I would like to understand a concept of homodyne architecture of I-Q demodulator and how the wave looks like at each step. It is a bandpass Signal: Step 1: multiplication with cos and sin wave. I represent cos and sin wave like Dirac function. For cos is even, sin is odd. Step 2: filter with LPF

Step 3: summation

Will be it correct?

You can also look at it in the time domain. Your received signal is

$$s(t)=s_I(t)\cos(\omega_c t)+s_Q(t)\sin(\omega_c t)\tag{1}$$

where $s_I(t)$ and $s_Q(t)$ are the in-phase and quadrature components, and $\omega_c$ is the carrier frequency (in rad). We have to assume that at the receiver the carrier frequency and phase are known (coherent demodulation), or can be retrieved somehow.

After demodulation we obtain two signals:

\begin{align}s(t)\cos(\omega_ct)&=s_I(t)\cos^2(\omega_ct)+s_Q(t)\sin(\omega_ct)\cos(\omega_ct)\\&=\frac12\big[s_I(t)(1+\cos(2\omega_ct))+s_Q(t)\sin(2\omega_ct)\big]\tag{2}\end{align}

and

\begin{align}s(t)\sin(\omega_ct)&=s_I(t)\cos(\omega_ct)\sin(\omega_ct)+s_Q(t)\sin^2(\omega_ct)\\&=\frac12\big[s_I(t)\sin(2\omega_ct)+s_Q(t)(1-\cos(2\omega_ct))\big]\tag{3}\end{align}

The low pass filters filter out all components at $2\omega_c$, so at the outputs of the filters we're left with

$$r_I(t)=\frac12s_I(t)\tag{4}$$

and

$$r_Q(t)=\frac12s_Q(t)\tag{5}$$

• thanks, will be my presentation(pictures) correct for each step? – user36610 Jul 19 '18 at 6:15
• @user36610: Well, sort of. Note that generally the spectra are complex valued, so it's not so straightforward to sketch them the way you did. – Matt L. Jul 20 '18 at 8:46