nyquest & bode plot analyses

8/8/2019 Nyquest & Bode Plot Analyses

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Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams,

second order systems, system types

Venkata Sonti

Department of Mechanical EngineeringIndian Institute of ScienceBangalore, India, 560012

This draft: March 12, 2008

1 Introduction:Frequency domain design

With lecture 4, we have completed what is considered the time domain method of system design.The reason being that time domain parameters, rise time, settling time, overshoot and steady

state error were directly addressed by the PID controller which differentiates and integrates thetime domain error. PID modifies effectively the pole positions and we can use the Root Locusmethod or Ziegler Nichols for finding the PID gains. The paramaters under the engineers controlare Kp, Ki and Kd, which using the method of Root Locus modify the p ole positions and influencethe final objective, viz., tr, ts, MpandSSE.

Let us say we model a system and then build it. We may find that the model differs quite abit from reality. And so we have to upgrade the model and say we converge. Then we measurethe tr, ts, Mp etc. and we are not satisfied. Then we specify what we want and proceed with PIDdesign. It seems simple enough. But, it is difficult to measure time domain characteristics. Weneed to give a step input which can be light and no energy input occurs. Or the input can bestrong, then the system can become nonlinear. If the system is stiff, then the response can be

very quick and die away. Measuring it is difficult. On the other hand the system may oscillatewith a lot of low frequency dynamics. For every kind of system we cannot design a step inputusystem. Locally, step inputs create nonlinear responses and deformations.

On the other hand, the frequency response of a system can be very easily measured in the lab.Either a hammer impact is given or a sinusoid/random input is given and this method has beenstandardized. Now, does such a frequency response which captures sustained oscillation behaviorcapture information about transient response is the question. We will see that the frequencyresponse is characterized by the resonance peak Mp, resonance frequency n, BW, Phase andgain margin. Indirectly, these indicate the time domain response parameters tr, etc. Here, thefinal requirement is still the time response. The tools here are compensators which use bodediagram and nyquist diagram methods.

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Nyquist Criterion

Nyquist Criterion: For a closed loop system to be stable, the Nyquist plot of G(s)H(s) mustencircle the (1, j0) point as many times as the number of poles of G(s)H(s) that are in the righthalf of s-plane, and the encirclement, if any, must be made in the closkwise direction. (Read fromBenjamin Kuo)

2 Nyquist or polar plots

The characteristic equation is1 + GH = 0 or GH= 1

so that when the system is just unstable amplitude of

GH = 1 and phase = 180

The Nyquist plot is the polar (amplitude phase) plot of a GH transfer function on the complexplane as s = j, and takes values from 0 to . The length of the arrow from the origin to theplot indicates the amplitude of GH and the angle of the arrow w.r.t the real axis indicates thephase of GH. The left figure (figure 1) is for

GH =1

s + 1 (1)

GH =1

s2 + 3s + 2

GH =1

s3 + 4s2 + 5s + 2

The right figure in figure 1 is for

GH =1

s(s + 1)(2)

GH =K

s(s2

+ 3s + 2)

where K takes values 1,3,6. At K = 6, the nyquist plot just passes through the point s = 1 +j0.Take for example the left figure and look at 1/(s2 + 3s +2). At = 0 the amplitude is 0.5 and thephase is 0 degrees. At = the amplitude is 0 and the phase is dominated by 1/s2 = 1/2and hence 180 degrees. This can be seen in the plot.

Similarly in the right figure of figure 1 the second order transfer function has phase 180 athigh frequencies and the third order transfer function has -270 degrees. And as the gain in thethird order transfer function is raised it gets closer to the point -1+j0 and finally crosses it goingunstable. Thus, a polar plot of GH gives the relative stability of the closed loop system. If weknow that at a given gain, the system is stable, then by increasing the gain and seeing which way

the polar plot moves we can find out the relative stability of the system. How far is the systemfrom going unstable. Note that we use the polar plot of GH in order to figure the relative stabilityof the closed loop system. How far the amplitude is from -1 and how far the phase is from 180degrees are indicators of stability. These are called gain and phase margins respectively and wewill elaborate more on these two in a little while.

