ee362-ch8

Frequency Response Analysis

Assoc. Prof. Enver Tatlicioglu

Department of Electrical & Electronics EngineeringIzmir Institute of Technology

Chapter 8

Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control Systems Chapter 8 1 / 129

Frequency response analysisAdvantages and disadvantages of root locus design approach

Advantages of root locus design approach:

Good indicator of transient response.

Explicitly shows location of closedloop poles.

Tradeoffs are clear.

Disadvantages of root locus design approach:

Requires the transfer function of the plant to be known.

Difficult to infer all performance values.

Hard to extract steady state response (for sinusoidal inputs).

Frequency response methods can be used to supplement root locus:

Can infer performance and stability from same plot.

Can use measured data when no model is available.

Design process is independent of system order (# poles).

Time delays are handled correctly (exp (s)). Graphical techniques (analysis/synthesis) are quite simple.


Frequency response

We want to know how a linear system responds to a sinusoidal inputin steady state.

Consider the systemY (s)

U (s)= G (s) .

Consider the input

u (t) = u0 sin (t) U (s) =u0

s2 + 2.

For zero initial conditions, we obtain

Y (s) = G (s)u0

s2 + 2.


Frequency response

After doing a partial fraction expansion (assume distinct roots)

Y (s) =1

s a1+ + n

s an+

0s + j

+0

s jy (t) = 1 exp (a1t) + + n exp (ant)

if stable, these decay to zero

+2 |0| sin (t + )

from which, we have

yss = 2 |0| sin (t + )= Au0 sin (t + )

where A is the amplitude gain and is the phase shift.

Important LTI system fact:

If the input to a stable LTI system is a sinusoid, then the steady stateoutput is a sinusoid of the same frequency but with differentamplitude and phase.


Frequency response

Notice that, 0 is found using partial fraction expansion method, andA and may be found from 0 as

A = |G (s)| |s=j= |G (j)|

=

[R (G (j))]2 + [I (G (j))]2

and

= G (s) |s=j= G (j)

= tan1( I (G (j))R (G (j))

)

.

Transfer function along jaxis tells us response to a sinusoid.

But, it also tells us about the stability since jaxis is the stabilityboundary!


Plotting frequency response

There are two common ways to plot a frequency response (i.e.,plotting the magnitude and phase for all frequencies).

Consider the following simple circuit.

The transfer function is found as

G (s) =1

1 + RCs.

After letting RC = 1 and setting s = j, we obtain

G (j) =1

1 + j.


Plotting frequency responsePlot method #1: Polar plot in splane

Evaluate G (j) at each frequency for 0 +.Result will be a complex number in the form a+ jb or A exp (j) foreach frequency.

Plot each point a+ jb or A exp (j) on the complex plane obtainedfor each frequency.

Result is a polar plot called a Nyquist plot.



For the RC circuit example, we obtain:

G (j)

0.0 1.0000.0

0.5 0.894 26.61.0 0.707 45.01.5 0.555 56.32.0 0.447 63.43.0 0.316 71.65.0 0.196 78.710.0 0.100 84.3+ 0.000 90.0



The polar plot is parametric in , so it is hard to read the frequencyresponse for a specific frequency from the plot.

Figure: Polar plot in splane

We will later see that the polar plot will help us determine stabilityproperties of the plant and the closedloop system.


Plotting frequency responsePlot method #2: Magnitude and phase plots

We separate magnitude and phase information from rationalpolynomials that are functions of j

Magnitude =magnitude of numerator

magnitude of denominator

=

[R (num)]2 + [I (num)]2

[R (den)]2 + [I (den)]2

and

Phase = phase of numerator phase of denominator

= tan1( I (num)R (num)

)

tan1( I (den)R (den)

)

.

For the RC circuit example, we obtain

G (j) =1

1 + 2 tan1 () .



We can plot the data by separating magnitude and phase plots.

Figure: Magnitude versus frequency plot for the RC circuit example



Figure: Phase versus frequency plot for the RC circuit example

These plots are in natural scale, but usually a loglog plot is preferred.

These are called Bode plots or Bode diagrams.


Bode plots

Simplest way to display the frequency response of a rationalpolynomial transfer function is to use Bode plots.

Bode magnitude plot is the plot of logarithmic |G (j)| versuslogarithmic .

Bode phase plot is the plot of G (j) versus logarithmic .

Since

log10

(ab

cd

)

= log10 (a) + log10 (b) log10 (c) log10 (d)

the factors of the transfer function can be split up and theircontributions can be evaluated separately.


