# Circuit Analysis-Converting a square wave to a sine wave

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1. Mar 16, 2016

### Forcefedglas

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.

$$v_i(t)= (-1)^n), nT_0<t\leq(n+1)T_0, n=...-2,-1,0,1,2..., T_0=\frac{1}{100}secs$$

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. Mar 16, 2016

### Staff: Mentor

Hi Forcefedglas.

You can look at it as a low-pass filter, and locate its "cut-off" where you wish.

3. Mar 16, 2016

### Forcefedglas

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

4. Mar 16, 2016

### Staff: Mentor

Cut-off is the frequency at which the filter starts to cause significant attenuation.

5. Mar 17, 2016

### LvW

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.

6. Mar 18, 2016

### Tom.G

Just as a FYI.
In practice, three cascaded stages are 'generally assumed' to be sufficient.