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How to prove a cosine signal with irrational period and another cosine signal with a rational period are orthogonal?

For example, $\cos(4t)$ and $\cos(4\pi t)$ sum is aperiodic. Hence it is said that these two signals are orthogonal over infinite limits? But how to prove this? Normal integration of product of the two signals cannot be used because we get $\cos(\infty)$ which is unknown.

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  • $\begingroup$ Can you please provide a reference for " these two signals are orthogonal over infinite limits?" $\endgroup$ – Oliver Jun 25 '15 at 4:37
  • $\begingroup$ I actually heard this from my professor,but wasn't satisfied with the answer. $\endgroup$ – skt9 Jun 25 '15 at 14:49
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Sinusoids are power signals, and in the case of power signals, cross-correlation between two signals $x(t)$ and $y(t)$ is defined by

$$R_{xy}(t)=\lim_{T\rightarrow\infty}\frac{1}{2T}\int_{-T}^{T}x(\tau)y(t+\tau)d\tau\tag{1}$$

If $x(t)=\cos(\omega_1 t)$ and $y(t)=\cos(\omega_2 t)$ then for $\omega_1\neq\omega_2$ the cross-correlation at $t=0$ is

$$\begin{align}R_{xy}(0)&=\lim_{T\rightarrow\infty}\frac{1}{2T}\int_{-T}^{T}\cos(\omega_1 t)\cos(\omega_2 t)dt\\ &=\lim_{T\rightarrow\infty}\frac{1}{2T}\left[\frac{\sin((\omega_1-\omega_2)T)}{\omega_1-\omega_2}+\frac{\sin((\omega_1+\omega_2)T)}{\omega_1+\omega_2}\right]\\&=0 \end{align}$$

Note that the definition (1) avoids all convergence problems that you would encounter otherwise.

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Using some trigonometry it can be shown that the product can be written as the following sum,

$$ \cos(4 t) \cos(4\pi t) = \frac{\cos(4(\pi+1)t) + \cos(4(\pi-1)t)}{2}. $$

So if we can show that the integral of just a cosine over infinite bounds is zero, then the infinite integral of this product will be zero as well.

The integral of the cosine will be split up into the following parts,

$$ \int_{-\infty}^{\infty}\cos(x)dx = \int_{-\infty}^{-\pi/2}\cos(x)dx+\int_{-\pi/2}^{\pi/2}\cos(x)dx+\int_{\pi/2}^{\infty}\cos(x)dx $$

The left and right part will be equal to each other because the cosine is an even function and the middle part is equal to 2. The left/right integral can be split up into infinitely many intervals defined by each zero crossing, such that the integral of each interval will be either $-2$ or $2$. Therefore it would be equivalent to write this integral as the following infinite summation,

$$ 2 \sum_{n=1}^\infty (-1)^n. $$

A similar summation is known as Grandi's series, which has the solution $1/2$. In our case the series is multiplied by $2$ and starts with $-1$ instead of $+1$, so the sum will be equal to $-1$. Now by adding the left, middle and right part of the integral it can be shown that those add up to zero, thus the two cosines are orthogonal.

If you would argue that the Grandi's series does not converges, then you should agree that a finite integral of a cosine will have a limited range of $[-2,2]$. This can be used when calculating the Fourier coefficient, which is defined as,

$$ a_n = \frac{2}{P} \int_{x_0}^{x_0+P} s(x)\ \cos\left(\frac{2\pi nx}{P}\right) dx. $$

If you choose the period $P$ to be an integer, define $s(x)$ as $\cos(4x)$, then for $n=2P$ you will find the Fourier coefficient of the $\cos(4\pi x)$ component of $\cos(4x)$. For this you do have to take the limit of $P\to\infty$. As shown before, this integral has a limited range, thus as $P$ gets bigger, then $a_n$ gets smaller and with the limit $a_n$ will go to zero.

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  • $\begingroup$ I don't agree: the integral diverges, and the sum is 1/2 only in a limited sense; it doesn't have an actual numerical value. $\endgroup$ – MBaz Jun 25 '15 at 14:46
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Orthogonality requires the definition of an inner product. An inner product definition typically looks like this. $$\langle f, g\rangle_w = \int_a^b f(x)g(x)w(x)\,dx. $$ sometimes with a lim thrown in there as well. The choice of weighting function and integral boundaries impact this a lot.

If you use the inner product definition from Matt L, they are orthogonal, if you use the inner product definition from fibonatic, they are not (despite fibonatic's answer, the integral does not converge). So if you treat them as "power signals" they are orthogonal and if you treat them as "energy signals" they are not. Both are actually valid approaches depending on the context you want to use the orthogonal properties for.

You can't really answer a orthogonality question without a definition of an inner product.

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