Topic: Continuous-time Fourier transform of a complex exponential

## Question

What is the Fourier transform of

$x(t)= e^{j \pi t}$?

Guess: $X(f)=\delta (f-\frac{1}{2})$

Proof:

$x(t)=\int_{-\infty}^{\infty} X(f)e^{j2\pi ft} df = \int_{-\infty}^{\infty} \delta (f-\frac{1}{2})e^{j2\pi ft} df = \int_{-\infty}^{\infty} \delta (f-\frac{1}{2})e^{j\pi t} df = e^{j\pi t} \int_{-\infty}^{\infty} \delta (f-\frac{1}{2}) df = e^{j\pi t}$

using the fact that $\delta (t-T)f(t) = \delta (t-T)f(T)$

Instructor's comments: Nice and clear solution! One can also justify the answer using the shifting property directly, which would save a couple of steps.-pm

$x(t) = \int_{-\infty}^{\infty} X(f)e^{j2\pi ft} df$

In order for the following to be true, $x(t)= e^{j \pi t}$

$X(f) = \delta(f - \frac{1}{2})$

because

$x(t) = \int_{-\infty}^{\infty} \delta(f - \frac{1}{2})e^{j2\pi ft} df = e^{j \pi t}$ with careful inspection.

$x(t)=e^{j2\pi 1/2t}=e^{j\omega_0 t},where \omega_0=1/2. F(e^{j\omega_0 t})=2\pi \delta(\omega-\omega_0),also C\delta(Cn)=\delta(n). so, X(f)=\delta (f-\frac{1}{2})$

\begin{align} \mathcal{F}[e^{j\pi t}]=\int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt \\=\int_{-\infty}^{\infty} e^{j\pi t}e^{-j2\pi ft} dt \\=\int_{-\infty}^{\infty} e^{-j2\pi (f-\frac{1}{2})t} dt \\=\delta (f-\frac{1}{2}) \end{align}

TA's comments: This is an infeasible solution! You cannot integrate a complex exponential over the range from -infinity to infinity. See the first solution for reference.

Using the inverse fourier transform definition,

$\, x(t)=e^{j \pi t}= \int_{-\infty}^{\infty}\mathcal{X}(f)e^{j2\pi f t} d f\,$

and the sifting property, we can see that an $X(f)$ that works is

$\delta (f-\frac{1}{2}) = X(f)$

\begin{align} \mathcal{X}(f)&=\int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt \\ &=\int_{-\infty}^{\infty} e^{j\pi t}e^{-j2\pi ft} dt \\ &=\delta \left (f-\frac{1}{2} \right) \end{align}

From the inverse Fourier Transform Definition:

$\, x(t)=e^{j \pi t}= \int_{-\infty}^{\infty}\mathcal{X}(f)e^{j2\pi f t} d f\,$

After inspection, we can see that need to pluck out only the portion of $e^{j 2\pi f t}$ where f = $1/2$

The sifting property will sift that portion out if a $\delta (f-\frac{1}{2})$ is used as X(f), so this is the FT of $e^{j \pi t}$

\begin{align} \mathcal{F}[e^{j\pi t}]=\int_{-\infty}^{\infty} e^{j\pi t}e^{-j2\pi ft} dt \\=\int_{-\infty}^{\infty} e^{-j2\pi (f-\frac{1}{2})t} dt \\=\delta (f-\frac{1}{2}) \end{align} 