Low pass RC filter circuit

Discussion in 'Homework Help' started by Hello, Mar 20, 2009.

  1. Hello

    Thread Starter Active Member

    Dec 18, 2008
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    Can you please have a look at Q3 part (iv) in the attached file, as i am having difficulty answering the question.

    Any help is appreciated.
     
    Last edited: Mar 20, 2009
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    If you've done part (iii) of the same question you should be able to determine the value of Vout(ω)/Vin(ω) - the "gain" or transfer function.

    As to the rest - this is a simple low pass filter [LPF].

    Are you able to draw the LPF filter curve from the transfer function ?

    When does the response fall to half its maximum - i.e. at which frequency (ω) does the 1μF capacitor's reactance equal the 1kΩ resistor value?
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Sorry this is a high pass filter [HPF].

    I'm half asleep!

    My apologies.

    Same ideas apply except the output response will be that of a HPF and the out will reach half its maximum under the same conditions.
     
  4. Hello

    Thread Starter Active Member

    Dec 18, 2008
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    The ratio for the gain, β=(10^3+10^-6)/10^3=1 not sure if it is correct!

    I have drawn a sketch of the frequency response curve for a real HPF but not sure how i am to calculate the frequency at which the gain is half the maximum value (if that is what the last part of the question is asking).
     
    Last edited: Mar 20, 2009
  5. t_n_k

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    Mar 6, 2009
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    At high frequencies >> the cut-off frequency the gain will indeed be 1.

    However, the question is really asking for an equation which shows how the gain varies as the frequency ω changes.

    If you consider when the frequency ω is zero (0), the gain will be zero since nothing can pass from input to output once "steady state" conditions are reached [another story for later in your course].

    So the gain can vary from 0 at DC (zero ω) to 1 at ω >> ωc, where ωc is the cut-off frequency. And ωc is that same frequency at which the gain has fallen (or risen) to half its maximum value.

    SO your graph would look like one in which the gain rises from zero at frequencies << ωc to 1 at frequencies >> ωc and beyond.

    If I were to give you the gain function for the earlier circuit in part (iii) it would have the form :-

    G(ω)=Vout(ω)/Vin(ω)=(1/jωC)/(R+(1/jω))

    {hope you have done complex numbers!}

    Can you now try to work out an equation with a similar form but for the circuit in part (iv)? Think about re-arranging things to match the circuit of part (iv) then plot the function you derive - i.e. G(ω) vs ω

    For the half gain case (ωc) you have to find out at what value of ω, 1/ωC=1000.

    See how you go with that.

    :)
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Hope the ωc and ωC terms aren't confusing .....

    I mean

    ωc = cutoff frequency ( perhaps I should have used ω0)

    and

    ωC = ω x C

    Cheers!
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    In case you're unsure about the complex number notations in the function for the solution to Q3 part (iii) LPF, the magnitude of the gain function would be :-

    |G(ω)|= (1/ωC)/(√(R^2 + (1/ωC)^2))

    which after simplification would be

    |G(ω)|=1/(√((ωRC)^2+1))

    So your solution for the gain (magnitude) in Q3 part (iv) would reduce to an equation without any complex operators, but with a different form to that above - the high pass filter case.
     
  8. thatoneguy

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    Feb 19, 2009
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    Nice work, tnk!

    Hello!: "Gain is half the maximum value", or half power, is known as the "3dB cutoff", and is the standard way to describe filter ranges.
     
  9. Hello

    Thread Starter Active Member

    Dec 18, 2008
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    For HPF,

    G(ω) = R/(Xc + R)= R/[(1/jωC) + R]= 1/[(R/jωC) + 1]

    When:

    ω=0: G(ω)=0
    ω=∞: G(ω)=1 (Hence, calculated ratio!)

    I am unsure whether i should draw the ideal frequency response or the real frequency response and i am having difficulty in understanding your description of the requency response!, because in the case of the real frequency response i thought as the frequency increased the gain increased to above 0, and in the case of the ideal response the gain was constant at 1 until it passed the cut-off requency, which than reached zero.
    Hope you can understand what i'm trying to say.

    For the last part of the question:

    Calculating cut-off frequency, ωo = 1/RC = 1/(1000x10^-6) = 1000 rad/s - I think the frequency it is asking for is the cut-off frequency since it is at ''3dB cut-off''.
     
    Last edited: Mar 21, 2009
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