Help understanding how to solve this problem - Nyquist, Fourier?

[dwverzwy]

This is from an exam prep that provides the solution. I don't understand how they came up with 2*9. To me the solution should be 2*10 to reflect the +-5f0 = 10..
Ugh I am too far out of school to remember all this..Anyone?

 

[dwverzwy]

Original Problem

 

[VEBill]

9+9=18, .: D) 20

I don't think it's much more complicated than that; provided you understand that square waves are made of the odd harmonics and you understand how AM sidebands work (nobody understands how FM sidebands work <- joke alert).

Beware that homework questions are explicitly not permitted. This one barely slips under the posting guidelines as it is an exam prep question.

 

[dwverzwy]

Maybe I understood this more thoroughly at one time, I just am not sure where the 9 comes from.
Not sure why you are warning me because like you said, it is not a homework problem and I am not in school and don't do homework at this time.
Anyway, I don't work in commmunications I work in electrical power distribution and transmission so understandably this stuff is not fresh to me. I am reaching out to those individuals who have a good understanding of this while I search for understanding myself through books and online. Sometimes it helps to have a real person to aid in understading, like on a forum for example.

 

[GregLocock]

The ninth harmonic is at 9 kHz, the maximum frequency of interest is hence 10Mhz+9 kHz, so the bandwidth of a filter centered on 10 MHz is 9*2=18 kHz, call it 20

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[dwverzwy]

Thanks that is helpful, but the problem states that the first 5 non zero components, where n=5. To me, that would be +-5fo = +-5 kHz which is a range of 10 kHZ... Not 9 .

 

[GregLocock]

no, as explained by VE1BLL, 2nd 4th etc are zero amplitude for a square wave.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[LiteYear]

If it helps, draw a number line centred at 10MHz. Then mark the first 5 non-zero components to the right of that - 10M+1kHz, +3kHz, +5kHz, +7kHz and +9kHz. Do the same to the left for the negative frequency components. Then count up the total bandwidth: 9+9=18kHz.

 

[VEBill]

Here's an image on Murata's webpage that shows how the odd harmonics contribute to an increasingly accurate square wave. This type of image can help to make this knowledge tidbit instinctive.

PS: Thank you for the background of your question.

 

[dwverzwy]

Hey guys I genuinely appreciate your insight in this problem. With your explanation, it makes a great deal more sense.. Thanks again..