# RC in changing voltage input

Discussion in 'Electronics Homework Help' started by perchick, Nov 2, 2012.

1. ### perchick

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0
Nov 2, 2012
Hey

this is very stupid, but i dont really know how to analyze this circuit. here's the input voltage signal: http://www.israup.net/images/58e1f2c937d9ea34bb56b3b1eb62d7f3.png
in this circuit τ=10^-6sec.
well, the Vout for changing voltage in rc circuit is Vin*1/[√(1+(τω)^2] but im not really sure i can do that in this case. im talking about the first part of the graph where the voltage input is 5t for 10^-6sec.im looking for the output over time but im having trouble with frequency in the Vout expression. i mean, can I use the normal methods for just one period? if not, how do i get the output voltage? should i just calculate the DC input and take it from there?
I know it's a pretty basic but it seems that there's no examples or explanation on line for this (that i found in 2 hours). maybe it's so basic no one thought it worth explaining
thanks

2. ### Harald KappModeratorModerator

12,454
2,984
Nov 17, 2011
This equation is of no use for this problem. The equation is suited for sinusoidal, repetitive input signals aka a sine wave. In other words: it is valid in the frequency domain.
That's not what you see in the image you linked.

Instead you have to find the equation in the time domain.
Draw the circuit and label all components, voltages and currents.
Then set up the equations using Kirchhoff's laws and the relationships between current and voltage for a resistor and a capacitor:
- V(R) = R*I(R)
- I(C) = C* dV(C)/dt
Then you solve the resulting differential equation.

Once you have drawn the circuit and set up the equations, you're welcome to come back to this forum for verification - to ensure a correct starting point before you start solving the differential equation.

To give you an idea what the result should look like: Regards,
Harald

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3. ### perchick

8
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Nov 2, 2012
sorry it took me so long to replay. thx for the help, don't know why i didn't try that
I'm pretty sure its the right answer.
first we have to get the differential equation of Vc. Vin=Vc+Vr=>5t=Vc+CR*dVc/dt
the answer for the equation is 5t-5RC(1-exp(-t/RC)) with the initial value of 0 (meaning Vc(0+)=0. this is for the first part. in time 1microSec we get Vc=1.84V and that is the initial value for the second part. for DC voltage we can get the equation from
Vc(t)=V(inf)+[V(0)-V(inf)]*exp(-t/RC) were V(inf) is the value in steady stat and Vc(0) is the initial value. the steady state value is just the DC and the initial value is 1.84 as calculated. in the third part we do the as in the second part.
can you please tell me how did you get the plot for this? im using multisim 2011 but its so hard to configure your own voltage signal and I didnt really wanted to waste so much time on that. thank you for the help

4. ### Harald KappModeratorModerator

12,454
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Nov 17, 2011
Good thing you evaluated the equations yourself. Do they match the curve form of the simulation?
I used LTSpice and a piecewise linear voltage source (PWL).

Harald

5. ### perchick

8
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Nov 2, 2012
thanks. my answer did match the curve. that's a nice program. how did you make your input voltage go up to 10 volt instantly in 1microSec (same in -5 volt for 2)? I cant seem to get how to config the source to give me a slope till 1 micro and then give me DC

6. ### Harald KappModeratorModerator

12,454
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Nov 17, 2011
Easy, like this: Harald

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7. ### perchick

8
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Nov 2, 2012
yeah, that's what i did. either way its a pretty cool simulator. and very easy to use. looks a bit old but still, much easier to handle than multisim with stuff like that
thank you for all your help  