Bode plotsExample

For

G (s) =s + 1

s/10 + 1 G (s) |s=j = G (j) =

j + 1

j/10 + 1

we obtain

|G (j)| = |j + 1||j/10 + 1|and thus

log10 |G (j)| = log10

1 + 2 log10

1 +(

10

)2.


Bode plots

Consider the following expression

log10

1 +

(

n

)2

.

For n,

log10

1 +

(

n

)2

log10 1 = 0.

For n,

log10

1 +

(

n

)2

log10(

n

)

.

So, we obtain two straight lines on log scale intersecting at = n.


Bode plots

Typically, we plot 20 log10 |G (j)|; which is in dB.A transfer function is made up of:

first order zeros and poles, complex zeros and poles, constant gains, delays.We will see how to make straight line approximations for magnitudeand phase plots, and combine them to form Bode plots.

Figure:Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control Systems Chapter 8 16 / 129

Bode magnitude plotsConstant gain

We have 20 log10 |K |dB which is not a function of frequency.So, it is a horizontal straight line.If |K | < 1, then 20 log10 |K | is negative.If |K | > 1, then 20 log10 |K | is positive.

Figure: Bode magnitude plot for constant gainAssoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control Systems Chapter 8 17 / 129

Bode magnitude plotsZero at origin

For a zero at the origin, G (s) = s.

So, we obtain

20 log10 |G (j)| = 20 log10 |j| = 20 log10 dB.

Figure: Bode magnitude plot for a zero at origin


Bode magnitude plotsPole at origin

For a pole at the origin, G (s) = 1/s.

So, we obtain

20 log10 |G (j)| = 20 log10 |j| = 20 log10 dB.

Figure: Bode magnitude plot for a pole at origin


Bode magnitude plotsZero or pole at origin

For a zero or pole at the origin,

Both are straight lines with a slope of 20dB per decade of frequency. Line intersects axis at = 1 (since log10 1 = 0).

For an nth order pole or zero at the origin, we obtain

20 log10 |(j)n| = 20 log10 n = 20n log10

which are still straight lines,

which still intersect axis at = 1,

which have a slope of 20n per decade.


Bode magnitude plotsZero on real axis, but not at origin

For a zero on the real axis (LHP or RHP), the standard Bode form is

G (s) =s

n 1

which ensures unity DC gain.

If we are given something like

G (s) = s n

then we factor the DC gain n as

G (s) = n

(s

n 1

)

and draw the constant gain n separately from the term (s/n 1).



An LHP or RHP zero has the following standard Bode form

G (s) =s

n 1 G (j) = j

(

n

)

1.

So, we obtain

20 log10 |G (j)| = 20 log10

1 +

(

n

)2

.

For n,

20 log10

1 +

(

n

)2

20 log101 = 0.

For n,

20 log10

1 +

(

n

)2

20 log10(

n

)

.




Figure: Bode magnitude plot for a zero on real axis, but not at origin


Bode magnitude plotsPole on real axis, but not at origin

For a pole on the real axis (LHP or RHP), standard Bode form is

G (s) =

(s

n 1

)1 G (j) =

(

j

(

n

)

1)1

.

So, we obtain

20 log10 |G (j)| = 20 log10

1 +

(

n

)2

1

= 20 log10

1 +

(

n

)2

which is same as zero on real axis, but not at origin except for aminus sign.


Bode magnitude plotsPole on real axis, but not at origin


Figure: Bode magnitude plot for a pole on real axis, but not at origin


Bode magnitude plotsComplex zero pair

For a complex zero pair (LHP or RHP), standard Bode form is

G (s) =

(s

n

)2

2(

s

n

)

+ 1

which ensures unity DC gain.

If we are given something like

G (s) = s2 2ns + 2nthen we factor the DC gain 2n as

G (s) = 2n

[(s

n

)2

2(

s

n

)

+ 1

]

and draw the constant gain 2n separately from the term[(sn

)2 2

(sn

)

+ 1

]

.



Complex zero pair does not lend themselves very well to straight lineapproximation.

We write complex zero pair (LHP or RHP) as

G (s) =

[(s

n

)2

2(

s

n

)

+ 1

]

.

At = n, magnitude value = 20 log10 (2) dB.

For n, magnitude value 0 dB. For n, magnitude value 40 dB.



Figure: Bode magnitude plots for complex zero pairs


Bode magnitude plotsComplex pole pair

We write complex pole pair (LHP or RHP) as

G (s) =

[(s

n

)2

2(

s

n

)

+ 1

]1

.

At = n, magnitude value = 20 log10 (2) dB. For n, magnitude value 0 dB. For n, magnitude value 40 dB.


