How Does Fourier Transform Analyze Beats in Signals?

[Salmone]

What is the Fourier transform of a beat? For example, I want to calculate the Fourier transform of the function $f(t)=\cos((\omega_p+\omega_v) t)+\cos((\omega_p-\omega_v)t),$ where $\omega_p+\omega_v=\Omega,\space\omega_p-\omega_v=\omega$ and $\Omega\simeq\omega.$

I think it is equal to $\frac{1}{2}(\delta(\omega_p+\omega_v-\omega)+\delta(\omega_p-\omega_v-\omega)+\delta(\omega_v-\omega_p-\omega)+\delta(-\omega_p-\omega_v-\omega))$, is it right?

 

[vela]

What is the Fourier transform of a beat? For example, I want to calculate the Fourier transform of the function $f(t)=\cos((\omega_p+\omega_v) t)+\cos((\omega_p-\omega_v)t),$ where $\omega_p+\omega_v=\Omega,\space\omega_p-\omega_v=\omega$ and $\Omega\simeq\omega.$

I think it is equal to $\frac{1}{2}(\delta(\omega_p+\omega_v-\omega)+\delta(\omega_p-\omega_v-\omega)+\delta(\omega_v-\omega_p-\omega)+\delta(-\omega_p-\omega_v-\omega))$, is it right?

Do you have some reason to think your answer is not correct?

 

[Salmone]

Do you have some reason to think your answer is not correct?

No, I just want to be sure.

 

[martinbn]

What does $\Omega\simeq\omega$ mean?

 

[Salmone]

Means that $\Omega$ has a similar value of $\omega$, for example: $\Omega=30Hz$ and $\omega=28Hz$

 

[Delta2]

Only slight problem I see is with the chosen symbols, since you chose $\omega$ for the constant $\omega_p-\omega_v=\omega$ you must use another symbol for the variable of the Fourier transform (that is the omega inside the dirac functions). I know we usually say the Fourier transform of $f(t)$ is $$\hat f(\omega)=...$$ but now you have already chosen $\omega$ to denote something else.

 

[Salmone]

Only slight problem I see is with the chosen symbols, since you chose $\omega$ for the constant $\omega_p-\omega_v=\omega$ you must use another symbol for the variable of the Fourier transform (that is the omega inside the dirac functions). I know we usually say the Fourier transform of $f(t)$ is $$\hat f(\omega)=...$$ but now you have already chosen $\omega$ to denote something else.

I can't edit the post no longer, but let's say $\Omega=\Omega_1$ and $\omega=\omega_p-\omega_v=\Omega_2$.

 

[Delta2]

I can't edit the post no longer, but let's say $\Omega=\Omega_1$ and $\omega=\omega_p-\omega_v=\Omega_2$.

ok fine if you put $\Omega_2$ instead of $\omega$ then I guess you are free to use $\omega$ as the variable of the Fourier transform