I am confused about how to give the expression of noise. Can someone tell me if my equations are correct?

If the received RF signal is expressed as:

enter image description here


enter image description here

If I want to add a AWGN, can I give the expression of noise as the following?

enter image description here

If not, what is the correct equation of the noise? Also, in the MATLAB simulation, I often add noise to a baseband signal by using: sqrt(NoisePower) * randn(). What should I do in the case of bandpass signal? Thanks guys!


Yes, you can represent your bandpass noise as:

$ n(t) =n_I(t)\cos(2\pi f_c t)-n_Q(t)\sin(2\pi f_c t),$

where $n_I(t)$ and $n_Q(t)$ represent the inphase and quadrature components of the noise signal. In matlab you can generate this sequence as

n = sqrt(Npwr/2) *(randn(1,len)+j*randn(1,len))

The scaling factor is different than in the baseband case because both the inphase and quadrature components have unit power (before scaling), so when added together the total power is equal to 2 and therefore the $\sqrt{2}$ factor is needed in the scaling factor to get unit power.

| improve this answer | |
  • $\begingroup$ Thanks! If the signal is a BPSK signal, the quadrature component should be considered as zero. And what is the noise then? Does the noise still has two components? $\endgroup$ – Berman Song Oct 2 '18 at 5:15
  • $\begingroup$ Yes the noise will have both inphase and quadrature components. You might want to look these notes - see slide VI-6. The modulator shifts the signal and noise by $\pm f_c$ and then the lowpass filter removes one of the shifts. $\endgroup$ – David Oct 2 '18 at 12:04
  • $\begingroup$ I just got another question. In your answer, you show me how to generate the noise in matlab. But, isn't this noise a broad band noise? If I want to generate a bandpass noise, do I need a bandpass filter? $\endgroup$ – Berman Song Dec 11 '18 at 3:59
  • $\begingroup$ No, the noise I showed you how to generate, generates a baseband version of the bandpass noise. This is so you don't have to generate the noise at high frequency (lots of samples) and then baseband. The noise generated does occupy the whole of the baseband frequency range. $\endgroup$ – David Dec 12 '18 at 20:13

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.