Designing Tuner: LRC Circuit w/377Ω Impedance @1MHz

[gfd43tg]

Homework Statement

Using the circuit layout shown in Fig. 9-15 (Standard LRC circuit), design a tuner that uses a variable inductor, a capacitor, and a resistor. The input impedance of the tuner should be 377 Ω at 1 MHz an its bandwidth should be 2%.

Homework Equations

The Attempt at a Solution

For this problem, I am assuming that a bandwidth of 2% means that B = .02. For an RLC circuit in series, B = R/L. Since the impedance is purely real at 1 MHz, and I know that 1 MHz is the resonant frequency, and I set it equal to 1/√LC. I choose R = 377 Ω, and the calculations that follow suit produce an inductor and capacitor value that doesn't sound right, the inductor is way too big and the capacitor is way too small. What is going on here?

 

[gfd43tg]

Here's a table showing the circuit and equations

And here is my attempt
https://www.physicsforums.com/attachments/69202

 

[gneill]

Bandwidth has units of frequency, not a unitless value (like a percentage).

So, what might a 2% bandwidth imply for a frequency of 1MHz?

 

[gfd43tg]

Thanks, how does this look? Still quite a small capacitor, but at least I know what letter to put in front of the F for this order of magnitude ;)


Assuming that this answer is correct, what was the point in specifying that the inductor was variable?

Edit: is Bandwidth the frequency or the angular frequency? Either B= .02*ω0, or B = .02*f?

 

[gneill]

Okay, looking better. But the bandwidth B = R/L yields a value in radians per second, not Hz. You want to work with values of ω.

 

[gfd43tg]

Ok great, thank you! I changed it to omega, but the exact same procedure.