag

nitu

de (

dB)

10-1

100

101

102

-260

-240

-220

-200

-180

-160

-140

-120

-100

-80Bode Plot Phase

Frequency (rad/sec)

Pha

se (

degr

ees)

Page 19

Improving Upon an Asymptotic Bode Plot

2nd Order Responses Near The Resonance Frequency.

( )22

21

1

21

1

−

⋅⋅⋅+

=

+

⋅⋅+

==

nn

n

nn

n

ww

ww

jws

ws

jwsTζζ

Magnitude:

⋅⋅+

−⋅−=

222

10 21log10n

n

n ww

ww

LogMagζ

Let nww =

Magnitude: ( )210 4log10 nLogMag ζ⋅⋅−=

Specific Points for nww = :

For 1=nζ : ( ) =⋅−= 4log10 10LogMag -6.02

For 5.025.0 ==nζ : ( ) 01log10 10 =⋅−=LogMag

For 25.0=nζ : ( ) 02.625.0log10 10 +=⋅−=LogMag

For 10.0=nζ : ( ) 3.98104.0log10 10 +=⋅−=LogMag

For 05.0=nζ : ( ) 00.2001.0log10 10 +=⋅−=LogMag

Page 20

What is the peak frequency of the magnitude? Notice how it is moving to the left as ζ decreases.

For nw

wu = ( ) 221

1uuj

uTn −⋅⋅⋅+

=ζ

The maximum is at ( ) ( )

⋅⋅+−∂∂

=222 21

10

uuunζ

( ) ( )( )

( ) ( ) ( )( )uuuuu

n

n

⋅⋅⋅+⋅−⋅−⋅⋅⋅⋅+−

⋅−= 22

23

22222212

21

121

0 ζζ

( ) ( )( ) 124210 2222 −⋅+=⋅⋅+⋅−⋅−= nn uuuu ζζ

221 nu ζ⋅−= or 221 nnww ζ⋅−⋅=

Condition: 021 2 >⋅− nζ or 707.02

10 =<< nζ

Note: for 707.02

1=≥nζ , there is no peak!

The peak at this point is

( ) ( ) ( )( ) ( ) ( )22

222222212211

1

21

1

nnnn uuMag

ζζζζ ⋅−⋅⋅+⋅−−=

⋅⋅+−=

2424 12

1

844

1

nnnnn

Magζζζζζ −⋅

=⋅−⋅+⋅

=

Page 21

Figure 8.11 showing the maximum of the frequency response and the percent shift in the frequency due to the damping factor, zeta.

Page 22

2nd Order Responses Near The Resonance Frequency.

Phase:

−

⋅⋅

−=∠ 2

1

2tan

n

n

n

ww

ww

a

ζ

Phase: deg902

tan −=

±⋅

−=∠εζ na

How steep is the slope at nww = ?

Compute the Phase Slope at nww = :

Using ( )[ ]xy

yya

x ∂∂

⋅+

−=−∂∂

211

tan

[ ]

−

⋅−⋅

⋅⋅

−

−

⋅⋅

−

⋅⋅+

−=∠

∂∂

22

2

222

1

22

1

2

121

1

n

nn

n

n

n

n

nn

n

ww

ww

ww

ww

w

ww

www

ζζ

ζ

[ ][ ] nnn

n

n

n

n

n

nn

n

nn

n

ww

ww

ww

ww

ww

ww

w

w ζζ

ζ

ζ

ζζ

12

4

21

412

2222

2

2

−=⋅

⋅⋅−=

⋅⋅+

−

⋅

⋅⋅+

−⋅⋅

−=∠∂∂

Page 23

Alternate derivation

For our case ( )[ ]wx

xy

yya

w ∂∂

⋅∂∂

⋅+

−=−∂∂

211

tan

[ ] [ ]( )( )

[ ]

−

⋅−⋅⋅⋅−

−⋅

⋅−⋅⋅+

−=∠

∂∂

22222 1

2212

121

1

u

uuuuuu

nn

n

ζζ

ζ

[ ] ( )[ ] [ ] [ ] nn

n

n

nn

uu

uuu ζζ

ζ

ζ

ζζ 12

4

21

4122222

22

−=⋅

⋅−=

⋅⋅+−

⋅⋅+−⋅⋅−=∠

∂∂

Therefore [ ]nnnnn ww

www

uw ⋅

−=

∂∂

⋅−=∂∂

⋅−=∠∂∂

ζζζ111

In what frequency region is this important?