Bode magnitude plotsComplex pole pair

Figure: Bode magnitude plots for complex pole pairs


Bode magnitude plotsTime delay

The transfer function for time delay is given by

G (s) = exp (s) |G (j)| = |exp (j)| = 1.

So, we obtain

20 log10 |G (j)| = 20 log10 1 = 0 dB

which obviously does not change the magnitude response.


Bode magnitude plotsExample

Draw the Bode magnitude response for

G (s) =2000 (s + 0.5)

s (s + 10) (s + 50).

First, rewrite in standard Bode form

G (s) =2000 0.510 50

(s0.5 + 1

)

s(

s10 + 1

) (s50 + 1

)

=2(

s0.5 + 1

)

s(

s10 + 1

) (s50 + 1

) .

Next, make the change s = j

G (j) = 2

(j0.5 + 1

)

j(j10 + 1

)(j50 + 1

) .



The components of the Bode magnitude plot may be obtainedseparately.

For the DC gain of 2:

20 log10 2 = 6 dB.

For the real zero not at origin (n = 0.5):

For 0.5, a straight line with a slope of 0 dB. For 0.5, a straight line with a slope of 20 dB. So, we obtain two straight lines on log scale intersecting at = 0.5.

For the pole at origin:

A straight line with a slope of 20 dB per decade of frequency andintersecting the axis at = 1.



Two real poles not at origin (n = 10 and n = 50):

For n = 10:

For 10, a straight line with a slope of 0 dB. For 10, a straight line with a slope of 20 dB. So, we obtain two straight lines on log scale intersecting at = 10.

For n = 50:

For 50, a straight line with a slope of 0 dB. For 50, a straight line with a slope of 20 dB. So, we obtain two straight lines on log scale intersecting at = 50.



Figure: Bode magnitude plot for G (s) = 2000(s+0.5)s(s+10)(s+50)


Bode magnitude plotsSteady state errors

Recall our discussion of steady state errors to step/ramp/parabolicinputs versus system type.

Consider a unity feedback system.

If the openloop plant transfer function has N poles at s = 0, thenthe system is type N.

Also recall that,

Kp is the error constant for type 0 systems,

Kv is the error constant for type 1 systems,

Ka is the error constant for type 2 systems.



For a unity feedback system, Kp = lims0 G (s) is nonzero for a type0 system.

At low frequencies, a type 0 system will have G (s) = Kp.

We can read this off from the Bode magnitude plot directly.

Kp is equal to the horizontal yintercept at low frequencies.So, we will have ess = 1/ (1 + Kp) for a step input.



Consider the below Bode magnitude diagram:

Horizontal as 0, so we know this system is type 0.Intercept the yaxis at 6 dB, so Kp = 6 dB or Kp = 2 in linear units.



For a unity feedback system, Kv = lims0 sG (s) is nonzero for atype 1 system.

At low frequencies, a type 1 system will have G (s) Kv/s.At low frequencies, |G (j)| Kv/ results in a slope of 20dB/decade.

Use the above approximation to extend the low frequency asymptoteto = 1.

The asymptote (not the original |G (j)|) evaluated at = 1 is Kv .So, we will have ess = 1/Kv for a ramp input.



Consider the below Bode magnitude diagram:

Slope is 20 dB/decade as 0, so we know the system is type 1.Extend the slope at low frequency to = 1, and read the magnitudeas 20 dB, so Kv = 20 dB or Kv = 10 in linear units.



For a unity feedback system, Ka = lims0 s2G (s) is nonzero for atype 2 system.

At low frequencies, a type 2 system will have G (s) Ka/s2.At low frequencies, |G (j)| Ka/2 results in a slope of 40dB/decade.

Use the above approximation to extend the low frequency asymptoteto = 1.

The asymptote (not the original |G (j)|) evaluated at = 1 is Ka.So, we will have ess = 1/Ka for a parabolic input.

Similar for higher order systems.


Bode phase plots

Bode plots we have seen so far consist of only the magnitude plots.

There are also phase plots.

They differ depending on whether the dynamics (the poles and thezeros) are at the RHP or the LHP.


Bode phase plotsFinding the phase of a complex number

Plot the location of the number as a vector in the complex plane.

Use trigonometry to find the phase.


Bode phase plotsFinding the phase of a complex number

Finding the phase of a complex function of is the same as findingthe phase of a complex number.

For complex numbers with positive real part,

(Complex number) = tan1( I (Complex number)R (Complex number)

)

.

For complex numbers with negative real part,

(Complex number) = 180 + tan1( I (Complex number)R (Complex number)

)

.

Note that

(ab

cd

)

= (a) + (b) (c) (d) .


