am i applying the property correctly??












0












$begingroup$


The time shifting property states that:



$$ mathcal{L}(u(t-t_0)f(t-t_0))=e^{-st_0}F(s) $$



Now I am given a question which is:



$$ mathcal{L}(e^{-4t}u(t-3)) $$



Can I do like I have done here? That the time shift is of 3 seconds, so using the property I get final answer as:



$$ frac{e^{-3s}}{s+4} $$



Edit: if we are asked to find unilateral Laplace Transform then and only then do I get the above answer?






laplace-transform







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    The time shifting property states that:



    $$ mathcal{L}(u(t-t_0)f(t-t_0))=e^{-st_0}F(s) $$



    Now I am given a question which is:



    $$ mathcal{L}(e^{-4t}u(t-3)) $$



    Can I do like I have done here? That the time shift is of 3 seconds, so using the property I get final answer as:



    $$ frac{e^{-3s}}{s+4} $$



    Edit: if we are asked to find unilateral Laplace Transform then and only then do I get the above answer?






    laplace-transform







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      The time shifting property states that:



      $$ mathcal{L}(u(t-t_0)f(t-t_0))=e^{-st_0}F(s) $$



      Now I am given a question which is:



      $$ mathcal{L}(e^{-4t}u(t-3)) $$



      Can I do like I have done here? That the time shift is of 3 seconds, so using the property I get final answer as:



      $$ frac{e^{-3s}}{s+4} $$



      Edit: if we are asked to find unilateral Laplace Transform then and only then do I get the above answer?






      laplace-transform







      share|cite|improve this question











      $endgroup$




      The time shifting property states that:



      $$ mathcal{L}(u(t-t_0)f(t-t_0))=e^{-st_0}F(s) $$



      Now I am given a question which is:



      $$ mathcal{L}(e^{-4t}u(t-3)) $$



      Can I do like I have done here? That the time shift is of 3 seconds, so using the property I get final answer as:



      $$ frac{e^{-3s}}{s+4} $$



      Edit: if we are asked to find unilateral Laplace Transform then and only then do I get the above answer?






      laplace-transform




      laplace-transform






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 30 '18 at 22:10









      Paul Enta

      5,18111334




      5,18111334










      asked Dec 20 '18 at 15:52









      Ahmad QayyumAhmad Qayyum

      677




      677






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          It's wrong!
          Go step by step.



          Step $1$: Find frequency shift due to $e^{-4t}$ using:



          If $mathcal{L}(f(t))=F(s)$, then
          $mathcal{L}(e^{-4t}f(t))=F(s+4)$. (which is the answer)



          Step $2$: Find $mathcal{L}(f(t))$ by the property of time shift as follows:
          $$mathcal{L}(u(t-3).1)=frac{e^{-3s}}{s}$$
          where $f(t)=u(t-3).1$



          So that finally $$F(s+4)=frac{e^{-3(s+4)}}{s+4}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            can you tell me why can't apply the property over here? what are the cases where we can apply this property? does those cases involve the unshifted version of unit step functions?
            $endgroup$
            – Ahmad Qayyum
            Dec 25 '18 at 18:19











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          It's wrong!
          Go step by step.



          Step $1$: Find frequency shift due to $e^{-4t}$ using:



          If $mathcal{L}(f(t))=F(s)$, then
          $mathcal{L}(e^{-4t}f(t))=F(s+4)$. (which is the answer)



          Step $2$: Find $mathcal{L}(f(t))$ by the property of time shift as follows:
          $$mathcal{L}(u(t-3).1)=frac{e^{-3s}}{s}$$
          where $f(t)=u(t-3).1$



          So that finally $$F(s+4)=frac{e^{-3(s+4)}}{s+4}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            can you tell me why can't apply the property over here? what are the cases where we can apply this property? does those cases involve the unshifted version of unit step functions?
            $endgroup$
            – Ahmad Qayyum
            Dec 25 '18 at 18:19
















          1












          $begingroup$

          It's wrong!
          Go step by step.



          Step $1$: Find frequency shift due to $e^{-4t}$ using:



          If $mathcal{L}(f(t))=F(s)$, then
          $mathcal{L}(e^{-4t}f(t))=F(s+4)$. (which is the answer)



          Step $2$: Find $mathcal{L}(f(t))$ by the property of time shift as follows:
          $$mathcal{L}(u(t-3).1)=frac{e^{-3s}}{s}$$
          where $f(t)=u(t-3).1$



          So that finally $$F(s+4)=frac{e^{-3(s+4)}}{s+4}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            can you tell me why can't apply the property over here? what are the cases where we can apply this property? does those cases involve the unshifted version of unit step functions?
            $endgroup$
            – Ahmad Qayyum
            Dec 25 '18 at 18:19














          1












          1








          1





          $begingroup$

          It's wrong!
          Go step by step.



          Step $1$: Find frequency shift due to $e^{-4t}$ using:



          If $mathcal{L}(f(t))=F(s)$, then
          $mathcal{L}(e^{-4t}f(t))=F(s+4)$. (which is the answer)



          Step $2$: Find $mathcal{L}(f(t))$ by the property of time shift as follows:
          $$mathcal{L}(u(t-3).1)=frac{e^{-3s}}{s}$$
          where $f(t)=u(t-3).1$



          So that finally $$F(s+4)=frac{e^{-3(s+4)}}{s+4}$$






          share|cite|improve this answer











          $endgroup$



          It's wrong!
          Go step by step.



          Step $1$: Find frequency shift due to $e^{-4t}$ using:



          If $mathcal{L}(f(t))=F(s)$, then
          $mathcal{L}(e^{-4t}f(t))=F(s+4)$. (which is the answer)



          Step $2$: Find $mathcal{L}(f(t))$ by the property of time shift as follows:
          $$mathcal{L}(u(t-3).1)=frac{e^{-3s}}{s}$$
          where $f(t)=u(t-3).1$



          So that finally $$F(s+4)=frac{e^{-3(s+4)}}{s+4}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 20 '18 at 18:19

























          answered Dec 20 '18 at 17:54









          Sameer BahetiSameer Baheti

          5718




          5718












          • $begingroup$
            can you tell me why can't apply the property over here? what are the cases where we can apply this property? does those cases involve the unshifted version of unit step functions?
            $endgroup$
            – Ahmad Qayyum
            Dec 25 '18 at 18:19


















          • $begingroup$
            can you tell me why can't apply the property over here? what are the cases where we can apply this property? does those cases involve the unshifted version of unit step functions?
            $endgroup$
            – Ahmad Qayyum
            Dec 25 '18 at 18:19
















          $begingroup$
          can you tell me why can't apply the property over here? what are the cases where we can apply this property? does those cases involve the unshifted version of unit step functions?
          $endgroup$
          – Ahmad Qayyum
          Dec 25 '18 at 18:19




          $begingroup$
          can you tell me why can't apply the property over here? what are the cases where we can apply this property? does those cases involve the unshifted version of unit step functions?
          $endgroup$
          – Ahmad Qayyum
          Dec 25 '18 at 18:19


















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