Question

A signal is described by the function $\mathbf{D}\left(t\right)\mathbf{=}{\mathbf{Ce}}^{\mathbf{-}\left|t\right|\mathbf{/}\mathbf{t}}$.

(a) Calculate the Fourier transform $\mathbf{A}\left(\omega \right)$. Sketch and interpret your result.

(b) How are $\mathit{D}\left(t\right)$and $\mathit{A}\left(\omega \right)$affected by a change in t ?

Short answer

(a) The Fourier transform is $A\left(\omega \right)=\frac{Ct}{2\mathrm{\pi }}\left(\frac{2}{1+{\left(t\omega \right)}^2}\right)$.

(b) The Fourier transform $A\left(\omega \right)$is inversely proportional to t and $D\left(t\right)$is directly proportional to t.

Step-by-step solution

Fourier transform:

The Fourier transform is a mathematical function that decomposes a waveform that is a function of time into the frequencies that make it up. The result produced by the Fourier transform is a complex valued function of frequency.

The Fourier transform is given by$\mathit{A}\left(k\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi }}{\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\mathit{\psi }\left(x\right){\mathit{e}}^{\mathbf{-}\mathbf{i}\mathbf{k}\mathbf{x}}\mathit{d}\mathit{x}$.

(a) Find Fourier transform:

The Fourier transform of the function$D\left(t\right)$is given by:

$$\begin{gathered} A\left(\omega \right)=\frac{1}{2\mathrm{\pi }}{\int }_{-\infty }^{\infty }D\left(t\right)e^{-i\omega t}dt \\ =\frac{1}{2\mathrm{\pi }}{\int }_{-\infty }^{\infty }C{e}^{-i\omega t}dt \\ =\frac{C}{2\mathrm{\pi }}{\int }_{-\infty }^0{e}^{\frac{t}{t}}{e}^{-i\omega t}dt+\frac{C}{2\mathrm{\pi }}{\int }_0^{\infty }{e}^{\frac{t}{t}}{e}^{-i\omega t}dt \\ =\frac{C}{2\mathrm{\pi }}{\int }_{-\infty }^0{e}^{t\left(\frac{1}{t}-i\omega \right)}dt+\frac{C}{2\mathrm{\pi }}{\int }_0^{\infty }{e}^{t\left(\frac{1}{t}-i\omega \right)}dt \end{gathered}$$

Solve the above equation further:

$$\begin{gathered} A\left(\omega \right)==\frac{C}{2\mathrm{\pi }}\left(\frac{1}{{\displaystyle \frac{1}{t}}-i\omega }+\frac{1}{{\displaystyle \frac{1}{t}}+i\omega }\right) \\ =\frac{C}{2\mathrm{\pi }}\left(\frac{2}{1+{\left(t\omega \right)}^2}\right) \end{gathered}$$

Thus, the Fourier transform is $A\left(\omega \right)=\frac{Ct}{2\mathrm{\pi }}\left(\frac{2}{1+{\left(t\omega \right)}^2}\right)$.

This is not an oscillatory function but an even function whose maximum occurs at$\omega =0$such that$A\left(\omega \right)=\frac{Ct}{\mathrm{\pi }}$. As there is an inverse relation between$∆\omega$and time interval$∆t$. So, the value of$A\left(\omega \right)$approaches 0 as$t\to \pm \infty$. So, the graph can be obtained as:

(b) Define how D(t) and A(ω) affected by a charge in t: 

The width of the function $A\left(\omega \right)$is approximately equal to the distance between the points of half maximum. At $\omega t=\pm 1$, the half maximum occurs, which implies $\omega =\pm \frac{1}{t}$. So, the value of $A\left(w\right)$is directly proportional to $t$. The function $D\left(t\right)$is exponentially decaying function is direct proportionality to $t$.

Thus, $A\left(w\right)$is inversely proportional to $t$and $D\left(t\right)$is directly proportional to $t$.