nyquist justification

7/29/2019 nyquist justification

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Control Systems Analysis of Nyquist Plot 1

w t norris: Nyquist Analysis: 23 March 2007

Nyquist Criterion - an analysis.

The Nyquist plot is used first as another means of establishing stability but also allows amethod of assessing the degree of stability by introducing the notion of phase andamplitude margins.

We consider a feedback system as shown below.

X Y = XX Y

+-

G H

1+ GH

G 1+ GH

G

For stability G has no poles in the right hand plane. If so then (1+ GH) must have no zerosin the right hand plane. We assume that the poles of GH in the right hand plane areknown including those on the imaginary axis. We can of course find zeros by evaluating

(1+ GH) and seeking the roots of the numerator, which are the zeros of (1+ GH) and findif any of these are in the right hand plane, i.e., with positive real parts.

The Nyquist plot method is less direct but probably yields more insight.

C

Real

Real

Imaginary

s

Z

|s-Z|

Imaginary

M

Consider such a zero, Z in the right hand plane.Further consider a contour C surrounding the zero.

The value at a typical point has s given by:

| s Z| exp( j) = (real quantity) exp( j)

where is marked in the figure.. If we plot thesepoints we get a curve, M, encircling the origin in acounter-clockwise direction as increases. ismeasured in the normal way in a counter-clockwisedirection

The M curve is similar in shape to the C curve


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C

Real

Imaginary

s

|s-P|

P

Real

Imaginary

M

|s-P|

1

In the case of a pole the value of s on the contour isgiven by:

1| s P|

exp (-j)(real quantity) exp( j)

Now the plot of the values of the points on the curveas increases. This gives a contour M againencircling the origin but this time it goes round theorigin in a clockwise direction.

Consider a function (s2 2s +2)which has zeros at(1+j) and (1j) as shown in the left hand plot below.

In this case the M curve is not the same shape as thec curve, swelling where the C curve is close to theorigin and vice versa.

There is general theorem, attributed to Cauchy about these contours about poles and zeros.

If a closed contour C encircles in an anticlockwise sense Z zeros and P poles of a functionthen the map of the function on the C contour yields a contour that encircles the origin N

times in a clockwise sense where

N = Z P

The proof for a rational function, that is one which is a ratio of polynomials is straightforward based on what is just written. You may recognise a parallel with the method forsketching the Bode plot from a knowledge of the poles and zeros.

Let the function be F(s) =(sz1)(sz2)...(szZ)(sp1)(sp2)...(spP)

=

|sz1| |sz2|...|szZ| ej1ej2...ejZ

(sp1)(sp2)...(spP) ej1ej1....ejP

where is the angle between the line from the point s to the zero 1, etc. and the m are thesame angles for the poles. All as illustrated for the zero and pole.

F(s) =|sz1| |sz2|...|szZ||sp1| |sp2|...|spP|

exp( )j[(1+2+...Z)(1+2 . .+P)]

= ( Positive real radius) exp( )j[(1+2+...Z)(1+2 . .+P)]

Thus as the point so moves around a closed contour the P and the Z vary from 0 to 2.

In the M plot the total angle changes in total by2 ( ZP)


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and so encircles the origin N = (ZP) times.

Some authors use the clockwise direction for traversing the contours or speak of the'positive' direction. It does not matter as long as one remains consistent. I have chosenanticlockwise since that is the sense in which angles from an axis are usually measured.

Consider a function (s2 2s +2)which has zeros at (1+j) and (1j) as shown in the left handplot below.

-1 1 2 3

-1

1

2

3

-10 -5 5

-10

-5

5

10

C1

C2

M1

M2

Real

Real

Imaginary Imaginary

s-planeZeros at (1+j) and (1-j)

Mapping of

s2-2s+2

Now evaluate the function of s on the circular contour C1 in the left hand figure. Thiscontour C! encloses only one of the zeros. Plot the result as in the right hand figure andthe curve M1 results. The curve M1 circles the origin once in the same anticlockwise

sense as shown as the original contour. The second zero has a marked influence on theshape of the curve M as we see in the appearance of the small loop which arises becauseof a kink in the phase of the function due to the second pole. But the total encirclement ofthe origin is only once,

The contour C2 encircles both zeros and the mapped contour now encircles the origin inthe right hand figure twice.

