Problem seen in discrete transfer function with varable z^-1, when calc ztrans of x(n)=n*u(n)

Hi dears,

Who knows why X1 and X2 are not the same?

X2 should be (z^-1)/(1-z^-1)^2 or (z^-1)/(1 - 2 z^-1 + z^-2)

Thanks

sympref('HeavisideAtOrigin', 1); % by default u(0)=0.5 so we set U(0)=1
u = @(n) heaviside(n) ;  % change function name
u0=u(0)
syms n 
x(n)=n*u(n)
X1=ztrans(x)

[num, den] = numden(X1);
X2 = tf(sym2poly(num), sym2poly(den),-1, 'variable', 'z^-1')
X2_var=X2.variable

[Walter Roberson]

My tests show it is related to specifying the variable. If you let the variable default to 'z' then you get a z in the numerator

Answer 1

[Paul]

Code works exactly as advertised

u = @(n) heaviside(n) ;  % change function name
syms n 
x(n)=n*u(n)

x(n) = 

X1=ztrans(x)

X1 = 

[num, den] = numden(X1)

num = 

z

den = 

As documented in sym2poly, it returns the polynomial in descending powers of the variable, in this case z

sym2poly(num)

ans = 1×2

1 0

[1 0] is the poly representation of z.

sym2poly(den)

ans = 1×3

1 -2 1

Here, we are telling tf that sym2poly(num) is the poly representation with variable z^-1. But wrt to z^-1, [1 0] = 1 + 0*z^-1 = 1, which is exactly what we get.

X2 = tf(sym2poly(num), sym2poly(den),-1, 'variable', 'z^-1')

X2 =
 
          1
  -----------------
  1 - 2 z^-1 + z^-2
 
Sample time: unspecified
Discrete-time transfer function.

So we need two steps

X2 = tf(sym2poly(num),sym2poly(den),-1)

X2 =
 
        z
  -------------
  z^2 - 2 z + 1
 
Sample time: unspecified
Discrete-time transfer function.

X2.Variable = 'z^-1'

X2 =
 
        z^-1
  -----------------
  1 - 2 z^-1 + z^-2
 
Sample time: unspecified
Discrete-time transfer function.

Finally, be very careful using heaviside. The default value of heaviside(0) is 1/2, which is (almost?) never what you want for discrete-time problems. It didn't matter here becasue u(0) = 0. Use sympref to control the value of heaviside(0).

[Ahmad]

Thank you very much dear @Paul

Yes, but why we need do it by 2 step as you said?

What is the difference between one step and two step to define the 'variable'?

To my mind don't you think it's confusing.... !!!

See the Matlab doc for tf, section: "Specify Polynomial Ordering in Discrete-Time Transfer Function"

Another Example:

num=[1];
den=[1 0 0 0];
% Two step OK!
X1=tf(num,den,0.1)
X1.Variable='z^-1'
% One step Not OK?!
X2=tf(num,den,0.1,'variable','z^-1')

[Walter Roberson]

Sertting the Variable after creating the tf() just changes the recorded variables, without changing the matrices.

[Ahmad]

yeah @Walter Roberson

Nowhere in the Matlab doc does it say about the difference.

[Walter Roberson]

I just went through the code.

In the tf([1], [1 0 0 0], 0.1) case, when it comes time to make the numerator and denominator the same size, with the numerator being shorter than the denominator, the numerator is padded on the left so that you get [0 0 0 1], [1 0 0 0] .

When you then set the Variable to z^-1, it checks to see if the numerator and denominator are the same size, sees that they are and does not pad them. Leading and trailing positions that are zero in both numerator and denominator would be removed if they existed, but in this situation they do not.

When, though, you tf([1], [1 0 0 0], 0.1, 'Variable', 'z^-1') then the internal consistency checking for array sizes is postponed until after the point where the variable has been set, and then size consistency is checked. This time, with the variable being q^-1 or z^-1, the padding to make [1] the same size as [1 0 0 0] is done on the right -- making it [1 0 0 0], [1 0 0 0] . Remove leading and trailing zeros from both positions and you get [1], [1]

So the tf([1], [1 0 0 0], 0.1, 'Variable', 'z^-1') case is numerator 1*z^-1, denominator 1*z^-1 + 0*z^-2 + 0*z^-3 + 0*z^-4 which is just 1*z^-1 same as the numerator

Thus you should be using

 tf(fliplr(num), fliplr(den), 0.1, 'Variable', 'z^-1')

to get the system you want, or something like that.

