MHB Fourier Transform to find sidebands with 2 different frequencies

[bugatti79]

Hi Folks,

I need to evaluate the following function f(t)=A[1+B \cos(\omega_1 t+ \phi)] \cos(\omega_2 t+ \phi) to find f(\omega) using the Fourier transform.
Ie, the Fourier transform I use is

f(\omega)=\displaystyle \frac{1}{\sqrt {2 \pi}} \int^{\infty}_{-\infty} f(t) (\cos \omega t+ j \sin \omega t)dt

giving

f(\omega)=\displaystyle \frac{1}{\sqrt {2 \pi}} \int^{\infty}_{-\infty} A[1+B \cos(\omega_1 t+ \phi)] \cos(\omega_2 t+ \phi) (\cos \omega t+ j \sin \omega t)dt

We are integrating wrt t but we have 3 different frequencies. Not sure how to handle this...

Basically i want to plot the frequency response as a function of \omega_1 and \omega_2...any ideas?

 

[Euge]

Hi Folks,

I need to evaluate the following function f(t)=A[1+B \cos(\omega_1 t+ \phi)] \cos(\omega_2 t+ \phi) to find f(\omega) using the Fourier transform.
Ie, the Fourier transform I use is

f(\omega)=\displaystyle \frac{1}{\sqrt {2 \pi}} \int^{\infty}_{-\infty} f(t) (\cos \omega t+ j \sin \omega t)dt

giving

f(\omega)=\displaystyle \frac{1}{\sqrt {2 \pi}} \int^{\infty}_{-\infty} A[1+B \cos(\omega_1 t+ \phi)] \cos(\omega_2 t+ \phi) (\cos \omega t+ j \sin \omega t)dt

We are integrating wrt t but we have 3 different frequencies. Not sure how to handle this...

Basically i want to plot the frequency response as a function of \omega_1 and \omega_2...any ideas?

Hi bugatti79,

Your integral diverges unless $A = 0$.

 

[chisigma]

Hi Folks,

I need to evaluate the following function f(t)=A[1+B \cos(\omega_1 t+ \phi)] \cos(\omega_2 t+ \phi) to find f(\omega) using the Fourier transform.
Ie, the Fourier transform I use is

f(\omega)=\displaystyle \frac{1}{\sqrt {2 \pi}} \int^{\infty}_{-\infty} f(t) (\cos \omega t+ j \sin \omega t)dt

giving

f(\omega)=\displaystyle \frac{1}{\sqrt {2 \pi}} \int^{\infty}_{-\infty} A[1+B \cos(\omega_1 t+ \phi)] \cos(\omega_2 t+ \phi) (\cos \omega t+ j \sin \omega t)dt

We are integrating wrt t but we have 3 different frequencies. Not sure how to handle this...

Basically i want to plot the frequency response as a function of \omega_1 and \omega_2...any ideas?

From the formal point of view the integral You have written diverges ... unless you do not use the so called 'singularity functions' based on Dirac delta ... see the way ...

Examples of the Fourier Transform

Kind regards

$\chi$ $\sigma$

 

[bugatti79]

Hi Guys, thanks for the reply.

1) A and B are just constants, so can we at least pull A outside of the integral? Would this solve the divergence issue?

2) I am not sure if this divergence is related to my original query regarding 3 different frequencies \omega \omega_1 \omega_2 in the eqn?

Hi bugatti79,

Your integral diverges unless $A = 0$.

From the formal point of view the integral You have written diverges ... unless you do not use the so called 'singularity functions' based on Dirac delta ... see the way ...

Examples of the Fourier Transform

Kind regards

$\chi$ $\sigma$

 

[Euge]

Hi Guys, thanks for the reply.

1) A and B are just constants, so can we at least pull A outside of the integral? Would this solve the divergence issue?

2) I am not sure if this divergence is related to my original query regarding 3 different frequencies \omega \omega_1 \omega_2 in the eqn?

The answers to 1) and 2) are no and yes, respectively. For 2), the frequencies do not change the divergence of the integral in the usual sense. If you're viewing the Fourier transform distributionally then you may be able to see the interaction of $\omega$ with the two other frequencies frequencies $\omega_1$ and $\omega_2$.

