0
$\begingroup$

So I'm currently studying ASK (amplitude shift keying) and how it is modulated and demodulated. The modulation take out data signal, $M(t)$, and multiplies it with our carrier frequency, lets say $\cos(2\pi f t)$, and that's our modulated signal.

For the demodulation, we our taking our received signal and multiplying it with our carrier frequency, which in the frequency domain shifts our received signal to the left and right by our carrier frequency. So we end up with out signal both at the origin and at 2 times our carrier frequency. After that we can do a low pass filter to recover out message signal $M(t)$.

So my question is, what happened to our correlator? I thought in digital communications you take your received signal and pass it through a match filter, which can also be implemented as a correlator. In discrete time this is a multiplication with a signal and then a summation, right? What it seems like we're doing is just multiplying just a single point from the received signal with a single point of our carrier frequency. I've simulated this in MATLAB, and I can see what happens to the signal, I'm just confused why this simple multiplication be being called a correlator.

That being said, can this multiplication and low pass filter be done in the frequency domain by any chance? I know a convolution (or correlation) in the time domain is just a multiplication in the frequency domain, but i'm not sure how to do what we're doing for the ASK demodulation in the frequency domain.

Thanks for your time and input!

||improve this question
$\endgroup$
0
$\begingroup$

You are mixing different notions.

What happens in basic digital transmission is :

  1. Digital signal generation (0's and 1's)
  2. Bit to symbol association (digital modulation), in our example ASK.
  3. Shape filtering, which allows the signal's spectrum to be contained within a standardized shape, since we can't just throw a digital signal with very sharp transitions in the air, in would pollute the electromagnetic spectrum.
  4. Finally, we mix the signal with a carrier frequency in order to shifts it's spectrum around the appropriate transmitting frequency

What happens at the receptor is the same operations in anti-chronological order :

  1. Analog demodulation (the signal is mixed with a carrier of the same frequency)
  2. Matched filtering, which is the same operation as in the emitter, but with a specific filter, which is ideally the complex conjugate and temporally reverted of the filter used in emission ($g^*(-t)$). Why is it this one and not another ? See Nyquist ISI criterion.
  3. Symbol estimation
  4. Digital demodulation of estimed symbol

About the filtering operation specifically : time domain and frequency domain operations are equivalent. If you do a time domain filtering, or any mathematical operation (under trivial notations) : $$ y(t) = g(t)*x(t) $$ It is equivalent to frequency domain multiplication :

$$ Y(f) = G(f)X(f) $$

Analog amplitude modulation of a signal $r(t)$ is, like you say, the operation $r(t)\cos(2\pi f_0t)$ (multiplication) which in frequency domain is equivalent to $R(f)(1/2)*(\delta(f-f_0)+\delta(f+f_0))$ (convolution).

Demodulation takes $r(t)\cos(2\pi f_0t)$ and multiply it again : $$ s(t) = r(t)\cos(2\pi f_0t)\cos(2\pi f_0t) $$ Which gives after simplifications : $$ s(t) = r(t)\frac{1}{2}\left(1+\cos(4\pi f_0t)\right) $$ That's when your LPF comes into action : it filters the unwanted $2f_0$ component and keeps only $r(t)$ which is the demodulated signal, and to do this mathematically, we write $s(t)*lpf(t)$, $lpf(t)$ being the LPF's impulse response.

In short : analog demodulation is either a multiplication in time domain ($r(t)\cos(2\pi f_0t)\cos(2\pi f_0t)$) followed by a convolution in time domain ($s(t)*lpf(t)$)

OR

It is a convolution in frequency domain ($R(f)*M(f)*M(f)$) (with $M(f)=(1/2)*(\delta(f-f_0)+\delta(f+f_0))$ followed by a multiplication in frequency domain ($S(f)LPF(f)$)

BUT both methods outputs are the same, physically speaking. We work in frequency or time domain according to which domain is best suited to give easy calculations, as you may have noticed.

||improve this answer
$\endgroup$
  • $\begingroup$ My question might have been answered, but i'm not understanding it, sorry. I know in class we are saying the demod is $r(t) $$\cos(2 \pi f t)$ and then pass this through a LPF, but our general form for demodulation is to multiply our basis function with our receive signal and then integrate, which is a correlation. Is that correct? if so, then how is a multiplication and a LPF the same as a correlation? $\endgroup$ – gerrgheiser Feb 23 '16 at 14:50
  • $\begingroup$ I think I answered your question now. But I still don't get why you integrate something ? Are you calculating convolution in time domain ? $\endgroup$ – MaximGi Feb 23 '16 at 15:14
  • $\begingroup$ Yes ! I understand now, when you talk about integrating, you are calculating what I wrote as $s(t)*g(t)$ which is the low-pass filtering. Right ? $\endgroup$ – MaximGi Feb 23 '16 at 15:15
  • $\begingroup$ I think i understand. So our lowpass filter is our integration which completes our correlation, and also filters out the 2$f_0$ component we created with multiplying by our carrier frequency. I didn't know you would ever take the convolution in the frequency domain though. Thanks for your clarification! $\endgroup$ – gerrgheiser Feb 23 '16 at 15:38
  • $\begingroup$ @gerrgheiser: what MaximGi describes is step 1 in the receiver, which downcoverts the transmitted signal back to baseband. You still need to do step 2, where you multiply the signal by the basis functions and integrate (in other words, do matched filtering). I hope this helps to clarify. $\endgroup$ – MBaz Feb 23 '16 at 19:05

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.