Effective Bandwidth of an AM-FM DSB-SC Signal

Have fun.

Could someone deduce the formula of the effective bandwidth which is needed to transmit a suppressed carrier, Fc, whose amplitude and frequency are modulated by F1 and F2 respectively?

Let us assume that 'dFc' is the frequency deviation by F2.

B (Bandwidth) = ?

Isn’t it homework too... for the radio amateurs

Kerim

Almost 50 years ago now, back at college, we did the formula for an AM modulated RF wave - as I recall it went to a number of lines, and was by far the most complex formula we ever did at college. Bear in mind, this was pre-calculator, so all maths was done on pencil and paper, with a log book where needed.

There was no way to remember such a long formula, and no point in doing so, as there seemed little point in it, and wasn't something you were ever going to use.

Nigel Goodwin said:

Almost 50 years ago now, back at college, we did the formula for an AM modulated RF wave - as I recall it went to a number of lines, and was by far the most complex formula we ever did at college. Bear in mind, this was pre-calculator, so all maths was done on pencil and paper, with a log book where needed.

There was no way to remember such a long formula, and no point in doing so, as there seemed little point in it, and wasn't something you were ever going to use.

Click to expand...

Perhaps I wasn't clear on my first post, or you are just kidding

The bandwidth formula, in case of AM only, is actually one of the simplest formulas!
If Fc is amplitude modulated by F1, B= 2*F1. Am I missing something?

Kerim said:

Perhaps I wasn't clear on my first post, or you are just kidding

Click to expand...

No, the formula we were given was a number of lines long, and very complicated.

Nigel Goodwin said:

No, the formula we were given was a number of lines long, and very complicated.

Click to expand...

I don't know what to say. It seems we are talking about two different things.
V_carrier = A*cos(2*pi*Fc*t)
V_audio = B*cos(2*pi*Fa*t)
V_am = V_carrier * V_audio
= A*cos(2*pi*Fc*t) * B*cos(2*pi*Fa*t)
= A*B/2 * [cos(2*pi*Fc*t - 2*pi*Fa*t) + cos(2*pi*Fc*t + 2*pi*Fa*t)]
= A*B/2 * {cos[2*pi*(Fc-Fa)*t] + cos[2*pi*(Fc+Fa)*t]}
...which is the sum of two signals whose frequencies are (Fc-Fa) and (Fc+Fa).
Therefore, the bandwidth, in the radio spectrum, which is needed to transmit V_am is from (Fc-Fa) to (Fc+Fa) or 2*Fa,

This is what I am talking about in case of amplitude modulation only.

In school I took both engineering and technology.
Engineering class we would spend days on a six-line equation.
Technology we learned the bandwidth is 2x the modulation frequency. Then we would build a modulator and watch it work.

ronsimpson said:

In school I took both engineering and technology.
Engineering class we would spend days on a six-line equation.
Technology we learned the bandwidth is 2x the modulation frequency. Then we would build a modulator and watch it work.

Click to expand...

Yes, this is the bandwidth if only the amplitude of the suppressed carrier (Fc) is modulated by F1. So, B= 2*F1.

But what could be the bandwidth if, at the same time, the frequency of the suppressed carrier is also modulated by another signal, F2?

I am afraid, the extended formula of B, for simultaneous amplitude and frequency modulations, cannot be found by searching the internet though it is also simple... as we will see

Those who had the chance to study FM modulation likely heard of Carson's rule which is a good/practical approximation of the effective FM bandwidth. In our case, we can write the FM bandwidth as:

BW_fm = 2 * ( dFc + F2 ) , [line_A]
where:
dFc = the frequency deviation
F2 = the FM modulating frequency.

From [line_A], BW_fm is from Fc – (dFc + F2) to Fc + (dFc + F2) [line_B]

We saw earlier that AM modulation generates two sides. When F2 signal deviates the frequency of the carrier (suppressed or not), all frequencies in these two sides are deviated equally. Therefore, if we assume F1 is the highest AM modulating frequency, we can say that BW_am is from Fc-F1 (=F_lo) to Fc+F1 (=F_hi).
Applying [line_B] around F_lo:
BW_fm_lo is from F_lo - ( dFc + F2 ) to F_lo + ( dFc + F2 ) [line_C]
Similarly, we get around F_hi:
BW_fm_hi is from F_hi - ( dFc + F2 ) to F_hi + ( dFc + F2 ) [line_D]

I hope it is clear from [line_C] and [line_D] that the lowest frequency in case of FM-AM modulation is F_lo - ( dFc + F2 ) , and the highest one is F_hi + ( dFc + F2 )

Replacing F_lo and F_hi, we get:
BW_fm_am is from (Fc - F1 - dFc – F2) to (Fc + F1+ dFc + F2)
or
BW_fm_am = 2 * (F1 + F2 + dFc)
where:
F1 is the highest frequency of the AM modulating signal.
F2 is the highest frequency of the FM modulating signal.
dFc is the greatest frequency deviation.

Et voila, problem solved
Kerim

The origin is over 100 yrs ago https://patentimages.storage.googleapis.com/e7/ab/03/a180f375f467e4/US1449382.pdf
Some cell phones sound worse than AM-FM-SSB-SC and less reliable than the gold standard POTS with > 72 dB SNR

Tony Stewart said:

The origin is over 100 yrs ago https://patentimages.storage.googleapis.com/e7/ab/03/a180f375f467e4/US1449382.pdf
Some cell phones sound worse than AM-FM-SSB-SC and less reliable than the gold standard POTS with > 72 dB SNR

Click to expand...

Sorry, I couldn't get your point well.

I guess you know that even the basic AM DSB-SC is disregarded by FCC because it takes twice the bandwidth of SSB-SC. And it is also believed that implementing a DSB-SC demodulator (of the known topologies, not mine ) is not practical for general use, mainly to radio amateurs, for example.

The case becomes worse with FM-AM DSB-SC because the needed bandwidth is even wider, as it is clear from the above formula of BW_fm_am.

In brief, the topic here is simply academic though it is not discussed yet in any university around the world because even the simple reliable PLL demodulator for AM DSB-SC only is supposed not existent (while FM-AM DSB-SC exists in my reality in the least ).

But this is not the end of the story... If I have time, I will show how to use FM-AM DSB-SC as a scrambled voice modulation while it takes about 4KHz bandwidth only. I guess this bandwidth could be licensed.

Cheers,
Kerim