I have a quite basic question, after starting to re-read some RF theory. If a voltage source has a resistance of 50 ohms, and the equivalent impedance of a connected bandpass filter and its load are 50 ohms for a perfect match, won't there be significant loss - so much so that the passband will disappear entirely? If the bandwidth of a filter is defined as the portion where the gain is above 1/sqrt2 ~ 0.707 V, and the highest the output can ever get is 0.5 V due to the voltage divider rule, is it impossible to have an impedance matched band pass filter? Why would someone want to have a matched preselection filter between an antenna and an LNA if matching losses might cause the weak signal to be out of amplification range?
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2 Answers
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The \$\frac{1}{\sqrt{2}}\$ point is defined relative to the magnitude, so if your magnitude in the passband is 0.5, then the -3dB point is at \$\frac{0.5}{\sqrt{2}}\approx0.354\$.
answered May 7, 2018 at 6:57
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\$\begingroup\$ That does explain a little, yes. I'm still curious as to the extent to which loss caused by a matching circuit is acceptable prior to an LNA. \$\endgroup\$ Commented May 7, 2018 at 10:23
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1\$\begingroup\$ You're not transmitting a signal, but a power of a signal, which means having a high input impedance will draw close to zero current; that's why the matched impedance is desired. \$\endgroup\$ Commented May 7, 2018 at 11:51
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There are various methods to design "matched filters".
Most filters reject/reflect the undesired frequencies, and there needs to be a method (back at the signal source) to absorb the reflected energy.
Note the need to absorb-energy will set the attenuation-floor achievable.
answered May 7, 2018 at 2:22