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I'm trying to design a notch filter to suppress an oscillator signal coming through a mixer (namely, 53.125 MHz), at the same time I must get the best possible signal at 60 MHz.

I've been running some simulations - it seems lowering the capacitor value and increasing the inductor value makes the "notch" steeper. It would seem in this example design, I would be losing a lot of signal on 60 MHz?

Simulation

What confuses me, also, is the role of the value of the resistor here, I don't really understand what the lower blue line and the dotted green line represent.

As I lower the resistor value, the lower blue line moves up and the dotted green line curves.

Can someone give a basic idea of what's going on here?

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    \$\begingroup\$ Your signal source is feeding power to something. Can you show a load on that dangling wire? Is the load resistive at 53.125 MHz? at 60 MHz? \$\endgroup\$ – glen_geek Aug 8 '17 at 17:03
  • \$\begingroup\$ What does "best possible signal" mean? Unity gain? High gain? What is the bandwidth of the pass band at 60MHz? Is phase important to this signal? \$\endgroup\$ – Reinderien Aug 9 '17 at 4:07
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I think the left side shows solid Blue = 20dB which I believe means 20dBV relative to 1V from 10V source.

The dotted blue line shows phase ref for 0 deg

THe dotted Green line show Phase going from -90 to +90 deg thru the peak notch.

The X(s)/R value gives the Q value and a simple 2nd order Notch will not give you the optimal pass/reject ratio. First define your goal (specs) for impedance , Zi, Zo at fo then the SNR ratio of 60/53.125, then the tolerance of your components to see what is needed. I suggest at least a 3rd order or 5th order LC filter.

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From what I know about LTSpice, and this is weird how they represent it, the dotted green line you see is the phase. So that spike in the phase line is a phase shift of \$180°\$, which usually occurs at the cutoff frequency.

The resistor value helps with your \$ Q \$ value in your curve which will affect the bandwidth of the curve.

The two blue lines must represent a DC signal since there is no change in phase or magnitude.

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