IQ generation separation

Question

To generate I & Q samples from data samples S

I = S(1:2:length(S)-1) * cos(2*pi*f*t) ;
Q = S(2:2:length(S)  ) * sin(2*pi*f*t) ;

I & Q to integer arithmetic

I = I * (2^16) ;
Q = Q * (2^16) ;

I Q combined

y = bitshiftleft(Q, 16) ; 
y = bitor(y, I);

and mixed with carrier fc

y = y * cos(2*pi*fc*t) ;

To recover the I & Q, we do:

after mixing y with the carrier fc to get baseband y, we do

Q = bitand(y, 0xFFFF0000) ;
Q = bitshiftright(Q, 16) ;
I = bitand(y, 0x0000FFFF) ;

I am assuming dividing is the reverse of the above multiplication

I = I / cos(2*pi*f*t) ;  
Q = Q / sin(2*pi*f*t) ;

My question is the above correct especially dividing by cos and sin? I am not sure how to calculate t in all the above.

Please help.

Answer 1

Your approach is putting the I and Q samples on a "sub-carrier" at frequency $f$. If that is really what you wanted to do, the typical demodulation approach is to multiply by the sub-carrier again and low pass filter to get I or Q. Dividing by cosine or sine creates time samples that go to infinity since you will divide by zero with those functions. The multiplication operation produces a baseband component and a component at twice the carrier frequency; this is made clearer with a quick refresh of the following trigonometric identities:

\begin{align} \cos(\alpha)\cos(\beta) &= \frac{1}{2}\cos(\alpha-\beta)+\frac{1}{2}\cos(\alpha+\beta)\\ \sin(\alpha)\sin(\beta) &= \frac{1}{2}\cos(\alpha-\beta)-\frac{1}{2}\cos(\alpha+\beta)\\ \sin(\alpha)\cos(\beta) &= \frac{1}{2}\sin(\alpha-\beta)+\frac{1}{2}\sin(\alpha+\beta) \end{align}

So in your case:

$$ \cos(2\pi ft)\cos(2\pi ft) = \frac{1}{2}\cos(0)+\frac{1}{2}\cos(2\pi ft) = \frac{1}{2} +\frac{1}{2}\cos(4\pi ft) $$

The $\frac{1}{2}$ represents the amplitude of I and the $\frac{1}{2}\cos(4\pi ft)$ is the doubled frequency component that is removed with a low pass filter.

However I am not sure that you intended or need to have such a "sub-carrier"? If you were going to simulate QPSK modulation for example, you can modulate directly to your final carrier as in:

\begin{align} I_t &= I \cos(2\pi f_c t), \quad\text{where}\quad I = {1,-1}\\ Q_t &= Q \sin(2\pi f_c t), \quad\text{where}\quad Q = {1,-1} \end{align}

And further the I and Q samples are sent simultaneously by adding the modulated outputs $I_t$ and $Q_t$ above (hence the increased data rate; we send two bits in each "symbol).

To demodulate, assuming we have done the carrier recovery work to have a good estimate of $f_c$ with all phase offsets removed, we can multiply the received sequence by $2\cos(2\pi f_c t)$ and low pass filter to recover I, and multiply the received sequence by $2\sin(2\pi f_c t)$ and low pass filter to recover Q:

Noting the demodulation steps below where LPF() denotes removing the $2f_c$ frequency component with a low pass filter:

\begin{align} I &= LPF\big\{2 I_t\cos(2\pi f_c t)\big\}\\ Q &= LPF\big\{2 Q_t\sin(2\pi f_c t)\big\} \end{align}

The Q component goes to zero on the I output since the result is in the $\sin(\alpha)\cos(\beta)$ form resulting in $\sin(0) = 0$:

$$ 0 = LPF\big\{2 Q_t\cos(2\pi f_c t)\big\} $$

And similarly the I component goes to zero on the Q output:

$$ 0 = LPF\big\{2 I_t\sin(2\pi f_c t)\big\} $$

Therefore we can "multiplex" I and Q together given this ability to separate the two that operate simultaneously and independently in time.

Note I highly prefer to think and model using exponential frequency terms instead of sines and cosines - it is so much simpler! In fact this best explains implementations with I and Q paths as well, since to implement a complex signal you will need two real sequences $(e^{j\omega t} = \cos(\omega t) + j\sin(\omega t))$. Please see Frequency shifting of a quadrature mixed signal for more details on that.