Fdtxjr

Pre-mixing cryogenic fuels and using only one fuel tank

Should I stop contributing to retirement accounts?

Is this toilet slogan correct usage of the English language?

copy and scale one figure (wheel)

Which one is correct as adjective “protruding” or “protruded”?

Why does the Sun have different day lengths, but not the gas giants?

How can "mimic phobia" be cured or prevented?

Redundant comparison & "if" before assignment

Did arcade monitors have same pixel aspect ratio as TV sets?

On a tidally locked planet, would time be quantized?

Not using 's' for he/she/it

Count the occurrence of each unique word in the file

It grows, but water kills it

Is it possible to have a strip of cold climate in the middle of a planet?

How to indicate a cut out for a product window

A social experiment. What is the worst that can happen?

Is the U.S. Code copyrighted by the Government?

Open a doc from terminal, but not by its name

GraphicsGrid with a Label for each Column and Row

How should I respond when I lied about my education and the company finds out through background check?

Can I sign legal documents with a smiley face?

Melting point of aspirin, contradicting sources

"Spoil" vs "Ruin"

What if a revenant (monster) gains fire resistance?

Inverse $Z$-transform $X (z) = log left( fraczz-a right)$

How to perform Inverse Z transformInverse $z$ transform - contour integrationCan Inverse z Transform have more than one solution?Inverse Z-transform of a rational functionHow can I find the inverse Z-transform of $1/z$?Inverse Z transformInverse Z Transform with $2-z^-2$Simple Inverse Z-TransformPartial Fraction Expansion Methods for inverse z transforminverse z-transform of $frac1(1-z^-1)^2$

1

$begingroup$

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

$$X (z) = log left( fraczz-a right)$$

I get to the point

$$x[n] = fracdelta[n]n+fracepsilon[n]*a^nn$$

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

$$X (z) = log left( fraczz-a right)$$

I get to the point

$$x[n] = fracdelta[n]n+fracepsilon[n]*a^nn$$

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

$$X (z) = log left( fraczz-a right)$$

I get to the point

$$x[n] = fracdelta[n]n+fracepsilon[n]*a^nn$$

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 $log(1-x)=sum_n=1^infty-fracx^nn$, we have
$$log(z/(z-a))=-log(1-a/z)=sum_n=1^infty frac(a/z)^nn$$
for $left| fracaz right| < 1.$ Therefore, the inverse $z$-transform should be
$$
x[n] = begincasesfrac1na^n & n>0 \ 0 & textotherwiseendcases
$$

share|cite|improve this answer

edited Mar 15 at 18:32

answered Mar 15 at 16:46

Mike EarnestMike Earnest

25.6k22151

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3149495%2finverse-z-transform-x-z-log-left-fraczz-a-right%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

Using the Taylor series $log(1-x)=sum_n=1^infty-fracx^nn$, we have
$$log(z/(z-a))=-log(1-a/z)=sum_n=1^infty frac(a/z)^nn$$
for $left| fracaz right| < 1.$ Therefore, the inverse $z$-transform should be
$$
x[n] = begincasesfrac1na^n & n>0 \ 0 & textotherwiseendcases
$$

share|cite|improve this answer

edited Mar 15 at 18:32

answered Mar 15 at 16:46

Mike EarnestMike Earnest

25.6k22151

$endgroup$

add a comment | 

1

$begingroup$

Using the Taylor series $log(1-x)=sum_n=1^infty-fracx^nn$, we have
$$log(z/(z-a))=-log(1-a/z)=sum_n=1^infty frac(a/z)^nn$$
for $left| fracaz right| < 1.$ Therefore, the inverse $z$-transform should be
$$
x[n] = begincasesfrac1na^n & n>0 \ 0 & textotherwiseendcases
$$

share|cite|improve this answer

edited Mar 15 at 18:32

answered Mar 15 at 16:46

Mike EarnestMike Earnest

25.6k22151

$endgroup$

add a comment | 

$begingroup$

Using the Taylor series $log(1-x)=sum_n=1^infty-fracx^nn$, we have
$$log(z/(z-a))=-log(1-a/z)=sum_n=1^infty frac(a/z)^nn$$
for $left| fracaz right| < 1.$ Therefore, the inverse $z$-transform should be
$$
x[n] = begincasesfrac1na^n & n>0 \ 0 & textotherwiseendcases
$$

share|cite|improve this answer

edited Mar 15 at 18:32

answered Mar 15 at 16:46

Mike EarnestMike Earnest

25.6k22151

$endgroup$

share|cite|improve this answer

edited Mar 15 at 18:32

answered Mar 15 at 16:46

Mike EarnestMike Earnest

25.6k22151

answered Mar 15 at 16:46

Mike EarnestMike Earnest

25.6k22151

answered Mar 15 at 16:46

Mike EarnestMike Earnest

25.6k22151