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“True” Digital Control

“True” Digital Control. .. or using the fact that we are sampling the signals to our advantge. More on the z-transform. We need to examine the z-transform in more detail.

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“True” Digital Control

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  1. “True” Digital Control .. or using the fact that we are sampling the signals to our advantge

  2. More on the z-transform • We need to examine the z-transform in more detail. • First, we look at two of the elements we have considered before -- SAMPLERS and the ZERO-ORDER HOLD (often referred to just as a HOLD).

  3. The sampler • When the switch (sampler) closes, the signal at A proceeds to B. • No signal reaches B except when the sampler closes. • This idea represents the action of a digital controller in that it only calculates its output at the sampling instants. A B

  4. The Hold • But the ‘plant’ will need to be given an input all the time -- just a ‘punch on the nose’ at the sampling instants is unlikely to work well. • This is where the Zero-Order Hold comes into action ...

  5. The Zero-Order Hold • This is the idea: • The value of the signal at C is ‘caught’ by the hold and held constant until the sampler closes again. C A Hold B

  6. Why a ‘Zero-Order Hold’ ? • Because it just catches the value of the signal at C. • A ‘First-Order Hold” would also catch the rate of change of the signal ... • ... and keep it changing at that rate. • But that is complicated, so the Zero-Order Hold is invariably used !

  7. What will it consist of ? • Normally a latching-type output port on the control computer attached to a D-A converter.

  8. Now for the z-transform • It only tells us what happens at sampling intervals • Consider this diagram: y y(4Ts) y(Ts) Time ( t) Ts 3Ts 5Ts etc

  9. The z-transform -- Continued • Suppose the switch closes for a time Tc at each sampling instant • We have a series of impulses • Each has a Laplace Transform equal to its area (‘strength’) times e-nTs to allow for the time at which it occurs • So the Laplace Transform of the complete sampled signal will be:

  10. ..and continues further ... • Tc(y(0) + y(Ts)e-sTs + y(2Ts)e-2sTs + ...) • or • We now put z = esTs and it simplifies ...

  11. So working them out looks heavy going ... • But fortunately, like Laplace Transforms, they can be looked up in tables ... • ... or converted by MATLAB using the ‘c2dm’ function. • This time, we leave the ‘tustin’ out... • [npd,dpd]=c2dm(num,den,ts,’zoh’) • The ‘zoh’ can be omitted.

  12. Rules for converting Block Diagrams to z • If blocks are separated by a sampler, convert the TFs to z and then combine the blocks. • If they are not, combine them in s and then convert to z. • This sounds complicated but it is actually easier than it looks ...

  13. An Example • The samplers close at each sampling instant • We must first combine the zero-order hold and the plant in s ... • ... and then convert to z. Plant Zero-order + -- hold G(s)

  14. Send for MATLAB ! • ‘c2dm’ will convert the hold-plant combination to z for us. • In order to determine the required D(z), we then specify the required closed-loop performance as a C.L.T.F. in z ... • ... and work out what D(z) will have to be in order to provide it. • We will find that a problem arises but it is not insurmountable.

  15. An example situation • We will re-examine our plant of transfer function • 40/(s2 + 10s + 20) • and we will try to work out a controller D(z) which will produce a unit step-response following a nice gentle first-order exponential. • We will use a time constant of 0.4 second.

  16. Our intended step-response 1 0.8 0.6 Amplitude 0.4 0.2 0 0 0.5 1 1.5 2 Time (secs)

  17. Converting the plant T.F. to z • We enlist the aid of MATLAB and ‘c2dm’, entering • num=40; • den=[1 10 20]; • [npd,dpd]=c2dm(num,den,.1) • and we obtain

  18. The digitised plant T.F. • npd = • 0 0.1449 0.1038 • dpd = • 1.0000 -1.2435 0.3679 • So the TF of plant + hold is • (0.1449z + 0.1038)/(z2 - 1.2435z + 0.3679)

  19. The digitised plant TF - Continued • or since it is (apart from MATLAB) usually more convenient to use negative powers of z • (0.1449z-1 + 0.1038z-2)/(1 - 1.2435z-1 + 0.3679z-2)

  20. The required closed-loop T.F. • Input = unit step converting to z/(z - 1) (from tables) • Output = 1 - e-t/0.4 • Another ‘bout’ with the tables produces • Output(z) =

