Laplace transform of piecewise function - making it to become heaviside unitstep function

Clash Royale CLAN TAG#URR8PPP

up vote
0
down vote

favorite

The piecewise function of $f(t)$ is as follows-

$f(t) = t , 0le t < 1 $

$f(t) = 2-t , 1le t le 2$

$f(t) = 0 , t>2$

I made a quick sketch -

now, I need help on making this into a formula f(t) = something.

I am not too sure on this shape of the graph. The function is ‘ON’ from 0 to 2.
If I am not wrong, it is called the heaviside unitstep function

I need to get a function of f(t) before I can apply the laplace transform of second shifting to get the answer for Laplace transform of that function.. thanks for the help!!

laplace-transform laplace-method

share|cite|improve this question

asked Jul 22 at 14:28

user185692

1045

add a comment | 

up vote
0
down vote

favorite

The piecewise function of $f(t)$ is as follows-

$f(t) = t , 0le t < 1 $

$f(t) = 2-t , 1le t le 2$

$f(t) = 0 , t>2$

I made a quick sketch -

now, I need help on making this into a formula f(t) = something.

I am not too sure on this shape of the graph. The function is ‘ON’ from 0 to 2.
If I am not wrong, it is called the heaviside unitstep function

I need to get a function of f(t) before I can apply the laplace transform of second shifting to get the answer for Laplace transform of that function.. thanks for the help!!

laplace-transform laplace-method

share|cite|improve this question

asked Jul 22 at 14:28

user185692

1045

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

The piecewise function of $f(t)$ is as follows-

$f(t) = t , 0le t < 1 $

$f(t) = 2-t , 1le t le 2$

$f(t) = 0 , t>2$

I made a quick sketch -

now, I need help on making this into a formula f(t) = something.

I am not too sure on this shape of the graph. The function is ‘ON’ from 0 to 2.
If I am not wrong, it is called the heaviside unitstep function

I need to get a function of f(t) before I can apply the laplace transform of second shifting to get the answer for Laplace transform of that function.. thanks for the help!!

laplace-transform laplace-method

share|cite|improve this question

asked Jul 22 at 14:28

user185692

1045

The piecewise function of $f(t)$ is as follows-

$f(t) = t , 0le t < 1 $

$f(t) = 2-t , 1le t le 2$

$f(t) = 0 , t>2$

I made a quick sketch -

now, I need help on making this into a formula f(t) = something.

I am not too sure on this shape of the graph. The function is ‘ON’ from 0 to 2.
If I am not wrong, it is called the heaviside unitstep function

I need to get a function of f(t) before I can apply the laplace transform of second shifting to get the answer for Laplace transform of that function.. thanks for the help!!

laplace-transform laplace-method

share|cite|improve this question

asked Jul 22 at 14:28

user185692

1045

share|cite|improve this question

share|cite|improve this question

asked Jul 22 at 14:28

user185692

1045

asked Jul 22 at 14:28

user185692

1045

asked Jul 22 at 14:28

user185692

1045

1045

add a comment | 

add a comment | 

5 Answers
5

active

oldest

votes

up vote
1
down vote



accepted

With Heaviside 's function..

$$f(t)=t-tH(t-1)+(2-t)(H(t-1)-H(t-2))+0H(t-2)$$
$$f(t)=t-tH(t-1)+(2-t)(H(t-1)-H(t-2))$$
$$f(t)=t+(2-2t)H(t-1)-(2-t)H(t-2))$$
$$f(t)=t+2(1-t)H(t-1)-(2-t)H(t-2))$$
$$f(t)=t-2(t-1)H(t-1)+(t-2)H(t-2))$$

