Applying Heaviside Function definition

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For the Heaviside function: e^(-cs)*F(s)=u(t-c)*f(t-c) I have 2 questions.

1. How is f(t-c) evaluated? If for instance my f(s) is 1/(s-2) and my e^(-cs) was e^(-3s) the answer should be u(t-3)*e^(2(t-3)). Apart from seeing that constant of 2 being left outside and a 3 being subtracted from the "t" I'm not getting what is happening. Is it f(t) - f(t) evaluated at c?

I'm reading that there is a shift of (3 units) happening here but that does not help me understand.

2. Is that value of C from the equation always positive?
Thank You.

laplace-transform

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asked Jul 15 at 11:08

MisterOH

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For the Heaviside function: e^(-cs)*F(s)=u(t-c)*f(t-c) I have 2 questions.

1. How is f(t-c) evaluated? If for instance my f(s) is 1/(s-2) and my e^(-cs) was e^(-3s) the answer should be u(t-3)*e^(2(t-3)). Apart from seeing that constant of 2 being left outside and a 3 being subtracted from the "t" I'm not getting what is happening. Is it f(t) - f(t) evaluated at c?

I'm reading that there is a shift of (3 units) happening here but that does not help me understand.

2. Is that value of C from the equation always positive?
Thank You.

laplace-transform

share|cite|improve this question

asked Jul 15 at 11:08

MisterOH

193

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

For the Heaviside function: e^(-cs)*F(s)=u(t-c)*f(t-c) I have 2 questions.

1. How is f(t-c) evaluated? If for instance my f(s) is 1/(s-2) and my e^(-cs) was e^(-3s) the answer should be u(t-3)*e^(2(t-3)). Apart from seeing that constant of 2 being left outside and a 3 being subtracted from the "t" I'm not getting what is happening. Is it f(t) - f(t) evaluated at c?

I'm reading that there is a shift of (3 units) happening here but that does not help me understand.

2. Is that value of C from the equation always positive?
Thank You.

laplace-transform

share|cite|improve this question

asked Jul 15 at 11:08

MisterOH

193

For the Heaviside function: e^(-cs)*F(s)=u(t-c)*f(t-c) I have 2 questions.

1. How is f(t-c) evaluated? If for instance my f(s) is 1/(s-2) and my e^(-cs) was e^(-3s) the answer should be u(t-3)*e^(2(t-3)). Apart from seeing that constant of 2 being left outside and a 3 being subtracted from the "t" I'm not getting what is happening. Is it f(t) - f(t) evaluated at c?

I'm reading that there is a shift of (3 units) happening here but that does not help me understand.

2. Is that value of C from the equation always positive?
Thank You.

laplace-transform

share|cite|improve this question

asked Jul 15 at 11:08

MisterOH

193

share|cite|improve this question

share|cite|improve this question

asked Jul 15 at 11:08

MisterOH

193

asked Jul 15 at 11:08

MisterOH

193

asked Jul 15 at 11:08

MisterOH

193

193

add a comment | 

add a comment | 

1 Answer
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0
down vote



accepted

The inverse Laplace transform of $$e^-csF(s)$$ is $$ u(t-c)f(t-c)$$ Thus$$frac e^-3ss-2$$ will change to $$ u(t-3)e^2(t-3)$$

The answer to your second question is yes, we shift functions to the right so c is positive.

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answered Jul 15 at 11:22

Mohammad Riazi-Kermani

27.6k41852

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Firstly thank you for the quick response. To expand on my question, why would writing the solution as u(t-3) e^(2t-3) be wrong? (aside from the fact that it is wrong). From the examples of this type i'm seeing a pattern of simply tacking on a value of (-c) to the value of the exponent. Does the 2 on the outside necessitate a need for parentheses?
  – MisterOH
  Jul 15 at 11:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your attention . You need to change $t$ to $t-3$ which results in $e^2(t-3)$
  – Mohammad Riazi-Kermani
  Jul 15 at 11:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've spent the last 3 hours toying around with different values of "t" in order to make sense of this. You have no idea how much your last remark just helped. Thanks.
  – MisterOH
  Jul 15 at 11:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MisterOH Thanks, I am glad that you figured it out.
  – Mohammad Riazi-Kermani
  Jul 15 at 11:58
  

  
  
  
  

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

up vote
0
down vote



accepted

The inverse Laplace transform of $$e^-csF(s)$$ is $$ u(t-c)f(t-c)$$ Thus$$frac e^-3ss-2$$ will change to $$ u(t-3)e^2(t-3)$$

The answer to your second question is yes, we shift functions to the right so c is positive.

share|cite|improve this answer

answered Jul 15 at 11:22

Mohammad Riazi-Kermani

27.6k41852

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Firstly thank you for the quick response. To expand on my question, why would writing the solution as u(t-3) e^(2t-3) be wrong? (aside from the fact that it is wrong). From the examples of this type i'm seeing a pattern of simply tacking on a value of (-c) to the value of the exponent. Does the 2 on the outside necessitate a need for parentheses?
  – MisterOH
  Jul 15 at 11:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your attention . You need to change $t$ to $t-3$ which results in $e^2(t-3)$
  – Mohammad Riazi-Kermani
  Jul 15 at 11:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've spent the last 3 hours toying around with different values of "t" in order to make sense of this. You have no idea how much your last remark just helped. Thanks.
  – MisterOH
  Jul 15 at 11:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MisterOH Thanks, I am glad that you figured it out.
  – Mohammad Riazi-Kermani
  Jul 15 at 11:58
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

