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Low pass filter: cutoff and peak output

  1. Mar 9, 2015 #1
    1. The problem statement, all variables and given/known data
    (a) If R1 = 4.7kΩ, what value for C1 will give a cutoff frequency of 50kHz?
    (b) If the input voltage is described by the equation Vin = 3cos(2π(4000)t), what will the peak output voltage be?

    669es3.png

    2. Relevant equations
    (a) fc = 1/(2πRC)
    (b) Vout/Vin = 1/√(1 + (2πfRC)2)
    3. The attempt at a solution
    (a) I rearranged the equation to C = 1/(2πfcR). I got 71.2 pF but am uncertain.
    (b) Not sure if this is the relevant equation. If it is, than I am uncertain how to get the frequency of the input voltage.
     
  2. jcsd
  3. Mar 9, 2015 #2
    The equation is fine, but you got the wrong number out of it. Can't tell you what went wrong if you don't show your calculation.

    Recall the general expression ##A\cos(\omega t + \phi)##, where ##A##, ##\omega##, ##\phi## is the amplitude, angular frequency and phase, respectively, of the sinusoid. Match that to what you're given.

    After you've done that, think about what the purpose of a low-pass filter is and how it affects signals below and above its cut-off frequency.
     
  4. Mar 9, 2015 #3
    (a) I rechecked my work and got 677.3 pF
    (b) So if without the low pass filter the peak voltage would be 3V. Would the frequency be 2π4000 = 25132.7? What about phase? If I'm assuming the values from part (a), then wouldn't the output voltage match the input voltage as 25.1 kHz < 50kHz and this is a low pass filter? If this is wrong and the frequency is really higher than the cutoff, I know that output lessens as frequency increases, but I'm not sure of the equation that would give me the voltage output.
     
  5. Mar 9, 2015 #4
    Great. :smile:

    You have ω = 2π4000 rad/s, which is the angular frequency. What is the relationship between frequency and angular frequency?

    Approximately, if the frequency of the input waveform was lower. There'll be a difference in phase.
     
    Last edited: Mar 9, 2015
  6. Mar 9, 2015 #5
    Ah right, so the frequency is simply 4 kHz, which means Vout would be approximately equal to Vin.
     
  7. Mar 9, 2015 #6

    LvW

    User Avatar

    Yes - approximately. However, do you think this answers the queston? Why not do an exact calculation?
     
  8. Mar 9, 2015 #7
    Well, I don't know how to do the calculation. This graph makes is seem as though they'd be exactly equivalent:
    lowpassfreqdomain.png
     
  9. Mar 9, 2015 #8

    LvW

    User Avatar

    No - it`s impossible. Because at 4 kHz the capacitor has an impedance other than infinite, you have a (small) voltage division between R and C.
    In addition, there is a (small) phase shift, Thus, you have to use the complex transfer function for separately calculating magnitude and phase.
    Don`t you remember - in your first post you have given already the correct magnitude expression. Insert the frequency - and you have the result.

    (Remark: The shown graph is not very realistic for a first-order lowpass)
     
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