This around the natural resonance … the curve is the same, except for near nw

Page 24

Matlab % % close all clear all wref=logspace(-1,1,1000)'; num=1; ii=0; for zeta=(.1:.1:1) ii=ii+1; den=[1 2*zeta 1]; sys=tf(num,den); [magii,phaseii]=bode(sys,wref); magw=squeeze(magii); phasew(:,ii)=squeeze(phaseii); magwdB(:,ii)=20*log10(magw); end figure(1) semilogx(wref,magwdB) title('Bode Plot Power') xlabel('Frequency (rad/sec)') ylabel('Magnitude (dB)'); grid axis([10^-1 10^1 -20 20]) figure(2) semilogx(wref,phasew) title('Bode Plot Phase') xlabel('Frequency (rad/sec)') ylabel('Phase (degrees)'); grid axis([10^-1 10^1 -180 0]) pause figure(1) axis([10^-1 10^1 -5 5])

Results in: …

Page 25

Zeta ranges from 0.1 to 1.0 in steps of 0.1 (blue, green, red, etc.)

Notice that the zeta=1 plot is about –6dB for later …

10-1

100

101-20

-15

-10

-5

0

5

10

15

20Bode Plot Power

Frequency (rad/sec)

Mag

nitu

de (d

B)

The values of zeta that are within 3dB of the desired resonance frequency are

0.4, 0.5, 0.6 and 0.7

Filters must have complex roots or else the roll off before you get to the –3dB point!

10-1

100

101-5

-4

-3

-2

-1

0

1

2

3

4

5Bode Plot Power

Frequency (rad/sec)

Mag

nitu

de (d

B)

At the resonance frequency the phase is changing dramatically.

This could be a problem for a frequency-modulated signal!

(additional changes in phase in time? But, frequency is a change in phase in time!)

10-1

100

101-180

-160

-140

-120

-100

-80

-60

-40

-20

0Bode Plot Phase

Frequency (rad/sec)

Pha

se (d

egre

es)

Page 26

The graphical derivation of the 2nd order system (or any function using the s-plane):

( )( ) ( )*

11

2

22

2

2 ssss

w

wsws

wsT n

nnn

n

+⋅+=

+⋅⋅⋅+=

ζ

( )22*11 , nnnnn wwjwss ⋅−⋅±⋅−= ζζ

( )( ) ( ) ( ) ( )*

11

2

22

2

2 sjwsjw

w

wjwwjw

wjwT n

nnn

n

+⋅+=

+⋅⋅⋅+=

ζ

( )( ) ( ) ( ) ( )*

11

2

22

2

2 sjwsjw

w

wjwwjw

wjwT n

nnn

n

+⋅+=

+⋅⋅⋅+=

ζ

Or ( ) ( )*11

exp*11

2

ss

n

jjsjwsjw

wjwT

θθ +⋅+⋅+=

Also note that:

( ) ( )

⋅−⋅−⋅−⋅

⋅−⋅+⋅−=⋅ 2222*

11 nnnnnnnnnn wwjwwwjwss ζζζζ

( ) ( )( ) ( ) ( )( )

⋅−+⋅⋅

⋅−+⋅=⋅ 222222*

11 nnnnnnnnnn wwwwwwss ζζζζ

222*11 nnn wwwss =

⋅

=⋅

Therefore ( ) ( )*11

exp*11

*11

ss jjsjwsjw

ssjwT

θθ +⋅+⋅+

⋅=

Using graphical computations to define the gain and phase:

