0
$\begingroup$

I'd like to know how one is supposed to show that the set $ \{ e^{jwt} \}$, where $\omega \in \mathbb{R}$ , is an ortho-normal basis?

So actually, how do I show that for every $w_1 \neq w_2 $: $\int_{-\infty}^{\infty}e^{jw_1t}e^{-jw2t}dt=0$ ?

Kindly

Sammy

$\endgroup$
  • $\begingroup$ Welcome to Maths SX! Are you sure the bounds are $+\infty$ and $-\infty$? $\endgroup$ – Bernard Oct 26 at 21:10
  • $\begingroup$ Not entirely actually. How do I know which boundaries to choose in order to show orthogonality for this specific set of functions? I have a feeling that its the [-$\pi$,$\pi$] interval... $\endgroup$ – Sammy Apsel Oct 26 at 21:13
  • $\begingroup$ Yes, I would compute it over a period. $\endgroup$ – Bernard Oct 26 at 21:15
0
$\begingroup$

Note first that if $R\in\Bbb R^+$ and $\omega\ne0$,$$\int_{-R}^R\exp(j\omega t)dt=\left[\frac{1}{j\omega}\exp(j\omega t)\right]_{-R}^R=\frac{2\sin(\omega R)}{\omega}=2R\operatorname{sinc}(\omega R).$$The final expression is also valid if $\omega=0$. Since $\frac{1}{\pi}\operatorname{sinc}(\omega)$ is a nascent delta function,$$\int_{-\infty}^\infty\exp(j\omega t)dt=2\pi\delta(\omega).$$In particular,$$\int_{-\infty}^\infty\overline{\exp(j\omega_1t)}\exp(j\omega_2t)dt=2\pi\delta(\omega_1-\omega_2)$$(in your question, you forgot one factor in the integrand should be complex-conjugated). This is our orthonormality condition (well, if we divide each basis element by $\sqrt{2\pi}$).

$\endgroup$
  • $\begingroup$ Thank you for the answer' but would you be kind enough to explain the right hand side of the following equation: $\int_{-\infty}^\infty\exp(j\omega t)dt=2\pi\delta(\omega)$ ? $\endgroup$ – Sammy Apsel Oct 26 at 21:17
  • $\begingroup$ @SammyAspel Are you familiar with nascent delta functions? If so, I'm setting $\epsilon=1/R$. $\endgroup$ – J.G. Oct 26 at 21:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.