What is the inverse $z$-transform of $frac{z}{z+1}$?
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When it's $frac z{z-1}$ the answer in the discrete domain can be found in the general tables = $u[n]$, but for $frac z{z+1}$ I can't find the rule. Some examples say it's $(−1)^nu[n] $ or $a^kcos(kpi)$, which one is it and why?
complex-analysis signal-processing inverse-function z-transform
edited Dec 19 '18 at 1:17
anomaly
17.8k42666
asked Feb 22 '17 at 16:55
quoquo
325
$endgroup$
add a comment |
$begingroup$
When it's $frac z{z-1}$ the answer in the discrete domain can be found in the general tables = $u[n]$, but for $frac z{z+1}$ I can't find the rule. Some examples say it's $(−1)^nu[n] $ or $a^kcos(kpi)$, which one is it and why?
complex-analysis signal-processing inverse-function z-transform
edited Dec 19 '18 at 1:17
anomaly
17.8k42666
asked Feb 22 '17 at 16:55
quoquo
325
$endgroup$
add a comment |
$begingroup$
When it's $frac z{z-1}$ the answer in the discrete domain can be found in the general tables = $u[n]$, but for $frac z{z+1}$ I can't find the rule. Some examples say it's $(−1)^nu[n] $ or $a^kcos(kpi)$, which one is it and why?
complex-analysis signal-processing inverse-function z-transform
edited Dec 19 '18 at 1:17
anomaly
17.8k42666
asked Feb 22 '17 at 16:55
quoquo
325
$endgroup$
When it's $frac z{z-1}$ the answer in the discrete domain can be found in the general tables = $u[n]$, but for $frac z{z+1}$ I can't find the rule. Some examples say it's $(−1)^nu[n] $ or $a^kcos(kpi)$, which one is it and why?
complex-analysis signal-processing inverse-function z-transform
complex-analysis signal-processing inverse-function z-transform
edited Dec 19 '18 at 1:17
anomaly
17.8k42666
asked Feb 22 '17 at 16:55
quoquo
325
edited Dec 19 '18 at 1:17
anomaly
17.8k42666
asked Feb 22 '17 at 16:55
quoquo
325
edited Dec 19 '18 at 1:17
anomaly
17.8k42666
edited Dec 19 '18 at 1:17
anomaly
17.8k42666
edited Dec 19 '18 at 1:17
anomaly
17.8k42666
17.8k42666
asked Feb 22 '17 at 16:55
quoquo
325
asked Feb 22 '17 at 16:55
quoquo
325
asked Feb 22 '17 at 16:55
quoquo
325
325
add a comment |
add a comment |
2 Answers
2
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To be well defined, the $z$ transform must include the ROC (radius of convergence). You can see that in this example.
You can write:
$$frac{z}{z+1}=frac{1}{1+z^{-1}}=sum_{n=0}^{infty}(-z^{-1})^n tag{1}$$
which is the $Z$ transform of $(−1)^nu[n]$
but also you can write
$$frac{z}{z+1}=z frac{1}{1+z}=z sum_{m=0}^{infty}(-z)^m = sum_{m=0}^{infty} (-1)^m z^{m+1} tag{2}$$
which, setting $n=-(m+1)$ gives
$$sum_{n=-infty}^{-1} (-1)^{n+1} z^{-n}$$
which is is $Z$ transform of $(-1)^{n+1}u(-n+1)$
So which one is the correct inverse transform? It depends on the ROC. Eq $(1)$ only converges when $|z^{-1}|<1$ or $|z|>1$ (outside the unit circle), while eq $(2)$ only converges for $|z|<1$
Hence, to do the anti-transform it's not enough to have the $Z$ transform as a mere formula, you need to have also the ROC. Or at least the data of the original sequence being causal or anticausal.
answered Feb 25 '17 at 15:51
leonbloyleonbloy
42k647109
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$$frac{z}{z+1}=frac{1}{1-(-z^{-1})}=sum_{n=0}^{infty}(-z^{-1})^n=sum_{n=0}^{infty}(-1)^nz^{-n}=mathcal{Z}{(-1)^nu[n]},quad |z|>1$$
answered Feb 25 '17 at 15:14
Matt L.Matt L.
9,034822
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
To be well defined, the $z$ transform must include the ROC (radius of convergence). You can see that in this example.
You can write:
$$frac{z}{z+1}=frac{1}{1+z^{-1}}=sum_{n=0}^{infty}(-z^{-1})^n tag{1}$$
which is the $Z$ transform of $(−1)^nu[n]$
but also you can write
$$frac{z}{z+1}=z frac{1}{1+z}=z sum_{m=0}^{infty}(-z)^m = sum_{m=0}^{infty} (-1)^m z^{m+1} tag{2}$$
which, setting $n=-(m+1)$ gives
$$sum_{n=-infty}^{-1} (-1)^{n+1} z^{-n}$$
which is is $Z$ transform of $(-1)^{n+1}u(-n+1)$
So which one is the correct inverse transform? It depends on the ROC. Eq $(1)$ only converges when $|z^{-1}|<1$ or $|z|>1$ (outside the unit circle), while eq $(2)$ only converges for $|z|<1$
Hence, to do the anti-transform it's not enough to have the $Z$ transform as a mere formula, you need to have also the ROC. Or at least the data of the original sequence being causal or anticausal.
answered Feb 25 '17 at 15:51
leonbloyleonbloy
42k647109
$endgroup$
add a comment |
$begingroup$
To be well defined, the $z$ transform must include the ROC (radius of convergence). You can see that in this example.
