# Why the bandwidth in frequency spectrum of AM wave is 2fm and not 2fc?

We know, the instantaneous voltage of AM wave is:

e_AM = Ec * cos(2*pi*fc*t) + ((m * Ec) / 2) * cos(2*pi*(fc + fs)*t) + ((m * Ec) / 2) * cos(2*pi*(fc - fs)*t)

Here:
Ec = Amplitude of the carrier
fc = frequency of the carrier
Es = Amplitude of the message signal
fs = frequency of the message signal
m  = modulation index

Here we get the bandwidth = (fc + fs) - (fc - fs) = 2*fs

But in the equation in the lower side-band's term if we take cos(2*pi*(fs - fc)*t) instead of cos(2*pi*(fc - fs)*t) therefore the bandwidth becomes = (fc + fs) - (fs - fc) that is 2*fc

That is instead of getting the 2 times of the frequency of the message signal (fs) we are getting the 2 times of the frequency of carrier signal (fc).

As cos(-x) = cos(x) hence if we swap the position of fc and fs inside the lower side-band term as the equation is mathematically valid. But it is clear that fc and fs are not the same thing. Hence we are getting two possible values of bandwidth. Although so far I've only seen 2*fs to be used as the value of the bandwidth. So why 2*fc is not used? Isn't it valid? If not, then why mathematically we are getting it to be an alternate solution?

Again if the alternate one is also a valid solution then what is the significance of it?

• $f_s-f_c < 0$ and so you are measuring bandwidth as the difference between the rightmost tip of the sidelobe on the negative axis to the rightmost tip of the sidelobe on the positive axis. This corresponds to no definition of bandwidth with which I am familiar. – Dilip Sarwate Jul 26 '20 at 21:53

Your "instantaneous voltage of AM wave" is a product of carrier and modulation waves decomposed into a sum of pure trig functions. Let us write down this product in its original form: $$e_{AM} = E_c·cos(\omega_c t) · (1 + m·cos(\omega_s t))$$ where angular frequency ω = 2πf.
A Fourier transform of this "voltage of AM wave" is $$\sqrt{\pi/2}·E_c·(\delta(\omega + \omega_c) + \delta(\omega - \omega_c)) +$$ $$\sqrt{\pi/2}·m·E_c·(\delta(\omega + \omega_c + \omega_s) + \delta(\omega + \omega_c - \omega_s) +$$ $$\delta(\omega - \omega_c + \omega_s) + \delta(\omega - \omega_c - \omega_s))/2$$ Mathematically, the bandwith of your "instantaneous voltage of AM wave" is $$(\omega_c + \omega_s) - (-\omega_c - \omega_s) = 2·(\omega_c + \omega_s).$$ TL;DR:
A Fourier transform of your "instantaneous voltage of AM wave" is a set of infinitely narrow peaks at frequencies $$-\omega_c - \omega_s, -\omega_c, -\omega_c + \omega_s, \omega_c - \omega_s, \omega_c, \omega_c + \omega_s.$$ There are negative frequencies, and those are not a pure mathematical peculiarity. These negative frequencies reveal itself in practical designs. Consider a design of superheterodyne transmitter. Two upconverters translate a source signal to a RF range: first to intermediate frequency and second to radio frequency. Two passband filters follow the upconverter mixers. Without filtering, a pure low-frequence tone would translate into six frequencies: $$(\omega_{RF} - \omega_{IF} - \omega_s), (\omega_{RF} - \omega_{IF}), (\omega_{RF} - \omega_{IF} + \omega_s), (\omega_{RF} + \omega_{IF} - \omega_s), (\omega_{RF} + \omega_{IF}), (\omega_{RF} + \omega_{IF} + \omega_s).$$ Filtering of "negative frequencies" in superheterodyne transceivers is called "image rejection".