Bode phase plotsConstant gain

For G (s) = K ,

(K ) =

{0 for K 0

180 for K < 0

which has a constant phase of 0 or 180.


Bode phase plotsZero or pole at origin

For a zero at origin, we have G (s) = s, so

G (j) = j = 90.

For a pole at origin, we have G (s) = 1/s, so

G (j) =1

j= j

=

1

90.

So both of them have constant phase of 90.


Bode phase plotsReal LHP zero

For a real LHP zero, we have G (s) =(

si

+ 1)

.

So,

G (j) =

(j

i+ 1

)

= tan1(

i

)

.

Figure: Bode phase plot of a real LHP zero


Bode phase plotsReal LHP pole

For a real LHP pole, we have G (s) = 1(si

+1) .

So,

G (j) = (1) (j

i+ 1

)

= tan1(

i

)

.

Figure: Bode phase plot of a real LHP pole


Bode phase plotsReal RHP zero

For a real RHP zero, we have G (s) =(

si

1)

.

So,

G (j) =

(j

i 1

)

= 180+tan1(

i

)

= 180tan1(

i

)

.

Figure: Bode phase plot of a real RHP zeroAssoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control Systems Chapter 8 49 / 129

Bode phase plotsReal RHP pole

For a real RHP pole, we have G (s) = 1(si

1) .

So,

G (j) = (1) (j

i 1

)

= [

180 + tan1(

i

)]

.

Figure: Bode phase plot of a real RHP pole


Bode phase plotsComplex LHP zero pair

Complex LHP zero pair causes phase to go from 0 to 180.

Figure: Bode phase plots for complex LHP zero pairs


Bode phase plotsComplex LHP pole pair

Complex LHP pole pair causes phase to go from 0 to 180.

Figure: Bode phase plots for complex LHP pole pairs


Bode phase plotsComplex RHP zero pair

Complex RHP zero pair causes phase to go from 360 to 180.

Figure: Bode phase plots for complex RHP zero pairs


Bode phase plotsComplex RHP pole pair

Complex RHP pole pair causes phase to go from 360 to 180.

Figure: Bode phase plots for complex RHP pole pairs


Bode phase plotsTime delay

The transfer function for time delay is given by

G (s) = exp (s) G (j) = exp (j) = 1 so

G (j) = in radians = 56.3 in degrees.

Figure: Bode phase plot for time delay

Notice that, a line becomes a curve in log scale.Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control Systems Chapter 8 55 / 129

Bode phase plotsExample

Draw the Bode phase response for

G (s) =2000 (s + 0.5)

s (s + 10) (s + 50).

First, rewrite in standard Bode form

G (s) =2000 0.510 50

(s0.5 + 1

)

s(

s10 + 1

) (s50 + 1

)

=2(

s0.5 + 1

)

s(

s10 + 1

) (s50 + 1

) .

Next, make the change s = j

G (j) = 2

(j0.5 + 1

)

j(j10 + 1

)(j50 + 1

) .



The components of the phase plot can be obtained separately.

For the DC gain of 2:

Zero phase contribution.

One real LHP zero (i = 0.5):

Phase changes from 0 at 0.1i = 0.05 to 90 at 10i = 5.

Pole at origin:

Constant phase contribution of 90.Two real LHP poles (i = 10 and i = 50):

For i = 10:

Phase changes from 0 at 0.1i = 1 to 90 at 10i = 100.For i = 50:

Phase changes from 0 at 0.1i = 5 to 90 at 10i = 500.



Figure: Bode phase plot for G (s) = 2000(s+0.5)s(s+10)(s+50)


Nonminimum phase systems

A system is called nonminimum phase if it has pole(s) or zero(s) inthe RHP.

Consider

G1 (s) = 10s+1s+10

} zero at 1pole at 10

}minimumphase

G2 (s) = 10s1s+10

} zero at + 1pole at 10

}nonminimum

phase

The magnitude responses of these two systems are found as

|G1 (j)| = 10 |j+1||j+10| = 102+1

2+100

|G2 (j)| = 10 |j1||j+10| = 102+1

2+100

which are the same.



Figure: Bode magnitude plot for G1 (s) = 10s+1s+10 and G2 (s) = 10

s1s+10



However, the phase responses are very different:

Figure: Bode phase plots for G1 (s) = 10s+1s+10 and G2 (s) = 10

s1s+10

Note that, the change in phase of G1 is much smaller than the changeof phase in G2.Hence G1 is minimum phase and G2 is nonminimum phase.



Nonminimum phase is usually associated with delay.

