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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

write it here.


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