If L = 10 micro Henries, C=1Nf, then calculate the value of the resistor when the resonant frequency is 1.4 Mhz
Correct me if I'm wrong, but resonance frequency doesn't care about the resistance, does it? Resistance affects the Q of the circuit.
It's been a while for me.
No tricks. The natural (undamped) resonant frequency is indeed Fr=1/(2pi*sqrt(LC)), which in this case is 1.59MHz. The series resistor "pulls" the resonant frequency (defined as the frequency where the impedance is highest) to a value lower than the natural resonant frequency. It is possible, using the equations on the page I referenced previously, to solve for the resistance value that causes damped resonance to occur at 1.4MHz.One suspects it's a trick question. The reactances may not be equal at the stated frequency, so the value of R will make that equality while degrading performance (poor Q).
Well, I agree that the Bode plot (attached) of the solution to his problem looks quite different than that of a high-Q tank circuit, but the peak is at 1.4MHz, and the circuit is the one he posted, with the resistor value set to yield maximum impedance at that frequency. If you don't want to call that the resonant frequency, what name would you give it? The page I referenced calls it the damped resonance frequency, and, a I said, 1 / (2pi(LC)^.5) is the undamped resonant frequency.I just don't agree with calling the 1.4 MHz the "resonant" frequency.
I agree a 1.4 MHz signal applied to a circuit with resonance at 1.59 MHz would pass a damped 1.4 MHz signal, and you can reduce the damping by lowering the Q of the circuit.
How much loss is acceptable before shifting the circuit's resonance to be close to the frequency of interest?
1.4 MHz is not a resonant frequency for the two components listed. In the sence of resonant frequencies, or 1 / (2pi(LC)^.5), 1.4 isn't in the ball park, unless your willing to accept a much lower signal level or a damped signal.
by Jake Hertz
by Jake Hertz
by Jeff Child