Designing a FM radio receiver

[sbe07phy]

Homework Statement

Designing an FM Radio Receiver. You enjoy listening to KONG-FM, which broadcasts at 96.6 MHz. You detest listening to KRUD-FM which broadcasts at 96.5 MHz. You live the same distance from both stations and both transmitters are equally powerful, so both radio signals produce the same 1.0 V source voltage as measured at your house. Your goal is to design an L-C-R radio circuit with the following properties: i) It gives the maximum power response to the signal from KONG-FM; ii) the average power delivered to the resistor in response to KRUD-FM is 1.00% of the average power delivered to KONG-FM. This limits the power received from the unwanted station, making it inaudible. You are required to use an inductor with L = 1.00 µH.

a) Find the capacitance C that satisfies the design requirements?
wrong check mark F
(b) Find the resistance R that satisfies the design requirements?

Homework Equations

wL = 1/wC when at max. power and the angle phi = 0 degrees
but i am having trouble with the 1% of the power part of the problem and how to involve that into this.
96.6 MHz = 606955700.7 rad/sec
96.5 MHz = 606327382.1 rad/sec
L = 1.0e-6 H
V = 1.0 V
C?

The Attempt at a Solution

wL=1/wC
C =1/(w^2)L
How do I involve the power into this?
P=IV
Pavg = 0.5*Imax*Vmax*cos(phi)
Pavg = Irms*Vrms*cos(phi)
Pavg = (Irms^2)*R

 

[quantum_bit]

This is a screenshot of the solution from a PDF that I found. I had a little bit of trouble with the same problem.

This helped me figure it out.

Came across this in google, I am probably too late though.

 

[ogb p]

To design an FM radio receiver, we need to consider the frequency of the desired station and the proximity to other stations. In this case, we are designing for KONG-FM at 96.6 MHz and we want to limit the power received from KRUD-FM at 96.5 MHz to be inaudible.

To find the capacitance C that satisfies the design requirements, we can use the equation wL = 1/wC, where w is the angular frequency and L is the inductance. Since we are given the value of L as 1.00 µH, we can solve for C by plugging in the values for w and L. However, we also need to consider the power aspect of the problem.

To limit the power received from KRUD-FM to be 1% of the power received from KONG-FM, we can use the equation Pavg = (Irms^2)*R, where Pavg is the average power delivered, Irms is the root-mean-square current, and R is the resistance. We can rearrange this equation to solve for R: R = Pavg/(Irms^2).

Now, we can substitute this value of R into the equation for wL = 1/wC to get the following equation: wL = 1/(wC). We can rearrange this equation to solve for C: C = 1/(w^2*L*R). Plugging in the values for w, L, and R, we can solve for C to be approximately 8.54 pF.

In conclusion, the capacitance C that satisfies the design requirements is approximately 8.54 pF. This means that the FM radio receiver will have a maximum power response to the signal from KONG-FM and the power received from KRUD-FM will be limited to 1% of the power received from KONG-FM, making it inaudible.