Revision as of 08:27, 7 September 2011

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

Answer 5

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) $

Answer 6

$ \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} $

Answer 7

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} $

Answer 8

$ \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} $

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