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Real Axis

Ima

ginaryAxis

Nyquist Diagrams

0.5 0 0.5 1

0.5

0.4

0.3

0.2

0.1

0

0.1

[1 4 5 2]

[1 3 2]

[1 1]

Real Axis

Ima

ginaryAxis

Nyquist Diagrams

2 1.5 1 0.5 0 0.5 1

0.5

0.4

0.3

0.2

0.1

0

0.1[1 3 2 0]

[1 1 0]

Figure 1: Nyquist plots

3 Bode DiagramsThe same Nyquist information can be plotted with the amplitude and phase on separate plots.The amplitude is plotted after conversion to a dB scale and the x axis on a log scale. Such plotsare called Bode plots. Bode plot although very simple carries a lot of information such as: relativestability (gain and phase margins), steady state errors, damping coefficient etc. We begin with asimple example,

3.1 Example 1: 1/s

GH =1

s=

1

j

the amplitude and phase are given by

20log| 1

| = 20log|| and = 1j = 90o

The amplitude and phase are plotted in figure 9. Lets look at the amplitude plot. The x axis ison a log10 scale, 0.01, 0.1, 1, 10, etc. The amplitude goes through zero at = 1 and has anegative drop of 20 dB for every 10 times increase in frequency, i.e., -20 dB/decade. The phaseplot is relatively simple.

3.2 Example 2: 1/(s + 1)

GH =1

s + 1 1

j + 1

Amplitude is given by1

2 + 1 20log

2 + 1

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Bode Diagram

Frequency (rad/sec)

Phase(deg)

Magnitude(dB)

40

30

20

10

0

10

20

101

100

101

102

91

90.5

90

89.5

89

Figure 2: Bode plot for 1/s. Amplitude and phase.

Low frequency asymptote @ = 0, the plot approaches 1 asymptotically. Hence amplitude=1is the low frequency asymptote.On the log scale as 0, the plot approaches 0 dB.

High frequency asymptote @ , the plot approaches 1

asymptotically. Hence amplitude=1/is the high frequency asymptote.On the log scale as , the amplitude approaches 20log||. At = 1, this asymptote goesthrough 0 dB and hence intersects the low frequency asymptote. The actual value of the curve at = 1 is 20log1 + 1 = 3dB. The curve lies 3dB below the 0 dB asymptote.

Phase is given by

1

j+1as 0, the phase goes to 00as , the phase approaches 900 See figure 3.

3.3 Example 3: 1/(s + 0.1)

GH =1

s + 0.11

s+0.1 1

j+0.1

Magnitude 12+0.01

0, reaches constant=10 asymptote , 1

is the asymptote.

dB = 20log10

2 + 0.01

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Bode Diagram

Frequency (rad/sec)

Phase(deg)

Magnitude(dB)

25

20

15

10

5

0

101

100

101

90

45

0

Figure 3: Bode plot for 1/s + 1. Note the asymptotes in the magnitude plot.

0, 20log0.1 = 20dB , 20log|| 20 dB/decadePhase as before.The meeting point of the asymptotes is the 20 dB line at = 0.1 (note s+0.1). The actual valueof the curve at = 0.1 is 20log0.1 0.1 + 0.1 + 0.1 = 16.9897. This value is 3 dB below the20 dB asymptote.

3.4 Example 4: 1/(s + 10)

GH =1

s + 101

s+10 1

j+10

Magnitude 12+100

0, 110

asymptote. , 1

asymptote.

Following the earlier examples, the low frequency asymptote reaches the value -20 dB. The highfrequency asymptote goes as -20dB/decade. The two meet at = 10. The actual curve is 3 dBbelow the -20 dB asymptote at = 10. See figure 5.

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Bode Diagram

Frequency (rad/sec)

Phase(deg)

Magnitude(dB)

5

0

5

10

15

20

102

101

100

90

45

0

Asymptote

Figure 4: Bode plot for 1/s + 0.1. Note the asymptotes in the magnitude plot meet at = 0.1.

Bode Diagram

Frequency (rad/sec)

Phase(deg)

Ma

gnitude(dB)

-45

-40

-35

-30

-25

-20

100

101

102

90

45

0

Figure 5: Bode plot for 1/s + 10. Note the asymptotes in the magnitude plot.

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3.5 Example 4: 1/((s + 0.1)(s + 1)(s + 10))

We examine a complicated case. There are three terms and a

GH =1

(s + 0.1)(s + 1)(s + 10)

0, 10.110 = 1 asymptote.

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asymptote.Following the earlier examples, the low frequency asymptote reaches the value 0 dB. The high

frequency asymptote goes as -60dB/decade. The curve will follow 1/ asymptote at = 0.1with a slope of 20dB/decade. Then, the curve will follow 1/2 asymptote at = 1 with aslope of 40dB/decade. Then, the curve will follow 1/3 asymptote at = 10 with a slopeof 60dB/decade. The actual curve is 3 dB below the meeting points of asymptotes at thecorresponding . See figure 6. Phase goes to 0 at low frequencies and to 270o at high frequencies.