Then take the Lpalace transform

Note that :
$$mathcalL(f(t-c)H(t-c))=e^-csF(s)$$
Therefore we have :
$$boxed F(s)=frac 1s^2-2frac e^-ss^2 +frac e^-2ss^2$$

share|cite|improve this answer

edited Jul 22 at 16:02

answered Jul 22 at 14:58

Isham

10.6k3829

add a comment | 

up vote
0
down vote

I think it's easier to calculate the laplace transform from the definition:
$$F(s)=int_0^inftyf(t)e^-stmathrmdt$$
$$F(s)=int_0^1te^-stmathrmdt+int_1^2(2-t)e^-stmathrmdt$$

share|cite|improve this answer

answered Jul 22 at 14:31

Botond

3,8432632

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I’ve been thought to form the heaviside function first and need to use that in practices and exam, is there a way to that ? I am not too sure of that.
  – user185692
  Jul 22 at 14:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user185692 Check Ian's answer.
  – Botond
  Jul 22 at 14:46
  

  
  
  
  

add a comment | 

up vote
0
down vote

Don't worry about Heaviside, just break it into two integrals and evaluate it.

Botond has it right. $$ F(s)=int_0^1te^-stmathrmdt+int_1^2(2-t)e^-stmathrmdt$$

Now you have to do integration by parts which I hope you still remember from your calculus class.

share|cite|improve this answer

edited Jul 22 at 14:50

answered Jul 22 at 14:35

Mohammad Riazi-Kermani

27.5k41852

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I’ve been thought to form the heaviside function first and need to use that in practices and exam, is there a way to that ? I’m not too sure of that.
  – user185692
  Jul 22 at 14:45
  

  
  
  
  

add a comment | 

up vote
0
down vote

Note that $$ t(H(0)-H(1)) + (2-t)(H(1)-H(2)) =tH(0)-2(t-1)H(1)+(t-2)H(2)$$

You should be able to finish it.

share|cite|improve this answer

answered Jul 22 at 15:00

Mohammad Riazi-Kermani

27.5k41852

add a comment | 

up vote
0
down vote

If you want to use the step function then you take a case which applies on $(a,b)$ and multiply it by $H(t-a)-H(t-b)$ (note the order, and use the convention $H(t-infty)=0$) and then sum over cases. This is useful because it matches the form used for Laplace transform tables.

share|cite|improve this answer

edited Jul 22 at 19:02

answered Jul 22 at 14:42

Ian

65k24681

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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2859456%2flaplace-transform-of-piecewise-function-making-it-to-become-heaviside-unitstep%23new-answer', 'question_page');

);

Post as a guest

Name Email

5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

With Heaviside 's function..

$$f(t)=t-tH(t-1)+(2-t)(H(t-1)-H(t-2))+0H(t-2)$$
$$f(t)=t-tH(t-1)+(2-t)(H(t-1)-H(t-2))$$
$$f(t)=t+(2-2t)H(t-1)-(2-t)H(t-2))$$
$$f(t)=t+2(1-t)H(t-1)-(2-t)H(t-2))$$
$$f(t)=t-2(t-1)H(t-1)+(t-2)H(t-2))$$

Then take the Lpalace transform

Note that :
$$mathcalL(f(t-c)H(t-c))=e^-csF(s)$$
Therefore we have :
$$boxed F(s)=frac 1s^2-2frac e^-ss^2 +frac e^-2ss^2$$

share|cite|improve this answer

edited Jul 22 at 16:02

answered Jul 22 at 14:58

Isham

10.6k3829

add a comment | 

up vote
1
down vote



accepted

With Heaviside 's function..

$$f(t)=t-tH(t-1)+(2-t)(H(t-1)-H(t-2))+0H(t-2)$$
$$f(t)=t-tH(t-1)+(2-t)(H(t-1)-H(t-2))$$
$$f(t)=t+(2-2t)H(t-1)-(2-t)H(t-2))$$
$$f(t)=t+2(1-t)H(t-1)-(2-t)H(t-2))$$
$$f(t)=t-2(t-1)H(t-1)+(t-2)H(t-2))$$

Then take the Lpalace transform

Note that :
$$mathcalL(f(t-c)H(t-c))=e^-csF(s)$$
Therefore we have :
$$boxed F(s)=frac 1s^2-2frac e^-ss^2 +frac e^-2ss^2$$

share|cite|improve this answer

edited Jul 22 at 16:02

answered Jul 22 at 14:58

Isham

10.6k3829

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

With Heaviside 's function..

