Determine the bandwidth of the following functions, assume f1>>>f2:

a) \(x_{2}(t) = sin(2\pi f_{1}t) + cos(2\pi f_{2}t)\)

Here the bandwidth is zero, because the frequency spectrum of these sinusoids consists of 4 impulses, all of which have no width.

b) \(sin(2\pi f_{1}t)cos^{2}(2\pi f_{2}t)\)

This one I'm not sure about. f1 would be acting as the carrier for the second signal since f1>>>f2 correct?

 

Determine the bandwidth of the following functions, assume f1>>>f2:

a) \(x_{2}(t) = sin(2\pi f_{1}t) + cos(2\pi f_{2}t)\)

Here the bandwidth is zero, because the frequency spectrum of these sinusoids consists of 4 impulses, all of which have no width.

From what you show, that is not true. I can only assume there is other information not offered. As shown, the signal will have two distinct frequencies f1 and f2 which would imply a band width of f(1)-f(2)

b) \(sin(2\pi f_{1}t)cos^{2}(2\pi f_{2}t)\)

This one I'm not sure about. f1 would be acting as the carrier for the second signal since f1>>>f2 correct?

This looks like amplitude modulation and the bandwidth will depend on the bandwidth of the modulating function. Side bands are produced by the modulation both above and below the carrier. If the modulating function is a single frequency of f(2), then the sidebands will be at f(1)+f(2) and f(1)-f(2) such that the total bandwidth would be 2*f(2)

 

From what you show, that is not true. I can only assume there is other information not offered. As shown, the signal will have two distinct frequencies f1 and f2 which would imply a band width of f(1)-f(2)
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I don't agree with that. If you have two steady-state sinewave signals then each occupies a single frequency in the spectrum which occupies zero bandwidth. The frequencies in between those two are unused and available for other signals.

 

You need to be able to assimilate both at the same time. For instance, if f(1)=10mHz and f(2)=1kHz, then any transmitter or receiver or other device that needs produce them or to take those signals in and process them needs to be able to pass both 1kHz and 10mHz and would therefore need a bandwidth that encompasses both frequencies.

 

You need to be able to assimilate both at the same time. For instance, if f(1)=10mHz and f(2)=1kHz, then any transmitter or receiver or other device that needs produce them or to take those signals in and process them needs to be able to pass both 1kHz and 10mHz and would therefore need a bandwidth that encompasses both frequencies.

Not necessarily. It could have a narrow bandwidth centered at each frequency with no response in between.

 

Not necessarily. It could have a narrow bandwidth centered at each frequency with no response in between.

Well, granted, but this would be a far more complicated situation.

There is nothing preventing other signals in the same network/system to inhabit the unused parts of the bandwidth. Narrow band filtering could be employed at the right points in the system as required, and is easy to accomplish. This sort of thing is done all the time.

If the two frequencies must be carried by even the same cable, antenna, or other simple devices, those devices need to pass the entire bandwidth.

The only real world example of total seperation I can think of that might be remotely practical is radio. At the least though, you'd need 2 transmitters and two receivers to totally isolate the signals. In the end however, if the two signals are related, they will have to be re-combined in some fashion, even if it is after processing.

I guess the question becomes what is most practical, or even possible.

 

for part b), the bandwidth is still zero right? The modulating function would be the two cos terms multiplying.

 

for part b), the bandwidth is still zero right? The modulating function would be the two cos terms multiplying.

No, the bandwidth in an AM signal is twice the highest frequency component in the modulating message signal as previously mentioned.

If you have a 100 MHz carrier, with a amplitude modulating audio message signal with the highest audio frequency being around 20 kHz, then there will be two side bands (USB and LSB) located around the carrier frequency in the frequency spectrum each of 20 kHz width. So the total bandwidth is 2*20 kHz = 40 kHz.

Later,
Michael