 

[bugatti79]

Hi Euge,

What if we changed the limits of infinity and make it a proper Integral by defining some definite limits?
Does this allow us to have A \ne 0 for successful evaluation?

The answers to 1) and 2) are no and yes, respectively. For 2), the frequencies do not change the divergence of the integral in the usual sense. If you're viewing the Fourier transform distributionally then you may be able to see the interaction of $\omega$ with the two other frequencies frequencies $\omega_1$ and $\omega_2$.

 

[Euge]

Hi Euge,

What if we changed the limits of infinity and make it a proper Integral by defining some definite limits?
Does this allow us to have A \ne 0 for successful evaluation?

You wouldn't need to do that if $f(t)$ is multiplied by a smooth cut-off function. Then the corresponding Fourier transform integral can be computed over a finite interval, regardless the value of $A$.

 

[bugatti79]

What would be a cutoff function that i could use?

 

[Euge]

At the moment you can't use them, because you haven't made sense of this Fourier transform -- is the transform a tempered distribution, or is it a usual integral?

 

[bugatti79]

Not sure I understand your question.

I already have a plot of f(t) against time. I want to convert it to f(w) to see how \omega_1 and \omega_2 interact with each other, ie potential sideband frequencies about the main frequency.

THe f(t) function is supposedly is a modulated waveform with the unaffected fundamental along with an upper and lower sideband. (This can be seen when f(t) is expanded out)

At the moment you can't use them, because you haven't made sense of this Fourier transform -- is the transform a tempered distribution, or is it a usual integral?

 

[Euge]

Ok, I see that you're viewing the transform in the distributional sense. In that case, we express the result in terms of delta functions. First express $f(t)$ as a linear combination of cosine waveforms by using the product-to-sum identity

$$\cos(x)\cos(y) = \frac{1}{2}(\cos(x + y) + \cos(x - y)).$$

You should get

$$f(t) = A\cos(\omega_2 t + \phi) + \frac{AB}{2}\cos[(\omega_1 + \omega_2)t + 2\phi] + \frac{AB}{2}\cos[(\omega_1 - \omega_2)t].$$

Next, using the cosine addition formula

$$\cos(x + y) = \cos(x)\cos(y) - \sin(x)\sin(y),$$

we "separate" the phases from the frequencies in $f(t)$, resulting in

$$(1) \quad f(t) = A\cos(\omega_2t)\cos(\phi) - A\sin(\omega_2 t)\sin(\phi) + \frac{AB}{2}\cos[(\omega_1 + \omega_2)t]\cos(2\phi) - \frac{AB}{2}\sin[(\omega_1 + \omega_2)t]\sin(2\phi) + \frac{AB}{2}\cos[(\omega_1 - \omega_2)t].$$

For convenience I'll let $\mathcal{F}(g(t))(\omega)$ denote the Fourier transform (your version) of a function $g(t)$. Then for all frequencies $\omega_0$,

$$(2) \quad \mathcal{F}(\cos(\omega_0 t))(\omega) = \sqrt{\frac{\pi}{2}}[\delta(\omega_0 + \omega) + \delta(\omega_0 - \omega)]$$

and

$$(3) \quad \mathcal{F}(\sin(\omega_0 t))(\omega) = -j\sqrt{\frac{\pi}{2}}[\delta(\omega_0 + \omega) - \delta(\omega_0 - \omega)].$$

Use $(1)$, $(2)$, $(3)$, and linearity of the Fourier transform to show that

$$\mathcal{F}(f(t))(\omega) = A\sqrt{\frac{\pi}{2}}e^{j\phi}\delta(\omega_2 + \omega) + A\sqrt{\frac{\pi}{2}}e^{-j\phi}\delta(\omega_2 - \omega) + \frac{AB}{2}\sqrt{\frac{\pi}{2}}e^{2j\phi}\delta(\omega_1 + \omega_2 + \omega) + \frac{AB}{2}\sqrt{\frac{\pi}{2}}e^{-2j\phi}\delta(\omega_1 + \omega_2 - \omega) + \frac{AB}{2}\sqrt{\frac{\pi}{2}}\delta(\omega_1 - \omega_2 + \omega) + \frac{AB}{2}\sqrt{\frac{\pi}{2}}\delta(\omega_1 - \omega_2 - \omega).$$

 

[bugatti79]

Hi Euge,
Thanks for you reply. Just working through what you have done.

