Calculating Bandwidth of FM Signal: Angle Modulation Q1

[lazyaditya]

Q1. A device with input x(t) and output y(t) is characterized by y(t) = x(t). If a FM signal with frequency deviation 90 Khz and modulating frequency 5 Khz is applied to the input terminals of the device then what will be the bandwidth of the output signal received ?



What i did was calculated the modulating index i.e by dividing frequency deviation by the modulating frequency and got the value equal to 18. Then by using carson rule calculated the bandwidth of the FM signal to be B.W = 2(18 + 1)(5) = 190 Khz.

But answer is not this , how should i do the question then ?

 

[CWatters]

Decades since I did this but isn't the modulation index the other way up eg

5/90 = very low

so only one pair of side bands?

B.W = 2(v.small + 1)(5) = 10 Khz

 

[rude man]

I got the same answer you did: 2(90 + 5) the way I did it.

Maybe your source wasn't satisfied with 98% of the energy contained within the Carson bandwidth. After all, the real answer is infinity! The Bessel function expansion of the modulated carrier extends without limit ...

 

[lazyaditya]

The answer given to me was 380 Khz that is 2 * 190 and i dnt know why , do it have to do something with the transfer function but h(t) = 1 since y(t)=x(t) , the FM signal is having bandwidth equal to 190 Khz at the time of input but how and why will it double when passed through the device having relationship y(t) = x(t), please help ?

 

[lazyaditya]

I got the same answer you did: 2(90 + 5) the way I did it.

Maybe your source wasn't satisfied with 98% of the energy contained within the Carson bandwidth. After all, the real answer is infinity! The Bessel function expansion of the modulated carrier extends without limit ...

why is the real answer infinity ?

 

[lazyaditya]

Decades since I did this but isn't the modulation index the other way up eg

5/90 = very low

so only one pair of side bands?

B.W = 2(v.small + 1)(5) = 10 Khz

Modulation index is 90/5 , i gave frequency deviation to be 90 and modulating frequency to be 5

 

[rude man]

why is the real answer infinity ?

Because if you expand the modulated signal in a series, that series has an infinite number of harmonics. In this case it's a Bessel series. You get Bessel series whenever you get functions like sin(a + bsin(x)) etc. This should be available somewhere in Wikipedia, if your textbook doesn't cover the subject rigorously.

E.g. http://en.wikipedia.org/wiki/Frequency_modulation

 

[CWatters]

Deleted

 

[lazyaditya]

Because if you expand the modulated signal in a series, that series has an infinite number of harmonics. In this case it's a Bessel series. You get Bessel series whenever you get functions like sin(a + bsin(x)) etc. This should be available somewhere in Wikipedia, if your textbook doesn't cover the subject rigorously.

E.g. http://en.wikipedia.org/wiki/Frequency_modulation

Ya total bandwidth is infinity but we don't consider the low powered components right? so they are neglected. But what about 380 Khz ?

 

[NascentOxygen]

Modulation index is 90/5 , i gave frequency deviation to be 90 and modulating frequency to be 5

With a modulation index of 18 it would be considered broadband FM, I think. But I can't see how bandwidth can be much different from 2*(18+1)*5 kHz.

 

[rude man]

Ya total bandwidth is infinity but we don't consider the low powered components right? so they are neglected. But what about 380 Khz ?

Right on the first part.

As for the second: I have no idea where the 380KHz comes from. It's obviously twice the answer we're getting. Are you sure you wrote the question down right? Seems funny they gave you y(t) = x(t). That's just a straight feedtru from input to output! If they had given you y(t) = x²(t) then you would have generated the 2nd harmonic of 190 KHz = 380 KHz at the output.

 

[NascentOxygen]

If they had given you y(t) = x²(t) then you would have generated the 2nd harmonic of 190 KHz = 380 KHz at the output.

Ding! I believe we have the winner.

 

[lazyaditya]

Right on the first part.

As for the second: I have no idea where the 380KHz comes from. It's obviously twice the answer we're getting. Are you sure you wrote the question down right? Seems funny they gave you y(t) = x(t). That's just a straight feedtru from input to output! If they had given you y(t) = x²(t) then you would have generated the 2nd harmonic of 190 KHz = 380 KHz at the output.

Thanks for this , i think then the question in book would have been wrong and answer would have been this "380 Khz" when y(t)= x^2(t)