Page 27

Using ( ) ( )*11

exp*11

*11

ss jjsjwsjw

ssjwT

θθ +⋅+⋅+

⋅=

0⋅= jjw

220 RXjjwj −⋅<<⋅

XjjwXRj ⋅<<−⋅ 22

jwXj <⋅

Page 28

P2.35 The suspension system for one wheel of an old-fashioned pickup truck is illustrated in Fig. P2.35. The mass of the vehicle is m1 and the mass of the wheel is m2. The suspension spring has a spring constant k1, and the tire has a spring constant k2.The damping constant of the shock absorber is b. Obtain the transfer function Y1(s)/X(s), which represents the vehicle response to bumps in the road.

x(t)

M1

M2

y1(t)

y2(t)

k1

b

k2

( ) ( ) 021121

21

2

1 =−+−

+ yykdt

yydb

dtyd

M

( ) ( ) ( ) 02212112

22

2

2 =−+−+−

+ xykyykdt

yydb

dtyd

M

xky

y

kkbssMkbskbskbssM

⋅

=

⋅

+++−−−−++

22

1

212

21

112

1 0

xkkkbssMkbs

kbskbssMyy

⋅

⋅

+++−−−−++

=

−

2

1

212

21

112

1

2

1 0

( ) ( ) ( )x

kkbskbssMkkbssM

kbssMkbskbskkbssM

yy

⋅

⋅

+−++⋅+++

+++++++

=

22

112

1212

2

12

11

1212

2

2

1 0

Page 29

( ) ( ) ( )( ) ( ) ( )2

112

1212

2

121

kbskbssMkkbssM

sXkbsksY

+−++⋅+++

⋅+⋅=

( ) ( ) ( )( ) ( ) ( ) 212122111

221

321

412

1 kkbksMkMkMksbMbMsMMssXkbsk

sY+⋅+++⋅++⋅+⋅

⋅+⋅=

( )( )

( )sX

kb

skM

kM

kM

skkbMbM

skkMM

s

skb

sY ⋅+⋅+

++⋅+

+⋅+⋅

+

=1

1

12

2

1

1

2

12

21

213

21

214

11

( ) ( )sX

ws

ws

ps

ps

zs

sY

nn

⋅

+

⋅⋅+⋅

+⋅

+

+

=2

21

1

2111

1

ζ

( ) ( )sXssss

s

sY ⋅

+

⋅⋅+⋅

+⋅

+

+

=21

1121

1001

000,11

101

ζ

0dB/decade

-40dB/decade

-20dB/decade

-40dB/decade

-60dB/decade

w-axis

Lo

gM

ag-a

xis

If the other roots don’t come into play soon enough, the car may bounce at 1 Hz!

Page 30

Twin-T Network

( ) ( ) ( )

⋅+

⋅=

⋅⋅++⋅

RsV

RsVCs

RRsV oin

112

112

( ) ( ) ( ) ( )sVsVRCssV oin +=⋅⋅⋅+⋅ 222

( ) ( ) ( ) CssVCssVCsCsR

sV oin ⋅⋅+⋅⋅=

⋅+⋅+⋅

23

( ) ( ) ( )sVsVRCs

RCssV oin +=

⋅⋅⋅⋅⋅+

⋅22

3

( ) ( ) ( ) CssVR

sVCsR

sVo ⋅⋅+⋅=

⋅+⋅ 32

11

( ) ( ) ( ) ( )( ) ( ) ( )( ) RCsRCs

RCssVsV

RCssVsVRCssV oinoino ⋅⋅⋅

⋅⋅⋅+⋅⋅

⋅++⋅⋅⋅+

⋅+=⋅⋅+⋅2222

11

( ) ( ) ( ) ( )RCs

RCssV

RCsRCs

RCssV ino ⋅⋅⋅+⋅⋅+

⋅=

⋅⋅⋅+⋅⋅+

−⋅⋅+⋅22

122

11

22

( )( )

( )( ) ( ) ( )( )2

2

11221

RCsRCsRCsRCs

sVsV

in

o

⋅⋅+−⋅⋅+⋅⋅⋅⋅+⋅⋅+

=

Laplace ( )( )

( )( )2

2

411

RCsRCsRCs

sVsV

in

o

⋅⋅+⋅⋅⋅+⋅⋅+

=

Frequency Response

( )( )

( ) ( )( ) ( )

( )( ) RCwjRCw

RCw

RCwjRCwj

RCwjjwVjwV

in

o

⋅⋅⋅⋅+⋅⋅−

⋅⋅−=

⋅⋅⋅+⋅⋅⋅⋅+

⋅⋅⋅+=

41

1

41

12

2

22

22

Page 31

num=[1 0 1]; den=[1 4 1]; freqs(num,den) or

num=[1 0 1]; den=[1 4 1]; bode(tf(num,den))

MATLAB Bode Plots:

Use freqs if you have the numerator (num) and denominator (den). If you want to derive the numerator and denominator of a transfer function use [sysnum,sysden]=tfdata(sys,’v’).

