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

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To convert a square wave to a sine wave using a circuit, the design must focus on selecting appropriate resistor and capacitor values. A low-pass filter approach is suggested, where the cut-off frequency is crucial for attenuating higher harmonics while preserving the fundamental frequency. The goal is to minimize the attenuation of the first harmonic while significantly reducing the third harmonic. It is noted that a simple first-order low-pass filter alone cannot achieve a perfect sine wave, and typically, three cascaded stages are recommended for better results. Understanding the cut-off frequency and its relation to the circuit's time constant is essential for effective design.
Forcefedglas
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Homework Statement


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

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

Homework Equations


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

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!
 
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Hi Forcefedglas. :welcome:

You can look at it as a low-pass filter, and locate its "cut-off" where you wish.
 
NascentOxygen said:
Hi Forcefedglas. :welcome:

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

Haven't learned 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
 
Cut-off is the frequency at which the filter starts to cause significant attenuation.
 
Forcefedglas said:

Homework Statement


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

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