After writing G2 (s) in terms of G1 (s), we obtain the followingexpression

G2 (s) = G1 (s)s 1s + 1

Delay

where we notice that s1s+1 is very similar to a first order Pade

approximation of a delay.


Bode plotsStability revisited

If we know the closedloop transfer function of a system in rationalpolynomial form, we can use Routh test to find stable ranges for K.

What if we only have openloop frequency response?

Consider the following system (for now, we assume that we know thetransfer function of the system, so that we can plot the root locus)

where

D (s) = K ,G (s) =1

s (s + 1)2.



5 4 3 2 1 0 1 23

2

1

0

1

2

3

Root Locus

Real Axis

Imag

inar

y A

xis

Figure: Root locus plot for G (s) = 1s(s+1)2

with D (s) = KAssoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control Systems Chapter 8 64 / 129


We can find that there is neutral stability at K = 2 (i.e., the system isstable for K < 2 and unstable for K > 2).

Recall that a point is on the root locus if |KG (s)| = 1 andG (s) = 180.If system is neutrally stable, jaxis will have a point (or points)where |KG (j)| = 1 and G (j) = 180.



Consider the Bode magnitude plots of KG (s) for K = 0.1, K = 2 andK = 10.



Consider the Bode phase plot of KG (s).



A neutral stability condition from Bode plots is obtained when|KG (j0)| = 1 and G (j0) = 180 at the same frequency 0.In this case, increasing K results in instability (which also means thatstability is obtained when |KG (j)| < 1 at G (j0) = 180).In some cases, decreasing K results in instability (which also meansthat stability is obtained when |KG (j)| > 1 at G (j0) = 180).We can find neutral stability point on Bode plot, but we dont (yet)have a way of determining if the system is stable or not.

Nyquist found a frequency domain method to do so.


Nyquist stability

If the poles of a closedloop transfer function are in RHP, then thesystem is unstable.

Nyquist found a way to count the closedloop poles in RHP.

If the count is greater than zero, then the system is unstable.

Idea:

First, find a way to count closedloop poles inside a contour.

Second, make the contour equal to the RHP.

Nyquists technique is a graphical method to determine:

system stability,

regions of stability,

margins of stability.

It involves graphing complex functions of s as a polar plot.

Counting is related to complex functional mapping.


Nyquist stabilityComplex functional mapping

Plotting a real function f (x) of a real variable x .

This can be done.Plotting a complex function F (s) of a complex variable s.

No, this is wrong!Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control Systems Chapter 8 70 / 129


We must draw mapping of points or lines from splane to F (s)plane.



Map points A, B, C, D by F (s) = 2s + 1.



Map a square contour (closed path) by F (s) = s/ (s + 2).

Idea:

By drawing maps of a specific contour, using a mapping functionrelated to the plant openloop frequency response, we will be able todetermine closedloop stability of systems.


Nyquist stabilityMapping function: Poles of the function

When we map a contour containing (encircling) poles and zeros ofthe mapping function, this map will give us information about howmany poles and zeros are encircled by the contour.

We practice drawing maps when we know poles and zeros.

Notice that, when we evaluate G (s) at s0

G (s) |s=s0 = G (s0)= |G (s0)| exp (j)

where =

zeros

poles.



Consider the below plots

where there are no zeros or poles inside the contour c1.

The phase increases and decreases, but never undergoes a netchange of 360 (i.e., F (c1) does not encircle the origin).




where there is one pole inside the contour c2.

Resulting map undergoes 360 net phase change (i.e., F (c2) encirclesthe origin).




where there are two poles inside the contour c3.

The map F (c3) encircles the origin twice.

These examples give heuristic evidence of the general rule known asCauchys theorem.


Nyquist stabilityCauchys Theorem

Theorem

Let F (s) be the ratio of two polynomials in s.Let the closed curve C in the splane be mapped into the complex plane

through the mapping F (s).If the curve C does not pass through any zeros or poles of F (s) as it istraversed in the CW direction, the corresponding map in the F (s)planeencircles the origin N = Z P times in the CW direction, where

Z = # of zeros of F (s) in C ,

P = # of poles of F (s) in C .



Consider the system

where the transfer function is

T (s) =D (s)G (s)

1 + D (s)G (s)H (s).

For closedloop stability, we want no poles of T (s) to be in RHP.

This is equivalent to saying, we want no zeros of 1 +D (s)G (s)H (s)to be in RHP.



Let F (s) = 1 + D (s)G (s)H (s), and count the number of zeros ofF (s) in RHP using Cauchys theorem where the contour is the entireRHP.



Note that, for F (s) = 1 + D (s)G (s)H (s), from Cauchys Theorem,N = # of encirclements of origin.


