Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

2

Lect

ure

3: B

ode

Plot

s

Prof

. Nik

neja

d

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Get

to k

now

you

r log

s!dB

ratio

dBra

tio-2

00.

100

2010

.000

-10

0.31

610

3.16

2-5

0.56

25

1.77

8-3

0.70

83

1.41

3-2

0.79

42

1.25

9-1

0.89

11

1.12

2En

gine

ers a

re v

ery

cons

erva

tive.

A “

mar

gin”

of

3dB

is a

fact

or o

f 2 (p

ower

)!K

now

ing

a fe

w lo

gs b

y m

emor

y ca

n he

lp y

ou

calc

ulat

e lo

gs o

f diff

eren

t rat

ios b

y em

ploy

ing

prop

ertie

s of l

og.

For i

nsta

nce,

kno

win

g th

at th

e ra

tio o

f 2 is

3 d

B, w

hat’s

the

ratio

of 4

?

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Bod

e Pl

ot O

verv

iew

Tech

niqu

e fo

r est

imat

ing

a co

mpl

icat

ed tr

ansf

er

func

tion

(sev

eral

pol

es a

nd z

eros

) qui

ckly

Bre

ak fr

eque

ncie

s :

)1(

)1)(

1()

1()

1)(1(

)(

)(

22

21

0pm

pp

znz

zK

jj

jj

jj

jG

Hωτ

ωτ

ωτ

ωτ

ωτ

ωτ

ωω

++

++

++

=

ii

τω

1=

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Sum

mar

y of

Indi

vidu

al F

acto

rs

Sim

ple

Pole

:

Sim

ple

Zero

:

DC

Zer

o:

DC

Pol

e:ωτ

j+1

1 ωτ

j+1 ωτ

j ωτ

j1

τω

1=

dB0 dB0 dB0dB0

90−

90+

90−

90+

τω

1=

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Exam

ple

Con

side

r the

follo

win

g tra

nsfe

r fun

ctio

n

Bre

ak fr

eque

ncie

s: in

vert

time

cons

tant

sps

100ns

10ns

100

321

===

τττ)

1)(1(

)1(

10)

(3

1

25

ωτ

ωτ

ωτ

ωω

jj

jj

jH

++

+=

−

Gra

d/s

10M

rad/

s10

0M

rad/

s10

32

1=

==

ωω

ω

)1)(

1(

)1(

10)

(

31

25

ωωωω

ωωω

ωj

j

jj

jH

++

+=

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Bre

akin

g D

own

the

Mag

nitu

de

Rec

all l

og o

f pro

duct

s is s

um o

f log

s

Let’s

plo

t eac

h fa

ctor

sepa

rate

ly a

nd a

dd th

em

grap

hica

lly

)1)(

1(

)1(

10lo

g20

)(

31

25

dB

ωωωω

ωωω

ωj

j

jj

jH

++

+=

31

25

1lo

g20

1lo

g20

1lo

g20

10lo

g20

ωωωω

ωωω

jj

jj

+−

+−

++

=

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Bre

akin

g D

own

the

Phas

e

Sinc

e

Let’s

plo

t eac

h fa

ctor

sepa

rate

ly a

nd a

dd th

em

grap

hica

lly

ba

ba

∠+

∠=

⋅∠

)1)(

1()

1(10

)(

31

25

ωτ

ωτ

ωτ

ωω

jj

jj

jH

++

+∠

=∠

−

31

25 1

1

110

)(

ωωωω

ωωω

ω

jj

jj

jH

+∠

−+

∠−

+∠

+∠

=∠

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Mag

nitu

de B

ode

Plot

: DC

Zer

o80 2060 40 -2

0

-60

-80

-40

104

105

106

107

108

109

1010

1011

ω

510ωj

0 dB

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Phas

e B

ode

Plot

: DC

Zer

o

180

45135

90 -45

-135

-180

-90

104

105

106

107

108

109

1010

1011

ω

510ωj

∠

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Mag

nitu

de B

ode

Plot

: Add

Firs

t Pol

e

80 2060 40 -20

-60

-80

-40

104

105

106

107

108

109

1010

1011

ω

dB5

10ωj

dB7

101

1ω

j+

Mra

d/s

101=

ω

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Phas

e B

ode

Plot

: Add

Firs

t Pol

e

180

45135

90 -45

-135

-180

-90

104

105

106

107

108

109

1010

1011

ω

510ωj

∠

710

1

1ω

j+

∠

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Mag

nitu

de B

ode

Plot

: Add

2nd

Zero

80 2060 40 -20

-60

-80

-40

104

105

106

107

108

109

1010

1011

ωdB8

101

ωj

+M

rad/

s10

02=

ω

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Phas

e B

ode

Plot

: Add

2nd

Zero

180

45135

90 -45

-135

-180

-90

104

105

106

107

108

109

1010

1011

ω

810

1ω

j+

∠

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Mag

nitu

de B

ode

Plot

: Add

2nd

Pole

80 2060 40 -20

-60

-80

-40

104

105

106

107

108

109

1010

1011

ω

dB10

101

1ω

j+

Gra

d/s

103=

ω

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Phas

e B

ode

Plot

: Add

2nd

Pole

180

45135

90 -45

-135

-180

-90

104

105

106

107

108

109

1010

1011

ω

1010

1ω

j+

∠−

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Com

paris

on to

“A

ctua

l”M

agPl

ot

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Com

paris

on to

“A

ctua

l”Ph

ase

Plot

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Why

do

I say

“ac

tual

”?

