How do you integrate a square wave?

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Discussion Overview

The discussion revolves around the integration of a square wave signal in the context of an op amp integrator circuit. Participants explore methods for representing the output signal mathematically, considering both theoretical and practical aspects of integration.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant seeks assistance in integrating a square wave signal for a homework problem involving an op amp integrator circuit, mentioning the use of the equation -1/RC * ∫Vin(t).
  • Another participant references a Wikipedia quote about definite integrals representing signed areas, suggesting it may be relevant to the integration of the square wave.
  • A participant suggests that integrating a discontinuous function like a square wave is easier when done in sections, emphasizing the need to account for initial conditions at each interval.
  • Further clarification is provided that each half cycle of the square wave should be integrated separately, leading to a triangle wave output, with the integration expressed as Vout(t) = (-1/RC)∫Vin(t) dt.
  • A participant shares their calculation for the output based on a 2V peak square wave input, arriving at a formula for the triangle wave output over a specified interval, while expressing uncertainty about the correctness of their solution.

Areas of Agreement / Disagreement

Participants generally agree on the approach of integrating the square wave in sections, but there is no consensus on the final output representation or the correctness of the calculations presented.

Contextual Notes

Participants mention the need to consider initial conditions and the specific intervals for integration, but do not resolve the implications of these factors on the final output.

Evilinside
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I'm doing homework where i have to find a function representation of the output signal from a simple op amp integrator circuit with a 4.7k resistor and a .01uf capacitor. I know I'm supposed to use the equation -1/rc * int vin(t) but the input is a square wave signal. I never learned how to integrate that. Can i anyone help me out? Do i have to use something called a Fourier series or Fourier transform?
 
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This quote from wikipedia also applies to a square wave:

A definite integral of a function can be represented as the signed area of the region bounded by its graph.
http://upload.wikimedia.org/wikipedia/commons/9/9f/Integral_example.svg
Does that help?
 
This is a discontinuous function, so it is easier to integrate in sections. You need to do integrals over specifc intervals and you need to account for the initial conditions at each interval.
 
As pointed out above, you have to integrate each half cycle of the input square wave in sections. The basic form of the integration is

V_{out}(t)=\frac{-1}{RC}\int_t V_{in}(t) dt

Each segment is a straight line. The line is continuous, but its derivative is not. The output of the square wave is a triangle wave.

Bob S
 
Yes I finally got the answer i think. I set the square wave input to 2V peak and i got an answer of (-4Vp/.000047)*t and since the frequency of the input square wave is 2000khz, the period is 500us. so (-4v/.00047)*t should be the graph of a triangle wave for the an interval of 0-500us and repeat itself every period after that. I think that's the answer anyway. Oh well whether I'm wrong or right, thanks for help.
 

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