Matlab is not giving the correct z transform of - (n*a^n)*u[- n - 1];

My code is:

syms n z a;
y = -(n*a*n)*heaviside(-n-1) ... heaviside is supposed to be the unit function but apparently it's not working.
yz = ztrans(y, n, z);

And the result I get everytime is 0... The correct answer must be (a*z^-1) / (1 - a*z^-1)^2

[Torsten]

syms n z a
y = n*a^n;
yz = ztrans(y, n, z)

yz = 

Now multiply numerator and denominator by z^-2.

Answer 1

[VBBV]

syms n z a;
y = (n*a^n)*heaviside(n+1) %... heaviside is supposed to be the unit function but apparently it's not working.

y = 

yz = ztrans(y, n,z)

yz = 

simplify(yz)

ans = 

[VBBV]

There is a typo error in the input function

y = (n*a^n)*heaviside(n+1)  % equivalent expression
%       ^ instead of power , you have a multiplier symbol

[juan]

I can't believe I didn't notice that typo...

Answer 2

[Paul]

In addition to what @VBBV said about a typo in the defintion of y[n], there are two other fundamental problem that needs to be addressed.

The first problem is that, by default, the heaviside is not the discrete-time unit step for integer values of its argument, because for n = 0 with default sym preferences

sympref('default');
heaviside(sym(0))

ans = 

So the first thing we need to do is change that from the default

sympref('HeavisideAtOrigin',1);
heaviside(sym(0))

ans = 

1

Here's y, with the corrected typo

syms n z a
y(n) = -(n*a^n)*heaviside(-n-1) % heaviside is supposed to be the unit function but apparently it's not working.

y(n) = 

Evaluate y for some values of n

[-3:3; y(-3:3)]

ans = 

It's clear that y(n) = 0 for n >= 0.

However, the function ztrans is the unilateral z-transform and ignores all values of y for n < 0. So ztrans is basically computing the z-transform based only on n >=0, and in that range y(n) = 0 which is why ztrans returned zero. What you really need is the bilateral z-transform for a strictly left-sided (or non-causal) signal. You can get that using ztrans and one rule from a standard z-transform table.