Block diagram of a demodulator with a low-pass filter

Question

I am having some difficulty designing a block diagram of a demodulator with a low pass filter - I don't know how to add it to the schematic(apply the LP filter to the diagram). I have an input signal (sinusoidal) given to me, it then goes through a mixer which adds a signal that is also harmonic, to then mix both of these signals together to form the output signal.

This looks more or less like in the picture. Now I need a low pass filter to filter out the upper frequencies to create a multi-level rectangular wave.

I've been told that mixers in fact multiply signals but it only happens when the signal is in time domain. In this case it is based on Angular frequency/pulsatance domain and I've been taught that mixers on pulsatance add or subtract the signal.

I would also appreciate it if it was based on any literature

Answer 1

This really should be covered in your coursework somewhere.

Everyone has their own favorite block diagram language, so if you do exactly what I say, and this is for a class, you may displease your prof. So -- find out what they are telling you to do, and do it.

Basically, add one of these

In the middle of the block you either put a transfer function (i.e. $\frac{\omega_0}{s + \omega_0}$, or you call out a transfer function (In the Laplace domain, i.e. $H(s)$ if continuous time, in the $z$ domain if discrete), or you put in a little graphic -- folks will call out filters with three horizontal squiggly lines, with diagonal "block" marks through some of them. So a lowpass filter would have the top two lines "blocked", and the bottom one "passing".

Answer 2

Note that a mixer as a time domain multiplier will result in the sum and the difference of the input frequencies. You can derive this from the cosine product rule and consider a carrier frequency as $\cos(2\pi f_c t)$ and a local oscillator at $\cos(2\pi f_1 t)$.

The cosine product rule is given as:

$$\cos(\alpha)\cos(\beta) = \frac{1}{2}\cos(\alpha+\beta) + \frac{1}{2}\cos(\alpha-\beta) $$

And thus we see how we can get the sum and the difference of two frequencies when we multiply them in the time domain. To do a down-converter to translate a frequency from a high frequency carrier to low frequency baseband, we would only be interested in the difference term for the sum and difference that would appear at the output of the multiplier. To select that difference term we use a “low-pass filter”.