Fourier transform of a time-derivative signal post-operated by an exponential

Question

Is the Fourier transform of $e^{-at}\dot{x}(t)$ for $a>0$ given by $j\omega X(a+j\omega)$ where $X(j\omega)$ is the Fourier transform of $x(t)$? Can somebody please show me the steps how to derive it without using the shortcuts.

Answer 1

By the Convolution theorem, we have that the Fourier Transform of a product is given by the convolution of their Fourier Transforms:

$$\mathcal{F}[f(t)g(t)](\omega) = \sqrt{2\pi}\int_{-\infty}^\infty \mathcal{F}[f(t)](s -\omega)\mathcal{F}[g(t)](s)\, ds$$

The Fourier Transform of $e^{-at}$ is given by

\begin{align} \mathcal{F}[e^{-at}](\omega) &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-(a+i\omega)t}\, dt\\ &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-i(\omega - ia)t}\,dt\\ &= \sqrt{2\pi}\,\delta(\omega-ia) \end{align}

The Fourier Transform of $\dot x(t)$ can be obtained using integration by parts:

\begin{align} \mathcal{F}[\dot x(t)](\omega) &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \dot x(t)e^{-i\omega t}\,dt\\ &= \frac{-i\omega}{\sqrt{2\pi}}\int_{-\infty}^\infty x(t) e^{-i\omega t}\,dt\\ &= -i\omega\mathcal{F}[x(t)](\omega) =: -i\omega\xi(\omega) \end{align}

Plugging these results into the convolution integral from above, we have

\begin{align} \mathcal{F}\left[e^{-at}\dot x(t)\right](\omega) &= -2\pi i\int_{-\infty}^\infty \delta(s - (\omega + ia))\, s\,\xi(s)\\ &= - 2\pi i\left(\omega + ia\right)\xi(\omega + ia) \end{align}

As you can see, aside from some factors of $-2\pi$, which can be down to convention, your second result is nearly the correct one - you're just missing an $i$ in there as well.

Answer 2

I provide the following as an alternative that sticks a little more closely to your original notation.

Let $$y\left( t \right) = {e^{ - at}}\dot x\left( t \right){\rm{ }}a > 0$$ Since $$X\left( {j\omega } \right) = \int\limits_{ - \infty }^\infty {dt\,{e^{ - j\omega t}}} x\left( t \right){\rm{ }} \Rightarrow {\rm{ }}X\left( {j\omega + a} \right) = \int\limits_{ - \infty }^\infty {dt\,{e^{ - \left( {j\omega + a} \right)t}}x\left( t \right)}$$ using integration by parts $$\begin{array}{c} Y\left( {j\omega } \right) = \int\limits_{ - \infty }^\infty {dt\,{e^{ - j\omega t}}} y\left( t \right) = \int\limits_{ - \infty }^\infty {dt\,{e^{ - \left( {j\omega + a} \right)t}}\dot x\left( t \right)} \\ = - \left( {j\omega + a} \right)\left. {{e^{ - \left( {j\omega + a} \right)t}}x\left( t \right)} \right|_{ - \infty }^\infty + \left( {j\omega + a} \right)\int\limits_{ - \infty }^\infty {dt\,{e^{ - \left( {j\omega + a} \right)t}}x\left( t \right)} \\ = \left( {j\omega + a} \right)X\left( {j\omega + a} \right) \end{array}$$ This derivation assumes both Fourier Transforms actually exist and that the endpoints in the integration by parts vanish. As far as I can tell, this is equivalent to the previous answer, other than the normalization assumed and the integration by parts, where I think an additional minus sign got added.