Bandwidth calculation

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mstechca

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When I looked at equations for bandwidth, they state that resistance (or is it impedance) is a factor in the calculations.

Because the inductor and capacitor are the only two items that directly affect the frequency of the "tank circuit", (an inductor and capacitor appearing connected in parallel) is it possible that I can still obtain a reasonable bandwidth figure without factoring in the resistance?

it seems that now, the only way to calculate the resistance of the tank circuit is to measure it with a special DMM. My DMM's are junk and can't measure lower than 1 ohm. I'm 100% certain that a tank circuit measures less than 1 ohm.

So what is the absolute easiest way to calculate bandwidth if the only two values given were the inductor value and the capacitor value?

and if I need resistance, is it safe to assume 1 ohm?
 
You can do one of two things:

1. Find yourself an oscilloscope.

2. Find the transfer function and find the theoretical bandwidth. I would recommend the 2nd way. You should be able to get a Bode plot in a matter of minutes.
 
Or a 3rd choice. Use a program similar to Pspice and compute the frequency response.

I'm not sure I can adequetly explain how to find a transfer function, I'm not sure how good your circuit theory is. However, if you post the circuit (assuming its not terribly complicated) I'll find the transfer function and compute the frequency response.
 
A capacitor and inductor in parallel have a high impedance and narrow bandwidth at resonance. The impedance can be reduced and the bandwidth increased by adding some resistance in their loop, or loading down the tank with a lower impedance load.
Real FM radios use ceramic filters in their IF amplifier/limiter circuits which have a wide bandwidth at resonance for good FM stereo and very steep sides for good selectivity.
 

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