Mixing
inverse z-transform of $frac{1}{(1-z^{-1})^{2}}$
$begingroup$
how to find inverse-z-transform of:
$$
x[n] = Z-Transformleft{ frac{1}{(1-z^{-1})^{2}} right}
$$
I could convert it to convolution... but then i'm stuck with solving convolution...which i'm not really sure how to perform either.
A few z-transform table definitions I'm looking at to solve the problem:
$$
begin{aligned}
X(z) &= sum_{n=-infty}^{infty} x[n] z^{-n} \
\
alpha^{n} u[n] &<=> frac{1}{1-alpha z^{-1}} \
\
x[n]*y[n] &<=> X(z) Y(z) \
\
x[n-k] &<=> z^{-k}X(z)
end{aligned}
$$
z-transform
asked Jan 5 at 1:56
JoJoJoJo
1
$endgroup$
add a comment |
$begingroup$
how to find inverse-z-transform of:
$$
x[n] = Z-Transformleft{ frac{1}{(1-z^{-1})^{2}} right}
$$
I could convert it to convolution... but then i'm stuck with solving convolution...which i'm not really sure how to perform either.
A few z-transform table definitions I'm looking at to solve the problem:
$$
begin{aligned}
X(z) &= sum_{n=-infty}^{infty} x[n] z^{-n} \
\
alpha^{n} u[n] &<=> frac{1}{1-alpha z^{-1}} \
\
x[n]*y[n] &<=> X(z) Y(z) \
\
x[n-k] &<=> z^{-k}X(z)
end{aligned}
$$
z-transform
asked Jan 5 at 1:56
JoJoJoJo
1
$endgroup$
$begingroup$
Use $3,4,5,6$ of en.m.wikipedia.org/wiki/…
$endgroup$
– lab bhattacharjee
Jan 5 at 2:06
add a comment |
$begingroup$
how to find inverse-z-transform of:
$$
x[n] = Z-Transformleft{ frac{1}{(1-z^{-1})^{2}} right}
$$
I could convert it to convolution... but then i'm stuck with solving convolution...which i'm not really sure how to perform either.
A few z-transform table definitions I'm looking at to solve the problem:
$$
begin{aligned}
X(z) &= sum_{n=-infty}^{infty} x[n] z^{-n} \
\
alpha^{n} u[n] &<=> frac{1}{1-alpha z^{-1}} \
\
x[n]*y[n] &<=> X(z) Y(z) \
\
x[n-k] &<=> z^{-k}X(z)
end{aligned}
$$
z-transform
asked Jan 5 at 1:56
JoJoJoJo
1
$endgroup$
how to find inverse-z-transform of:
$$
x[n] = Z-Transformleft{ frac{1}{(1-z^{-1})^{2}} right}
$$
I could convert it to convolution... but then i'm stuck with solving convolution...which i'm not really sure how to perform either.
A few z-transform table definitions I'm looking at to solve the problem:
$$
begin{aligned}
X(z) &= sum_{n=-infty}^{infty} x[n] z^{-n} \
\
alpha^{n} u[n] &<=> frac{1}{1-alpha z^{-1}} \
\
x[n]*y[n] &<=> X(z) Y(z) \
\
x[n-k] &<=> z^{-k}X(z)
end{aligned}
$$
z-transform
z-transform
asked Jan 5 at 1:56
JoJoJoJo
1
asked Jan 5 at 1:56
JoJoJoJo
1
asked Jan 5 at 1:56
JoJoJoJo
1
asked Jan 5 at 1:56
JoJoJoJo
1
asked Jan 5 at 1:56
JoJoJoJo
1
1
$begingroup$
Use $3,4,5,6$ of en.m.wikipedia.org/wiki/…
$endgroup$
– lab bhattacharjee
Jan 5 at 2:06
add a comment |
$begingroup$
Use $3,4,5,6$ of en.m.wikipedia.org/wiki/…
$endgroup$
– lab bhattacharjee
Jan 5 at 2:06
$begingroup$
Use $3,4,5,6$ of en.m.wikipedia.org/wiki/…
$endgroup$
– lab bhattacharjee
Jan 5 at 2:06
$begingroup$
Use $3,4,5,6$ of en.m.wikipedia.org/wiki/…
$endgroup$
– lab bhattacharjee
Jan 5 at 2:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
for positive m:
$$
left(begin{array}{c} n + m - 1 \ m - 1 end{array} right) a^n u[n] <=> frac{1}{(1-a z^{-1})^m}
$$
$$
left(begin{matrix}n\k\end{matrix}right)=frac{nleft(n-1right)ldotsleft(n-k+1right)}{kleft(k-1right)ldots1}
$$
for m=2:
$$
(n+1) a^{n} u[n] <=> frac{1}{(1-az^{-1})^2} ROC: |z| > |a|
$$
answered Jan 5 at 1:59
JoJoJoJo
1
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
for positive m:
$$
left(begin{array}{c} n + m - 1 \ m - 1 end{array} right) a^n u[n] <=> frac{1}{(1-a z^{-1})^m}
$$
$$
left(begin{matrix}n\k\end{matrix}right)=frac{nleft(n-1right)ldotsleft(n-k+1right)}{kleft(k-1right)ldots1}
$$
for m=2:
$$
(n+1) a^{n} u[n] <=> frac{1}{(1-az^{-1})^2} ROC: |z| > |a|
$$
answered Jan 5 at 1:59
JoJoJoJo
1
$endgroup$
add a comment |
$begingroup$
for positive m:
$$
left(begin{array}{c} n + m - 1 \ m - 1 end{array} right) a^n u[n] <=> frac{1}{(1-a z^{-1})^m}
$$
$$
left(begin{matrix}n\k\end{matrix}right)=frac{nleft(n-1right)ldotsleft(n-k+1right)}{kleft(k-1right)ldots1}
$$
for m=2:
$$
(n+1) a^{n} u[n] <=> frac{1}{(1-az^{-1})^2} ROC: |z| > |a|
$$
answered Jan 5 at 1:59
JoJoJoJo
1
$endgroup$
add a comment |
$begingroup$
for positive m:
$$
left(begin{array}{c} n + m - 1 \ m - 1 end{array} right) a^n u[n] <=> frac{1}{(1-a z^{-1})^m}
$$
$$
left(begin{matrix}n\k\end{matrix}right)=frac{nleft(n-1right)ldotsleft(n-k+1right)}{kleft(k-1right)ldots1}
$$
for m=2:
$$
(n+1) a^{n} u[n] <=> frac{1}{(1-az^{-1})^2} ROC: |z| > |a|
$$
answered Jan 5 at 1:59
JoJoJoJo
1
$endgroup$
for positive m:
$$
left(begin{array}{c} n + m - 1 \ m - 1 end{array} right) a^n u[n] <=> frac{1}{(1-a z^{-1})^m}
$$
$$
left(begin{matrix}n\k\end{matrix}right)=frac{nleft(n-1right)ldotsleft(n-k+1right)}{kleft(k-1right)ldots1}
$$
for m=2:
$$
(n+1) a^{n} u[n] <=> frac{1}{(1-az^{-1})^2} ROC: |z| > |a|
$$
answered Jan 5 at 1:59
JoJoJoJo
1
answered Jan 5 at 1:59
JoJoJoJo
1
answered Jan 5 at 1:59
JoJoJoJo
1
answered Jan 5 at 1:59
JoJoJoJo
1
1
add a comment |
add a comment |
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$begingroup$
Use $3,4,5,6$ of en.m.wikipedia.org/wiki/…
$endgroup$
– lab bhattacharjee
Jan 5 at 2:06