Nyquist stabilityNyquist criterion

After a simple modification, Nyquist proposedF (s) = D (s)G (s)H (s) and N = # of encirclements of 1.

The Nyquist criterion simplifies Cauchys criterion for feedbacksystems of T (s) = D(s)G(s)1+D(s)G(s)H(s) form.


Nyquist stability test

We think of Nyquist path as four parts

I. The origin (which is sometimes a special case as we will see later),

II. +jaxis which is the frequency response of the openloop system(we will plot it in polar form),

III. For physical systems = 0,

IV. jaxis (i.e., complex conjugate of the region in part II).Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control Systems Chapter 8 83 / 129


So, for most physical systems, the Nyquist plot, used to determineclosedloop stability, is basicly a polar plot of loop frequency responseD (j)G (j)H (j).

We dont even need a mathematical model of the system. Measureddata of G (j) combined with known D (j) and H (j) are enoughto determine the closedloop stability.

N = # of encirclements of 1 for F (s) = D (s)G (s)H (s).P = # of poles of 1 + F (s) in RHP = # of openloop unstable poles(assuming that H (s) is stable which is a reasonable assumption).

Z = # of zeros of 1 + F (s) in RHP = # of closedloop unstablepoles

Z = N + P .

The system is stable iff Z = 0.



You need to be careful when counting encirclements.

A safe way is to draw a line from 1 in any direction, and to count #of crossings of line and diagram, and

N = # of clockwise crossings# of counterclockwise crossings.


Enhanced Nyquist stability test

Note that, changing the gain K of F (s) magnifies the entire plot.

This magnification property allows an enhanced test to obtain rangesof K for stability.

For the loop transfer function KD (s)G (s)H (s), N = # ofencirclements of 1/K point when F (s) = D (s)G (s)H (s) wherethe rest of test is the same.


Nyquist stabilityExample

Consider the system with

G (s) =5

(s + 1)2, D (s) = H (s) = 1.

We have:

I. For s = 0, G (s) = 5.

II. For s = j, G (j) = 5/ (j + 1)2.

III. For |s| = , G (s) = 0.IV. For s = j, G (j) = 5/ (j + 1)2 = G (j).



R (G (j)) I (G (j)) R (G (j)) I (G (j))0.0000 5.0000 0.0000 0.9253 0.2086 2.68560.0019 4.9999 0.0186 2.0108 0.5983 0.79060.0040 4.9998 0.0404 4.3697 0.2241 0.10820.0088 4.9988 0.0879 9.4957 0.0536 0.01140.0191 4.9945 0.1908 20.6351 0.0117 0.00110.0415 4.9742 0.4135 44.8420 0.0025 0.00010.0902 4.8797 0.8872 97.4460 0.0005 0.00000.1959 4.4590 1.8172 500.0000 0.0000 0.00000.4258 2.9333 3.0513



The Nyquist plot is obtained as:

No encirclements of 1: N = 0.No openloop unstable poles: P = 0.

So, Z = N + P = 0, thus, closedloop system is stable.

No encirclements of 1/K for any K > 0.So, the system is stable for any K > 0.



Lets confirm by checking Routh array.

Check the stability of a (s) = 1 + KG (s) = s2 + 2s + 1 + 5K .

The routh array is obtained as

s2 1 1 + 5Ks1 2

s0 1 + 5K

so the system is stable for any K > 0.




G (s) =50

(s + 1)2 (s + 10), D (s) = H (s) = 1.

We have:

I. For s = 0, G (s) = 50/10 = 5.

II. For s = j, G (j) = 50(j+1)2(j+10)

.

III. For |s| = , G (s) = 0.IV. For s = j, G (j) = 50

(j+1)2(j+10) = G (j).



R (G (j)) I (G (j))0.0 5.0000 0.0000

0.1 4.9053 0.80080.2 4.4492 1.86240.5 2.4428 3.27251.2 0.5621 2.02412.9 0.4764 0.19337.1 0.0737 0.026217.7 0.0046 0.006443.7 0.0002 0.0006100 0.0000 0.0000



Note the loop to the left of the origin

so the system is not stable for all K > 0.


Nyquist stabilityPole at the origin


G (s) =1

s (s + 1), D (s) = H (s) = 1.

Note that there is a pole at the origin.

Warning: We cannot blindly follow procedure.

Nyquist path goes through a pole at the origin.

Remember from Cauchys theorem that the path cannot pass directlythrough a pole or a zero.

Remember that we want to count the closedloop poles inside acontour that encompasses the RHP.



So, we use a slightly modified Nyquist path

where the bump at the origin makes a detour around the offendingpole.



