Null-to-Null Bandwidth of AM Signal (10ms, 1kHz)

[Jammin_James]

1. What is the null-to-null bandwidth of an AM signal with a single-tone message whose duration is 10ms and whose frequency is 1kHz?
2. Relevant transforms (sorry, I don't know how to include special formula functions that are on the forum here):

x(t)cos(ft) <-> 1/2[X(f+f0)+X(f-f0)]

Also the pulse transform, I don't know how I'd type that out, sorry. : /
3. The Attempt at a Solution .

In my mind I've pictured this in the time domain as a cosine wave being multiplied by a pulse 10ms long, centered around 0s. So translated into the frequency domain this would become a Sinc function split by the cosine tone at -1kHz and +1kHz. Normally, since only the positive frequencies are considered, only 1/T of the Sinc would be involved, but because of the shift (from the cosine) both zero crossings of the Sinc function now come into play making the width of the zero crossing 2/T.

So my final answer came to 200Hz.

Does this seem right?

 

[RFengineer75]

James,
I hope this will help you.
First of all a single tone message is just a sinusoid at a constant frequency such
as: m(t) = Acos(2πfmt) where m(t) is the message function, fm is the frequency
of the cosine and A is the amplitude of the message in this case the amplitude of
the a cosine. The bandwidth of this is simply B = fm = (1000 - 0)Hz, from 0 to
the highest frequency in the m(t). This scheme is called tone modulation.
I will switch between f and ω, where ω = 2πf and make reference where needed.

If the frequency of this sinusoid is 1kHz then in the frequency domain it is a pair of pulses
(delta function) at ±fm = ±1kHz in a 2 sided spectrum. The Fourier Transform is
F(ω) = π[δ(ω + ωm)t + δ(ω - ωm)t], where ωm = 2πfm.

When you say AM modulation I assume you mean DSB-SC. If so then the modulated
signal would be:
x(t) = m(t)cos(wct) = Acos(ωmt)cos(ωct) = A/2[cos(ωc + ωm)t + cos(ωc - ωm)t]
In the frequency domain the modulated signal is 2 pulses of half the amplitude of m(t)
at ± ωm on either side of ±ωc for a total bandwidth of 2ωm = 4πB = 4000π .
The bandwidth in Hz of the modulated signal is 2B = 2000 Hz. This is always the case
for DSB-SC.

I see where you refer to the pulse width of 10 ms, so I know that I am not answering exactly what you want, but for a single tone as you put it, the above is correct in DSB-SC
and the duration of the tone would simply be how long the carrier would be modulated.
Like when CW (Morse Code) is being used, and the pulse width would not effect the bandwidth in the frequency domain of M(ω) in this case.
If it is conventional AM = DSB with a carrier then everything would be the same except
at the center freq of ±ωc there would be another pulse which would be the carrier.
With DSB-SC you would use coherent demodulation and with DSB with carrier you could use
envelope detection. If you need more help just let me know. I hope this helps you some.
If you can be more specific I can more than likely explain exactly what your solution should be.
Rob