How to get z transfer function from difference equation?

I have the difference equation

y(k) == (4*y(k - 1))/5 + (2*u(k))/5

and would like to get the transfer function

          0.4*z
  Gz(z)= -------
          z-0.8

There are two issues. If I use vpa() to get

y(k) == 0.8*y(k - 1) + 0.4*u(k)

ztrans throws this error:

Error using symengine
The 'List' option is not allowed for input of this type.

Error in transform (line 74)
F = mupadmex('symobj::vectorizeSpecfunc', f.s, x.s, w.s, trans_func,
'infinity');

Error in sym/ztrans (line 28)
F = transform('ztrans', 'n', 'z', 'w', f, varargin{:});

Error in drt (line 31)
sys = ztrans(discrete_eqn,k,z)

but even if I don't use vpa(), ztrans(eqn,k,z) just gives the following output:

ztrans(y(k) == (4*y(k - 1))/5 + (2*u(k))/5, k, z)

Why doesn't it replace y(k-1) with 1/z*ztrans(y(k)) etc?

Answer 1

[Vidya Viswanathan]

The function "ztrans" returns the Z-transform of a symbolic expression/symbolic function with respect to the transformation index at a specified point. For example, the line of code

ztrans(f,trans_index,eval_point)

computes the Z-transform of f with respect to trans_index at point eval_point. For a given difference equation, say, y(n)=0.8y(n-1)+0.4u(n), the Z-transform can be computed as follows:

syms y(n) z;
eq=y(n)-0.8*y(n-1)-0.4*heaviside(n);
Zeq=ztrans(eq,n,z)

The output of the above is as follows:

Zeq =

ztrans(y(n), n, z) - 2/(5*(z - 1)) - (4*ztrans(y(n), n, z))/(5*z) - (4*y(-1))/5 - 1/5

In this case, the Z-transform of y(n-1) is correctly replaced by (1/z)*ztrans(y(n)).

Refer to the following link for more information about the computation of Z-Transforms using MATLAB:

http://www.mathworks.com/help/symbolic/compute-z-transforms-and-inverse-z-transforms.html

Answer 2

[Yash Sharma]

Zeq =

ztrans(y(n), n, z) - 2/(5*(z - 1)) - (4*ztrans(y(n), n, z))/(5*z) - (4*y(-1))/5 - 1/5