0
$\begingroup$

Carrier is 2 * pi * 25, modulating signal is 2 * pi, all the amplitudes are 1s and zero phase.

I realized that I could add a sine wave right at the end. I get a carrier in frequency domain, but my signal in time domain looks strange. Spectogram Simulation

$\endgroup$
  • $\begingroup$ what's the frequency of your carrier and of your modulator ? $\endgroup$ – Hilmar Mar 18 at 20:50
  • $\begingroup$ @Hilmar, I have added information in the post. $\endgroup$ – Alex Mar 18 at 20:53
  • $\begingroup$ Can you confirm that the frequencies you gave are in herz? If your carrier has frequency 25 Hz, I would sample much slower than your 5 kHz -- maybe at 250 Hz. $\endgroup$ – MBaz Mar 18 at 21:46
  • $\begingroup$ @MBaz Frequencies are in radians/second. I changed sampling frequency to 250 HZ for both signals. There is no change. I think it should not affect signal anyway unless I do undersampling. $\endgroup$ – Alex Mar 18 at 22:51
  • $\begingroup$ Can you see the carrier using the scope? $\endgroup$ – MBaz Mar 19 at 0:20
1
$\begingroup$

I see how you added the carrier by multipying by $1 + 0.5cos(2\pi f t)$ (or you used sine, wouldn't change it), and that approach seems fine to me.

It looks like you do actually see the carrier in your plot! What I see from your plot does appear to be a signal at +/-25 Hz which is what we would expect to see for $cos(2\pi 25 t)$ (or sine if you used that)

However also note that your plot claims the resolution bandwidth is 4.88 Hz. Your modulating signal of 1 Hz would not be visible unless you decrease that resolution bandwidth! Right now all signals (25 Hz carrier and +/-1 Hz modulation sidebands) are within that bandwidth so would appear as one signal which is what you see.

To increase the resolution bandwidth, you need to increase the total simulation time (which would increase the number of points in the FFT if you keep the same sampling rate, or you could decrease the sampling rate with the same number of samples or do both). Resolution bandwidth is approximately the inverse of the simulation time, but that bandwidth would be increased further than $1/T$ if you are using windowing in the FFT. From the plot it does not look like windowing is used, in which case your simulation duration would be predicted to be $1/4.88$ Hz or $204.9$ ms. With your sampling rate of $5$ KHz this would suggest $1024$ samples. (It may be a sliding FFT block size over 1024 samples as I also see T = 10s). Thankfully your plot tells you what the bandwidth is so just increase the total FFT size, decrease your sampling rate or both --to see your $1$ Hz sidebands clearly you should make the resolution bandwitdth on the order of $0.1$ Hz.

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Good catch, Dan! $\endgroup$ – MBaz Mar 19 at 16:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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