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Continuous-time Fourier transform of a complex exponential
What is the Fourier transform of $ x(t)= e^{j \pi t} $? Justify your answer.
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Answer 1
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
Answer 2
$ 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.
Answer 3
$ 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}) $
Answer 4
$ \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 \\=\int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt \\ = e^{j\pi t} \end{align} $