Bump is defined by the curve

s = lim0

exp (j)

where 0 90.From above,

G (s) |s= exp(j) =1

exp (j) ( exp (j) + 1)

where 0 90.



Magnitude when 0 is

lim0

G (s) |s= exp(j)

=

1

| exp (j) + 1| 1

.

Phase when 0 is

lim0

G (s) |s= exp(j) = ( exp (j) + 1) .

So,

lim0

G ( exp (j)) lim0

(1

+

)

.

This is an arc of infinite radius, sweeping from 0 to 90+ (a littlemore than 90 because of contribution from 1/ (s + 1) term).

We cannot draw this to scale!




N = # of encirclements of 1: N = 0.P = # of loop transfer function poles inside the modified contour:P = 0.

So, Z = N + P = 0, thus, the closedloop system is stable.




G (s) =1

s2 (s + 1), D (s) = H (s) = 1.

We will use the modified Nyquist path again.

Near the origin, we will have a bump defined by curve

s = lim0

exp (j)

where 0 90.From above,

G (s) |s= exp(j) =1

2 exp (j2) ( exp (j) + 1)

where 0 90.Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control Systems Chapter 8 100 / 129


This is an arc of infinite radius, sweeping from 0 to 180+ (a littlemore than 180 because of contribution from 1/ (s + 1) term).

N = # of encirclements of 1: N = 2.P = # of loop transfer function poles inside the modified contour:P = 0.

So, Z = N + P = 2, thus, closedloop system is unstable for K = 1.

In fact, the system is unstable for any K > 0.


Stability margins

A large fraction of systems to be controlled are stable for small gainbut become unstable if gain is increased beyond a certain point.

The distance between the current (stable) system and an unstablesystem is called a stability margin.

We can have a gain margin and a phase margin.

Gain Margin: Factor by which the gain is less than the neutralstability value.

Gain margin measures How much can we increase the gain of theloop transfer function D (s)G (s)H (s) and still have a stablesystem?

Phase Margin: Phase factor by which phase is greater than neutralstability value.

Phase margin measures How much delay can we add to the looptransfer function and still have a stable system?


Stability margins

Many Nyquist plots are like this one:

Increasing loop gain magnifies the plot.

GM = 1/(distance between origin and place where Nyquist mapcrosses real axis).

If we increase gain, Nyquist map stretches and we may encircle 1.For a stable system, GM > 1 (linear units) or GM > 0 dB.


Stability margins

Many Nyquist plots are like this one:

PM = Angle to rotate Nyquist plot to achieve neutral stability =intersection of Nyquist with circle of radius 1.

If we increase openloop delay, Nyquist map rotates and we mayencircle 1.For a stable system, PM > 0.


Stability margins

This is usually easier to check on Bode plot, even though derived onNyquist plot.

Define gain crossover as frequency where Bode magnitude is 0 dB.

Define phase crossover as frequency where Bode phase is 180.PM = Bode phase at gain crossover (180).GM = 1/(Bode gain at phase crossover frequency) if Bode gain ismeasured in linear units = (Bode gain at phase crossover frequency)[dB] if Bode gain measured in dB.


Stability margins

We can also determine stability as K changes.

Instead of defining gain crossover where |G (j)| = 1, use thefrequency |KG (j)| = 1.You need to be careful using this test.

It works only if the system is minimum phase.

Nyquist plot is the safest way for this test.


Bode compensator design

The frequency response methods we have seen so far largely tell usabout stability and stability margins of a closedloop system based onopenloop response.

Now, we look at frequency response based design methods whichprimarily aim at improving stability margins.

Also, given the relationship between PM and performance, we havesome idea of transient response as well.

Start thinking of Bode magnitude and Bode phase plots as lego tomake the frequency response we want.


Bode compensator designP control

Consider the system

whereD (s) = K .

Similar to the root locus method, we use the characteristic equationin the form

cl (s) = 1 + KG (s)

to examine the stability and the performance of Bode plot parameters.



Consider the Bode plots

Gain crossover frequency is denoted by p and is equal to thefrequency where |KG (j)| = 0 dB.Phase crossover frequency is denoted by and is equal to thefrequency where KG (j) = 180.Recall that PM = 180 + G (jp).If PM < 0, the closedloop system is unstable.Recall that GM = 1/ |KG (j)| (linear units).If GM < 1 (linear units), the closedloop system is unstable.



A bonus of computing PM from the openloop frequency responsegraph is that it can help us predict the closedloop performance ofthe system.

