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Circuit Analysis-Converting a square wave to a sine wave

  1. Mar 16, 2016 #1
    1. The problem statement, all variables and given/known data
    For an upcoming lab I've been asked to build a circuit to convert a square wave (vi(t))e into a sine wave (v0(t)) by selecting appropriate resistor/capacitor values for the circuit below (from what I know, it's impossible to produce an accurate sine wave with just this, I assume that I just have to do the best I can). Searching around online has only given me some qualitative explanations so I'm looking for a circuit analysis based explanation on how this is supposed to work.
    24x2geg.png
    $$v_i(t)= (-1)^n), nT_0<t\leq(n+1)T_0, n=...-2,-1,0,1,2..., T_0=\frac{1}{100}secs$$

    2. Relevant equations
    $$a_0=\frac{1}{T}\int_{0}^{T}f(t)dt$$
    $$a_n=\frac{2}{T}\int_{0}^{T}cos(nw_0t)f(t)dt$$
    $$b_n=\frac{2}{T}\int_{0}^{T}sin(nw_0t)f(t)dt$$
    $$V=IZ$$

    3. The attempt at a solution
    I started by calculating the fourier series, which I believe works out to be $$-\sum_{k=1}^{\infty}\frac{8}{(2k-1)\pi}sin(\pi (2k-1)t)$$

    Then I attempted to get an equation for v0(t) in terms of vi(t). Simplifying the resistor and capacitor in parallel then applying voltage division gave:

    $$v_0(t)=\frac{R_2}{R_1+R_2+jR_1R_2\pi nC}v_i(t)$$

    May have made a mistake in there somewhere but either way from this point on I don't have a clue on how to proceed, I thought about making the denominator real but I'm not seeing how that would help. Any tips will be appreciated, thanks!
     
    Last edited: Mar 16, 2016
  2. jcsd
  3. Mar 16, 2016 #2

    NascentOxygen

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    Staff: Mentor

    Hi Forcefedglas. :welcome:

    You can look at it as a low-pass filter, and locate its "cut-off" where you wish.
     
  4. Mar 16, 2016 #3
    Haven't learnt about low pass filters yet but I'll go read up on it now. What do you mean by cut-off?

    EDIT: Ah I see it's to do with resonance which we get to in a few weeks...My course was recently restructured to teach fourier series at the start instead of at the end but I guess the lab questions didn't change
     
  5. Mar 16, 2016 #4

    NascentOxygen

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    Staff: Mentor

    Cut-off is the frequency at which the filter starts to cause significant attenuation.
     
  6. Mar 17, 2016 #5

    LvW

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    You are right. With this simple first-order lowpass it is not possible to convert squarewave into a sinewave.
    Your goal should be to attenuate the 3rd harmonic as much as possible (the 2nd harmonic does not exist in a poor squarewave) without attenuation the first harmonic too much. That means: The "cut-off frequency" must be located somewhere between the first and the third harmonic.
    This cut-off frequency ωo=2πfo is the inverse of the time constant of the circuit.
     
  7. Mar 18, 2016 #6
    Just as a FYI.
    In practice, three cascaded stages are 'generally assumed' to be sufficient.
     
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