I plo

tted

the

trans

fer c

hara

cter

istic

s with

M

athe

mat

ica

The

rang

e of

freq

uenc

y fo

r the

plo

t is 6

ord

ers o

f m

agni

tude

. Th

e pr

ogra

m h

as to

find

the

“hot

sp

ots”

in o

rder

to p

lot t

he fu

nctio

n. N

ear t

he h

ot

spot

s, m

ore

poin

ts a

re p

lotte

d. I

n be

twee

n ho

t sp

ots,

the

func

tion

is in

terp

olat

ed.

If y

ou p

ick

the

wro

ng p

oint

s, yo

u’ll

end

up w

ith th

e w

rong

plo

t:mag

= LogLinearPlot[20*Log[10, Abs[H[x]]], {x, 10^4,

10^11},PlotPoints -> 10000, Frame -> True,PlotStyle

->

Thickness[.005], ImageSize

-> 600,GridLines -> Automatic,

PlotRange

-> {{10^4, 10^11}, {-20, 100}} ]

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Don

’t al

way

s be

lieve

a c

ompu

ter!

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Seco

nd O

rder

Tra

nsfe

r Fun

ctio

n

The

serie

s res

onan

t circ

uit i

s one

of t

he m

ost

impo

rtant

ele

men

tary

circ

uits

:

The

phys

ics d

escr

ibes

not

onl

y ph

ysic

al L

CR

ci

rcui

ts, b

ut a

lso

appr

oxim

ates

mec

hani

cal

reso

nanc

e (m

ass-

sprin

g, p

endu

lum

, mol

ecul

ar

reso

nanc

e, m

icro

wav

e ca

vitie

s, tra

nsm

issi

on li

nes,

build

ings

, brid

ges,

…)

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Serie

s LC

R A

naly

sis

With

pha

sora

naly

sis,

this

circ

uit i

s rea

dily

an

alyz

ed

RIC

jI

LjI

V s+

+=

ωω

1

RR

Cj

Lj

VRI

V

RC

jL

jI

V

s

s

++

==

++

=

ωω

ωω

1

1

0

+ V o −

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Seco

nd O

rder

Tra

nsfe

r Fun

ctio

nSo

we

have

:

To fi

nd th

e po

les/

zero

s, le

t’s p

ut th

e H

in c

anon

ical

fo

rm:

One

zer

o at

DC

freq

uenc

y ca

n’t c

ondu

ct D

C d

ue

to c

apac

itor

RC

jL

j

RVV

jH

s+

+=

=

ωω

ω1

)(

0+ V o −

RCj

LCCR

jVV

jH

sω

ωω

ω+

−=

=2

0

1)

(

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Pole

s of

2nd

Ord

er T

rans

fer F

unct

ion

Den

omin

ator

is a

qua

drat

ic p

olyn

omia

l:

LRj

jLC

LRj

RCj

LCCR

jVV

jH

sω

ωω

ωω

ωω

++

=+

−=

=2

20

)(

11

)(

LRj

jLR

jj

Hω

ωω

ωω

++

=2

2 0)

()

(LC1

2 0≡

ω

Qj

j

Qj

jH

02

2 0

0

)(

)(

ωω

ωω

ωω

ω+

+=

RLQ

0ω

≡

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Find

ing

the

pole

s…

Let’s

fact

or th

e de

nom

inat

or:

Pole

s are

com

plex

con

juga

te fr

eque

ncie

sTh

e Q

para

met

er is

cal

led

the

“qua

lity-

fact

or”

or Q

-fac

tor

This

par

amet

ers i

s an

impo

rtant

para

met

er:

Re

Im

0)

(2 0

02

=+

+ω

ωω

ωQ

jj

22

−±

−=

−±

−=

Qj

QQ

Q41

12

42

00

2 0

2 00

ωω

ωω

ωω

∞→

→

0R

Q

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Res

onan

ce w

ithou

t Los

s

The

trans

fer f

unct

ion

can

para

met

eriz

ed in

term

s of

loss

. Fi

rst,

take

the

loss

less

cas

e, R=

0:

Whe

n th

e ci

rcui

t is l

ossl

ess,

the

pole

s are

at real

freq

uenc

ies,

so th

e tra

nsfe

r fun

ctio

n bl

ows u

p!A

t thi

s resonance

freq

uenc

y, th

e ci

rcui

t has

zer

o im

agin

ary

impe

danc

eEv

en if

we

set t

he so

urce

equ

al to

zer

o, th

e ci

rcui

t ca

n ha

ve a

stea

dy-s

tate

resp

onse

Re

Im

02 0

2 00

42

ωω

ωω

ωj

QQ

Q

±=

−±

−=

∞→

2

Dep

artm

ent o

f EEC

SU

nive

rsity

of C

alifo

rnia

, Ber

kele

y

EE

CS

105

Fall

2003

, Lec

ture

3P

rof.

A. N

ikne

jad

Mag

nitu

de R

espo

nse

The

resp

onse

pea

kine

ssde

pend

s on Q

QjQ

j

LRjLR

jj

H0

22 0

0

002

2 0

00

)(

ωω

ωω

ωω

ωωω

ωω

ωωω

ω+

−=

+−

=

1=

Q

10=

Q

100

=Q

0ω

1)

(0

02 0

2 0

2 0

0=

+−

=

Qj

Qj

jH

ωω

ωω

ω

ω

1)

(0=

ωjH

0)0(=

H