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Minimum frequency spacing for having orthogonal sinusoidals

Posted By Krishna Sankar On December 31, 2007 @ 4:46 am In Uncategorized | 8 Comments

In this post, the objective is to figure out the minimum separation between two sinusoidals having frequencies $f_1$, $f_2$ of duration $T$each to be orthogonal. Let the phase difference between the sinusoidals is $\phi$ where $\phi$ can take any value from $0$ to $2\pi$ (Refer Example 4.3 [DIG-COMM-SKLAR] [1]).

For the two sinuosidals to be orthogonal,

$\int_0^Tcos(2\pi f_1 t+\phi)cos(2 \pi f_2t)dt = 0$

Integrating and applying the limits, the above equation simplifies to (thanks to the detailed simplification in Example 4.3 [DIG-COMM-SKLAR] [1]) ,

$cos(\phi)\left[\frac{sin(2\pi (f_1+f_2)T)}{2\pi(f_1+f_2)} + \frac{sin(2\pi (f_1-f_2)T)}{2\pi(f_1-f_2)} \right] \\ + sin(\phi)\left[\frac{cos(2\pi (f_1+f_2)T)-1}{2\pi(f_1+f_2)} + \frac{cos(2\pi (f_1-f_2)T)-1}{2\pi(f_1-f_2)} \right]=0$.

Note:

$sin(n\pi)=0$ and $cos(2n\pi)=1$ where $n$ is an integer.

Let as assume that $(f_1+f_2)T$ is an integer. Then two terms in the above equation vanishes as

$sin(\2\pi (f_1+f_2)T)=0$ and $cos(\2\pi (f_1+f_2)T)=1$.

The above equation simplifies to,

$cos(\phi)\frac{sin(2\pi (f_1-f_2)T)}{2\pi(f_1-f_2)} + sin(\phi)\frac{cos(2\pi (f_1-f_2)T)-1}{2\pi(f_1-f_2)} =0$.

For an arbitrary value of $\phi$from $0$ to $2\pi$

In a such a case, for the above equation to be zero, then the cosine term to be equal to 1 and the sine term need to be equal to 0 for making the above equation zero. To satisfy that requirement, need to have,

$2\pi (f_1-f_2)T=2n\pi$

$f_1-f_2=\frac{n}{T}$.

Ofcourse, the minimum value of $n$is 1, then

$f_1-f_2=\frac{1}{T}$.

For $\phi$= $0$

When $\phi$= $0$, then cosine term in the equation is already zero. To make the eqution 0, the sine term need to be equal to be zero. To satisfy that requirement, need to have,

$2\pi (f_1-f_2)T=n\pi$

$f_1-f_2=\frac{n}{2T}$.

Ofcourse, the minimum value of $n$is 1, then

$f_1-f_2=\frac{1}{2T}$.

% Simple Matlab/Octave code

% Minimum frequency seperation between two sinusoidals

T = 1;
fs = 100;
t = 0:1/fs:T;
t = t(1:end-1);

% with random phase
f1 = 1;
f2 = 2;
phi = 2*pi*rand; % uniformly distributed from 0 tp 2pi
s1 = cos(2*pi*f1*t+phi);
s2 = cos(2*pi*f2*t);
sum_with_phi_random = sum(s1.*s2)

% with zero phase difference
f3 = 3/4;
f4 = 5/4;
s3 = cos(2*pi*f3*t);
s4 = cos(2*pi*f4*t);
sum_with_phi_zero = sum(s3.*s4)

close all
figure
plot(t,s1,’b.-’)
hold on
plot(t,s2,’rx-’)
legend(‘s1′,’s2′)
title(‘Minimum frequency seperation for random phase’)
grid on
xlabel(‘time’)
ylabel(‘amplitude’)

figure
plot(t,s3,’b.-’)
hold on
plot(t,s4,’rx-’)
legend(‘s3′,’s4′)
title(‘Minimum frequency seperation for zero phase’)
grid on
xlabel(‘time’)
ylabel(‘amplitude’)

Figure: Two sinusoidals with frequency difference = 1/T

Figure: Two sinusoidals with frequency difference = 1/2T

Summary

1. When the phase difference between two sinuosidals is not known, then the minimum frequency separation between them is $\frac{1}{T}$ for the sinusoidals to be orthogonal.

2. When the phase difference between two sinuosidals is zero, then the minimum frequency separation between them is $\frac{1}{2T}$ for the sinusoidals to be orthogonal.

3. In the above Matlab code snippet, with $\frac{1}{2T}$seperation, the sum of the product of two sinusoidals is only nearly equal to zero (and not zero). Need to think more and revert.

References

Hope this helps,

Krishna