The circle C1 has a radius 15. On C1 at s = 1+j + ej/2 M1 point = 525On C1 at s = 1+j + ej(/2+.3) M1 point = 4.72j 216.

Further points can be checked.

An example of a pole and a zero comes next. Note that these are a very artificial pole andzero. ( Why ?)


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-1 1 2 3

-1

1

2

3

C1

C2

M1

M2

Real

Real

Imaginary

Imaginary

s-planeZeros at (1+j)

and Pole at (1-j)

Mapping of

s-1-j

s-1+j

-2 2 4

-2

-1

12

Contour C1 on the left encloses the zero and gives the curve M1 which encircles theorigin. Both C1 and M1 are counter clockwise.

The contour C2 encloses both a zero and a pole and the corresponding curve M2, does notencircle the origin.

Application to Feedback Control Systems

Coming back to the feedback circuit

X Y = XX Y

+-

G H

1+ GH

G 1+ GH

G

To write it again. The transfer function isG

1+ GH. For stability (1+ GH) must have no

zeros. If it has poles then multiplication by the denominator of GH will remove these inthe final working. Nyquist's idea was to count the number of poles by using the Cauchytheorem just discussed. We wish to count the number of zeros in (1+ GH) in the righthand plane. Choose as the contour C the imaginary axis and a return in a 'circle' at infinity

embracing the whole of the right hand plane.


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C

Real

Imaginary, j

s-plane

Return at

infinity

=+

=

We evaluate (1+ GH) on the contour. Wework with GH since this will normallyhave fewer zeros than poles and so thevalue for s = will be zero. This part ofthe contour loop does not have to be

computed.

Taking an anticlockwise contour we needto evaluate GH along the imaginary axisfrom = + to = .

Counting the number of times, N, that thismap encircles the origin will give the value

N = ZP.

The number of poles, P, in the right hand

plane, that is with positive real parts, isthat in GH. Since these are usually easilyfound we may write

Z = N + P

The final step is, instead of evaluating F = 1 + GH to evaluate F'= F-1 = GH. Then wehave the number N as the number of encirclements of the point (1,j 0 ).

Thus the Nyquist criterion for stability becomes

Plot GH from = to = . The return circle at infinity is all at zero.

If N is the number ofclockwise encirclements of the point (1, 0) and

if P is the number of poles with positive real parts in GH

then the number of zeros of (1+ GH) in the right hand plane is Z = N+P.

If Z > 0 then the system is unstable.

Notes

1. Clearly if there are poles with positive real parts then for stability the mapped contourmust encircle the -1 point in a clockwise direction so that N is negative.

2. Since normally GH has fewer zeros than poles at = the value of GH will be zero.

3. Further GH(j) is the complex conjugate of GH(j) so the plot for will be themirror image in the real axis of the plot for GH(j) with < < 0.

4. A wrinkle occurs if there are poles on the imaginary axis: we deal with these cases with,indeed, a wrinkle as would be seen on closer examination of the case which we do not dohere.


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Examples

1) Set GH =1

(s+1)(s+2).

The plot is shown on the right and clearlydoes not encircle ( -1, 0) and so therepresented system is stable. Indeed all 2ndorder functions GH are stable.

0.2 0.4

-0.3

0.3

=

=0

=Real

Imaginary

2) Set GH =1

s(s+1).

The Nyquist plot is shown opposite. The

great circle at infinity has been shown onlyroughly.

Note that it starts at (1, ) and not on theimaginary axis. This reflects theapproximation in the wrinkle computation.

Terms like1s

occur quite often in

mechanical systems where there isintegration of a velocity to get position.

This, being second order is stable.

-1

-20

-10

10

20

=

=

=0+

=0

3) Set GH =8

s(s+1)(s+2).

In this case the contour encloses the point(1, 0) twice. There are two poles, apartfrom the pole at = 0 which we haveexcluded. The system is unstable.

Plotting the diagrams involves numericalwork at a range of frequencies. FortunatelyStudent MATLAB can do this.

nyquist([8],[1 3 2 0],{1,100}]

gets the loop for you. Note the curlybrackets around the frequency range in thestatement above.

Stability can be achieved by reducing the 8in the numerator to a lower value.

