# Simple Miller Integrator Question

Discussion in 'Homework Help' started by jegues, Apr 20, 2011.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
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I think I have the first few portions of this question correct, but I am unsure if I have the correct answer for the portion with the sin wave.

Is what I'm doing correct?

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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It's kind of obvious. The point is that you have already calculated the values of R & C that give unity gain at 1krad/sec. So the output magnitude at 1000 rad/sec will be 2V. There will be a phase difference.

Like all things electrical there will always be strange behavior that crops up from time to time. Depending at what point (time-wise) in the AC cycle you apply the input to the integrator you get a different result with respect to the transient behavior. Do you know what the difference might be?

Mar 6, 2009
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4. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
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Here's the solution they are giving in the manual.

It looks like my answer would be equivalent to theirs if I hadn't had that -2 trailing my cosine term. Is that -2 accounted for somewhere or did I make a mistake?

Actually know that I think about it more, if I apply the signal at a time t1= pi and go to time t2 = t then the -2 goes away, right? Is this correct?

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Last edited: Apr 20, 2011
5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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The constant term in the solution is a DC offset in the output. As you observed, this offset can be removed by changing the phase of the input. In fact if you make the input a simple cosine, rather than a sine function, the offset vanishes.

6. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
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So having there negative 2 there is okay, right?

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