  21. The required C.L.T.F. -- Continued • Which becomes with Ts = 0.1 second • And we divide by the input z/(z - 1) to give the CLTF • 0.2212/(z - 0.7788)

  22. Now for the required D(z) • We do a DG/(1 + DG) act again: • DG/(1 + DG) = 0.2212/(z - 0.7788) • and by rearranging • DGz - 0.7788DG = 0.2212 + 0.2212DG • and • D = 0.2212/[(z - 1)G]

  23. And now for D(z) ... • G(z) was • (0.1449z + 0.1038)/(z2 - 1.2435z + 0.3679) • so by rearranging, D(z) becomes • 1.5267 - 1.8985z-1 + 0.5616z-2 • ------------------------------------- • 1.0000 - 0.2833z-1 - 0.7167z-2 • We will try this in SIMULINK ....

  24. The resulting step-response 1.5 1 0.5 0 1 2 3 4 Time (second)

  25. Not quite as intended ! • Oh dear ! There is a slight wobble on the plant output ... • ... and a much worse one from the controller. • Let us examine D(z) again.

  26. The Ringing Pole • The denominator of the controller transfer function was calculated by: • (z - 1)(0.1449z-1 + 0.1038z-2) • The second bracket is zero when z = - 0.72 • So the controller has a pole at z = - 0.72 • This is a Ringing Pole .. a Bad Thing

  27. The s- and z-planes • We remember that in s ... • ...poles with positive real parts meant unstable systems, and .. • ...poles with non-zero imaginary parts meant systems with overshoot. • What is the situation in z ?

  28. This one may be ignored by non-mathematicians ... • We think of s as a + jw • Normally a > 0 means instability • For z, e(a + jw)Ts = eaTs x ejwTs • De Moivre: ejwTs = cos(wTs) + j sin(wTs) • so its magnitude is 1. • And |z| > 1 if a > 0 • So for stability |z| < 1.

  29. Rules of the z-plane • Systems with all their poles inside a circle of radius 1 unit centred on the origin -- the Unit Circle -- are stable. • Again, poles off the real axis indicate step overshoot. • Additionally, poles with negative REAL parts indicate oscillatory behaviour at half the sampling frequency. • We avoid them at all costs !!!

  30. “Oh, my controller is ringing !” • What we need to do to stop it is to replace the offending bracket by an equivalent static gain. • Not as difficult as it sounds • Calculate the gain by putting z = 1 ... • ... and keep the most positive (or least negative) power of z.

  31. “The bell is removed ... ” • We start from 0.1449z-1 + 0.1038z-2 • Putting z = 1 gives 0.2487 • Reinstating the z-1 gives 0.2487z-1 • The complete controller T.F. becomes • 0.8894 - 1.1062z-1 + 0.3273z-2 • ------------------------------------ • 1 - z-1 • This one works much better

  32. Response - no ringing pole 1 0.8 0.6 0.4 0.2 0 1 2 3 4 Time (second)

  33. The Kalman Controller • “Back to basics !” • nth-order systems can be made to settle in n sampling intervals • For a second-order, we could regard the first interval controller action as the “accelerator” and the second one as the “brakes”.

  34. Producing a Kalman Controller • We start with the plant transfer function in z ... in our example: • (0.1449z + 0.1038)/(z2 - 1.2435z + 0.3679) • and we make the coefficients in the numerator add up to 1 (known as ‘normalising’ the T.F.) • We do this by adding the 0.1449 and the 0.1038 together, giving 0.2487 ...

  35. The Kalman -- Continued • ... and dividing top and bottom of our T.F. by that number. • The plant T.F. becomes • 0.5825z-1 + 0.4175z-2 • ------------------------------------ • 4.020 - 4.9995z-1 + 1.4790z-2 • and we regard it as P(z)/Q(z).

  36. Nearly to the Kalman ! • It turns out that the controller TF should be Q(z)/[1 - P(z)] ! • So it will be • 4.020 - 4.9995z-1 + 1.4790z-2 • ----------------------------------- • 1 - 0.5825z-1 - 0.4175z-2 • and we duly test with SIMULINK ...

  37. The response with the Kalman 4 3 2 1 0 -1 0.2 0.4 0.6 0.8 1 Time (second)

  38. Why don’t we always use the Kalman ? • It depends on a good plant model being available (and the plant dynamics not changing) in order to work well • A heavy controller action is usual unless we sample slowly enough to encounter aliasing problems -- we will revisit this problem in the lecture on ‘Practicalities’

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