1) I am not familiar with your notation \mathcal{F}(\cos(\omega_0 t))(\omega) where you have added on the omega term...?
I know that \mathcal{F}(f(t))= \hat f(\omega)

I see you have added the delta function but I am not sure how you arrived at those final terms (2) and (3)?.

For convenience I'll let $\mathcal{F}(g(t))(\omega)$ denote the Fourier transform (your version) of a function $g(t)$. Then for all frequencies $\omega_0$,

$$(2) \quad \mathcal{F}(\cos(\omega_0 t))(\omega) = \sqrt{\frac{\pi}{2}}[\delta(\omega_0 + \omega) + \delta(\omega_0 - \omega)]$$

$$(3) \quad \mathcal{F}(\sin(\omega_0 t))(\omega) = -j\sqrt{\frac{\pi}{2}}[\delta(\omega_0 + \omega) - \delta(\omega_0 - \omega)].$$<br>

Use $(1)$, $(2)$, $(3)$, and linearity of the Fourier transform to show that

$$\mathcal{F}(f(t))(\omega) = A\sqrt{\frac{\pi}{2}}e^{j\phi}\delta(\omega_2 + \omega) + A\sqrt{\frac{\pi}{2}}e^{-j\phi}\delta(\omega_2 - \omega) + \frac{AB}{2}\sqrt{\frac{\pi}{2}}e^{2j\phi}\delta(\omega_1 + \omega_2 + \omega) + \frac{AB}{2}\sqrt{\frac{\pi}{2}}e^{-2j\phi}\delta(\omega_1 + \omega_2 - \omega) + \frac{AB}{2}\sqrt{\frac{\pi}{2}}\delta(\omega_1 - \omega_2 + \omega) + \frac{AB}{2}\sqrt{\frac{\pi}{2}}\delta(\omega_1 - \omega_2 - \omega).$$

 

[Euge]

Hi Euge,
Thanks for you reply. Just working through what you have done.

1) I am not familiar with your notation \mathcal{F}(\cos(\omega_0 t))(\omega) where you have added on the omega term...?
I know that \mathcal{F}(f(t))= \hat f(\omega)

I see you have added the delta function but I am not sure how you arrived at those final terms (2) and (3)?.

Your version of the Fourier transform of a function $f(t)$ is

$$\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(t)e^{j\omega t}\, dt$$

I let $\mathcal{F}(f(t))(\omega)$ represent this integral (since $\omega$ varies I put $\omega$ in the notation to be definite). Given a frequency $\omega_0$,

$$\cos(\omega_0 t) = \frac{e^{j\omega_0 t} + e^{-j\omega_0 t}}{2}.$$

Therefore

$$\mathcal{F}(\cos(\omega_0 t))(\omega)$$

$$ = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \cos(\omega_0 t)e^{j\omega t} \, dt$$

$$ = \frac{1}{2\sqrt{2\pi}}\int_{-\infty}^\infty (e^{j\omega_0 t} + e^{-j\omega_0 t})e^{j\omega t}\, dt$$

$$= \sqrt{\frac{\pi}{2}}\cdot \frac{1}{2\pi}\int_{-\infty}^\infty [e^{j(\omega_0 + \omega)t} + e^{j(\omega - \omega_0)t}]\, dt = \sqrt{\frac{\pi}{2}}[\delta(\omega + \omega_0) + \delta(\omega - \omega_0)].$$

Since $\delta$ is even, $\delta(\omega - \omega_0) = \delta(\omega_0 - \omega)$. So the result follows. Compute $\mathcal{F}(\sin(\omega_0 t))(\omega)$ using the same method.

 

[bugatti79]

Use $(1)$, $(2)$, $(3)$, and linearity of the Fourier transform to show that

$$\mathcal{F}(f(t))(\omega) = A\sqrt{\frac{\pi}{2}}e^{j\phi}\delta(\omega_2 + \omega) + A\sqrt{\frac{\pi}{2}}e^{-j\phi}\delta(\omega_2 - \omega) + \frac{AB}{2}\sqrt{\frac{\pi}{2}}e^{2j\phi}\delta(\omega_1 + \omega_2 + \omega) + \frac{AB}{2}\sqrt{\frac{\pi}{2}}e^{-2j\phi}\delta(\omega_1 + \omega_2 - \omega) + \frac{AB}{2}\sqrt{\frac{\pi}{2}}\delta(\omega_1 - \omega_2 + \omega) + \frac{AB}{2}\sqrt{\frac{\pi}{2}}\delta(\omega_1 - \omega_2 - \omega).$$

1) I presume you used the linearity of \mathcal{F}(a*f+b*g)=a \mathcal{F}(f)+b \mathcal{F}(g)?

