convergence of Fourier transform of $e^{-t}\sin(2\pi ft)u(t)$

As you see Fourier transform function is being divergent for the first statement but it seems to converge. What is my fault?
\begin{align} \int\limits_{-\infty }^{+\infty }{{{e}^{-t}}\sin(2\pi ft)u(t){{e}^{-j2\pi ft}}\,dt}\\ &=\int\limits_{0}^{+\infty }{{{e}^{-t}}\sin(2\pi {{f}_{c}}t){{e}^{-j2\pi ft}}\,dt}\\ &=\int\limits_{0}^{+\infty }{{{e}^{-t}}\frac{{{e}^{j2\pi {{f}_{c}}t}}-{{e}^{-j2\pi {{f}_{c}}t}}}{2j}{{e}^{-j2\pi ft}}\,dt} \\ & =\int\limits_{0}^{+\infty }{\frac{{{e}^{j2\pi t({{f}_{c}}-f)-t}}-{{e}^{-j2\pi t({{f}_{c}}-f){-t}}}}{2j}\,dt}\\ &=\int\limits_{0}^{+\infty }{\frac{{{e}^{t(j2\pi ({{f}_{c}}-f)-1)}}-{{e}^{-t(j2\pi ({{f}_{c}}-f){+1)}}}}{2j}\,dt} \\ & =\left(\frac{{{e}^{t(j2\pi ({{f}_{c}}-f)-1)}}}{2j(j2\pi ({{f}_{c}}-f)-1)}+\frac{{{e}^{-t(j2\pi ({{f}_{c}}-f){+1)}}}}{2j(j2\pi ({{f}_{c}}-f)+1)}\right)_{0}^{+\infty } \\ \end{align}

• Check the formula for Fourier transform..It should be -j2πft not j2πft.. – Navin Prashath Nov 20 '16 at 14:21
• What you wrote immediately after the third = sign is incorrect and the error propagates from there to give you nonsense at the end – Dilip Sarwate Nov 20 '16 at 14:32
• these are three totally unrealted questions. Don't ask three totally unrelated questions in one question! I'm removing the second and the third, please ask them separately (you can still find your original question in the edit history). – Marcus Müller Nov 20 '16 at 14:38
• Thanks for your response and Thank you Marcus for editing. I've edited the equation. – Ehsan Zakeri Nov 20 '16 at 15:26
• Actually The $e^t(...)$ converges when you set t to infinity – Ehsan Zakeri Nov 20 '16 at 15:59

2 Answers

Let $$x(t)={{e}^{-t}}\sin(2\pi f_ct)u(t)$$

Its Fourier transform denoted by $X(f)$ is calculated as follows \begin{align} X(f)&=\int\limits_{-\infty }^{+\infty }{{{e}^{-t}}\sin(2\pi f_ct)u(t){{e}^{-j2\pi ft}}\,dt}\\ &=\int\limits_{0}^{+\infty }{{{e}^{-t}}\sin(2\pi {{f}_{c}}t){{e}^{-j2\pi ft}}\,dt}\\ &=\int\limits_{0}^{+\infty }{{{e}^{-t}}\left(\frac{{{e}^{j2\pi {{f}_{c}}t}}-{{e}^{-j2\pi {{f}_{c}}t}}}{2j}\right){{e}^{-j2\pi ft}}\,dt} \\ & =\frac{1}{2j}\int\limits_{0}^{+\infty }{{{e}^{-(1+j2\pi (f-f_c))t}}-{{e}^{-(1+j2\pi (f+f_c))t}}\,dt}\\ & = \frac{1}{2j}\left( \frac{-e^{-(1+j2\pi (f-f_c))t}}{1+j2\pi (f-f_c)}-\frac{-e^{-(1+j2\pi (f+f_c))t}}{1+j2\pi (f+f_c)} \right)\biggr\vert^{t=+\infty}_{t=0}\\ &=\frac{1}{2j}\left(\frac{1}{1+j2\pi (f-f_c)}-\frac{1}{1+j2\pi (f+f_c)} \right)\\ &=\frac{j}{2}\left(\frac{1}{1+j2\pi (f+f_c)}-\frac{1}{1+j2\pi (f-f_c)} \right) \end{align}

Notice that this result can also be found using the convolution theorem:

$$\mathcal{F}\{f(t)g(t)\}=F(\omega)*G(\omega)$$

where $f(t)=e^{-t}u(t)$ and $g(t)=\sin(2\pi f_c t)$

• Dear msm , I appreciate you for your help and spend much time to solve this problem.Thanks – Ehsan Zakeri Nov 21 '16 at 6:35
• My pleasure. I'm glad it helped. – msm Nov 21 '16 at 8:51

Thanks to $u(t)$, \begin{align} \int\limits_{-\infty }^{+\infty }{{{e}^{-t}}\sin(2\pi ft)u(t){{e}^{-j2\pi ft}}\,dt}=\int\limits_{0}^{+\infty }{{{e}^{-t}}\sin(2\pi {{f}_{c}}t){{e}^{-j2\pi ft}}\,dt} \end{align} Now $$0 \le | e^{-t}\sin(2\pi {{f}_{c}}t){{e}^{-j2\pi ft}}| \le | e^{-t}|$$ so your function is absolutely integrable.

The trouble comes for the exponential product rule:

$${{{e}^{-t}}{{e}^{j2\pi {\pm{f}_{c}}t}}{{e}^{-j2\pi ft}}} = e^{-t\left(1\mp j2\pi {{f}_{c}}+j2\pi f\right)}$$ which integrates quite well.

• Very true, my mistake. I changed the $\pm$ into $\mp$ too. – Laurent Duval Nov 21 '16 at 9:44