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I have this signal $ s(t) = e^{- \frac{t}{RC} } [ u(t) - u(t - T_c) ] $ and I have to calculate the Fourier transform. I obtained $$ S(f) = \frac{ RC( 1 - e^{-i2 \pi f T_c } )}{1 + i 2 \pi f RC } $$ but I should obtain $$ S(f) = \frac{ RC( 1 - e^{-i2 \pi f T_c } e^{\frac{-T_c}{RC}}) }{1 + i 2 \pi f RC } $$

The Fourier transform of the first part gave me $ \frac{1}{ \frac{1}{RC} + iw } $ and the Fourier transform of the second part of the subtraction gave me $ \frac{1}{ \frac{1}{RC} + iw } e^{-i 2 \pi f T_c } $. I this the error is in the second part but I don’t know how to solve it.

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  • $\begingroup$ You should indicate what is $u$ in your post, this is not a very standard notation. $\endgroup$
    – LL 3.14
    Aug 3, 2020 at 8:46
  • $\begingroup$ $u(\cdot)$ is understood to be the unit step function. $\endgroup$
    – mjw
    Aug 3, 2020 at 11:38

1 Answer 1

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Basically, we just need a repeated application of this Fourier transform:

$$\bbox[5px, border: 1pt solid blue]{{\cal{F}}\left\{e^{-at}u(t-b)\right\} = \int_b^\infty e^{-at} e^{-2\pi i f t}\,dt=\frac{e^{-b(a+2\pi i f)}}{a+2\pi i f}.}$$

Applying it twice:

$$\bbox[5px, border: 1pt solid green]{\begin{aligned}{\cal{F}}\left\{e^{-at}[u(t)-u(t-b)]\right\} &= \int_0^\infty e^{-at} e^{-2\pi i f t}\,dt-\int_b^\infty e^{-at} e^{-2\pi i f t}\,dt \\ &=\frac{1}{a+2\pi i f}-\frac{e^{-b(a+2\pi i f)}}{a+2\pi i f}.\end{aligned}}$$

Equivalently, this can be written as:

$$\bbox[5px, border: 1pt solid purple]{\begin{aligned}{\cal{F}}\left\{e^{-at}[u(t)-u(t-b)]\right\} &= {\cal{F}}\left\{e^{-at}I_{[0,b]}\right\} \\\ &= \int_0^b e^{-at} e^{-2\pi i f t}\,dt\\ &=\frac{1-e^{-b(a+2\pi i f)}}{a+2\pi i f}.\end{aligned}}$$

We have $a=\frac{1}{RC}$ and $b=T_c$:

$$\bbox[5px, border: 1pt solid green]{ \begin{aligned}{\cal{F}}\left\{e^{-\frac{t}{RC}}[u(t)-u(t-T_c)]\right\} &= \int_0^\infty e^{-\frac{t}{RC}} e^{-2\pi i f t}\,dt-\int_{T_c}^\infty e^{-\frac{t}{RC}} e^{-2\pi i f t}\,dt\\ &=\frac{1}{\frac{1}{RC}+2\pi i f}-\frac{e^{-T_c\left(\frac{1}{RC}+2\pi i f\right)}}{\frac{1}{RC}+2\pi i f}.\end{aligned}}$$

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  • $\begingroup$ Thank you so so much !!!! Now i understand perfectly where’s my mistake :) $\endgroup$ Aug 3, 2020 at 11:13
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    $\begingroup$ You are welcome! $\endgroup$
    – mjw
    Aug 3, 2020 at 11:37

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