Use bode if you have a transfer function ( sys=tf(num,den) ).

Page 32

Estimating the system transfer function based on the frequency response.

There are many known elements about the shape of a transfer function. For example, 20dB/decade defines a first order pole or zero. Therefore, where we see +/-20dB/decade the magnitude response has one more zero/pole “active” at that frequency.

Page 33

Starting with the magnitude plot

As the curve changes from flat to +/-20dB/decade the frequency of the transition is +/- 3dB from the expected curve.

Using this, there is a pole at 300 that causes the system to transition to –20dB/decade.

Therefore

+

3001

1s

This stops and turns around at 2450 becoming +20dB/decade

Therefore 1

2450245021

2

+⋅⋅+

ssζ

If you want to make a guess at the damping factor, measure the overshoot from the corner of the asymptotes. If we say 10 dB, then the curve from Fig. 8.10 approximates that 15.0=ζ .

Therefore 1

2450245015.021

2

+⋅⋅+

ss

The rising signa l returns to a constant coefficient moving from +20dB/decade at the -3dB point and frequency 20,000.

Therefore

+000,20

1

1s

Page 34

The composite transfer function is then:

( )

+

⋅

+⋅⋅+

⋅

+

=

000,201

11

2450245015.021

3001

1

2

s

ss

ssT

Verifying the Equation

We can verify the proposed asymptotic changes be checking the phase plot for values of –90, -45, 0, 45, and 90 degrees.

For a pole we should have –45 degrees at the pole.

For a zero we should have +45 degrees.

For a second order, we should have +/-90 degrees depending on if they are zeros or poles.

Summing from frequency nearer 0 going toward infinity for our estimate would mean that the first frequency is at –45 degrees (going to –90 deg.), the second is at (-90+90=) 0 degrees (going to –90+180=+90 deg.), and the third is at (-90+180-45=) 45 degrees go to (-90+180-90=0 deg). The frequencies in the curve verify those selected.

Page 35

The composite transfer function is verified as:

( )

+

⋅

+⋅⋅+

⋅

+

=

000,201

11

2450245015.021

3001

1

2

s

ss

ssT

or in a compact form as:

( )

+⋅

+

+⋅⋅+

=

000,201

3001

2450245015.021

2

ss

ss

sT

The MATLAB plot is

Note: This is an example of the Twin-T response previously described!

Page 36

Minimum Phase 9th ed. - p. 424-426 or 10th ed. - p. 450-452

Start with a basic transfer function.

( ) ( ) ( )( )jwTjwT

ps

zs

KsTM

m m

Q

i i ∠⋅=

+

+

⋅=

∏

∏

=

= exp1

1

1

11

What if: ( ) ( ) ( )( )jwTjwT

ps

zs

KsTM

m m

Q

i i ∠⋅=

+

+−

⋅=

∏

∏

=

= exp1

1

1

12

For example, compare:

( )

+⋅

+

+

⋅=

21

11

11

1

ps

ps

zs

KsT and ( )

+⋅

+

+−

⋅=

21

2

11

1

ps

ps

zs

KsT i

The magnitude is:

( )[ ]2

2

2

1

2

11

11

1

+⋅

+

+

⋅=

pw

pw

zw

KjwTmag and ( )2

2

2

1

2

12

11

1

+⋅

+

+

⋅=

pw

pw

zw

KsT

The phase is:

( )[ ]

∠−

∠−

∠+∠=∠

2111 tantantan

pw

apw

azw

aKjwT

and ( )

∠−

∠−

−∠+∠=∠

2112 tantantan

pw

apw

azw

aKsT

or ( )

∠−

∠−

∠−+∠=∠

2112 tantantan

pw

apw

azw

aKsT π

Page 37

Matlab Example

Let z=+/-10 and p1, p2=-2 –5.

