bode plots of transfer functions

8/13/2019 Bode Plots of Transfer Functions

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Lecture 45Bode Plots of Transfer Functions

A. Goals : Design Oriented Analysis1. Design Oriented Analysis Overview2. Inspection of T(s) in normalized form3. Five points in design oriented analysis

B. Review of Bode Plots1. db values2. Log - Log Slopes vs f

3. Normalized Form 1 + s/wa. Regular Poleb. Right Half Plane Zero

flyback / Buck - Boost

c. Inverted Forms 1 + w/s1. Pole

d. Complex T(s) Plots versus Frequency

4. Low-Pass Filter Resonant Circuit5. Series Resonant R-L-C Circuit


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Lecture 45Bode Plots of Transfer Functions

A. Goal : Design Oriented Analysis1. Design Oriented Analysis

Our aim in design is to always keep the whole system viewat hand so that we can alter some parts of the system andimmediately see the results of the change in other parts ofthe system. The keystone will be the Bode plots of each partof the system that comprise the open loop response. Fromthe open loop response Bode plots various design changesmay be explored. We seek simple intuitive understanding ofa transfer function via Bode Plots vs f

2. Inspection of T(s) in normalized form to:a. Guess / estimate pole and zero location

b. Determine asymptotic behaviors


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3. The five points in design oriented analysis wewill emphasize are given below:

Before we can do this we need to review how to constructand make Bode plots as they are the key to our designoriented approach. Asymptotic approximations to the fullBode plots are key to rapid design and analysis. Dependingon whether or not we know the high frequency or lowfrequency behavior of the transfer function we may choose

either normal pole/zero from or inverted pole/zero forms aswe will discuss below. Each has its advocates and we needto be familiar with both. What we dont want to do is to mixthe forms in one T(s) expression, if at all possible.

For those students to whom Bode plots are hazy oreven new we will give a brief review.


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B. Bode Plot Review

Usually in a transfer function Vo/Vinhas a value at eachapplied frequency. We use db for the transfer functionmagnitudes, as it will allow for easy asymptoticapproximations to the curves.

1. db values 20 log10G

To employ a db scale we always need a BASE value. For

example 50kon a base of 10 k, is considered as 14 db.20 db

is 20 db greater than a base of 10k

, or 100k

60 db A is a current 60 db greater than a base of 1mA orone Ampere. Do you get the new way of thinking in db?

Actual Magnitude Magnitude in dB

-6 dB

1 0 dB

2 6 dB

5 = 10/2 20 dB - 6 dB = 14 dB

10 20dB1000 = 103

3 * 20dB = 60 dB

What does -3db mean? What does +3 db mean? Can youtell the value of the T(s) ratio that gives 3db? Three db willbe a useful rule of thumb to place the crude amplitude T(s)plots closer to the real ones as we will see later.


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2. Log - log Plots and G vs f/foslopes in db units

focorresponds to acharacteristic pole or zero

of the transfer function

3. Normalized Forms for Transfer Functionsa. Single Isolated Pole G(s) = 1/(1+s/wp)

wpPole break frequency in radians/sec.Bode Plots for Single Pole

Crude Amplitude Plot Accurate Amplitude

fois the characteristic frequencyWhen f is fo/2 or 2fowe go off the crude plot by

only 1db to achieve the actual values. Likewise @fowe areoff the crude plot by 3dbas shown above.


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Crude Phase Plot Accurate Phase Plot

Real phase crosses Real phase crossesasymptote only once asymptote 3 times

These break frequencies above and below f0will be useful.


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b. Isolated Right Half Plane ZeroG(s) = 1 - s/wz

Bode Plot Right Half Plane Zero vs. Left Half Plane Zero

G(s) = 1+ s/wz G(s) = 1 - s/wzVery usual in T(s) Flyback / Buck-Boostfor many converters T(s) have this unique

right half zero featureOn page 8 we compare and contrast the right and left planezeros behavior versus applied frequency. While the amplitudeplots cannot tell the difference between the two, the phase plotscertainly do. What is the difference?


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Gain up vs. f Gain up vs. f

Phase up away from -180 Phase down toward -180

These asymptotic plots of phase for left and right planezeroes tell us the whole story.

c. Inverted G(s) forms Have Unique Bode Plots

When we focus on high f response of T(s) or G(s) wesometimes utilize w/s forms for the poles or zeros.

1. Inverted pole

G(s) -1

1+ w / sp

Bode plot of inverted pole has some unique properties:

Low f amplitude vs w decreases to -at low f, f < fo, unlikeRHP zero where low f response saturates at some f, f < fo.


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For an inverted pole we the gain up and the phase down vs.f for f < fo2. Inverted Zero

On page 10 we plot the frequency response of an invertedzero, with all the details of an asymptotic approach to bothgain and phase plots.


