c2d function 'zoh' method formula

Hi!

My following question would be, say i have a continuous time PI controller:

and i apply the function c2d(,,'zoh'), then it appears that matlab uses he substitution:

.

Which is completly fine, however if i have a transfer function like:

then when applying the c2d function c2d(,,'zoh'), it appears that the previous substitution in this case changes (s!=(z-1)/Ts)depending on the value of . I would like to know why Matlab does this, and what is its algebraic formula if possible?

Thank you!

Answer 1

[Paul]

I suspect that in your first case for C the substitution you cite is only applicable because of the form of C. In general, the ZOH approximation does not use that substitution. Though probably not implemented this way, the genaral form for the ZOH approximation can be implemented as shown below, and compared to what Matlab produces

P=5;Ti=6;Ts=.1;  % example data
C=tf(P*[Ti 1],[Ti 0]);
Cz=c2d(C,Ts,'zoh');
Cznew = minreal(c2d(C*tf(1,[1 0]),Ts,'impulse')*tf([1 -1],[1 0],Ts)/Ts);
[Cz Cznew]

ans =
 
  From input 1 to output:
  5 z - 4.917
  -----------
     z - 1
 
  From input 2 to output:
  5 z - 4.917
  -----------
     z - 1
 
Sample time: 0.1 seconds
Discrete-time transfer function.

R=10;Te = 5; W=tf(1,[R*Te 1]);
Wz=c2d(W,Ts,'zoh');
Wznew = minreal(c2d(W*tf(1,[1 0]),Ts,'impulse')*tf([1 -1],[1 0],Ts)/Ts);
[Wz Wznew]

ans =
 
  From input 1 to output:
  0.001998
  ---------
  z - 0.998
 
  From input 2 to output:
  0.001998
  ---------
  z - 0.998
 
Sample time: 0.1 seconds
Discrete-time transfer function.

The substitution you cite, s = (z -1)/Ts, the forward rectangular rule, which appears to be an allowable, albeit undocumented, method input to c2d.

Edit: The forward rectangular rule is not implemented. When specifying the method input as a character string, only the first character matters, so 'forward' is actually 'foh'

H = tf(1,[1 1 1]);
Ts = 0.1;
H1 = c2d(H,Ts,'foh');
H2 = c2d(H,Ts,'forward');
H3 = c2d(H,Ts,'fu');
H1 - H2

ans =
 
  0
 
Static gain.

H1 - H3

ans =
 
  0
 
Static gain.

[Tamas Toth-K]

Splendid! Thank you!

[Tamas Toth-K]

Although i believed i would be able to go on from this, i must say i am stuck once again, i somewhat understand what is happening, but it is not clear to me what 'impulse' does in Matlab as in mathematica no method within function ToDiscreteTimeModel gives quite the results back as matlab does with 'zoh'.

Furthermore i am studying control theory but no combination of BWD FWD LSR RSR or tustin methods or even using step response equvilent method give what matlab does, (which uses something like 'impulse' as you discribed), i am trying to find an algebraic equation how the Matlab function works as i would need it in mathematica, but i had no success, do you possibly know the answer?

[Paul]

I don't know anything about Mathematica, but it seems to give the same results as Matlab as far as I can tell. Using the example on this page:

sys=tf([1 0],[1 1 1]);
sysd=c2d(sys,0.4,'tustin')

sysd =
 
   0.1613 z^2 - 0.1613
  ----------------------
  z^2 - 1.548 z + 0.6774
 
Sample time: 0.4 seconds
Discrete-time transfer function.

Note that Mathematica doesn't scale the coeffiecient of z^2 to unity. So multiply the top and bottom to match the Mathematica result:

>> sysd.num{:} = sysd.num{:}*4.96;sysd.den{:} = sysd.den{:}*4.96;
>> sysd

sysd =
 
       0.8 z^2 - 0.8
  ------------------------
  4.96 z^2 - 7.68 z + 3.36
 
Sample time: 0.4 seconds
Discrete-time transfer function.

Which is the same as the Mathematica result for Tustin (or blinear). Now do the same for ZOH:

>> sysd=c2d(sys,0.4,'zoh')

sysd =
 
     0.321 z - 0.321
  ---------------------
  z^2 - 1.54 z + 0.6703
 
Sample time: 0.4 seconds
Discrete-time transfer function.

Which is the same as the Mathematica result. Can you show an example where Mathemeatica and Matlab give different results?

In Matlab, c2d supports several methods. The 'impulse' method is the z-transform of the samples of the continuous time impulse response of the system. Based on its doc page, it appears that Mathematica does not support the 'impulse' method.

What exactly are you trying to do?

[Tamas Toth-K]

Thank you, very much! after you said it is he same I took another look and realized that out of the many substituted variables one was incorrect, after the fix it is the same :)

Answer 2

[raphael antunes]

n=1.501877347e+06;
d=[1 4.255319149e+03 125.1564456e0+03];
g=tf(n,d)

gz=c2d(g,0.000369,'zoh')
pole(gz)
zero(gz)

[Paul]

If this Answer is supposed to be a new Question, I suggest deleting this answer and posting a new Question (with more detail on what the question actually is, in addition to the code).