Using the pole-zero map, estimate the phases. Start at w=0 and move toward w=inf.

At 0=w ( ) 000001 =−−+=∠ jwT

and ( ) ππ =−−+=∠ 0002 sT

At ∞=w ( )2222

01ππππ

−=−−+=∠ jwT

and ( )2222

02ππππ

−=−−+=∠ sT

They both end up at the same angle, but T2 starts at π! As a result, the phase transition will be significantly larger!

Page 38

Using MATLAB: den=poly([-2 -5])/10 num1=[1 10]/10 num2 = [1 -10]/10 sys1=tf(num1,den) sys2=tf(num2,den) figure(1) pzmap(sys1,sys2) figure(2) bode(sys1,sys2) grid figure(3) step(sys1,-sys2) grid

For this example, ( )jwT1 represents the minimum phase angle curve for the same magnitude output and is called the minimum phase solution.

Definition: The minimum phase solution must have all zeros in the left-half of the s-plane. Therefore, all poles and zeros should be in the LHP!

A non-minimal-phase solution will have zeros in the right half plane .

Page 39

By the way, the step response to these two systems is also different:

Page 40

All Pass Network

The all-pass network is designed to pass all frequencies, but above the resonance, the signal will be phase shifted.

( ) ( )

⋅=

⋅+⋅

+⋅R

sVLCs

CsR

sV in1

11

22

( ) ( )

⋅+⋅

⋅=

⋅+⋅

+⋅LCs

CssV

LCsCs

RsV in 223 11

1

( ) ( )[ ] ( ) ( ) ( ) ( )

⋅+⋅+

⋅+⋅⋅

⋅+⋅

−

⋅+⋅+

⋅+⋅⋅⋅=⋅=−

CRsLCsLCsR

LCsCs

CRsLCsLCsR

RsVsVsVsV inout 2

2

22

2

32 11

1111

( ) ( ) ( )

⋅+⋅+⋅

−

⋅+⋅+

⋅+⋅=⋅

LCsCRsCRs

LCsCRsLCs

sVsV inout 22

2

111

( ) ( )

⋅+⋅+⋅+⋅−

⋅=⋅LCsCRsLCsCRs

sVsV inout 2

2

11

Page 41

The bode plot of the all-pass network is

Page 42

Matlab Example of setting the phase transition … for a=.1:.1:2 num=[1 -a 1] den=[ 1 a 1] sys=tf(num,den) figure(1) pzmap(sys) hold on wref=logspace(-2,2,1000); [magw,phasew]=bode(sys,wref); magw=squeeze(magw)'; phasew=squeeze(phasew)'; magwdB=dBv(magw); figure(2) semilogx(wref,magwdB) title('Bode Plot Power') xlabel('Frequency (rad/sec)') ylabel('Magnitude (dB)'); axis([ 0.0100 100.0000 -0.1000 0.1000]) grid hold on figure(3) semilogx(wref,phasew) title('Bode Plot Phase') xlabel('Frequency (rad/sec)') ylabel('Phase (degrees)'); grid hold on pause end

Page 43

Bandwidth p. 432

For this class, the bandwidth is the frequency at which the “passband” magnitude response has dropped 3 dB.

In doing filter design, the 3 dB bandwidth typically defines the filter passband.

When designing a filter, typically you are also interested in the stopband. The stopband is the bandwidth outside of which the magnitude is below some defined value.

Passband

Dynam

icR

ang

e

0 dB

-3 dB

-40 dBStopband

or

Passband

Dynam

icR

ang

e

0 dB

-3 dB

-40 dBStopband

Page 44

Dorf’s attempt to confuse a simple concept: 2nd order systems

Notice that the –3dB bandwidth value changes based on the damping factor. From the curve, the 3 dB bandwidth is below or at nw for 707.0>nζ while the 3 dB bandwidth frequency is at or above nw for 707.00 << nζ . Dorf Fig. 8.26 shows the actual offset.

