1
$\begingroup$

What is the relation between the AM modulation index and the SNR? (I plotted this and it came as the shape of the exponential graph)

Also what is the relation between the channel noise power and the SNR? I plotted this and I got the shape as this graph:

enter image description here

(ignore the labels of the axis)

Improve this question
$\endgroup$
1
  • $\begingroup$ This question is unanswerable because you give no indication of where the noise is and how it is affecting the signal. $\endgroup$
    – Dilip Sarwate
    Commented Apr 2, 2014 at 13:44

1 Answer 1

Reset to default
1
$\begingroup$

If $x(t)$ is your message signal, then you can write the AM signal as

$$s(t)=A[1+mx(t)]\cos\omega_0 t$$

where $A>0$ is a real-valued constant, $m$ is the modulation index, and $\omega_0$ is the carrier frequency (in radians). If you assume that the noise added by the channel is white with power spectral density $N_0$, then the SNR after demodulation is

$$SNR=\frac{A^2m^2\overline{x^2(t)}}{4N_0B_x}$$

where $B_x$ is the bandwidth of the message $x(t)$ and $\overline{x^2(t)}$ is its average power.

Improve this answer
$\endgroup$
8
  • $\begingroup$ But SNR is not an exponential function of modulation index $m$ as the OP claims his result shows, so the OP must be using a different definition of SNR. $\endgroup$
    – Dilip Sarwate
    Commented Apr 2, 2014 at 14:07
  • $\begingroup$ @DilipSarwate You're right, but this is the most basic case and if the OP is not happy with it, this will hopefully at least start a discussion revealing the exact details of the question. $\endgroup$
    – Matt L.
    Commented Apr 2, 2014 at 15:03
  • $\begingroup$ For SNR I used power of signal/power of noise $\endgroup$
    – user1930901
    Commented Apr 2, 2014 at 18:55
  • $\begingroup$ Yes, that's what's usually done. $\endgroup$
    – Matt L.
    Commented Apr 2, 2014 at 20:21
  • $\begingroup$ so what you posted is applicable or not? $\endgroup$
    – user1930901
    Commented Apr 3, 2014 at 6:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.