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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?

Adam Haun
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Padmal
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    No problem! Inline MathJAX on EE.SE needs backslashes before the dollars signs. – Adam Haun Jan 05 '16 at 04:04
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    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. – robert bristow-johnson Jan 05 '16 at 04:47
  • 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 – Mark Ch Jan 05 '16 at 06:08
  • 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. – Mark Ch Jan 05 '16 at 06:30
  • I still don't get the definition of this AM modulation and $\mu$ properly. Different people give different definitions and it's pretty confusing. What is the bad equation at the start? – Padmal Jan 05 '16 at 10:41
  • So the modulated signal should be something like $$x(t) = 5[1 + \frac{0.8}{2}x_m(t)]\cos(\omega_ct)$$ – Padmal Jan 05 '16 at 10:48
  • Hopefully if this question gets answered, I think I can manage with AM at the exam :) – Padmal Jan 05 '16 at 10:49
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    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 $$x(t) = 5[1 + 0.8x_m(t)]\cos(\omega_ct)$$ – Padmal Jan 05 '16 at 12:52
  • great, yes that makes perfect sense. Glad your problem is solved. Sorry, I didn't see your comments. It is good practice to put a username in the comment like @Blogger if you are expecting someone to read it (not meant to be patronising if you already know this, which I'm sure you already do!) – Mark Ch Jan 10 '16 at 15:37

1 Answers1

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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.

rdtsc
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dennis menace
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  • This is an examination question so it has to be possible :) – Padmal Jan 05 '16 at 10:43
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    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. – rdtsc Jan 05 '16 at 12:19
  • 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 – Padmal Jan 05 '16 at 12:55