Answer 3

[raphael antunes]

n=1.502e+06;
d=[1 4.255e+03 125.1e+03];
g=tf(n,d)
gz=c2d(g,0.0000125,'zoh')

z=tf('z');
c=(z-0.9998)/(z-1)

step(feedback(gz*c/(1+gz*c)))

Answer 4

[raphael antunes]

%|----------------Função de transferência em tempo continuo----------|
n=1.502e+06;
d=[1 4.255e+03 125.1e+03];
g=tf(n,d)

g =
 
        1.502e06
  ---------------------
  s^2 + 4255 s + 125100
 
Continuous-time transfer function.

rlocus(g)

%|-------------------Controlador em tempo continuo-------------------|
s=tf('s');
c=1.1*((s+32.5)/(s+1.2))

c =
 
  1.1 s + 35.75
  -------------
     s + 1.2
 
Continuous-time transfer function.

%|------------------------Analise de Malha fechada Contínua-----------------------------|
gf=g*c/(1+g*c)

gf =
 
           1.652e06 s^4 + 7.086e09 s^3 + 4.437e11 s^2 + 7.24e12 s + 8.061e12
  -----------------------------------------------------------------------------------
  s^6 + 8512 s^5 + 2.003e07 s^4 + 8.194e09 s^3 + 4.619e11 s^2 + 7.279e12 s + 8.083e12
 
Continuous-time transfer function.

pole(gf)

ans = 6×1

1.0e+03 * -4.2254 -3.7896 -0.4338 -0.0328 -0.0296 -0.0012

zero(gf)

ans = 4×1

1.0e+03 * -4.2254 -0.0325 -0.0296 -0.0012

step(gf,0.5)

%|----------------------------ZOH--------------------------------|
gz=c2d(g,0.0000125,'zoh')

gz =
 
  0.0001153 z + 0.0001133
  -----------------------
  z^2 - 1.948 z + 0.9482
 
Sample time: 1.25e-05 seconds
Discrete-time transfer function.

pole(gz)

ans = 2×1

0.9996 0.9486

zero(gz)

ans = -0.9824

%|----------------------------Discretização--------------------------------|
cz=c2d(c,0.0000125,'tustin')

cz =
 
  1.1 z - 1.1
  -----------
     z - 1
 
Sample time: 1.25e-05 seconds
Discrete-time transfer function.

cprewarping=c2d(c,0.0000125,c2dOptions('Method','tustin','PrewarpFrequency',1))

cprewarping =
 
  1.1 z - 1.1
  -----------
     z - 1
 
Sample time: 1.25e-05 seconds
Discrete-time transfer function.

cmpz=c2d(c,0.0000125,'matched')

cmpz =
 
  1.1 z - 1.1
  -----------
     z - 1
 
Sample time: 1.25e-05 seconds
Discrete-time transfer function.

%|------------------------Analise de Malha fechada discreta-----------------------------|
gfz=gz*cz/(1+gz*cz)

gfz =
 
  0.0001268 z^5 - 0.0003761 z^4 + 0.0002492 z^3 + 0.0002407 z^2 - 0.0003587 z + 0.0001181
  ---------------------------------------------------------------------------------------
          z^6 - 5.896 z^5 + 14.48 z^4 - 18.97 z^3 + 13.98 z^2 - 5.493 z + 0.8992
 
Sample time: 1.25e-05 seconds
Discrete-time transfer function.

pole(gfz)

ans = 
   1.0002 + 0.0000i
   0.9995 + 0.0004i
   0.9995 - 0.0004i
   0.9946 + 0.0000i
   0.9539 + 0.0000i
   0.9486 + 0.0000i

zero(gfz)

ans = 5×1

-0.9824 1.0000 0.9996 0.9996 0.9486

%|-------------------------Resposta ao degrau (aproximações discretas)------------------| 
step(feedback(cmpz*gz,1),'r',0.5)
hold on
step(feedback(cprewarping*gz,1),'b',0.5)
step(feedback(cmpz*gz,1),'G',0.5)
legend('Tustin','Tustin prewarping ','MPZ')

%|-------------------------continuo x discreto----------------------------|
step(feedback(cmpz*gz,1),'r',0.5)
hold on
step(gf,'b',0.5)

step(feedback(cmpz*gz,1),'r',0.05)
hold on
step(gf,'b',0.05)

Answer 5

[raphael antunes]

n=-20.27;
d=[1 0 -66.93];
g=tf(n,d)
s=tf('s');
pi1=1.3*s+6;
g1f=feedback(-13.05*g,pi1)
aux=(-0.1628*s^2+9.81)/s^2
pi2=0.8*s+0.7;
g2f=feedback(pi2*g1f*aux,1)
step(g2f)