-2 -1

-0.5

0.5=

=

=0+

=0


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4) Set GH =1

s2(s+1)

Since we have a double pole at s=0 we geta whole circle around infinity.

The branches fro finite stretch up andaway to infinity so that the circle is a circleand a half. We have drawn the branchesover a limited range but the whole curveencircles the (1,0) point twice and thesystem is unstable.

One needs to watch for double polesystems like this one.

-0.5

-0.4

0.4

=

=

=0+

=0

5) A pole of GH in the right hand planedoes not necessarily lead to instability.

Set GH =1

s(s1)

In this case however the contour doesencircle (1,0) in the anticlockwisedirection and the system is not stable.

Note that sGH is negative at s=0 and the

infinite half circles are in the left handplane.

-0.5

-0.4

0.4

=

=

=0+

=0

6) Systems can be stabilised byintroducing a derivative term as we sawfor the oven.

Set GH =1+2ss(s1)

Now the encirclement of (1, 0) is in theanticlockwise direction so N = 1. There isa pole in the right hand plane of GH soP = 1. And

Z = P + N = 1 + 1 = 0.

-2-1

-1.5

-1

1

1.5

=

=

=0+

=0

The system is stable. This is a rather unusual system. For reassurance let us look at theoverall closed loop transfer function.

Hclosed loop =1

1+GH=

1

1+1+2ss(s1)

=s(s1)

s2 + s +1.

The poles are at s = 12 j3

2; the system is indeed stable


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Bode Plots.

The Nyquist plot is of the magnitude and phase of the response of the open transferfunction. This can equally be shown as a Bode plot. It can also be shown as a Nichols plotwhich we deal later ( but not in this edition of the notes).

Consider example 3 of the above Nyquist plots. GH = 8s(s+1)(s+2)

. On the left below is

the Nyquist plot showing just part of the branch from = towards = 0. Indeed sincethere is mirror symmetry of the Nyquist plot in the real axis it is usual to draw only the partfor positive.

On the right is the Bode plot over the range of frequency ( in Hz, not in rad s1) . Spendsome time comparing these plots and in understanding why they show the sameinformation. and see how to draw the Nyquist plot given the Bode plot.

The range in the Bode plot covers the part of the Nyquist plot near the crossing of the realaxis.

0.15 0.2 0.3 0.5Frequency,Hz.

-180

-100

0

100

180

Phasedeg

rees

0.1 0.15 0.2 0.3 0.5 Frequency,Hz.

-10

-5

0

5

10

MagnitudedB

-2 -1

-0.5

0.5=

=0+

Amplitudemargin,+3 dB

Phasemargin,

-10

Nyquist Plot

Bode Plots

We have marked the amplitude margin which is the amplitude at the cross over of thereal axis on the Nyquist diagram where the phase is 180. We have also marked theplace where the amplitude is 0 dB, i.e., no amplification and shown the phase margin.

This system is unstable. The phase amplitude is too large at = 180 and the phasemargin at 0 dB lies beyond 180

If we reduce the 'K' coefficient and make GH =4

s(s+1)(s+2)we get a stable system.


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The Nyquist and Bode plots for the open loop transfer function GH =4

s(s+1)(s+2)are:

-1.2 -1

-0.4

-0.2

0.1=

=0+

Nyquist Plot

Bode Plot

0.15 0.2 0.3 0.5

Frequency,Hz.

-180

-90

0

90

180

Phasedegrees

0.1 0.15 0.2 0.3 0.5Frequency,Hz.

-20

-15

-10

-5

0

5

Magnitude

dB

Amplitudemargin,

3 dBAmplitudemargin,

3 dB

Phasemargin, 15

Phasemargin, 15

1

We can compare the critical quantities, the amplitude margin and the phase margin ineither plot but it is easier with the Bode plot.

For stability in the Bode plot where there are no poles in the open loop transferfunction are

a) the 0 dB value of frequency must be f or a phase angle of less than 180

The amplitude or gain margin is the gain ( usually in dB) at the frequency where the phaseangle is equal to 180. In this case the gain margin is 3 dB,. i.e., a factor of 14

b) the phase at 0 dB must be less than 180 The phase margin is the phasedifference between the phase where the amplitude is 0 dB and 180. IN this case the phasemargin is 15.


