Calculate Bandwidth of Filtered Signal

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In summary, the filtered signal is still a Fourier series, but the higher order terms have no amplitude. 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.
  • #1
marina87
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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?
 

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  • #2
marina87 said:

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! :smile:

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?
 
  • #3
I like Serena said:
Welcome to PF, marina87! :smile:

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?
 
  • #4
So what is the highest frequency represented in your Fourier series that is less than your cut-off frequency of 4 kHz?
 
  • #5
marina87 said:
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.
 
  • #6
I like Serena said:
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?
 
  • #7
I like Serena said:
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.
 
  • #8
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
 
  • #9
I like Serena said:
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?
 
  • #10
Which properties are you thinking of?
 
  • #11
I like Serena said:
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
 
  • #12
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.
 
  • #13
I like Serena said:
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.
 

1. What is the bandwidth of a filtered signal?

The bandwidth of a filtered signal refers to the range of frequencies that can pass through the filter without significant attenuation or distortion. It is typically measured in Hertz (Hz) or kilohertz (kHz).

2. How do you calculate the bandwidth of a filtered signal?

The bandwidth of a filtered signal can be calculated by taking the difference between the upper and lower cutoff frequencies of the filter. This can be determined by analyzing the filter's transfer function or frequency response curve.

3. What factors affect the bandwidth of a filtered signal?

The bandwidth of a filtered signal can be affected by several factors, including the type of filter used (e.g. low-pass, high-pass, band-pass), the order of the filter, and the specifications of the components used in the filter circuit.

4. Why is it important to know the bandwidth of a filtered signal?

Knowing the bandwidth of a filtered signal is important for understanding the frequency range of the signal and how it may be affected by the filter. This information is also crucial for designing and optimizing filter circuits for specific applications.

5. Can the bandwidth of a filtered signal be changed?

Yes, the bandwidth of a filtered signal can be changed by adjusting the cutoff frequencies or components in the filter circuit. Different filter designs and configurations can also affect the bandwidth of the filtered signal.

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