Question

apply z transform to x[n] = a^(n+1) u[n+1] step by step please

Answer 1

u[n] is a unit step function
it is used as a bounded input

Answer 2

Is this a bilateral Z transform?

Answer 3

it is bilaterial with out the unit step function

Answer 4

the unit step is for a input bound

Answer 5

the unit step changes the bounds

Answer 6

I really don't know what I'm doing. I'll leave this to someone else.

Answer 7

kk, ty for trying

Answer 8

Let the discrete function be a(n)
Z transform is defined as
\[\Large H(z)=\Sigma_{n=-\infty}^{n=\infty} a(n)z^{-n}\]
Do you get this?

Answer 9

yes

Answer 10

we have
\[a[n]=a^{n+1} U(n+1)\]
so the discrete function is defined from \(n=-1\ to\ n=\infty\)

Answer 11

ok, i understand there is a time shift
n+1 = 0
n=-1

Answer 12

yeah:)

Answer 13

Let's plugin the values

Answer 14

k

Answer 15

\[\Large H(z)=\Sigma_{n=-1}^{n=\infty} a^{n+1}z^{-n}\]
\[\Large H(z)=a\times \Sigma_{n=-1}^{n=\infty} a^{n}z^{-n}\]
or
\[\Large H(z)=a\times \Sigma_{n=-1}^{n=\infty} (az^{-1})^n\]
do you get till here?@mathman12

Answer 16

yes

Answer 17

i got to here and try to match a infinite geometric series and got something strange

Answer 18

if
\[|a|\le 1\]
then this z transform will exist, other wise it won't exist

Answer 19

i tried matching \[\sum_{k=0}^{oo} x^k = 1/(1-x)\]

Answer 20

you can do that
[\[\Large H(z)=a\times ((az^{-1})^{-1}+\Sigma_{n=0}^{n=\infty} (az^{-1})^n)\]
now we get
[\[\Large H(z)=a\times ((\frac z a+\Sigma_{n=0}^{n=\infty} (az^{-1})^n)\]
Now use the formula for the last part

Answer 21

I took out n=-1 from the sum and then the sum becam from n=0 to n= (\infty\)

Answer 22

n=\(\infty\)

Answer 23

can't you subtract it from a^(n+1)z^(-n) as a^(n+1-1)z^(-(n-1)) ?

Answer 24

to compensate for the adding to the bound you subtract from the exp?

Answer 25

let me think

Answer 26

that won't work

Answer 27

Read my post , if you don't understand. I'll help you:)