I have the equation \$5\cos(t)e^{-3t}u(t)\$ and the Fourier Transform of it is $$\frac{5(3+j\omega)}{(3+j\omega)^2 + 1}$$ I can't figure out how to arrive at this answer.

Using the FT pairs table, I have \$\cos(t)=\pi[\delta(\omega-1)+\delta(\omega+1)]\$ and $$e^{-3t}u(t)=\frac{1}{3+j\omega}$$

Using the property of multiplication I get: $$\frac{5}{2}\left(\frac{1}{3+j(w-1)}+\frac{1}{3+j(w+1)}\right)$$

I just can't figure out how we go from here to $$\frac{5(3+j\omega)}{(3+j\omega)^2 + 1}$$

  • 1
    \$\begingroup\$ You've lost a pi somewhere. Common denominator is [3+j(w-1)][3+j(w+1)], just add the fractions. \$\endgroup\$
    – Chu
    Jun 10, 2015 at 16:31
  • \$\begingroup\$ You're right, the \$pi\$ from \$FT(cos(t))\$ should cancel out the \$pi\$ from the multiplication property. I've changed it. \$\endgroup\$ Jun 10, 2015 at 16:33
  • \$\begingroup\$ I understand now, I was thinking I couldn't distribute the \$j(w-1)\$ and \$j(w+1)\$ \$\endgroup\$ Jun 10, 2015 at 16:45

1 Answer 1


As you mentioned: $$\operatorname{FT}(\cos(t)) = \pi[\delta (\omega-1) + \delta (\omega+1)]$$ $$\operatorname{FT}(e^{-3t}u(t))=\frac{1}{3+j\omega}$$

The multiplication in the time domain is the convolution in the frequency domain with factor \$\frac{1}{2\pi}\$: $$\frac{1}{2\pi}\pi[\delta (\omega-1) + \delta (\omega+1)] * \left(\frac{1}{3+j\omega}\right)=$$ $$=\frac{1}{2}[\delta (\omega-1) + \delta (\omega+1)] * \left(\frac{1}{3+j\omega}\right)=$$ As convolution of a function \$f(\omega)\$ with \$\delta (\omega-a)\$ is \$f(\omega-a)\$:

$$=\frac{1}{2} \left(\frac{1}{3+j(\omega-1)} + \frac{1}{3+j(\omega+1)}\right)=$$

$$=\frac{1}{2}\left( \frac{3+j(\omega+1)+3+j(\omega-1)}{(3+j(\omega+1))(3+j(\omega-1))} \right)=$$ $$=\frac{1}{2}\left( \frac{6+ 2j\omega}{ 9+3j(\omega+1+\omega-1)-(\omega+1)(\omega-1) }\right)=$$ $$=\frac{3+ j\omega}{ 9+6j\omega-\omega^2+1 }=$$ $$=\frac{3+ j\omega}{ (3+j\omega)^2+1 }$$ Add the factor \$5\$ we omitted in the beginning, and you will get your result.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.