Heterodyne Mixers

1 Mixing (a.k.a. Heterodyning, Multiplying)

First, you mix (multiply) your input signal with a sine wave of a carefully selected frequency. As we know from the convolution theorem, multiplying in the time domain is the same as convolving in the frequency domain, so if our input signal is f(t) with Fourier transform $\hat f(\omega )$, we have

$\mathcal{F}\left[ f(t) e^{-i\omega _0 t}\right] = \hat f(\omega ) * \delta (\omega -\omega _0). \,\!$

By substituting $\omega ^\prime \equiv \omega -\omega _0$, so that

$\mathcal{F}\left[ f(t) e^{-i\omega _0 t}\right] = \hat f(\omega ^\prime +\omega _0) * \delta (\omega ^\prime ), \,\!$

it should become apparent that we’ve shifted the spectrum of f(ω). We now measure at frequency $\omega ^\prime$ the spectrum a $f(\omega ^\prime +\omega _0)$.

The only thing to pay attention to is that when I said “sine wave”, I really meant $e^{i\omega _0 t}=\cos (\omega _0 t) + i\sin (\omega _0 t)$. That is, you actually need to multiply f(t) by both a cosine (to get the real part) and a sine (to get the imaginary part) to have truly mixed with a tone at frequency ω0. This is commonly known as I/Q mixing (I = cosine, Q = sine). One of the hardest parts of I/Q mixing is ensuring that your cosine and sine waves are in perfect quadrature (i.e. phase shifted by exactly $\frac\pi 2$). Failure to do so will result in false “images” of tones at corresponding positive/negative frequencies.

If you are too cheap to mix with both components, and instead only mix with the cosine (or just the sine, it doesn’t matter), you’ve actually mixed with two tones: one at ω0 and one at − ω0. This we know from our basic trig identities for sine and cosine. As a result, $\omega ^\prime$ will now measure $\hat f$ at $\hat f(\omega ^\prime \pm \omega _0)$. If you are careful with your filters so that you get rid of the $\hat f(\omega ^\prime + \omega _0)$ part, then your cheapness didn’t cost you performance, and may have saved you some money.