frequency response analysisabl.gtu.edu.tr/hebe/abldrive/70976026/w/storage/...phase margin is...

Frequency Response Analysis

Consider

let the input be in the form

Assume that the system is stable and the steady state response of the system to a sinusoidal inputdoes not depend on the initial conditions

We have PFEyields

Complex conjugate of

Inverse Laplace transform yields

At steady state (when ) we have

where

● Since is a complex quantity, we have

where

and similarly

Hence

That is a sinusoidal input generates a sinusoidal output

Note that

– Negative phase is called phase lag

– Positive phase is called phase lead

There are many ways representing these, We will concentrate on 3 common graphical representations

Amplitude ratios of the output sinusoid to the input

Phase shift of the output with respect to the input

Graphical Representations

1) Bode Plot

– A plot of magnitude in decibels (dB) vs in semi-logaritmic coordinates. The Phase angle is also plotted against in semilogarithmic coordinates

2) Polar Plot

– A plot of magnitude vs phase in polar coordinates as from zero to infinity

3) Magnitude vs Phase plots

– A plot of magnitude in dB vs the phase on the rectangular coordinates with as a varying parameter on the curve

Bode Plot Construction

Consider the transfer function of the system given in the form

where are positive real values.

In order to put the given tranfer function into “Bode“ form let and normalize

Magnitude PlotMagnitude Plot Phase PlotPhase Plot

Curve Breaks {up/down} by 20 dB/dec at {a,b / c,d,e}

Curve breaks {up/down} by 45 degrees one decade before {a,b / c,d,e} and breaks {down/up} by 45 degrees one decade after {a,b / c,d,e}

Initial Value,

● For n=0 Calculate and convert to dBusing

● For n>0 Calculate -intercept =

Calculate the initial slope -20n dB /dec Final magnitude slope -20 RD dB /dec

Initial and final values

● Plot a zero / pole plot for the original transfer function

● Calculate the angle from the figure at = 0

An illustrative example

The

Initial angle

Final angle

-10 -8 -6 -4 -2 0-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Real Axis

Imag

Axi

s

Bode Plot Example

Construct the Bode Plot for

Start with putting the transfer function into Bode form

Magnitude PlotMagnitude Plot

– Initial Magnitude (note that n=0)

– Use “up“ arrow to denote a slope of +20 dB/dec and “down“ arrow to denote a slope of -20 db/dec

110 110 210010

-20

-40Decade Decade

Initial Magnitude

RD = 0

Final slope is zero

Semi log

plot

– Start with the initial magnitude and use arrows to draw the asymptotic plot

110 110 210010

-20

-40

+ 20 dB/dec

dB

Phase plot :

Start with the pole / zero Plot

Initial angle :

Final angle :

-10 -8 -6 -4 -2 0-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Real AxisIm

ag A

xis

Use the negative of the poles and the zeros on the jw axis (zero at -1 and pole at -10)

Use an “up“ arrow to denote a slope of +45 deg/dec and a “down“ arrow to denote a slope of -45 deg/dec

Frequency (rad/sec)

110 110 210010

45

90Decade

Initial Phase Final Phase

- 45 deg/dec 45 deg/dec

deg

The actual Bode plot is obtain using the following code

% (s+1)/(s+10)

figure;

num = [1 1];

denum = [1 10];

SYS=tf(num,den);

bode(SYS);

Bode plot construction

Another example : This time sketch the Bode plot for

In Bode form

Magnitude PlotMagnitude Plot● Initial Magnitude (note that n=1) Calculate the intercept

Initial slope = -20.n dB/dec = -20 dB/dec

Plot the negative poles and zeros on the jw axis and use “up“ arrows to denote a slope of -20 dB/dec and “down“ arrows to denote a slope of +20 dB/dec

Finally use the initial value and the arrows to draw the asymptotic plot

Mag

nitu

de (

dB)

Frequency (rad/sec)

110 210010

-40

-20

-20 dB/dec

-20 dB/dec

-40 dB/dec

-40 dB/dec-60

0

20

Phase plot :

Start with the pole / zero Plot

Initial angle :

Final angle :

Use the negative of the poles and the zeros on the jw axis (zero at -10 and poles at 0,-1,-100)

Use an “up“ arrow to denote a slope of +45 deg/dec and a “down“ arrow to denote a slope of -45 deg/dec

