On the Use of I/Q Signals
- All signals are complex, that is they have the form of $$x(t)=A\exp(i\omega t)$$. I/Q presentation of a signal fully captures this by storing the real component in the I, the in-phase signal, and the complex component in Q, the quadrature signal.
- When we force signals to be purely real, e.g. \(\cos(\omega t)\), we taking the real part of \(exp(i\omega t)\). Because we ignore the imaginary component, we lose information. Specifically, \(\cos(\omega t)\) can be the real part of \(\exp(i\omega t)\) or \(\exp(-i\omega t)\). We don't know any more.
- This ambiguity is present in the exponential form of \(\cos\): $$\cos(\omega t) = \frac{\exp(i \omega t) + \exp(-i\omega t)}{2}$$. Note the presence of the two complex signals whose sum is always purely real as their imaginary parts cancel out. The real part of either complex signal will produce \(\cos(\omega t)\).
- This ambiguity is why when you multiply \(\cos(\omega_0 t)\) and \(\cos(\omega_1 t)\) you end up with \(\cos[(\omega_0 + \omega_1)t]\) and \(\cos[(\omega_0 - \omega_1)t]\). In other words, the frequency add as \(\pm \omega_0 \pm \omega_1\), which produces 4 unique combinations that reduces to 2 because \(\cos\) is even.
- The result of multiplying real cosines produces two peaks in the frequency spectrum.
- On the other hand, preserving the complex nature of a signal by presenting it as \(A\exp(i\omega t)\) means that multiplying two signals together produces a unique result: $$A_0\exp(i\omega_0 t) \times A_1\exp(i\omega_1 t) = A_0A_1\exp[i(\omega_0 + \omega_1)t]$$. This is because there is no ambiguity as we have not thrown away any information.
- The result of multiplying complex signals produces one peak in the frequency spectrum.
- This is one of the advantages of working with I/Q data, which preserves the complex nature of signals, and allows us to frequency shift a signal through multiplication without also generating an additional unwanted image of the signal.
Cheers,
Steve