What function do high-pass filters in a direct conversion receiver perform?

Question

Use case: I'm considering hacking into my SoftRock RXTX to bypass the filters, then add external filters. I'm wondering if I can use just the low-pass filters I've already built, or if there's a compelling reason I'd need to also build high-pass filters.

I know a low pass filter is necessary to attenuate harmonics in multiples of the LO. But what about high-pass filtering? Is it strictly necessary or no, and why? In other words, what reason(s) are there for removing frequencies below the band of interest prior to the mixer?

Answer 1

So, I don't know the Softrock personally, but from your description, it's a switching mixer architecture.

Receiver side

What that means is that you mix not by multiplying the input signal with a single harmonic oscillation (a tone), but by switching it on and of.

Pre-Mixer HPF

Mathematically, this description is omitting one interesting detail:

The operation done with the input signal $r(t)$ of the downconcerter is not actually a multiplication with a perfect square wave, i.e. a wave that has the shape

$$l'(t) =A \cdot \text{sgn}(\sin(2\pi f t))$$

but is actually just an "on- and off" switching; so in the half-periods where $l'$ would be negative, it's actually zero. This leads to an actual LO function

$$\begin{align} l(t) &= \max(l'(t), 0)\\ &=A \max( \text{sgn}(\sin(2\pi f t),0)\\ &=\frac12l'(t) +\frac A2 \end{align}$$

Whereas $l'$ has a time-average of zero, we can clearly see the DC offset of $\frac A2$ of $l(t)$!

Now, when multiplying

$$r(t) \cdot l(t) = \frac12 l'(t)r(t) + \frac A2 r(t)$$

we get the sub-$f$ content of $r(t)$ directly in baseband (to worsen it, amplified by $\frac A2$)!

So, with a switching down-mixer, you must have an input high-pass filter to avoid getting low-frequency input signal directly in your desired output band.

Post-Mixer HPF

On the output of the same mixer, you'd also want a DC-cancelling high pass filter – just because there's practically zero information in an arbitrary small bandwidth, but having a DC offset on your ADC will simply reduce your usable dynamic range. Anyway, that hpf is usually just the AC coupling you'd do anyways to get your signal onto the center of the $[0;V_\text{Full Scale}]$ range of the ADC.

Transmitter side

Pre-Mixer

You don't want DC offset from your DAC to reach the switching mixer – simply because that is typically simply a transistor, and imprinting a DC current would shift its operating point and possibly be detrimental to linearity.

Post-Mixer

The Mixer will introduce a DC current. You typically don't want that to reach your antenna, your HPA (same effect as for the switching transistor) or your baluns (which might be a DC short, anyway).

Answer 2

They prevent (or at least reduce) front end overload due to strong and/or messy nearby signal sources at lower frequencies. Think of the 80 meter station at Field Day blanking the 40 and 20m receivers every time it's keyed up.

Chances are good that you'll be ok without them if your environment is pretty RF quiet, but it may not take all that much to cause trouble. My RTL-SDR, modified for direct sampling (which has chicken wire for filtering I think), was wiped out on big chunks of 20m when I keyed up my hf rig on 40m, even at only 10 watts. The transmitter was a squeaky clean Elecraft, the transmit antenna was a resonant, well choked dipole more than 100 feet away, and the SDR antenna seemed to have little influence, as it occurred even with no antenna.

A good set of coaxial stubs on the transmitter, and a good band pass filter on the receiver helped a lot.

I definitely wouldnt try to use it for Field Day or SO2R without the high pass filtering, at least.