Practice Question on "Digital Signal Processing"
Topic: Continuous-time Fourier transform of a complex exponential
Contents
Question
What is the Fourier transform of
$ x(t)= e^{j \pi t} $?
Justify your answer.
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
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} $
- 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.
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} $