This is a nice mathematical demonstration of defining a 3dB bandwidth for a specific case. Painfully, there is so much math that you can miss the important part … 3dB!

Log Magnitude vs. Phase Diagrams

From the Bode Plot we showed curves of the LogMag vs LogFreq and Phase vs. LogFreq. We can also combine these two curves into a LogMag vs. Phase plot.

Page 45

( )

⋅+⋅⋅⋅+⋅

⋅+⋅

⋅+⋅

=2

50503.021

21

1015

wj

wj

wjjw

wj

jwT

The resulting plot is called a Nichols Chart that is addressed in Chap. 9.

den=conv([1 0],conv([.5 1],[(1/50)^2 (0.6/50) 1])) num=5*[1/10 1] sys=tf(num,den) figure(1) [smag,sphase,sw]=bode(sys); smag2=dBv(squeeze(smag)); sphase2=squeeze(sphase); plot(sphase2,smag2) grid figure(2) nichols(sys) grid

A Nichols Chart with grid

Page 46

A Plot of the dBv(Mag) vs Phase

Coming in the next chapter, Gain and Phase Margins to determine stability.

Notice: if the plot were of the open loop response or the characteristic equation elements ( ) ( )jwHjwG ⋅ from

( ) ( ) ( )jwHjwGjw ⋅+=∆ 1

The same gain magnitude and phase requirements that were defined when we studied the root locus must hold here.

Where, the magnitude of 1 is 0 dB and the phase requirement is still –180 degrees.

Once the gain is below 1, it won’t significantly affect the feedback. Therefore, gain greater than 1 will significantly change the closed loop response and “difference error”.

Once the phase changes by 180 degrees, you have positive feedback and not negative feedback.

From the open loop frequency plots, we can define design margins in gain and phase.

Gain margin: the gain when the phase is +/- 180 degrees (about 28 dB above)

Phase margin: the phase when the gain is 0 dB (about 180-132=48 degrees above)

Page 47

Matlab Function Support den=conv([1 0],conv([.5 1],[(1/50)^2 (0.6/50) 1])) num=5*[1/10 1] sys=tf(num,den) figure(1) [smag,sphase,sw]=bode(sys); smag2=dBv(squeeze(smag)); sphase2=squeeze(sphase); plot(sphase2,smag2) grid figure(2) nichols(sys) grid figure(3) margin(sys)

-150

-100

-50

0

50

Mag

nitu

de (

dB)

10-1

100

101

102

103

-270

-225

-180

-135

-90

Pha

se (

deg)

Bode DiagramGm = 28.6 dB (at 47.5 rad/sec) , Pm = 48.6 deg (at 2.94 rad/sec)

Frequency (rad/sec)

Page 48

Spectral Power Responses

For ( ) ( ) ( )( )jwTjwTjwT ∠⋅= exp

What is ( )jwT −

Magnitude (must be the same!):

( )

∏∏

∏

==

=

⋅⋅+

−⋅

+⋅

+

⋅=N

n n

n

n

M

m m

R

Q

i i

ww

ww

pw

w

zw

KjwT

1

222

1

2

1

2

211

1

ζ

Phase:

( ) ∑∑∑===

−

⋅⋅

−

−⋅−

=∠

N

n

n

n

nM

m m

Q

i i

ww

ww

apw

aRzw

ajwT1

211

1

2tantan

2tan

ζπ

( ) ∑∑∑===

−−

⋅⋅−

−

−−⋅+

−=−∠

N

n

n

n

nM

m m

Q

i i

ww

ww

ap

waR

zw

ajwT1

211

1

2tantan

2tan

ζπ

Therefore ( ) ( )jwTjwT =− and ( )( ) ( )( ) 0=∠−+−∠ jwTjwT

Then, what is ( ) ( ) ( ) ( )( ) ( ) ( )( )jwTjwTjwTjwTjwTjwT ∠−⋅⋅∠⋅=−⋅ expexp

or ( ) ( ) ( ) 2jwTjwTjwT =−⋅

This is a power term (notice the square), so we use 10*log to create decibels. It is the same result!! Note, this works for (s,-s) too!!

( ) ( ) ( ) 2sTsTsT =−⋅