The inverse Laplace transform of $$e^-csF(s)$$ is $$ u(t-c)f(t-c)$$ Thus$$frac e^-3ss-2$$ will change to $$ u(t-3)e^2(t-3)$$

The answer to your second question is yes, we shift functions to the right so c is positive.

share|cite|improve this answer

answered Jul 15 at 11:22

Mohammad Riazi-Kermani

27.6k41852

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Firstly thank you for the quick response. To expand on my question, why would writing the solution as u(t-3) e^(2t-3) be wrong? (aside from the fact that it is wrong). From the examples of this type i'm seeing a pattern of simply tacking on a value of (-c) to the value of the exponent. Does the 2 on the outside necessitate a need for parentheses?
  – MisterOH
  Jul 15 at 11:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your attention . You need to change $t$ to $t-3$ which results in $e^2(t-3)$
  – Mohammad Riazi-Kermani
  Jul 15 at 11:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've spent the last 3 hours toying around with different values of "t" in order to make sense of this. You have no idea how much your last remark just helped. Thanks.
  – MisterOH
  Jul 15 at 11:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MisterOH Thanks, I am glad that you figured it out.
  – Mohammad Riazi-Kermani
  Jul 15 at 11:58
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

The inverse Laplace transform of $$e^-csF(s)$$ is $$ u(t-c)f(t-c)$$ Thus$$frac e^-3ss-2$$ will change to $$ u(t-3)e^2(t-3)$$

The answer to your second question is yes, we shift functions to the right so c is positive.

share|cite|improve this answer

answered Jul 15 at 11:22

Mohammad Riazi-Kermani

27.6k41852

The inverse Laplace transform of $$e^-csF(s)$$ is $$ u(t-c)f(t-c)$$ Thus$$frac e^-3ss-2$$ will change to $$ u(t-3)e^2(t-3)$$

The answer to your second question is yes, we shift functions to the right so c is positive.

share|cite|improve this answer

answered Jul 15 at 11:22

Mohammad Riazi-Kermani

27.6k41852

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 11:22

Mohammad Riazi-Kermani

27.6k41852

answered Jul 15 at 11:22

Mohammad Riazi-Kermani

27.6k41852

answered Jul 15 at 11:22

Mohammad Riazi-Kermani

27.6k41852

27.6k41852

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Firstly thank you for the quick response. To expand on my question, why would writing the solution as u(t-3) e^(2t-3) be wrong? (aside from the fact that it is wrong). From the examples of this type i'm seeing a pattern of simply tacking on a value of (-c) to the value of the exponent. Does the 2 on the outside necessitate a need for parentheses?
  – MisterOH
  Jul 15 at 11:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your attention . You need to change $t$ to $t-3$ which results in $e^2(t-3)$
  – Mohammad Riazi-Kermani
  Jul 15 at 11:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've spent the last 3 hours toying around with different values of "t" in order to make sense of this. You have no idea how much your last remark just helped. Thanks.
  – MisterOH
  Jul 15 at 11:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MisterOH Thanks, I am glad that you figured it out.
  – Mohammad Riazi-Kermani
  Jul 15 at 11:58
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Firstly thank you for the quick response. To expand on my question, why would writing the solution as u(t-3) e^(2t-3) be wrong? (aside from the fact that it is wrong). From the examples of this type i'm seeing a pattern of simply tacking on a value of (-c) to the value of the exponent. Does the 2 on the outside necessitate a need for parentheses?
  – MisterOH
  Jul 15 at 11:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your attention . You need to change $t$ to $t-3$ which results in $e^2(t-3)$
  – Mohammad Riazi-Kermani
  Jul 15 at 11:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've spent the last 3 hours toying around with different values of "t" in order to make sense of this. You have no idea how much your last remark just helped. Thanks.
  – MisterOH
  Jul 15 at 11:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MisterOH Thanks, I am glad that you figured it out.
  – Mohammad Riazi-Kermani
  Jul 15 at 11:58
  

  
  
  
  

Firstly thank you for the quick response. To expand on my question, why would writing the solution as u(t-3) e^(2t-3) be wrong? (aside from the fact that it is wrong). From the examples of this type i'm seeing a pattern of simply tacking on a value of (-c) to the value of the exponent. Does the 2 on the outside necessitate a need for parentheses?
– MisterOH
Jul 15 at 11:38

Firstly thank you for the quick response. To expand on my question, why would writing the solution as u(t-3) e^(2t-3) be wrong? (aside from the fact that it is wrong). From the examples of this type i'm seeing a pattern of simply tacking on a value of (-c) to the value of the exponent. Does the 2 on the outside necessitate a need for parentheses?
– MisterOH
Jul 15 at 11:38

Thanks for your attention . You need to change $t$ to $t-3$ which results in $e^2(t-3)$
– Mohammad Riazi-Kermani
Jul 15 at 11:48

Thanks for your attention . You need to change $t$ to $t-3$ which results in $e^2(t-3)$
– Mohammad Riazi-Kermani
Jul 15 at 11:48

I've spent the last 3 hours toying around with different values of "t" in order to make sense of this. You have no idea how much your last remark just helped. Thanks.
– MisterOH
Jul 15 at 11:53

I've spent the last 3 hours toying around with different values of "t" in order to make sense of this. You have no idea how much your last remark just helped. Thanks.
– MisterOH
Jul 15 at 11:53

@MisterOH Thanks, I am glad that you figured it out.
– Mohammad Riazi-Kermani
Jul 15 at 11:58

@MisterOH Thanks, I am glad that you figured it out.
– Mohammad Riazi-Kermani
Jul 15 at 11:58

add a comment | 

 

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