z transform of a left handed sequence

Hello. I want to calculate z trnasform of the following sequence :

how can i do it without doing the math? The first part of sequence is easy I just need to subtract the elements of n=0 and n=1 from the but for the second part i need to use math and that's a bother to be honest. Is there a way to just calculate the z transform of a left handed sequence or a command for and ?

here's what I've got so far

clear; close; clc

syms n z

yp1 = ztrans((1/3)^n)
v   = (1/3)^n * z^-n
yp2 = symsum(v,n,0,1)
Yp  = yp1-yp2
YPs = simplify(Yp)

y1 = n*2^n*z^-n
v = symsum(y1,n,0,inf)
meq = simplifyFraction(children(v,1))

eqn = YPs + meq

[AlireaA Mirrabie]

Hello and thank you for your answer. I've tried the heaviside command but heviside is defined as

but i want the u(t) as

is there a command for that? i tried sympref and other ways to make that 0.5 at t=0, 1 but it didn't work for z transform. For example look at the code below. I want to calculate the z transform of

syms a n

y = a^n*heaviside(-n-2)

y = 

ztrans(y)

ans = 

0

and I'm sure that, this answer is incorrect.

Answer 1

[Paul]

Hi AlireaA

You are on the right track by breaking this problem into two parts. Let's look at the first part

syms n integer
syms z
yp1 = ztrans((1/3)^n)

yp1 = 

v   = (1/3)^n * z^-n;
yp2 = symsum(v,n,0,1);
Yp  = yp1-yp2

Yp = 

YPs = simplify(Yp)

YPs = 

That result is correct, and could also have been obtained as follows:

sympref('HeavisideAtOrigin',1);
u(n) = heaviside(n);
simplify(ztrans((1/3)^n*u(n-2)))

ans = 

We can use ztrans here because the sequence in question is zero for n < 0.

Also, it's very important to note that the Region of Convergence (ROC) for Yps is abs(z) > 1/3.

Now, moving onto the second component .... it looks like you're trying to use symsum to compute the z-transform explicitly. However, because the sequence is left-sided we need to use the bi-lateral z-transform (which I'm sure is the point of the exercise). Furthermore, because the sequence is zero for n > -1, we can take the sum from -inf to -1 (instead of -inf to inf),

y1 = n*2^n*z^-n;  % no need to include u(-n-1) when summing from -inf to -1, because y1[n>=0] = 0
%v = symsum(y1,n,0,inf)
v = symsum(y1,n,-inf,-1)

v = 

The z-transform of this sequence only coverges over the ROC abs(z) < 2. Combined with result above, the ROC for the z-transform of the sum of the two sequences is

assume(1/3 < abs(z) < 2);

Recompute v with this assumption to get a compact form

v = simplify(symsum(y1,n,-inf,-1))

v = 

BTW, you could also have obtained this result from a standard, z-transform table.

Consequently, the z-transform of x[n] is

X(z) = YPs + v

X(z) = 

with ROC

assumptions(z)

ans = 

[AlireaA Mirrabie]

Thank you Paul for your answer. That is exactly what I was looking for you're a life saver.