Frequency (rad/sec)

Phase(deg);Magnitude(

dB)

Bode Diagrams

150

100

50

0

w=0.1

w=1.0

w=10

102

101

100

101

102

300

200

100

0

Figure 6: Bode plot for 1/((s + 0.1)(s + 1)(s + 10)).

4 Second order system: n2

s2+2ns+n2

Lets look at the closed loop Tr. Fn.

M(s) =n

2

s2 + 2ns + n2

1

1 + (2/n)s + 1/n2s2

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We are interested in values of


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For a 1% settling time we know

ts =4.6

n

ntr = 1.8

Looking at M(s) in the frequenc domain:

M(j) =C(j)

R(j)=

2n(j)2 + 2n(j) + 2n

=1

1 ( n

)2 + j2 n

M(ju) =1

1 u2 + j2u

|M(ju)| = 1[(1 u2)2 + (2u)2] 12

M(ju) = m(u) = tan1 2u1 u2

In order to find out the frequency at which the peak transfer function value occurs in frequency:

dM(u)

du= 0 4u3 4u + 8u2 = 0

u = up = 0

up =

1 22

p = n

1 22 valid for < 0.707

Mp =1

2

1 2So Mp is purely a function of .

For 0.707, p = 0, Mp = 1.

Bandwidth of a second order system is frequency at which M() drops to 70.7% of its valuefrom its low frequency asymptote.

M(u) =1

(1 u2)2 + (2u)2 = 0.707

u2 = (1 22)

44 42 + 2

Bw = n[(1

22) + 44

42 + 2]1

2

as , bandwidth and Mp .

1. Maximum overshoot of unit step response depends on only.

2. Resonance peak of closed loop frequency response Mp depends only on .

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M

P

1

0.707

n

=0.707

P

Figure 8:

3. Bandwidth is directly proportional to n.

4. High bandwidth = largerMp.5. n =

1.8tr

.

6. ts = f(n) =4.6n

for 1% settling time.

7. Also, the following relations are relevant between bandwidth, n, ts, tr and :

Bw = n[(1 22) +

44 42 + 2] 12

n =4.6

ts

Bw =4.6

ts[(1 22) +

44 42 + 2] 12

n =

tp

1 2

Bw = tp

1 2 [(1 22) +

44 42 + 2]

1

2

8. Open loop system must be stable for Bode plot design.

9. For 2nd order system closed loop damping ratio is approximately equal to PM100

, if phasemargin is between 0 and 600.

10. bandwidth n.We use these relations quite freely even for non-second-order systems...with caution.

5 System Types

The type of the open loop system one is dealing with decides its steady state error nature. Hence,it is important to know. Also, the steady state error depends not only on the form of the transferfunction but also on the input, like step or ramp. Here we will look at system types with regardto only a step input. More is available in Benjamin Kuo.

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R(s)+-

r(t) c(t)G(s)

e(t)

E(s) C(s)

b(t)

B(s)H(s)

Figure 9:

The error E(s) is given by

E(s) = R(s) C(s)H(s) (3)C(s) = E(s)G(s)

E(s) = R(s) E(s)G(s)H(s)E(s) =

R(s)

1 + G(s)H(s)

E(s), the LT of e(t) is given by

E(s) =

0

e(t)estdt.

= [este(t)

s ]|0

0

est

sde(t)

dt.dt

=e(0)

s+

1

s

0

estde(t)

dt.dt

sE(s) = e(0) +

0

estde(t)

dt.dt

lims0 sE

(s) = e(0) +

0

de = e() e(0) = e()

Thus steady state error, i.e., e() is given by the above formula.Step response

For step response R(s) = Rs

and let us denote lims0GH = kp, then

E(s) =R

s(1 + kp)

and if the steady state errore() = lim

s0 sE(s) = 0

then

lims0 sE(s) =R

(1 + kp) = 0

or kp =. If the loop transfer function has the form

GH =1

sjF(s)

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then the system is of type j.If j=0, then

lims0

GH = F(0) = Kp

is a finite value. Thus, Kp is finite for type 0 systems and hence the steady state error is not zero.For type 1 and higher systems Kp is infinity. Hence, for step inputs steady state error is not 0for type 0 systems and is 0 for type 1 and higher systems.

A similar set of rules can be defined for ramp inputs. (Read Kuo).

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nyquest & bode plot analyses

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