How to cancel same pole-zero pair for Z-function ?

With matlab, How to cancel the same pole-zero pair for Z-function , for example : minreal for s-function?

Answer 1

[Star Strider]

The minreal function works with discrete systems.

Example —

n = [1    -9    26   -24];
d = [1   -12    41   -42];

tfs = tf(n,d);
figure
bode(tfs)
[ps,zs] = pzmap(tfs)
rs = minreal(tfs)

tfz = tf(n, d, 0.1)
figure
bode(tfz)
[pz1,zz1] = pzmap(tfz)
rz = minreal(tfz)

For the continuous system:

rs =
  s - 4
  -----
  s - 7

and for the discrete system:

rz = 
  z - 4
  -----
  z - 7

In both, it cancels the poles-zeros at 2 and 3.

[tan zhang]

Script in Matlab as follows , NTF1=NTF2 by math, but run result : NTF1 <> NTF2, why????

fs=250e3;

z=tf([1 0],[1],1.0/fs,'Variable','z');

K1=1.0/2,K2=1.0/3,K3=1.0/2;

I1=1.0/(z-1),I2=z/(z-1);

NTF1=(1+K1*I1*(K1*I1+K1*I1*K2*I2))^-1; % this input mode is Convenience and liked

NTF2=tf([1 -3 3 -1],[1 -3 3+K1^2+K1^2*K2 -1-K1^2],1.0/fs,'Variable','z'); % this input mode isn't Convenience and liked

bode(NTF1,NTF2);

matlab运行结果：

NTF1 =

z^4 - 4 z^3 + 6 z^2 - 4 z + 1

----------------------------------------

z^4 - 4 z^3 + 6.333 z^2 - 4.583 z + 1.25

NTF2 =

z^3 - 3 z^2 + 3 z - 1

----------------------------

z^3 - 3 z^2 + 3.333 z - 1.25

[Star Strider]

The poles and zeros are not exactly the same in ‘NTF1’ and ‘NTF2’, so they are not exactly equal, and in ‘NTF1’ the pole and zero near 1+0j are not exactly the same, so minreal will not cancel them:

NTF1p = pole(NTF1)
NTF1z = zero(NTF1)

NTF2p = pole(NTF2)
NTF2z = zero(NTF2)

produces:

NTF1p =
  1.109321664527059 + 0.607607613792185i
  1.109321664527059 - 0.607607613792185i
  1.000000000000006 + 0.000000000000000i
  0.781356670945876 + 0.000000000000000i

NTF1z =
  1.000219151659574 + 0.000000000000000i
  0.999999983229428 + 0.000219134884851i
  0.999999983229428 - 0.000219134884851i
  0.999780881881574 + 0.000000000000000i


NTF2p =
  1.109321664527060 + 0.607607613792186i
  1.109321664527060 - 0.607607613792186i
  0.781356670945880 + 0.000000000000000i

NTF2z =
  1.000005497150145 + 0.000000000000000i
  0.999997251424926 + 0.000004760605284i
  0.999997251424926 - 0.000004760605284i

If you specify a tolerance, minreal will cancel the pole and zero in ‘NTF1’ at 1+j0:

NTF1m = minreal(NTF1, 0.001)
NFT2m = minreal(NTF2)

and the result is as expected:

NTF1m =
 
     z^3 - 3 z^2 + 3 z - 1
  ----------------------------
  z^3 - 3 z^2 + 3.333 z - 1.25
 
Sample time: 4e-06 seconds
Discrete-time transfer function.


NFT2m =
 
     z^3 - 3 z^2 + 3 z - 1
  ----------------------------
  z^3 - 3 z^2 + 3.333 z - 1.25
 
Sample time: 4e-06 seconds
Discrete-time transfer function.

so with the pole-zero cancellation, and the bode plot shows they are now the same:

bode(NTF1m, NFT2m)

[tan zhang]

Star Strider：

Thank you !

I have get your thought ,but select appropriate tolerance for minreal isn't easy in reality!

Thanks again!

The Learner

[Star Strider]

My pleasure!

If my Answer helped you solve your problem, please Accept it!