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I'm taking a basis course in DSP and in class we had an exercise on amplitude modulating signal. we were given a signal \$x(t)\$ with fourier transform \$X^F (f)\$ with magnitude which looks like this:

enter image description here

now, the modulated signal, \$x_{AM}(t)=2x(t)\cos (2\pi f_c t)\$ has fourier transform: \$X^F _{AM}=X^F(f-f_c)+X^F(f-f_c)\$.

then, I saw in class that the magnitude of this fourier transform looks like this:

enter image description here

so it looks to me as if they implied that the relation: $$|X^F _{AM}|=|X^F(f-f_c)|+|X^F(f-f_c)|$$

holds. I think that the correct relation should be with an additional interference term.

Are there hidden assumptions for this signal which makes the above relation correct?

edit: to be more specific, if I write the magnitude of the fourier transform of the AM signal I get: $$\scriptsize{|X^F _{AM}(f)|=\sqrt{|X^F (f-f_c)|^2+|X^F (f+f_c)|^2+2|X^F (f-f_c)|\cdot |X^F (f+f_c)|\cdot \cos(\angle X^F (f-f_c)-\angle X^F (f+f_c))}}$$ this means that the above relation holds iff \$\angle X^F (f-f_c)=\angle X^F (f+f_c)\$ so, why is it correct generally?

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  • \$\begingroup\$ This isn't really a question; it's more a suspicion with a request for hidden assumptions. Can you be more specific? \$\endgroup\$
    – Andy aka
    Jun 16, 2016 at 9:20
  • \$\begingroup\$ more specifically: first, I'm looking for an approval that this relation is not true for general signal \$x(t)\$ with the attached magnitude of it's fourier transform. second, I'm asking what are the assumptions that one has to make on \$x(t)\$ in order for this relation to hold. \$\endgroup\$
    – Vegetal605
    Jun 16, 2016 at 9:25

2 Answers 2

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Your computation of the cross-term is wrong (at least it was, before you edited). The squared magnitude of the amplitude modulated spectrum is

$$|X_{AM}(f)|^2=|X(f-f_c)|^2+|X(f+f_c)|^2+2|X(f-f_c)||X(f+f_c|\cos(\Delta\phi)\tag{1}$$

where \$\Delta\phi\$ is the phase difference between \$X(f-f_c)\$ and \$X(f+f_c)\$. If \$f_c>f_m\$ (which is usually the case in practice), the two shifted spectra do not overlap and, consequently, the cross-term is zero. For taking the square root a similar reasoning applies: since \$X(f-f_c)\$ and \$X(f+f_c)\$ do not overlap, the square root of the sum is simply the sum of the square roots. Consequently, you get

$$|X_{AM}(f)|=|X(f-f_c)|+|X(f+f_c)|\tag{2}$$

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  • \$\begingroup\$ thanks! the minus sign in the cross-term was a typo. I edited it. \$\endgroup\$
    – Vegetal605
    Jun 16, 2016 at 11:11
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    \$\begingroup\$ @dorsh605: Welcome! By the way, if you have more DSP questions you could ask them on Signal Processing(dsp.stackexchange.com). \$\endgroup\$
    – Matt L.
    Jun 16, 2016 at 11:12
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You are both right and wrong.

The AM process is linear, that means it obeys superposition, which means your course is correct. Linearity is both necessary and sufficient.

In real life, nothing is ever perfectly linear, and there will be distortion, aka interference terms. However, for 'good' hardware, say lab instrumentation, these terms will be very suppressed. For 'cheap' hardware, say commercial radio equipment, these terms will be suppressed enough for the system to work adequately, but maybe no more

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  • \$\begingroup\$ can you elaborate on the mathematical aspect? if I write the magnitude of the fourier transform of the AM signal I get: $$\scriptsize{|X^F _{AM}(f)|=\sqrt{|X^F (f-f_c)|^2+|X^F (f+f_c)|^2+2|X^F (f-f_c)|\cdot |X^F (f-f_c)|\cdot \cos(\angle X^F (f-f_c)-\angle X^F (f+f_c))}}$$ this means that the above relation holds iff \$\angle X^F (f-f_c)=\angle X^F (f+f_c)\$ so, why does it is correct generally? \$\endgroup\$
    – Vegetal605
    Jun 16, 2016 at 10:05

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