2) How did you justify or transform the phases into a complex number? Can you show? I understand that they don't effect the amplitude which is fine for me.

3) Not sure how I can plot this for viewing. I intend to use an interactive math package to show how the frequencies vary etc by assigning slider bars for \omega_1, \omega_2, A and B while having \omega as the abscissa value. Not sure how to handle the \delta or the complex terms?

 

[Euge]

You're right about the linearity. I didn't transform the phases into complex numbers. I used Euler's identity

$$e^{j\theta} = \cos(\theta) + j\sin(\theta)$$

to get those complex exponentials.

 

[bugatti79]

To understand the overall approach:

1)what is the reason we use the dirac function for? Is it because a single sine wave acts as a single impulse where all the power is concentrated?

2)Ie, can the Dirac approach be avoided if we changed the limits of the integral to upper and lower real values?

3)Is this Dirac approach a separate issue from the "cut-off point" that was mentioned in an earlier approach.

Thanks

 

[Euge]

To understand the overall approach:

1)what is the reason we use the dirac function for? Is it because a single sine wave acts as a single impulse where all the power is concentrated?

No, the Dirac functions come out directly from calculation of the Fourier transfrom. The sine and cosine waves are linear combinations of complex exponentials, and complex exponentials translate to (constant multiples of) deltas in Fourier space.

2)Ie, can the Dirac approach be avoided if we changed the limits of the integral to upper and lower real values?

If both limits are finite, the integral will exist in the Riemann sense, so there will be no deltas in the calculation. You'll just have sines in your final answer.

3)Is this Dirac approach a separate issue from the "cut-off point" that was mentioned in an earlier approach.

There was no "Dirac approach" but Dirac delta functions came out from direct calculation. It's a separate from cut-off functions.

 

[Opalg]

Hi Folks,

I need to evaluate the following function f(t)=A[1+B \cos(\omega_1 t+ \phi)] \cos(\omega_2 t+ \phi) to find f(\omega) using the Fourier transform.
Ie, the Fourier transform I use is

f(\omega)=\displaystyle \frac{1}{\sqrt {2 \pi}} \int^{\infty}_{-\infty} f(t) (\cos \omega t+ j \sin \omega t)dt

giving

f(\omega)=\displaystyle \frac{1}{\sqrt {2 \pi}} \int^{\infty}_{-\infty} A[1+B \cos(\omega_1 t+ \phi)] \cos(\omega_2 t+ \phi) (\cos \omega t+ j \sin \omega t)dt

We are integrating wrt t but we have 3 different frequencies. Not sure how to handle this...

Basically i want to plot the frequency response as a function of \omega_1 and \omega_2...any ideas?

Maybe I am completely misunderstanding the problem, but it seems to me that it would be easier to use some elementary trigonometry, rather than the Fourier transform, to analyse this waveform, because of of the divergence problems noted by Euge and chisigma. In fact, $$\begin{aligned} f(t)\ &=A\bigl(1+B \cos(\omega_1 t+ \phi)\bigr) \cos(\omega_2 t+ \phi) \\ &= A\cos(\omega_2 t+ \phi) + AB\cos(\omega_1 t+ \phi) \cos(\omega_2 t+ \phi) \\ &= A\cos(\omega_2 t+ \phi) + \tfrac12AB\bigl( \cos[(\omega_1 - \omega_2)t] - \cos[(\omega_1 + \omega_2)t + 2\phi]\bigr).\end{aligned}$$ That is a combination of three sinusoidal waveforms, one with frequency $\omega_2$, one with frequency $\omega_1 - \omega_2$ and phase difference $-\phi$, and one with frequency $\omega_1 + \omega_2$ and phase difference $\pi + \phi$, the amplitudes being respectively $A$, $\frac12AB$ and $\frac12AB$. That pretty much agrees with the Fourier transform description obtained by Euge in comment #11.