Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is a part of two Passband Signals. My question involves inner product. Is
$$ \left < \cos(2\pi ft + \phi) , \sin(2\pi ft + \phi) \right> $$ where $\phi \in [-\pi, \pi)$ equal to zero, i.e. that functions are orthogonal? Thank you in advance.

share|cite|improve this question
@Kaster: Thank you for editing. – karl71 Apr 10 '13 at 8:29
Which inner product are you using? – Dennis Gulko Apr 10 '13 at 8:35
@DennisGulko: imagine <u,v>, so the inner product will be the result of integrating from -infinite to infinite the product of (u)(v*) – karl71 Apr 10 '13 at 8:41
You have to integrate over 1 period, not from $-\infty$ to $\infty$. – Matt L. Apr 10 '13 at 8:42
+1 for posting interesting question – dato datuashvili Apr 10 '13 at 8:43

Yes, they are orthogonal. Consider the interval $[0,T]$ with $T=1/f$ and integrate, then you'll see it. In this case the inner product is defined by

$$\int_{0}^T \sin\left(\frac{2\pi t}{T}+\phi\right) \cos\left(\frac{2\pi t}{T}+\phi\right)dt$$

share|cite|improve this answer
That's true. But my question is if that inner product equals zero for any value of ϕ when ϕ∈[−π,π) – karl71 Apr 10 '13 at 8:39
The value of $\phi$ doesn't matter as long as it shifts both function values equally. You always integrate over one period $T$, and it doesn't matter where you start. – Matt L. Apr 10 '13 at 8:40
+1 for show your efford.i was interesting for me too – dato datuashvili Apr 10 '13 at 8:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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