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I've this RC filter circuit (r is connect to earth on the bottom):

enter image description here

My Vin is a square wave:

enter image description here

With Laplace transform I derived that:

$$Vout(s)=\frac{r}{r+\frac{1}{cs}}\cdot Vin(s)$$

and this square wave has a laplace transform of $$Vin(s)=\frac{\text{A}\tanh\left(\frac{\text{T}s}{4}\right)}{s}$$

So, when we substitute that in,in the equation for Vout(s) and calculate the inverse laplace transform, and plot that function I got this:

enter image description here

Q: Why do I get (from the theory) this strange output voltage?

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  • \$\begingroup\$ If I rember my old classes well - and I probably don't - doesn't 's' in the laplace have something to do with an impulse? That looks like a series of impulses to me, so I don't think that it is totally valid in this context... \$\endgroup\$ – slightlynybbled May 13 '16 at 23:18
  • \$\begingroup\$ How did you calculate the inverse laplace transform? \$\endgroup\$ – Magic Smoke May 13 '16 at 23:29
  • \$\begingroup\$ How did you choose T? You're missing the positive part of the graph - the shapes you can see are exponentials \$\endgroup\$ – Chu May 13 '16 at 23:55
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You should get a crude saw-tooth shaped output. Only the rising edge from -A to A should go through this hi-pass filter. A real-world setup and an oscilloscope would show a cleaner image, including the negative going return spike.

They would be visible on an oscilloscope, which can draw pulses or any wave shape with very fast rise and fall times, as the signal draws directly on the display without any software intervention.

I do not know why it is not on your plot, unless it did not have time for the drawing engine to do them, which is using software to calculate how to draw the image. The other reason for missing parts of the waveform would be a calculation with an irrational or out-of range t value. Where is your t value for Vout?

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