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d. Various complex T(s) Plots versus Frequency

1. G(s) = 32 db or 40 on a linear scale

(1+s

2 100Hz)(1+

s

2 2 Hz)

kWe plot this below by asymptotes. For 1+s/wo, wois the breakpoint.

Note phase break frequencies are 10 times off from


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amplitude break frequencies.

2. Well separated pole and zero means by a factor >100

A(s) = A 1+s/ w

1+s/ wo

z

p

wp> 100 wz fz= 100f1(below)

When f 0 we obtain A(DC) = Ao. Why is A(f = ) higher invalue? By exactly how much?

limit s 0 A01/1 limit s o zp

op

z

A s/ w

s/ w = A

w

w=

Aof

fz

1

Phase starts early at fz/10 with 45/decade slope.Phase ends late at fp*10 with 45/decade slope.


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3. Given log - log plots Find T(s)Two possible answers occur depending on the form chosen forT(s): normal pole and zero form or inverted poles and zeros.Consider the two T(f) amplitude plots below

-20 db/dec

-20 db/decw3

Inverted

zero at

Inverted

pole at w1

Pole at

Midband Gain

Gm

w1

w2+20 db/dec -20 db/dec

Gm

w1w0

Midband Gain

Inverted

pole at w0Pole at w1

Low f saturationidentifies inverted pole

Consider the inverted pole/zero form first and then normal

form.

G(s)G (1

w

s)

(1w

s)(1

s

w)

m2

1

3

=+

+ + G(s)

G

(1w

s)(1

s

w)

m

o

1

=+ +

OR:w G (s w )

(s w )(s w )3 m 2

1 3

++ +

OR:w G s

(s w )(s w )1 m

o 1+ +

4. Low-Pass Filter Resonant CircuitThese Bode plots are unique and different from all priorones. The circuit diagram is at the bottom of the page:

2

1

o o

2

v (s)

v (s) =

1

1 +s

Qw+ (

s

w)

Q Q factor of the resonant circuit. For Q 1/2 roots arereal while for Q 1/2 roots are complex. Q is in linear units.

C R

L

v1(s) v2(s)

+

-


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woCorner radian Frequency =1

LCor f =

1

2 LC

We will find below that Q(low pass filter) RL

C.

This may seem wrong as higher R values mean higher Q forthe low-pass filter. Q for the series R-L-C circuit, as distinctfrom the low-pass filter, differs in its R dependence asshown below in section 5. In the low-pass filter case

L 0, Q reallymeans we have only a RC filter. The


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transfer function of the low-pass filter is easily found:

|V (s)

V (s)| =

1

[1-(w

w

) ] +(w/ w )

Q

2

1

o

2 2 o2

2

For w > wo:V

V

w

w2

1 o

2

We plot this response including the resonant bump below.The resonant bump near f=f0is asymmetric in shape.

Crude Plot Accurate Plot with Q Bump

|Q|db

fo

slow

-40 db/dec

sharp||G||

0 db

f

0 db

||G(jw)||db

0db

-20db

-40db

-60db

0.1fo fo 10fo

-40 db/dec

(f/fo)-2

2

1 L 2

o

2

o2

V (s)

V (s) =

1

1+ sR C+ s LC =

1

1+s

Q w+

s

w

=1

(1+s/ w )(1+ s/ w )1 2For the resonant bump to occur w1, w0are complex, not real!We do not have a simple Bode plot with only straight line

asymptotes, as this is a resonant circuit with wo=1

LC

!

Rather we have to learn the proper way to treat resonantcircuits which involves the linear asymptotes at frequenciesfar from f0and a resonant bump near f=f0.


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1

Qw = R

o

lC Q =1

RCw

=

L

C

Rl o

In series R-L-C RQ

1

LC

V(s)

V (s) = tan |

1

Q

w

w

1- (w

w)

|2

1

-1 o

o

2 = (f)

In summary for the low-pass filter:

We now need to see some trends of T(s) phase plots for thelow-pass filter near f=f0as well as far from resonance. Thebehavior is very different from the Bode plots we coveredbefore that did not have a resonance.

Trend of phase shift of low-pass T(s),, vs. f withthe varying Q factor as chosen by circuit componentsTotal change of 180 occurs over two decades of frequencyaround fofor low Q and even more rapidly for high Q. Thelocation of the TWO phase angle break frequencies is thekey.


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Rule of thumb: f lower= 10-1/2Qfo when phase shift begins

fupper= 10+1/2Qfowhen phase shift ends

This has a special pair of break frequencies for the case

For Q = 1/2 (minimum value) or beginning of low Q casefL= fo10 and fu= 10 fu

Phase slope is 90/ decade 180 over 2 decades.For high Q conditions in the circuit values the break

frequencies flowerand fhigherare closer together and for Q = ,fL= fH= fo.Below we give the full-blown amplitude (with Q peaking) and

phase plots for T(s) for the low-pass filter. On page 17 wegive the phase plots only from Erickson which emphasizethe way to spot high Q via the rate of change of phase.