[Walter Roberson]

sysz = tf([1 2], [3 4 5], 0.1)

sysz =
 
       z + 2
  ---------------
  3 z^2 + 4 z + 5
 
Sample time: 0.1 seconds
Discrete-time transfer function.

sysz.Variable = 'z^-1'

sysz =
 
     z^-1 + 2 z^-2
  -------------------
  3 + 4 z^-1 + 5 z^-2
 
Sample time: 0.1 seconds
Discrete-time transfer function.

sysz1 = tf([1 2], [3 4 5], 0.1, 'Variable', 'z^-1')

sysz1 =
 
      1 + 2 z^-1
  -------------------
  3 + 4 z^-1 + 5 z^-2
 
Sample time: 0.1 seconds
Discrete-time transfer function.

sysz1 is z times larger than sysz

[Ahmad]

Thank you so much for your code tracing @Walter Roberson it seems as you said above. :)

numerator and denominator vectors must be the same size!

[Paul]

At the risk of beating this further into the ground .....

tf interprets the num and den arguments differently depending on if the Variable is 'z,' which is the default, or 'z^-1'. (and similarly for the other options, q and q^-1 I think)

When the Variable is 'z', the num and den arguments are interpreted as coefficients in descending powers of z. The coeffients [1 2 3] are interpreted as

1*z^2 + 2*z^1 + 3*z^0 = z^2 + 2*z + 3

When the Variable is 'z^-1', the num and den arguments are interpreted as coefficients in ascending powers of z^-1. The coeffients [1 2 3] are interpreted as

1*(z^-1)^0 + 2*(z^-1)^1 + 3*(z^-1)^3 = 1 + 2*z^-1 + 3*z^-2

In other words, it's incumbent on the user to make sure the num and den inputs are consistent with the specified variable.

I'm sure these conventions are documnted somewhere.

It's because of this different meaning of num and den that the code does the zero padding differently depending on the Variable specification.

So if we want to construct a system that's given by

H(z) = (z + 2)/(3*z^2 + 4*z + 5)

the tf command is

H1 = tf([1 2],[3 4 5],-1,'Variable','z')

H1 =
 
       z + 2
  ---------------
  3 z^2 + 4 z + 5
 
Sample time: unspecified
Discrete-time transfer function.

If we mentally mulitply the top and bottom by z^-2, we get the equivalent system

H(z) = (z^-1 + z^-2)/(3 + 4*z^-1 + 5*z^-2)

so enter it like this (the leading zero in num is criitical)

H2 = tf([0 1 2],[3 4 5],-1,'Variable','z^-1')

H2 =
 
     z^-1 + 2 z^-2
  -------------------
  3 + 4 z^-1 + 5 z^-2
 
Sample time: unspecified
Discrete-time transfer function.

Verify equality

H1 - H2

ans =
 
  0
 
Static gain.

Not sure what @Walter Roberson was implying in this comment re: using fliplr. That is not correct.

[Ahmad]

Yes exactly numerator and denominator vectors must be the same size!

[Paul]

They only need to be the same size if that's what the problem requires. For an example of when it's not required

H(z) = (1 + z^-1) / (1 + 2*z^-1 + 3*z^-2)

H = tf([1 1],[1 2 3],-1,'Variable','z^-1')

H =
 
       1 + z^-1
  -------------------
  1 + 2 z^-1 + 3 z^-2
 
Sample time: unspecified
Discrete-time transfer function.

You can, of course, zero-pad the numerator if you wish (zero-pad to the right for z^-1)

H = tf([1 1 0],[1 2 3],-1,'Variable','z^-1')

H =
 
       1 + z^-1
  -------------------
  1 + 2 z^-1 + 3 z^-2
 
Sample time: unspecified
Discrete-time transfer function.

but you're not obligated to do so. The only reuqirement is that num and den represent the system for the Variable that's being used.