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Inverse $Z$-transform $X (z) = log left( frac{z}{z-a} right)$

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1

$begingroup$

I need help to find the inverse $Z$-transform of the following function

$$X (z) = log left( frac{z}{z-a} right)$$

I get to the point

$$x[n] = frac{delta[n]}{n}+frac{epsilon[n]*a^n}{n}$$

But how do I find the value when $n = 0$?

z-transform

share|cite|improve this question

edited Mar 15 at 17:39

Rodrigo de Azevedo

13.2k41960

asked Mar 15 at 16:24

Oskar LundinOskar Lundin

62

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is $epsilon[n]$?
  $endgroup$
  – Rodrigo de Azevedo
  Mar 15 at 17:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Heavyside step function, I don't know if everyone is using epsilon but in my class we do:/
  $endgroup$
  – Oskar Lundin
  Mar 15 at 17:49
  

  
  
  
  

add a comment | 

1

$begingroup$

I need help to find the inverse $Z$-transform of the following function

$$X (z) = log left( frac{z}{z-a} right)$$

I get to the point

$$x[n] = frac{delta[n]}{n}+frac{epsilon[n]*a^n}{n}$$

But how do I find the value when $n = 0$?

z-transform

share|cite|improve this question

edited Mar 15 at 17:39

Rodrigo de Azevedo

13.2k41960

asked Mar 15 at 16:24

Oskar LundinOskar Lundin

62

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is $epsilon[n]$?
  $endgroup$
  – Rodrigo de Azevedo
  Mar 15 at 17:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Heavyside step function, I don't know if everyone is using epsilon but in my class we do:/
  $endgroup$
  – Oskar Lundin
  Mar 15 at 17:49
  

  
  
  
  

add a comment | 

$begingroup$

I need help to find the inverse $Z$-transform of the following function

$$X (z) = log left( frac{z}{z-a} right)$$

I get to the point

$$x[n] = frac{delta[n]}{n}+frac{epsilon[n]*a^n}{n}$$

But how do I find the value when $n = 0$?

z-transform

share|cite|improve this question

edited Mar 15 at 17:39

Rodrigo de Azevedo

13.2k41960

asked Mar 15 at 16:24

Oskar LundinOskar Lundin

62

$endgroup$

share|cite|improve this question

edited Mar 15 at 17:39

Rodrigo de Azevedo

13.2k41960

asked Mar 15 at 16:24

Oskar LundinOskar Lundin

62

share|cite|improve this question

edited Mar 15 at 17:39

Rodrigo de Azevedo

13.2k41960

asked Mar 15 at 16:24

Oskar LundinOskar Lundin

62

edited Mar 15 at 17:39

Rodrigo de Azevedo

13.2k41960

edited Mar 15 at 17:39

Rodrigo de Azevedo

13.2k41960

asked Mar 15 at 16:24

Oskar LundinOskar Lundin

62

asked Mar 15 at 16:24

Oskar LundinOskar Lundin

62

1 Answer
1

active

oldest

votes

1

$begingroup$

Using the Taylor series $log(1-x)=sum_{n=1}^infty-frac{x^n}n$, we have
$$log(z/(z-a))=-log(1-a/z)=sum_{n=1}^infty frac{(a/z)^n}n$$
for $left| frac{a}{z} right| < 1.$ Therefore, the inverse $z$-transform should be
$$
x[n] = begin{cases}frac1na^{n} & n>0 \ 0 & text{otherwise}end{cases}
$$

share|cite|improve this answer

edited Mar 15 at 18:32

answered Mar 15 at 16:46

Mike EarnestMike Earnest

25.9k22151

$endgroup$

add a comment | 

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1

$begingroup$

Using the Taylor series $log(1-x)=sum_{n=1}^infty-frac{x^n}n$, we have
$$log(z/(z-a))=-log(1-a/z)=sum_{n=1}^infty frac{(a/z)^n}n$$
for $left| frac{a}{z} right| < 1.$ Therefore, the inverse $z$-transform should be
$$
x[n] = begin{cases}frac1na^{n} & n>0 \ 0 & text{otherwise}end{cases}
$$

share|cite|improve this answer

edited Mar 15 at 18:32

answered Mar 15 at 16:46

Mike EarnestMike Earnest

25.9k22151

$endgroup$

add a comment | 

1

$begingroup$

Using the Taylor series $log(1-x)=sum_{n=1}^infty-frac{x^n}n$, we have
$$log(z/(z-a))=-log(1-a/z)=sum_{n=1}^infty frac{(a/z)^n}n$$
for $left| frac{a}{z} right| < 1.$ Therefore, the inverse $z$-transform should be
$$
x[n] = begin{cases}frac1na^{n} & n>0 \ 0 & text{otherwise}end{cases}
$$

share|cite|improve this answer

edited Mar 15 at 18:32

answered Mar 15 at 16:46

Mike EarnestMike Earnest

25.9k22151

$endgroup$

add a comment | 

$begingroup$

Using the Taylor series $log(1-x)=sum_{n=1}^infty-frac{x^n}n$, we have
$$log(z/(z-a))=-log(1-a/z)=sum_{n=1}^infty frac{(a/z)^n}n$$
for $left| frac{a}{z} right| < 1.$ Therefore, the inverse $z$-transform should be
$$
x[n] = begin{cases}frac1na^{n} & n>0 \ 0 & text{otherwise}end{cases}
$$

share|cite|improve this answer

edited Mar 15 at 18:32

answered Mar 15 at 16:46

Mike EarnestMike Earnest

25.9k22151

$endgroup$

share|cite|improve this answer

edited Mar 15 at 18:32

answered Mar 15 at 16:46

Mike EarnestMike Earnest

25.9k22151

answered Mar 15 at 16:46

Mike EarnestMike Earnest

25.9k22151

answered Mar 15 at 16:46

Mike EarnestMike Earnest

25.9k22151