October 05, 2015

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