The Bode design parameters are related to the root locus designparameters according to

p = n

1 + 44 22

PM = tan1

2

1 + 44 22

where n is the natural frequency and is the damping ratio.

The phase margin is also approximately related to the damping ratioas follows

PM/100 for PM 60.Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control Systems Chapter 8 111 / 129


Remember that is a measure of relative stability:

Hence, PM is also a measure of relative stability.


Bode compensator designExample

Consider the system

where

G (s) =1

s(s2 + 1

) , D (s) = K .

Find K so that PM = 45 for the closedloop system.



Outline of the solution:

1. Find the closedloop denominator in the form cl (s) = 1 + KG (s).

2. Plot |KG (j)| and G (j) by assuming K = 1.3. From the given (desired) PM, calculate the desired G (jp) (i.e.,

desired G (jp) =PM180).4. From the desired G (jp) and the G (j) plot, find the desired p.

5. Raise or lower the magnitude plot so that it crosses the 0 dB level atthe desired value of p.

6. Find the distance between the desired magnitude plot and the originalmagnitude plot at the desired value of p (i.e., desired + original =shift in dB).

7. Calculate K by using the following formula derived from themagnitude condition

K = 10(shift/20).



1. The closedloop denominator is in the root locus form

cl (s) = 1 + K1

s(s2 + 1

) .

2. The Bode plot for KG (s) when K = 1:



3. Desired G (jp) = PM180 = 45 180 = 135.4. The desired p is found from the desired G (jp) and the G (j)

plot.

Also note that, the actual PM = 180 118 = 62 for K = 1.Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control Systems Chapter 8 116 / 129


5. Raise or lower magnitude plot by changing K :

6. The difference between the desired magnitude plot and the originalmagnitude plot is

shift = desired + original = 0 9 = 9 dB.




K = 10(shift/20) = 10(9/20) = 2.8.

Note that, for this problem, the below specifications are equivalent:

PM = 45,

= 0.45, Desired closedloop poles are at 1 j2.



Consider the system

where

G (s) =1

(s + 1)(

s10 + 1

) (s

100 + 1) , D (s) = K .

Find K so that PM = 45 for the closedloop system.



1. The closedloop denominator is in the root locus form

cl (s) = 1 + K1

(s + 1)(

s10 + 1

) (s

100 + 1) .

2. The Bode plot for KG (s) when K = 1:



3. Desired G (jp) = PM180 = 45 180 = 135.4. The desired p is found from the desired G (jp) and the G (j)

plot.

Also note that, the actual PM = 180 0 = 180 for K = 1.Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control Systems Chapter 8 121 / 129


5. Raise or lower magnitude plot by changing K :



6. The difference between the desired magnitude plot and the originalmagnitude plot is

shift = desired + original = 0 20 = 20 dB.


K = 10(shift/20) = 10(20/20) = 10.


Bode compensator designPD control

Controller has the form D (s) = K (1 + TDs).We have seen from root locus that this has a stabilizing effect.Magnitude and phase effects of a PD controller:

PD control increases phase for frequencies over 0.1/TD .Locate 1/TD < crossover so that phase margin at crossover is better.Problem: Magnitude response continues to increase as frequencyincreases. This amplifies high frequency sensor noise.


Bode compensator designLead compensator

Lead compensator has the form D (s) = K[

Ts+1Ts+1

]

where < 1.

This is an approximate of PD control for frequencies up to = 1/ (T ).

Magnitude and phase effects of a lead compensator:


Bode compensator designLead compensator

The phase contributed at frequency is

= tan1 (T) tan1 (T) .

If we need more phase improvement, we can use a doubleleadcompensator

D (s) = K

[Ts + 1

Ts + 1

]2

.

Need to compromise between good phase margin and good sensornoise rejection at high frequency.


Bode compensator designPI control

In many problems, it is important to keep bandwidth low, and alsoreduce steady state error.

PI controller has the form D (s) = K(

1 + 1TI s

)

.

Magnitude and phase effects of a PI controller:


Bode compensator designPI control

Infinite gain at zero frequency.

Reduces steady state error to step, ramp, etc reference inputs.

But also has integrator antiwindup problems.

Adds phase below break point.

We want to keep break point frequency very low to keep fromdestabilizing system

1

TI c .


Bode compensator designLag compensator

Approximates PI controller, but without integrator overflow.

Lag compensator has the form D (s) = K(

Ts+1Ts+1

)

where > 1.

Primary objective of lag compensator design is to add 20 log10 dBgain to low frequencies without changing PM.

Magnitude and phase effects of a lag compensator:

Steady state response improves with little effect on transient response.


Frequency Response AnalysisBode plotsNyquist plotsBode Compensator Design

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