Calculate Bandwidth of Filtered Signal

[marina87]

Homework Statement

My message signal is the Fourier Series (x(t)) of a bipolar pulse. This signal passes through the ideal low pass filter, h(f), which has a cutoff frequency of 4Khz. What the baseband bandwidth of the filtered signal?

Homework Equations

ƩAn*cos(2*pi*n*f0*t), An= [4*(-1)^((n-1)/2)]/(pi*n)

In other words: x(t)=(4/pi)*[cos(2*pi*f0*t)-(1/3)*cos(2*pi*3*f0*t)+(1/5)*cos(2*pi*5*f0*t)]

The Attempt at a Solution

I don't know if I should apply the Fourier Transform to the Fourier Series because this one is x(t) and my filter is h(f). What should I do?

 

[I like Serena]

Homework Statement

x(t) is my message signal and its a Fourier series. This signal passes through an ideal low pass filter, h(f), which has a cutoff frequency of 4Khz. What the baseband bandwidth of the filtered signal?

Homework Equations

Its in the attachment that is in
https://www.physicsforums.com/showthread.php?p=3978590#post3978590

The Attempt at a Solution

I have no idea how to work a Fourier series x(t) because I feel I am in the time domain and my filter is in the frequency domain. Should I apply the Fourier Transform to the Fourier Series?

Welcome to PF, marina87!

Your Fourier Series gives you the frequency spectrum.
For instance at $f_0$ the amplitude is ${4 \over \pi}$, and at $3f_0$ the amplitude is ${4 \over \pi}{1 \over 3}$.

If you apply a low pass filter, all the higher frequencies are cut off.
To find the baseband bandwidth, you need to know what $f_0$ is.
Do you?

 

[marina87]

Welcome to PF, marina87!

Your Fourier Series gives you the frequency spectrum.
For instance at $f_0$ the amplitude is ${4 \over \pi}$, and at $3f_0$ the amplitude is ${4 \over \pi}{1 \over 3}$.

If you apply a low pass filter, all the higher frequencies are cut off.
To find the baseband bandwidth, you need to know what $f_0$ is.
Do you?

I see so if my f0 is 1KHz then my bandwidth will be 3KHz.
But if I modulate this signal with a carrier signal cos(2*pi*fc*t) I still need to use the Fourier Transform , right? In other words I still need to apply the Fourier Transform to the series or wait because I filtered the signal is not a Fourier series anymore? Am I correct?

 

[I like Serena]

So what is the highest frequency represented in your Fourier series that is less than your cut-off frequency of 4 kHz?

 

[I like Serena]

I see so if my f0 is 1KHz then my bandwidth will be 3KHz.

Yep. :)

But if I modulate this signal with a carrier signal cos(2*pi*fc*t) I still need to use the Fourier Transform , right? In other words I still need to apply the Fourier Transform to the series or wait because I filtered the signal is not a Fourier series anymore? Am I correct?

There's no real need to apply a Fourier transform.
You can read off the frequency spectrum straight from the Fourier series.

But if you would apply a Fourier transform to you Fourier series, you'll find a transform that shows you spikes at exactly the aforementioned frequencies.Btw, the filtered series is still a Fourier series.
It's just that the higher order terms all have amplitude zero.

 

[marina87]

So what is the highest frequency represented in your Fourier series that is less than your cut-off frequency of 4 kHz?

3Khz but can you help me with my new analysis. I just need someone to tell me if I am in the right track.

After filtered this signal its not a Fourier Series anymore is just a "regular signal that add to cosines functions. Then I can apply the Fourier Transform right?

 

[marina87]

Yep. :)



There's no real need to apply a Fourier transform.
You can read off the frequency spectrum straight from the Fourier series.

But if you would apply a Fourier transform to you Fourier series, you'll find a transform that shows you spikes at exactly the aforementioned frequencies.


Btw, the filtered series is still a Fourier series.
It's just that the higher order terms all have amplitude zero.

Sorry I am confusing you. Let me explain a little more the problem.
First I have to filter my signal x(t) (The Fourier Series) with and ideal low pass filter which has a cut off frequency 4KHz. Afther that this "new signal x1(t)" is multiply by a cos(wct) to obtain a DSB-SC modulation. I am thinking apply the Fourier transform to the modulated signal to obtain the AM expression.

 

[I like Serena]

Hmm, the new signal x1(t) is just the sum of 2 cosine terms.
You can simply multiply them by your new cosine term, yielding your DSB-SC modulation.

If you want to find the frequency spectrum, the easiest way is to replace the cosine products by cosine sums, using the relevant goniometric identity.

See for instance the section "Product-to-sum and sum-to-product identities" in http://en.wikipedia.org/wiki/List_of_trigonometric_identities

 

[marina87]

Hmm, the new signal x1(t) is just the sum of 2 cosine terms.
You can simply multiply them by your new cosine term, yielding your DSB-SC modulation.

If you want to find the frequency spectrum, the easiest way is to replace the cosine products by cosine sums, using the relevant goniometric identity.

See for instance the section "Product-to-sum and sum-to-product identities" in http://en.wikipedia.org/wiki/List_of_trigonometric_identities

That is what I was thinking but ones I have the product of the cosines then I can use the properties of the Fourier Transform to obtain the spectrum?

 

[I like Serena]

Which properties are you thinking of?

 

[marina87]

Which properties are you thinking of?

x(t)*cos(2pif0t) ----> 0.5*(X(f+f0)+X(f-f0))

If you use the Eulers formula of a cos(x)= 0.5*(e^(jx) + e^(-jx)) and then multiply by a a function x(t) then you can apply the Frequency Shift property

 

[I like Serena]

Yes, that looks usable.

You do realize that:
$$\cos \theta \cos \varphi = {\cos(\theta - \varphi) + \cos(\theta + \varphi) \over 2}$$
Note that this looks a lot like your property.

 

[marina87]

Yes, that looks usable.

You do realize that:
$$\cos \theta \cos \varphi = {\cos(\theta - \varphi) + \cos(\theta + \varphi) \over 2}$$
Note that this looks a lot like your property.

Thanks for the help.