using Fourier Transforms to solve the question.












1












$begingroup$


I am given a question of Fourier Transform:



$$ e^{2(t-1)}u(t-1) $$
My teacher solved it by using the formula which I couldn't understand so I tried to apply the properties on it.
Now I have solved it by the following method:



$$ e^{2(t)}u(t) rightarrow frac{1}{2+jomega} $$



Now we know that:
$$delta(t-t_0) rightarrow e^{-jomega t_0} $$
So I used the above property on $u(t-1)$ and got the following answer which is same as my teacher got, which is:



$$ frac{e^{-jomega}}{2+jomega} $$



Is my method correct?






fourier-transform







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I am given a question of Fourier Transform:



    $$ e^{2(t-1)}u(t-1) $$
    My teacher solved it by using the formula which I couldn't understand so I tried to apply the properties on it.
    Now I have solved it by the following method:



    $$ e^{2(t)}u(t) rightarrow frac{1}{2+jomega} $$



    Now we know that:
    $$delta(t-t_0) rightarrow e^{-jomega t_0} $$
    So I used the above property on $u(t-1)$ and got the following answer which is same as my teacher got, which is:



    $$ frac{e^{-jomega}}{2+jomega} $$



    Is my method correct?






    fourier-transform







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am given a question of Fourier Transform:



      $$ e^{2(t-1)}u(t-1) $$
      My teacher solved it by using the formula which I couldn't understand so I tried to apply the properties on it.
      Now I have solved it by the following method:



      $$ e^{2(t)}u(t) rightarrow frac{1}{2+jomega} $$



      Now we know that:
      $$delta(t-t_0) rightarrow e^{-jomega t_0} $$
      So I used the above property on $u(t-1)$ and got the following answer which is same as my teacher got, which is:



      $$ frac{e^{-jomega}}{2+jomega} $$



      Is my method correct?






      fourier-transform







      share|cite|improve this question









      $endgroup$




      I am given a question of Fourier Transform:



      $$ e^{2(t-1)}u(t-1) $$
      My teacher solved it by using the formula which I couldn't understand so I tried to apply the properties on it.
      Now I have solved it by the following method:



      $$ e^{2(t)}u(t) rightarrow frac{1}{2+jomega} $$



      Now we know that:
      $$delta(t-t_0) rightarrow e^{-jomega t_0} $$
      So I used the above property on $u(t-1)$ and got the following answer which is same as my teacher got, which is:



      $$ frac{e^{-jomega}}{2+jomega} $$



      Is my method correct?






      fourier-transform




      fourier-transform






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 25 '18 at 17:39









      Ahmad QayyumAhmad Qayyum

      677




      677






















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          $begingroup$

          Assuming that your definition of the Fourier transform is
          $$
          hat f(omega) = int_{mathbb R} f(t)e^{jomega t} mathsf dt,
          $$
          then yes, your answer is correct. We can use a change of variables $s=t+1$ to compute
          begin{align}
          hat f(omega) &= int_1^infty e^{-2(t+1)}e^{jomega t} mathsf dt\
          &= e^{-jomega}int_0^infty e^{-2s}e^{jomega s} mathsf ds\
          &= e^{-jomega}cdotfrac{1}{2+jomega}.
          end{align}






          share|cite|improve this answer









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            $begingroup$

            Assuming that your definition of the Fourier transform is
            $$
            hat f(omega) = int_{mathbb R} f(t)e^{jomega t} mathsf dt,
            $$
            then yes, your answer is correct. We can use a change of variables $s=t+1$ to compute
            begin{align}
            hat f(omega) &= int_1^infty e^{-2(t+1)}e^{jomega t} mathsf dt\
            &= e^{-jomega}int_0^infty e^{-2s}e^{jomega s} mathsf ds\
            &= e^{-jomega}cdotfrac{1}{2+jomega}.
            end{align}






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Assuming that your definition of the Fourier transform is
              $$
              hat f(omega) = int_{mathbb R} f(t)e^{jomega t} mathsf dt,
              $$
              then yes, your answer is correct. We can use a change of variables $s=t+1$ to compute
              begin{align}
              hat f(omega) &= int_1^infty e^{-2(t+1)}e^{jomega t} mathsf dt\
              &= e^{-jomega}int_0^infty e^{-2s}e^{jomega s} mathsf ds\
              &= e^{-jomega}cdotfrac{1}{2+jomega}.
              end{align}






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Assuming that your definition of the Fourier transform is
                $$
                hat f(omega) = int_{mathbb R} f(t)e^{jomega t} mathsf dt,
                $$
                then yes, your answer is correct. We can use a change of variables $s=t+1$ to compute
                begin{align}
                hat f(omega) &= int_1^infty e^{-2(t+1)}e^{jomega t} mathsf dt\
                &= e^{-jomega}int_0^infty e^{-2s}e^{jomega s} mathsf ds\
                &= e^{-jomega}cdotfrac{1}{2+jomega}.
                end{align}






                share|cite|improve this answer









                $endgroup$



                Assuming that your definition of the Fourier transform is
                $$
                hat f(omega) = int_{mathbb R} f(t)e^{jomega t} mathsf dt,
                $$
                then yes, your answer is correct. We can use a change of variables $s=t+1$ to compute
                begin{align}
                hat f(omega) &= int_1^infty e^{-2(t+1)}e^{jomega t} mathsf dt\
                &= e^{-jomega}int_0^infty e^{-2s}e^{jomega s} mathsf ds\
                &= e^{-jomega}cdotfrac{1}{2+jomega}.
                end{align}







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 25 '18 at 22:26









                Math1000Math1000

                19.3k31745




                19.3k31745






























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