Can anyone help me approach this quesiton? I have to derive an expression for the frequency response. How do i start this question?

 

I would replace L and C with their impedances.

 

Am i heading in the right direction? How do i get the Qaulity factor and the gain?

 

Am i heading in the right direction? How do i get the Qaulity factor and the gain?

You made an error.

Think of the circuit as a voltage divider. So you have one resistor: R+Rs+jwL. Second resistor is -j/(wC). You want the voltage across the second resistor. What is the voltage divider equation looks like?

 

You made an error.

Think of the circuit as a voltage divider. So you have one resistor: R+Rs+jwL. Second resistor is -j/(wC). You want the voltage across the second resistor. What is the voltage divider equation looks like?

Oh am i missing (-j/ωC) on the bottom summation?

 

Oh am i missing (-j/ωC) on the bottom summation?

Yes, you do.

 

Alright after adding that, what is K? It says it is the DC gain where K = H(j0). If this is true then my H(jw) = 0 from the derivation i just made. Is this correct?

 

Ohhh disregard that last one please, it's the MAGNITUDE! I see now. which should be 1/sqrt(1+(W/W0)^2) if i'm not wrong.

 

Hi,

It is more clear to write it out in terms of the variable 's' where s=jw. This makes the two complex impedances:
sL and 1/sC in shorthand notation which means s*L and 1/(s*C).

Also, for a voltage divider it is always the impedance of the lower element divided by the sum of all the element impedances.

So we would end up with:
1. The lower element: 1/sC
2. The sum: Rx+sL+1/sC
3. The division: (1/sC)/(Rx+sL+1/sC)
Note Rx=R+Rs where R and Rs are the original two resistors.

Now all that is left to do is to simplify #3 and then rearrange it into the required form to make it easier to extract the required quantities like Q.