You can write:
$$frac{z}{z+1}=frac{1}{1+z^{-1}}=sum_{n=0}^{infty}(-z^{-1})^n tag{1}$$
which is the $Z$ transform of $(−1)^nu[n]$
but also you can write
$$frac{z}{z+1}=z frac{1}{1+z}=z sum_{m=0}^{infty}(-z)^m = sum_{m=0}^{infty} (-1)^m z^{m+1} tag{2}$$
which, setting $n=-(m+1)$ gives
$$sum_{n=-infty}^{-1} (-1)^{n+1} z^{-n}$$
which is is $Z$ transform of $(-1)^{n+1}u(-n+1)$
So which one is the correct inverse transform? It depends on the ROC. Eq $(1)$ only converges when $|z^{-1}|<1$ or $|z|>1$ (outside the unit circle), while eq $(2)$ only converges for $|z|<1$
Hence, to do the anti-transform it's not enough to have the $Z$ transform as a mere formula, you need to have also the ROC. Or at least the data of the original sequence being causal or anticausal.
answered Feb 25 '17 at 15:51
leonbloyleonbloy
42k647109
$endgroup$
add a comment |
$begingroup$
To be well defined, the $z$ transform must include the ROC (radius of convergence). You can see that in this example.
You can write:
$$frac{z}{z+1}=frac{1}{1+z^{-1}}=sum_{n=0}^{infty}(-z^{-1})^n tag{1}$$
which is the $Z$ transform of $(−1)^nu[n]$
but also you can write
$$frac{z}{z+1}=z frac{1}{1+z}=z sum_{m=0}^{infty}(-z)^m = sum_{m=0}^{infty} (-1)^m z^{m+1} tag{2}$$
which, setting $n=-(m+1)$ gives
$$sum_{n=-infty}^{-1} (-1)^{n+1} z^{-n}$$
which is is $Z$ transform of $(-1)^{n+1}u(-n+1)$
So which one is the correct inverse transform? It depends on the ROC. Eq $(1)$ only converges when $|z^{-1}|<1$ or $|z|>1$ (outside the unit circle), while eq $(2)$ only converges for $|z|<1$
Hence, to do the anti-transform it's not enough to have the $Z$ transform as a mere formula, you need to have also the ROC. Or at least the data of the original sequence being causal or anticausal.
answered Feb 25 '17 at 15:51
leonbloyleonbloy
42k647109
$endgroup$
To be well defined, the $z$ transform must include the ROC (radius of convergence). You can see that in this example.
You can write:
$$frac{z}{z+1}=frac{1}{1+z^{-1}}=sum_{n=0}^{infty}(-z^{-1})^n tag{1}$$
which is the $Z$ transform of $(−1)^nu[n]$
but also you can write
$$frac{z}{z+1}=z frac{1}{1+z}=z sum_{m=0}^{infty}(-z)^m = sum_{m=0}^{infty} (-1)^m z^{m+1} tag{2}$$
which, setting $n=-(m+1)$ gives
$$sum_{n=-infty}^{-1} (-1)^{n+1} z^{-n}$$
which is is $Z$ transform of $(-1)^{n+1}u(-n+1)$
So which one is the correct inverse transform? It depends on the ROC. Eq $(1)$ only converges when $|z^{-1}|<1$ or $|z|>1$ (outside the unit circle), while eq $(2)$ only converges for $|z|<1$
Hence, to do the anti-transform it's not enough to have the $Z$ transform as a mere formula, you need to have also the ROC. Or at least the data of the original sequence being causal or anticausal.
answered Feb 25 '17 at 15:51
leonbloyleonbloy
42k647109
answered Feb 25 '17 at 15:51
leonbloyleonbloy
42k647109
answered Feb 25 '17 at 15:51
leonbloyleonbloy
42k647109
answered Feb 25 '17 at 15:51
leonbloyleonbloy
42k647109
42k647109
add a comment |
add a comment |
$begingroup$
$$frac{z}{z+1}=frac{1}{1-(-z^{-1})}=sum_{n=0}^{infty}(-z^{-1})^n=sum_{n=0}^{infty}(-1)^nz^{-n}=mathcal{Z}{(-1)^nu[n]},quad |z|>1$$
answered Feb 25 '17 at 15:14
Matt L.Matt L.
9,034822
$endgroup$
add a comment |
$begingroup$
$$frac{z}{z+1}=frac{1}{1-(-z^{-1})}=sum_{n=0}^{infty}(-z^{-1})^n=sum_{n=0}^{infty}(-1)^nz^{-n}=mathcal{Z}{(-1)^nu[n]},quad |z|>1$$
answered Feb 25 '17 at 15:14
Matt L.Matt L.
9,034822
$endgroup$
add a comment |
$begingroup$
$$frac{z}{z+1}=frac{1}{1-(-z^{-1})}=sum_{n=0}^{infty}(-z^{-1})^n=sum_{n=0}^{infty}(-1)^nz^{-n}=mathcal{Z}{(-1)^nu[n]},quad |z|>1$$
answered Feb 25 '17 at 15:14
Matt L.Matt L.
9,034822
$endgroup$
$$frac{z}{z+1}=frac{1}{1-(-z^{-1})}=sum_{n=0}^{infty}(-z^{-1})^n=sum_{n=0}^{infty}(-1)^nz^{-n}=mathcal{Z}{(-1)^nu[n]},quad |z|>1$$
answered Feb 25 '17 at 15:14
Matt L.Matt L.
9,034822
answered Feb 25 '17 at 15:14
Matt L.Matt L.
9,034822
answered Feb 25 '17 at 15:14
Matt L.Matt L.
9,034822
answered Feb 25 '17 at 15:14
Matt L.Matt L.
9,034822
9,034822
add a comment |
add a comment |
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