“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.

• 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).

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

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 ...

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.

HoldA BC

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 !

What will it consist of ?

• Normally a latching-type output port on the control computer attached to a D-A converter.

Now for the z-transform

• It only tells us what happens at sampling intervals

• Consider this diagram:y

Time ( t)Ts 3Ts 5Ts etc

y(Ts)

y(4Ts)

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:

..and continues further ...

• Tc(y(0) + y(Ts)e-sTs + y(2Ts)e-2sTs + ...)

• or

• We now put z = esTs and it simplifies ...

T y rT ec s

srT

r

s( )

0

T y rT zc s

r

r

( )

0

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.

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 ...

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.

Zero-orderhold

PlantG(s)+--

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.

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.

Our intended step-response

0 0.5 1 1.5 20

0.2

0.4

0.6

0.8

1

Time (secs)

Am

plitu

de

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

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)

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)

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) = ( )

( )( )

.

.

1

1

0 4

0 4

e z

z z e

T

T

s

s

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)

0 22121 0 7788

.( )( . )

zz z

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]

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 ....

The resulting step-response

1 2 3 4

0

0.5

1

1.5

Time (second)

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.

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

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 ?

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.

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 !!!

“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.

“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

Response - no ringing pole

1 2 3 40

0.2

0.4

0.6

0.8

1

Time (second)

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”.

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 ...

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).

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 ...

The response with the Kalman

0.2 0.4 0.6 0.8 1-1

0

1

2

3

4

Time (second)

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’