# Can anyone help with frequency response?

Discussion in 'Homework Help' started by thisonedude, Aug 6, 2014.

1. ### thisonedude Thread Starter Member

Apr 20, 2014
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Can anyone help me approach this quesiton? I have to derive an expression for the frequency response. How do i start this question?

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2. ### shteii01 AAC Fanatic!

Feb 19, 2010
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I would replace L and C with their impedances.

3. ### thisonedude Thread Starter Member

Apr 20, 2014
52
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Am i heading in the right direction? How do i get the Qaulity factor and the gain?

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4. ### shteii01 AAC Fanatic!

Feb 19, 2010
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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?

5. ### thisonedude Thread Starter Member

Apr 20, 2014
52
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Oh am i missing (-j/ωC) on the bottom summation?

Feb 19, 2010
3,974
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Yes, you do.

7. ### thisonedude Thread Starter Member

Apr 20, 2014
52
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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?

8. ### thisonedude Thread Starter Member

Apr 20, 2014
52
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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.

9. ### MrAl Distinguished Member

Jun 17, 2014
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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.