The Q peaking in the amplitude plots can be estimated by

a set of rules we will develop later, so that we need not haveto plot out the curve in detail to see the major features of Qpeaking. This will be done in lectures 46-47.


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For a second look at resonant circuits and their T(s) behaviorvia Bode plots we will leave the low-pass filter and look atthe series resonance circuit. It will reveal a different Q

behavior, different phase behavior and unique dynamicalbehavior. We do this because in resonant converters thisseries R-L-C circuit is as crucial to operation as the low-passfilter is to switch mode buck, boost and buck-boost. It alsoshows us that Q is not a fixed concept the same for allresonant circuits.


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5. Series R-L-C Resonant circuit

C

LRL

v2(s)v1(s)

V1(s) = I(s)[RL+ sL + 1/sC]

The characteristic equation is:

s2+ Rs/L + 1/LC = 0

Solutions for the roots are:

s =

RL

RL

4LC

2

2

R

2Lis termed the damping parameter as given on page 13

of this lecture. w1

LCo= is the radian resonance frequency.

s = - wo

2 2 . We have several ways to solve

for the natural time response to a delta function input:

2 2 wo > 0 We have exponential decaying solutions .

2 2 wo < 0 We have decaying sinusoidal solutions . = 0 The step response solutions are pure sinusoids.

In the series resonance circuit the characteristic impedance

of an L-C pair is Zo=L

C

. also define the quality factor Q as:

Q =Z

R

o

With this set of definitions we find the characteristic equation

becomes: sw

Qs w2 o o

2+ + = 0 and the damping factor


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Is =w

2Qo .

Hence solutions are: s = ( )4 1 1 2QWe restate the division between different dynamicalsolutions depending on the chosen circuit Q. The case Q = is pivotal:

Q < two real roots damped exponential responseQ = two repeated real roots critically dampedQ > two complex roots decaying oscillatory

response

We find for a step voltage input the capacitor

voltage is:

Vc= Vin- Vine-t(

wSin wt + Cos wt)

This is plotted below

For long times the energy stored in the capacitor goes to CVIN

2. The inductor current has a different time response.

IL= CVine-t(

wSin wt + Cos wt) - wCVine

-t(

wSin wt - Cos

wt)


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As time goes on the inductor current goes to zero and theenergy stored in the inductor goes to zero. Inbetween wehave a roller coaster of energy exchange.

The changing transient energy storage between the

capacitor and the inductor versus time in the series R-L-Ccircuit for a step voltage input is shown below.

Note how inductor energy decays to zero with time.

Ultimately, = CVin2but during the transient response theenergy in the inductor has wide swings depending on the Q

chosen for the circuit before settling down to zero.If we analyze the response of a series R-L-C to a

periodic wave (square, triangle, etc.) and plot VR/Vinat eachharmonic we find the general T(s) relationship for the voltageacross the resistor, VR(f), as shown on the top of page 21:


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V

V=

RZ

jnw

w

jnw

w

RZ

R

in

o

o

o

c+ +

Considering only the fundamental at fr=1

LCthis T(s) for

the resistor voltagevaries as shown below in what manyconsider the more classic resonance shape:

For Q > 100 the response is very selective. High Q meanshigh selectivity about fr. This seems too good to be true asthere first appears no price is paid for high Q. This is anillusion we will remove next as there is a price to be paid.WHAT IS IT?Consider the capacitor voltage T(s) at each harmonicfrequency, fH=n fIN, even the case n=1

V

V=

1

1 nw

w jnw

w Q

C

in 22

o2

o +When the input frequency component nw = wo

V

VC

in

(nw = wo) = QVin

For Vinof 120V rms Vc= 12,000V rms for a Q of 100.


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Similarly for VLbut 180out of phase with Vcwe find another12,000 V level. Can L and C withstand voltages 100*Vin?

Consider a resonant filter at 60 Hz for a 120V, 1KWload. This implies we have a particular value load resistor.

RL= (120)2/P = 14.4load.To achieve a given Q for the circuit the individual

components must have Qs well in excess of the targetcircuit Q. For a filter with a desired Q = 10 at f = 60 Hz =

1

LC2we need L and C with 3 times higher Q levels. For

the chosen series R-L-C with Q=10 we can then find therequired characteristic impedance from the relation Q =

Zo/RLor 144 = Zo=L

C. We then find L = 382mH and C =

18.4 F as one possible solution set. Assuming Q(L) = 30implies that associated with the inductor is an equivalent

series resistance ,RL= 4.8. For Q(C) = 30 means ESR Rc= 4.8. The total R-L-C circuit with all associated equivalentseries resistances and the load resistance of 14.4 Ohms isshown below:

Note that the circuit Q is = Zo/R(total) = 144/24 = 6 waybelow expected values from L and C components each withQ = 30. To do better higher Q components are needed.

bode plots of transfer functions

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