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
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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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    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?
     
  5. thisonedude

    Thread Starter Member

    Apr 20, 2014
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    Oh am i missing (-j/ωC) on the bottom summation?
     
  6. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Yes, you do.
     
  7. thisonedude

    Thread Starter Member

    Apr 20, 2014
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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
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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

    Well-Known 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.
     
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