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How come there is no attenuation across a filter

Question asked by rulonv Employee on Feb 17, 2004
There are two kinds of loss produced by a filter 1) dissipative (resistive) and 2) impedance mismatch.  The dissipative loss of a filter is the same as its insertion loss.  For in-band frequencies the dissipative loss is typically more significant than the mismatch loss.  However, for out-of-band frequencies the mismatch loss it typically several times greater than the insertion loss.  Signals arriving at a filter input will either be reflected or transmitted.  All transmitted signals through the filter will undergo the loss due to dissipation. 

Dissipative loss is a function of the Q of the components that make up the filter.  Ideal components have infinite Q and will have no insertion loss.

When looking at out-of-band channel power or voltage along a cascade of stages it will appear that the stage prior to a filter has more loss than expected and the filter attenuation is much less than expected.  This seems counter intuitive at first but when we look at mismatch loss and transmitted and reflected power it all makes good sense.  The out-of-band impedance looking into a filter is either really high or really low. There is an out-of-band impedance mismatch between the filter and the stage prior to it.  The actual out-of-band voltage appearing at the filter input will be affected by mismatch impedance which automatically accounts for the out-of-band attenuation caused by the filter.  In the lab the only way to accurately measure the out-of-band input voltage to a filter is to break the input node connection and measure the forward voltage, transmitted voltage, and reflected voltage (Vforward = Vtransmitted + Vreflected).

In order to simplify the simulation process all SPECTRASYS measurements and spectrums only use transmitted signals not the forward signals assumed by spreadsheets.  Spreadsheets typically never account for mismatch loss so it is assumed that all forward traveling signals equal the transmitted signals ... which is really not the case!  

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