$$f(t)=t-tH(t-1)+(2-t)(H(t-1)-H(t-2))+0H(t-2)$$
$$f(t)=t-tH(t-1)+(2-t)(H(t-1)-H(t-2))$$
$$f(t)=t+(2-2t)H(t-1)-(2-t)H(t-2))$$
$$f(t)=t+2(1-t)H(t-1)-(2-t)H(t-2))$$
$$f(t)=t-2(t-1)H(t-1)+(t-2)H(t-2))$$

Then take the Lpalace transform

Note that :
$$mathcalL(f(t-c)H(t-c))=e^-csF(s)$$
Therefore we have :
$$boxed F(s)=frac 1s^2-2frac e^-ss^2 +frac e^-2ss^2$$

share|cite|improve this answer

edited Jul 22 at 16:02

answered Jul 22 at 14:58

Isham

10.6k3829

With Heaviside 's function..

$$f(t)=t-tH(t-1)+(2-t)(H(t-1)-H(t-2))+0H(t-2)$$
$$f(t)=t-tH(t-1)+(2-t)(H(t-1)-H(t-2))$$
$$f(t)=t+(2-2t)H(t-1)-(2-t)H(t-2))$$
$$f(t)=t+2(1-t)H(t-1)-(2-t)H(t-2))$$
$$f(t)=t-2(t-1)H(t-1)+(t-2)H(t-2))$$

Then take the Lpalace transform

Note that :
$$mathcalL(f(t-c)H(t-c))=e^-csF(s)$$
Therefore we have :
$$boxed F(s)=frac 1s^2-2frac e^-ss^2 +frac e^-2ss^2$$

share|cite|improve this answer

edited Jul 22 at 16:02

answered Jul 22 at 14:58

Isham

10.6k3829

share|cite|improve this answer

share|cite|improve this answer

edited Jul 22 at 16:02

edited Jul 22 at 16:02

edited Jul 22 at 16:02

answered Jul 22 at 14:58

Isham

10.6k3829

answered Jul 22 at 14:58

Isham

10.6k3829

answered Jul 22 at 14:58

Isham

10.6k3829

10.6k3829

add a comment | 

add a comment | 

up vote
0
down vote

I think it's easier to calculate the laplace transform from the definition:
$$F(s)=int_0^inftyf(t)e^-stmathrmdt$$
$$F(s)=int_0^1te^-stmathrmdt+int_1^2(2-t)e^-stmathrmdt$$

share|cite|improve this answer

answered Jul 22 at 14:31

Botond

3,8432632

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I’ve been thought to form the heaviside function first and need to use that in practices and exam, is there a way to that ? I am not too sure of that.
  – user185692
  Jul 22 at 14:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user185692 Check Ian's answer.
  – Botond
  Jul 22 at 14:46
  

  
  
  
  

add a comment | 

up vote
0
down vote

I think it's easier to calculate the laplace transform from the definition:
$$F(s)=int_0^inftyf(t)e^-stmathrmdt$$
$$F(s)=int_0^1te^-stmathrmdt+int_1^2(2-t)e^-stmathrmdt$$

share|cite|improve this answer

answered Jul 22 at 14:31

Botond

3,8432632

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I’ve been thought to form the heaviside function first and need to use that in practices and exam, is there a way to that ? I am not too sure of that.
  – user185692
  Jul 22 at 14:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user185692 Check Ian's answer.
  – Botond
  Jul 22 at 14:46
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

I think it's easier to calculate the laplace transform from the definition:
$$F(s)=int_0^inftyf(t)e^-stmathrmdt$$
$$F(s)=int_0^1te^-stmathrmdt+int_1^2(2-t)e^-stmathrmdt$$

share|cite|improve this answer

answered Jul 22 at 14:31

Botond

3,8432632

I think it's easier to calculate the laplace transform from the definition:
$$F(s)=int_0^inftyf(t)e^-stmathrmdt$$
$$F(s)=int_0^1te^-stmathrmdt+int_1^2(2-t)e^-stmathrmdt$$

