I don't agree. The selection of a harmonic is a
signal processing function and not one of power transfer.

"W3JDR" wrote in message
...
Absolutely not!!!!!!!!!!!!!!!!!!
He needs best possible loaded Q for maximum selectivity! Good unloaded Q
will help minimuze circuit losses, but won't help with selectivity.

"Airy R. Bean" wrote in message
...
Then it is unloaded Q that you are interested in.

"Paul Burridge" wrote in message
...
I am seeking to select for the 5th harmonic.

"The selection of a harmonic is a signal processing function and not one of power transfer."


Exactly Airy.
The output of the multiplier stage, which contains many frequencies, needs to be processed in order to select the desired component and reject all others to the best extent possible. This is an indirect way of saying you need a very selective tuned circuit.

The selectivity of the tuned circuit is defined by it's loaded Q; ie, the Q of the reactances in network loaded when by their operating environment. This means that for best selectivity, the reactances of the L and C should be as low as possible, which in turn means that the C should be as large as can be achieved, consistent with realizably small L and component (unloaded) Q. Unfortunately, as the ratio of unloaded component Q to loaded circuit Q gets small, the losses go up. This implies that there's a tradeoff to be made between selectivity and loss. Thus it is in the world of real components - in the world of DSP things can be different.

If you want high selectivity, you have to balance it against tolerable circuit losses, at least with real components.

Joe
W3JDR


"Airy R. Bean" wrote in message ...
I don't agree. The selection of a harmonic is a
signal processing function and not one of power transfer.

"W3JDR" wrote in message
...
Absolutely not!!!!!!!!!!!!!!!!!!
He needs best possible loaded Q for maximum selectivity! Good unloaded Q
will help minimuze circuit losses, but won't help with selectivity.

"Airy R. Bean" wrote in message
...
Then it is unloaded Q that you are interested in.

"Paul Burridge" wrote in message
...
I am seeking to select for the 5th harmonic.

"The selection of a harmonic is a signal processing function and not one of power transfer."


Exactly Airy.
The output of the multiplier stage, which contains many frequencies, needs to be processed in order to select the desired component and reject all others to the best extent possible. This is an indirect way of saying you need a very selective tuned circuit.

The selectivity of the tuned circuit is defined by it's loaded Q; ie, the Q of the reactances in network loaded when by their operating environment. This means that for best selectivity, the reactances of the L and C should be as low as possible, which in turn means that the C should be as large as can be achieved, consistent with realizably small L and component (unloaded) Q. Unfortunately, as the ratio of unloaded component Q to loaded circuit Q gets small, the losses go up. This implies that there's a tradeoff to be made between selectivity and loss. Thus it is in the world of real components - in the world of DSP things can be different.

If you want high selectivity, you have to balance it against tolerable circuit losses, at least with real components.

Joe
W3JDR


"Airy R. Bean" wrote in message ...
I don't agree. The selection of a harmonic is a
signal processing function and not one of power transfer.

"W3JDR" wrote in message
...
Absolutely not!!!!!!!!!!!!!!!!!!
He needs best possible loaded Q for maximum selectivity! Good unloaded Q
will help minimuze circuit losses, but won't help with selectivity.

"Airy R. Bean" wrote in message
...
Then it is unloaded Q that you are interested in.

"Paul Burridge" wrote in message
...
I am seeking to select for the 5th harmonic.