Solution LTI exam

7/24/2019 Solution LTI exam

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Andrea Bisoffi Solution to Ex. 5, July 20, 2015

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1.1 Requirements

Consider the control scheme below

yoR(s) G(s)

y

d

e u

n

and the following selections for R(s) and G(s):

R(s) =R(1 + sR)

(1 + sTR), G(s) = 20

es

s2(1 + s/30).

Set temporarily = 0 and tune the parameters of R(s) in such a waythat

m 35,

disturbance d is attenuated by a factor 10 on the output y for 5 rad/s.

When the parameters are tuned, (i) draw the Bode diagram ofL(s) (mag-nitude and phase), (ii) compute the corresponding c and m, (iii) draw(preferably with a different color) the magnitudeBode diagram ofS(s) =

1

1 + L(s), (iv) find the smallest value of for which any of the above re-

quirements is violated.

1.2 Solution

In view of parameter tuning, note the following.

It is completely intuitive from the geometric representation of the at-tenuation requirements for disturbances d and n that they set lowerand upper bounds on the crossover frequency c.

Compared to these bounds, the plant pole in = 30 rad/s is at highfrequency with respect to the realistically achieavable c.

Due to the double integrator in the plant, the initial phase in the phasediagram is 180. To obtain a phase margin m of 35

, the controllerzero has to be used to suitably increase the phase.

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150

100

50

0

50

100

c= 1.8895,

m= 40.5098, (min) = 0.050895

Magnitude(dB)

102

101

100

101

102

103

300

250

200

150

100

50

Phase(deg)

Frequency (rad/s)

Figure 1: Asymptotic (blue) and exact (red) Bode diagram ofL(s), for-bidden areas for disturbance attenuation (magenta boxes) and magnitudediagram for S(s) (black).

By tuningR and R, one can satisfy the requirements on attenuationof d and on m. Then, one can realize that the controller pole isneeeded to guarantee the required attenuation on n. Incidentally, ifone setTR= 0, the controller would not be physically realisable.

A possible choice of the three controller parameters is R = 1/20, R =1/0.46, TR= 1/3.

The resulting Bode diagrams forL(s) are in Figure 1, together with theresulting c and m.

The only requirement that is affected by a delay different from zero is the

phase margin, consistently with the fact that delays can reduce the phasemargin and also make the closed loop system unstable. Depending on thevalue of phase margin you obtained after the design, the delay contributionc

180

should be less than m 35

. Therefore, m35

c

180 . The

minimum value of for the current controller parameters can be found inFigure 1.

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2 Text 2

2.1 Requirements

Consider the control scheme below

yoR(s) G(s)

y

d

e u

n

and the following selections for R(s) and G(s):

R(s) =R(1 + sR)

(1 + sTR), G(s) = 20

1

s2(1 + s30

)

Tune the parameters ofR(s) in such a way that

m 35,

disturbance d is attenuated by a factor 10 on the output y for 5 rad/s.

When the parameters are tuned, (i) draw the Bode diagram ofL(s) (mag-nitude and phase) (ii) compute the corresponding c and m, (iii) draw the

magnitudeBode diagram ofS(s) = 1

1 + L(s)starting from the one ofL(s)

(preferably with a different color), (iv) give the steady state response for

y(t) when yo(t) = 2 ramp(t), d(t) = 0, n(t) = 0.

2.2 Solution

The solution is identical to the one in Section 1.2, apart from point (iv).

Since L(s) contains a doubleintegrator, the steady-state error to a setpointramp response is zero, which means that asymptotically y(t) = yo(t). Theanswer is then y(t) = 2ramp(t).

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Solution LTI exam