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For the relatively unusual case where there are poles in the open loop transfer functionwe get, as for case 6 above where

GH =1+2s

s(s1)we get

0.15 0.2 0.3 0.5

Frequency,Hz.

-180

-90

0

90

180

Phasedegre

es

0.1 0.15 0.2 0.3 0.5Frequency,Hz.

-4

-2

0

2

4

6

Magnitude

dB

-2-1

-1.5

-1

1

1.5

=

=0+

Nyquist plotBode plot

Amplitudemargin,

6 dB

Phasemargin,

45

Note that the phase angle lag is increasing as the frequency increases in contradistinction tothe case where there is no pole in the open loop transfer function. Thus we measure fromthe high frequency end of the Bode plot to determine the overall response.

The gain margin is 6 dB, i.e., a factor of 2. The phase margin is 45.

The plot curls round the (-1 + j 0) point in a clockwise direction so N = 1. But since thereis a pole of GH in the right hand plane P = 1. Thus the number of zeros in (1+ GH) is

Z = N + P = -1 +1 = 0The system is stable.

This example suggests another rule for the Nyquist plot:

If the Nyquist plot is such as when moving from = towards = 0 the point (1 + j 0)

lies on the right of the line then the system is stable.

But except in the clearest cases it is wisest to check the conclusion of the Nyquist plot.

In this case GH =1+2ss(s1)

. If we set H = 1 so we have a unity feed back system and a

forward transfer function, G =1+2s

s(s1)that is definitely unstable we nevertheless get

T =G

1+ G=

1+2s1+2s+s(s1)

=1+2s

s2 + s + 1

of which the poles are at s = 12

j 32 ; the system is stable.


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THE Rest of This text is still in preparation.

Nichols Plot

Nathaniel B Nichols contrived a graphical way of finding the closed loop transfer functionfrom the open loop transfer function when the feed back is unity, i.e., for a simplifiedsystem with H = 1. A plotting paper was used. With the advent of computers this methodis now less valuable but we show it here for the curious.

The Nichols plot has as axes the phase of G and its magnitude expressed in dB. The chartis marked with lines of constant M and constant , the magnitude and phase of the closedloop transfer function.

One may also make an Inverse Nichols plot with the magnitude in dB and the phase of theclosed loop transfer function as axes and lines of constant magnitude and phase of G theopen loop transfer function marked.

Setting the open loop transfer function, G = A ej, we get the closed loop transferfunction, T = Mej

T=Mej =Aej

1+Aejand Aej=

Mej

1Mej

Using these expressions we may draw charts of lines of constant M and on a plot ofA v. ( the Nichols Chart) and of constant A and on a plot of M v. ( the inverseNichols Chart).whence

|T| =A

1+A2+2Acosand

A =T2 |T|1T2sin2

T21Since A is real

1T2sin2 0 so |sin| 1

|T|The phase of T is

= arctan

sin

A+cos

whence A =sin()

sinPoles on the imaginary axis

Take a single pole on the imaginary axis at jo. Write GH =F'(s)sjo

.I

Make a small indent of radius in the contour to the right of the pole as shown in theupper left hand graph below. We may write

s= jo + ej

over the part of the contour that forms the indent. Since F'(s) has no singularities we canfor small values of set F'(s) = F'(jo) which will be approximately a constant, complexperhaps. Thus


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GH F'(jo)

jo + ej jo= F'(jo)

ej

C

Real

Real

Imaginary, jImaginary

s-plane

j

Radius,

Pole

at j

M

a

a

b

b

Real

Imaginary

M

a

b

Radius

Radius

Real

Imaginary

M

ab

F'(j)

F'(j)

F'(j)

=F'(j)

F'(j)

F'(j)

=F'(j

)

=0

=0 =0

=

2

=2

=

2

= 2

= 2

Since varies around the indent circle from2

t o 2

the mapping of this part of the

contour is a semi-circle of radius F'(s)/ . For F'(jo) real and positive the semicircle isas shown in the upper right hand graph above. Note the positions of the correspondingpoints, 'a' and 'b'.

If F'(jo) is complex one gets the situation as in the lower right hand graph. For a double

pole the mapped contour becomes a full circle. For triple and higher multiple poles thereader will need to be able to think for themselves.

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nyquist justification