share|cite|improve this answer

answered Jul 22 at 14:31

Botond

3,8432632

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 14:31

Botond

3,8432632

answered Jul 22 at 14:31

Botond

3,8432632

answered Jul 22 at 14:31

Botond

3,8432632

3,8432632

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I’ve been thought to form the heaviside function first and need to use that in practices and exam, is there a way to that ? I am not too sure of that.
  – user185692
  Jul 22 at 14:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user185692 Check Ian's answer.
  – Botond
  Jul 22 at 14:46
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I’ve been thought to form the heaviside function first and need to use that in practices and exam, is there a way to that ? I am not too sure of that.
  – user185692
  Jul 22 at 14:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user185692 Check Ian's answer.
  – Botond
  Jul 22 at 14:46
  

  
  
  
  

I’ve been thought to form the heaviside function first and need to use that in practices and exam, is there a way to that ? I am not too sure of that.
– user185692
Jul 22 at 14:44

I’ve been thought to form the heaviside function first and need to use that in practices and exam, is there a way to that ? I am not too sure of that.
– user185692
Jul 22 at 14:44

@user185692 Check Ian's answer.
– Botond
Jul 22 at 14:46

@user185692 Check Ian's answer.
– Botond
Jul 22 at 14:46

add a comment | 

up vote
0
down vote

Don't worry about Heaviside, just break it into two integrals and evaluate it.

Botond has it right. $$ F(s)=int_0^1te^-stmathrmdt+int_1^2(2-t)e^-stmathrmdt$$

Now you have to do integration by parts which I hope you still remember from your calculus class.

share|cite|improve this answer

edited Jul 22 at 14:50

answered Jul 22 at 14:35

Mohammad Riazi-Kermani

27.5k41852

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I’ve been thought to form the heaviside function first and need to use that in practices and exam, is there a way to that ? I’m not too sure of that.
  – user185692
  Jul 22 at 14:45
  

  
  
  
  

add a comment | 

up vote
0
down vote

Don't worry about Heaviside, just break it into two integrals and evaluate it.

Botond has it right. $$ F(s)=int_0^1te^-stmathrmdt+int_1^2(2-t)e^-stmathrmdt$$

Now you have to do integration by parts which I hope you still remember from your calculus class.

share|cite|improve this answer

edited Jul 22 at 14:50

answered Jul 22 at 14:35

Mohammad Riazi-Kermani

27.5k41852

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I’ve been thought to form the heaviside function first and need to use that in practices and exam, is there a way to that ? I’m not too sure of that.
  – user185692
  Jul 22 at 14:45
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Don't worry about Heaviside, just break it into two integrals and evaluate it.

Botond has it right. $$ F(s)=int_0^1te^-stmathrmdt+int_1^2(2-t)e^-stmathrmdt$$

Now you have to do integration by parts which I hope you still remember from your calculus class.

share|cite|improve this answer

edited Jul 22 at 14:50

answered Jul 22 at 14:35

Mohammad Riazi-Kermani

27.5k41852

Don't worry about Heaviside, just break it into two integrals and evaluate it.

Botond has it right. $$ F(s)=int_0^1te^-stmathrmdt+int_1^2(2-t)e^-stmathrmdt$$

Now you have to do integration by parts which I hope you still remember from your calculus class.

share|cite|improve this answer

edited Jul 22 at 14:50

answered Jul 22 at 14:35

Mohammad Riazi-Kermani

27.5k41852

share|cite|improve this answer

share|cite|improve this answer

edited Jul 22 at 14:50

edited Jul 22 at 14:50

edited Jul 22 at 14:50

answered Jul 22 at 14:35

Mohammad Riazi-Kermani

27.5k41852

answered Jul 22 at 14:35

Mohammad Riazi-Kermani

27.5k41852

answered Jul 22 at 14:35

Mohammad Riazi-Kermani

27.5k41852

27.5k41852

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I’ve been thought to form the heaviside function first and need to use that in practices and exam, is there a way to that ? I’m not too sure of that.
  – user185692
  Jul 22 at 14:45
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I’ve been thought to form the heaviside function first and need to use that in practices and exam, is there a way to that ? I’m not too sure of that.
  – user185692
  Jul 22 at 14:45
  

  
  
  
  

