I have a question as follows. This is not a homework question I just need to clarify my doubt on how this modulation index is defined.

Suppose a 2kHz audio tone having 2V amplitude is to be amplitude modulated on a carrier \$x_c(t) = 5\cos(6\pi 10^5t)\$ with a modulation index of 0.8. For the resulting AM signal

  1. Derive the mathematical expression $$x(t) = A_c[1+\mu x_m(t)]\cos(\omega_c t)$$ $$x(t) = 5[1 + \frac{0.8 * 2}{5}x_m(t)]\cos(2\pi*3 *10^{5}t)$$

is this correct? Isn't modulation index made from \$\frac{2}{5}\$ so do I have to use it like this?

$$x(t) = 5[1 + (0.8 * 2)x_m(t)]\cos(2\pi*3 *10^{5}t)$$

I carried on with the first formula

  1. Sketch the frequency spectrum

enter image description here

I calculated the amplitudes as follows
Carrier Amplitude = \$\frac{A_c}{2}\$ Side band amplitude = \$\frac{\mu A_c a}{4}\$ where a = 2

  1. Find the bandwidth

    \$2 \times f_m\$ = 4 kHz

  2. Find the power of the carrier frequency component

\$\frac{A_c^2}{2} = \frac{5^2}{2}\$ = 12.5 W

  1. Express the total sideband power as a ratio to the carrier power

\$(A_c[1+\mu x_m(t)]\cos(\omega_c t))^2\$ simplifies into \$\frac{A_c^2[1+\mu ^2 x_m^2(t)]}{2}\$

so the carrier power is \$\frac{A_c^2}{2}\$ and total sideband power is \$\frac{A_c^2\mu^2x_m^2(t)}{2}\$

so as a ratio to the carrier power, it is \$\mu^2x_m^2(t)\$ which simplifies as \$\frac{2^2*0.8^2}{2}\$ (Because amplitude of modulating signal is 2V)

Is this assumption correct?

  • 1
    \$\begingroup\$ No problem! Inline MathJAX on EE.SE needs backslashes before the dollars signs. \$\endgroup\$
    – Adam Haun
    Commented Jan 5, 2016 at 4:04
  • 2
    \$\begingroup\$ but not before the double dollar signs. and if you go to DSP.SE, or physics.se or math.se, the backslashes are not there. \$\endgroup\$ Commented Jan 5, 2016 at 4:47
  • \$\begingroup\$ I am willing to admit (0.8*2)/5 in equ 2 looks incorrect. It would be more accurate if at just (0.8/2) so that audio tone is scaled to same amplitude as carrier before applying the modulation index \$\endgroup\$
    – Mark Ch
    Commented Jan 5, 2016 at 6:08
  • \$\begingroup\$ It is difficult to continue with the question, based on the (possibly) bad equation at the start, but let's try... the spectrum simply must be incorrect because it is showing over 100% modulation. In voltage terms, when each sideband is 0.5 of carrier amplitude, you have 100% modulation. \$\endgroup\$
    – Mark Ch
    Commented Jan 5, 2016 at 6:30
  • 1
    \$\begingroup\$ Well, I kinda found a solution. As we are supposed to normalize the signal before modulating, the amplitude of the modulating signal doesn't involve in the equation. So the modulated signal will be $$x(t) = 5[1 + 0.8x_m(t)]\cos(\omega_ct)$$ \$\endgroup\$
    – Padmal
    Commented Jan 5, 2016 at 12:52

1 Answer 1


This formula seems to be misinterpreted by you:
$$x(t) = A_c[1+\mu x_m(t)]\cos(\omega_c t)$$ \$x_m(t)\$ is any message signal, not necessarily a sine wave. Therefore,
$$x(t)=[A_c + A_m x_m(t)]cos(\omega_c t)$$ \$A_c\$ : carrier amplitude
\$A_m\$ : message amplitude

Then the carrier amplitude \$A_c\$ term is taken common to yield:
$$x(t)=A_c[1+\mu x_m(t)]cos(\omega _ct)$$ Where \$\mu\$ = modulation index \$\dfrac{A_m}{A_c}\$ and \$0 \leq\mu \leq 1\$.

So according to this link you cannot give both the carrier and message amplitude and expect a signal with a modulation index within reality.

  • \$\begingroup\$ This is an examination question so it has to be possible :) \$\endgroup\$
    – Padmal
    Commented Jan 5, 2016 at 10:43
  • 1
    \$\begingroup\$ As a professor, it is entirely possible that a) an unintentional error has been made in the exam question, and/or b) it was intentionally introduced. :) Always check your answers. \$\endgroup\$
    – rdtsc
    Commented Jan 5, 2016 at 12:19
  • \$\begingroup\$ Yes this is a tricky question if we don't know the definition properly. I got to know that the modulating signal is normalized before modulating. So the modulating signal amplitude is 1. So we can use \$\mu\$ directly \$\endgroup\$
    – Padmal
    Commented Jan 5, 2016 at 12:55

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