10/09 - Fourier Transform

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\left \langle e^{j2\pi n t/T} \mid e^{j2\pi m t/T} \right \rangle =\int_{-\infty}^{\infty}e^{j2\pi n t/T} e^{-j2\pi m t/T}\,dt
=\int_{-\infty}^{\infty}e^{j2\pi (n-m) t/T}\,dt
=\int_{-T/2}^{T/2}e^{j2\pi (n-m) t/T}\,dt Assuming the function is perodic with the period T

Fourier Transform

Remember from 10/02 - Fourier Series

  •  \alpha_n = \frac{1}{T}\int_{-T/2}^{T/2} x(t) e^{-j\,2\,\pi \,n\,t/T}\, dt
  • x(t) = x(t+T) = \sum_{n=-\infty}^\infty \alpha_n e^{j\,2\pi \,n/T}

If we let  T \rightarrow \infty

\frac{1}{T} \rightarrow df
\frac{n}{T} \rightarrow f Remember f=\frac{2\pi n}{T} \,\!
T\,\! \rightarrow \infty
\sum_{n=-\infty}^{\infty} \frac{1}{T} \rightarrow \int_{-\infty}^{\infty}() df


F[x(t)]\,\! =X(f)\,\! =\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}dt =\left \langle x(t) \mid e^{j2\pi ft}\right \rangle_t
F^{-1}[X(f)]\,\! =x(t)\,\! =\int_{-\infty}^{\infty} X(f) e^{j2\pi ft}df =\left \langle X(f) \mid e^{-j2\pi ft}\right \rangle_f


\int_{-\infty}^{\infty}e^{j2\pi ft}e^{-j2\pi f\lambda}df =\left\langle e^{j2\pi ft}\mid e^{j2\pi ft}\right\rangle_f =\delta(t-\lambda)\,\!
\int_{-\infty}^{\infty}e^{j2\pi tf}e^{-j2\pi tf_0}dt =\left\langle e^{j2\pi tf}\mid e^{j2\pi tf_0}\right\rangle_t =\delta(f-f_0)\,\!
F^{-1}[F[x(t)]]\,\! =\int_{-\infty}^{\infty}\left [ \int_{-\infty}^{\infty} x(\lambda) e^{-j2\pi f\lambda}d\lambda\right ]e^{j2\pi ft} df =\int_{-\infty}^{\infty}X(f) e^{j2\pi ft} df =x(t)\,\!
=\int_{-\infty}^{\infty}x(\lambda) \int_{-\infty}^{\infty}e^{j2\pi f(t-\lambda)} df d\lambda =\int_{-\infty}^{\infty}x(\lambda) \delta(t-\lambda) d\lambda =x(t)\,\!
=\int_{-\infty}^{\infty}\left [ \int_{-\infty}^{\infty} x(\lambda) e^{-j\omega\lambda}d\lambda\right ]e^{j\omega t} \frac{1}{2\pi}d\omega =\int_{-\infty}^{\infty}x(\lambda) \left [ \frac{1}{2\pi} \int_{-\infty}^{\infty}  e^{j(t-\omega) \lambda}d\omega\right ] d\lambda =\int_{-\infty}^{\infty}x(\lambda) \delta(t-\omega) d\lambda =x(t)\,\!

Sifting property of the delta function

The dirac delta function is defined as any function, denoted as  \delta(t-u)\,\!, that works for all variables that makes the following equation true: x(t)=\int_{-\infty}^{\infty} x(u)\delta(t-u) du

  • When dealing with \omega\,\!, it behaves slightly different than dealing with f\,\!. When dealing with =\int_{-\infty}^{\infty}x(\lambda) \left [ \frac{1}{2\pi} \int_{-\infty}^{\infty}  e^{j(t-\omega) \lambda}d\omega\right ] d\lambda, note that the delta function is \frac{1}{2\pi} \int_{-\infty}^{\infty}  e^{j(t-\omega) \lambda}d\omega. The \frac{1}{2\pi} is tacked onto the front. Thus, when dealing with \omega\,\!, you will often need to multiply it by 2\pi\,\! to cancel out the \frac{1}{2\pi}.

More properties of the delta function

\delta(a\,t) = \frac{1}{\left | a \right |}

\delta(\omega)=\delta(2\pi f)=\frac{1}{2\pi}\delta(f)

\int_{-\infty}^{\infty}\delta(a\,t)\,dt =\int_{-\infty}^{\infty}\delta(u\,t)\,\frac{du}{\left|a\right|} Let a\,t=u and du=a\,dt