I’ve been thought to form the heaviside function first and need to use that in practices and exam, is there a way to that ? I’m not too sure of that.
– user185692
Jul 22 at 14:45

I’ve been thought to form the heaviside function first and need to use that in practices and exam, is there a way to that ? I’m not too sure of that.
– user185692
Jul 22 at 14:45

add a comment | 

up vote
0
down vote

Note that $$ t(H(0)-H(1)) + (2-t)(H(1)-H(2)) =tH(0)-2(t-1)H(1)+(t-2)H(2)$$

You should be able to finish it.

share|cite|improve this answer

answered Jul 22 at 15:00

Mohammad Riazi-Kermani

27.5k41852

add a comment | 

up vote
0
down vote

Note that $$ t(H(0)-H(1)) + (2-t)(H(1)-H(2)) =tH(0)-2(t-1)H(1)+(t-2)H(2)$$

You should be able to finish it.

share|cite|improve this answer

answered Jul 22 at 15:00

Mohammad Riazi-Kermani

27.5k41852

add a comment | 

up vote
0
down vote

up vote
0
down vote

Note that $$ t(H(0)-H(1)) + (2-t)(H(1)-H(2)) =tH(0)-2(t-1)H(1)+(t-2)H(2)$$

You should be able to finish it.

share|cite|improve this answer

answered Jul 22 at 15:00

Mohammad Riazi-Kermani

27.5k41852

Note that $$ t(H(0)-H(1)) + (2-t)(H(1)-H(2)) =tH(0)-2(t-1)H(1)+(t-2)H(2)$$

You should be able to finish it.

share|cite|improve this answer

answered Jul 22 at 15:00

Mohammad Riazi-Kermani

27.5k41852

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 15:00

Mohammad Riazi-Kermani

27.5k41852

answered Jul 22 at 15:00

Mohammad Riazi-Kermani

27.5k41852

answered Jul 22 at 15:00

Mohammad Riazi-Kermani

27.5k41852

27.5k41852

add a comment | 

add a comment | 

up vote
0
down vote

If you want to use the step function then you take a case which applies on $(a,b)$ and multiply it by $H(t-a)-H(t-b)$ (note the order, and use the convention $H(t-infty)=0$) and then sum over cases. This is useful because it matches the form used for Laplace transform tables.

share|cite|improve this answer

edited Jul 22 at 19:02

answered Jul 22 at 14:42

Ian

65k24681

add a comment | 

up vote
0
down vote

If you want to use the step function then you take a case which applies on $(a,b)$ and multiply it by $H(t-a)-H(t-b)$ (note the order, and use the convention $H(t-infty)=0$) and then sum over cases. This is useful because it matches the form used for Laplace transform tables.

share|cite|improve this answer

edited Jul 22 at 19:02

answered Jul 22 at 14:42

Ian

65k24681

add a comment | 

up vote
0
down vote

up vote
0
down vote

If you want to use the step function then you take a case which applies on $(a,b)$ and multiply it by $H(t-a)-H(t-b)$ (note the order, and use the convention $H(t-infty)=0$) and then sum over cases. This is useful because it matches the form used for Laplace transform tables.

share|cite|improve this answer

edited Jul 22 at 19:02

answered Jul 22 at 14:42

Ian

65k24681

If you want to use the step function then you take a case which applies on $(a,b)$ and multiply it by $H(t-a)-H(t-b)$ (note the order, and use the convention $H(t-infty)=0$) and then sum over cases. This is useful because it matches the form used for Laplace transform tables.

share|cite|improve this answer

edited Jul 22 at 19:02

answered Jul 22 at 14:42

Ian

65k24681

share|cite|improve this answer

share|cite|improve this answer

edited Jul 22 at 19:02

edited Jul 22 at 19:02

edited Jul 22 at 19:02

answered Jul 22 at 14:42

Ian

65k24681

answered Jul 22 at 14:42

Ian

65k24681

answered Jul 22 at 14:42

Ian

65k24681

65k24681

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2859456%2flaplace-transform-of-piecewise-function-making-it-to-become-heaviside-unitstep%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

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

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

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email