Broadband impedance matching question

It is the same - for impedance matching you need the conjugate on *all* frequencies within the band of interest. This is the reason why most broadband matching recipes apply to purely resistove loads.

It's the same, except that you'll have to work out a matching circuit that has the correct impedance over the whole band you want to match, and not just at one single frequency.

Sometimes it's easy. You can match a Butterworth filter by putting a dual of the filter in parallel at its input. It's a little harder for 'tame' filters, such as Bessels or Gaussians. You'd have to grind the math.

I've never successfully done it for more aggressive filters, Chebycheff, etc. A naive attempt got me a few negative values in my matching circuit. I believe there are ways, involving transformers for example, but I'd have to invest some time to work on it.

For some random measured impedance, it could get pretty hairy pretty fast. I've done it once or twice, but I don't recall all the details. The idea is always the same: From the polynomial approximation of the measured impedance, work out the expression for the impedance or admittance of the matching circuit and then develop the circuit for it by continued fraction expansion. You'll get a result in the form of a ladder circuit.

Jeroen Belleman

For RC loads, there's a theorem of Bode's to the effect that the integral of minus the log of the insertion gain of the network has to be less than 2 pi/RC, so you have to trade off match quality to gain bandwidth. Since it goes as the log, however, the tradeoff is pretty favourable. Proving things like that became a bit of an industry back in the '50s, with Fano, Darlington, and others publishing more complicated cases, but the Bode one is the simplest and (for what I do) generally the most useful.

For octave-band things you can usually get within about half a dB of the limit using a single tee network.

Cheers

Phil Hobbs

Or more precisely, resistive in the mid-band, where it necessarily must be resistive -- for gain to be constant, impedance also needs to be constant, and a constant impedance is a resistance.

At the band edges, impedance starts to fall over, and twists into capacitance or inductance, where you can compensate it (extending bandwidth maybe 2x) with conjugates, peaking, filter networks, whatever you want to call it.

Tim

IIRC, Butterworth is the only self-dual filter (that is to say, its dual is also the HPF version, and when combined, gives a resistive match), and the others break this by being too sharp, or too lumpy (e.g., a Bessel filter has a gentle rolloff, so its dual must have an even softer cutoff, that's not a Bessel prototype).

Or that a Cheb. filter has antinodes in the passband, so its dual must have zeroes in the stopband (a Cheb. II filter at least).

Oh hm, I wonder if it's that the poles of a Cheb or Bessel lie on an ellipse (translated or centered, respectively), and that the dual rotates the ellipse? That'd be neat, and illustrate graphically why it's wrong. (Butterworth, all poles lie on a circle so a Re/Im flip doesn't affect it.) Or more likely they're just discombobulated, but I know so little about pole-space relationships.

Tim