Pha

se (

deg)

Frequency (rad/sec)

110 110 210010

-90

-45

-135

-180

Initial Phase

Final Phase

- 45 deg/dec

- 45 deg/dec

310

-225

The actual Bode plot

figure;

num = [10 10];

denum = [1 101 100 0];

SYS=tf(num,den);

bode(SYS);

Bode Plot for repeated roots

Sketch the Bode plot of

In Bode form

Magnitude PlotMagnitude Plot● Initial Magnitude (note that n=0)

dB conversion

Plot the negative poles on jw axis (2 poles at )

use “up“ arrows to denote a slope of -20 dB/dec and “down“ arrows to denote a slope of +20 dB/dec

Finally use the initial value and the arrows to draw the asymptotic plot

Bode Diagramsn n1010/n

-20

-40

Decade20

0

n100

Initial Magnitude ( )

2

1log20

n

- 40 dB/dec

Phase plot :

Start with the pole / zero Plot

Initial angle :

Final angle :

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Real Axis

Imag

Axi

s

Use the negative of the poles the jw axis

Use an “up“ arrow to denote a slope of +45 deg/dec and a “down“ arrow to denote a slope of -45 deg/dec

n

-90

-180

90

0

-270

Initial PhaseFinal Phase

n 10 n 10010

n

The actual Bode plot when

figure;

num = [1];

denum = [1 2 1];

SYS=tf(num,den);

bode(SYS);

Example

Use the Bode plot to determine the stability of a closed loop system

-120 -100 -80 -60 -40 -20 0 20-100

-80

-60

-40

-20

0

20

40

60

80

100

Real Axis

Imag

Axi

s

dj

djc 100 10

Transfer Function

Root locus form of the denominator

For the closed loop system is marginally stable. At the closed loop poles are at (from Root Locus)

Angle Condition

Magnitude Condition

Bode Solution for finding and

Note that

Might not be easy to solve for and

● A Bode plot is the graph of magnitude and phase of the transfer function with respect to

– Hence we can directly obtain (read) the value of the desired frequency from the phase plot

– Using the desired frequency we can read the corresponding value of the magnitude of the transfer function and obtain the value of as

or in “normal“ units

in dB

● The Bode plot gives us a graphical way of finding the values of the and

● Specific to our problem, the closed loop system is defined by

Construct the Bode Plot

-120

-100

-80

-60

-40

-20

0

10-1

100

101

102

103

-250

-200

-150

-100

-50

0

Frequency (rad/sec)

Ph

ase

(deg

)M

agn

itu

de

(dB

)

- 40 dB

-180 deg

20d

Then using

That is for the stability of the system we need

Bode Compensator Design

● Similar to the root locus design we can use the characteristic equation to examine the stability and/or the performance of the closed loop system using Bode plot

10-1

100

101

102

103

Frequency (rad/sec)

Ph

ase

(deg

)M

agn

itu

de

(dB

)

0

-180

|KH ( jωφ )|

∠KH ( jωp )

ωφω p

: Gain cross over freq.● The value of frequency

when

: phase cross over freq.● The value of frequency

when

Gain margin and Phase margin

● These points on the Bode plot leads us to 2 new definitions

Phase Margin

Gain Margin

The system is unstableif this value is less than zero

Bode plot – Root locus● The Bode plot parameters are related to the root locus

parameters according to

● Phase margin is related to the damping ratio as follows

Note that damping ratio is a measure of relative stability

Root locus design parameters

Damping ratio

Natural frequency

for

Hence phase cross over frequency is a measure of relative stability as well

Bode Design Example

Given

with Find K so that the phase margin is 45 degrees for the closed loop system

Solution :

Design Steps

● Find the characteristic equation in the form

● Plot and asuming

● From calculate the desired

i.e.

● From the desired , and the plot, find the desired

● Raise or lower the magnitude plot so that it crosses the 0 dB level at the desired value of

● Find the shift between the desired plot and the original amgnitude plot at the desired value of

● Calculate the value of using

Desired + original = shift in dB

● Step 1 :

● Step 2 :

● Step 3 :

Desired

● Given the desired value find from the plot

● Step 5 :

● Step 6 :

The difference between the desired magnitude plot and the actual magnitude plot is -9 dB , thefore

● Step 7 :