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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 $$\frac{5(3+j\omega)}{(3+j\omega)^2 + 1}$$

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    \$\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

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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.

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