Low Pass RC filter with input transient

[jonnybgood]

A first order RC low pass filter has a first order differential equation.

C=22µF
R=1kΩ





V_out+ V_R= V_in




If the input voltage is a transient V_in=5e ^-100t , does this remain a first order differential equation?

How do you recommend solving it?

 

[albbg]

I can't see the pictures/equations.

 

[jonnybgood]

How about now.

 

[albbg]

OK, now the pictures and equations are OK.

The order of the system doesn't depend from the signal, but only from the components of the system itself. What do you want to find ? q/Vin ?

However the complete solution is given by homogeneous solution (ie. setting Vin = 0) + particular integral (in you case an exponential)

 

[jonnybgood]

I forgot q(0) = 0. So I have enough information to get a complete solution given the initial solution. Does the q(0)=0 specifiy the charge of the cap at time=0sec?

If so how do they make sense to find other variables at t=0? We know that by the Vin function we have a 5V input at t=0.

---------- Post added at 14:19 ---------- Previous post was at 13:54 ----------

Shall I use q and Vin as functions of time and work with normal integrating factor method?

 

[_Eduardo_]

.....
How do you recommend solving it?

The best way is one of those you saw in class, we cannot know which.

In the first stages that exercises are solved using differential equations resources.
Later, after having learned the basis of Laplace transform, they are solved by transform-antitransform method.
And later, is not solved in time domain (L antitransform), only study the properties in 's' domain except in special cases.

 

[jonnybgood]

The best way is one of those you saw in class, we cannot know which.

In the first stages that exercises are solved using differential equations resources.
Later, after having learned the basis of Laplace transform, they are solved by transform-antitransform method.
And later, is not solved in time domain (L antitransform), only study the properties in 's' domain except in special cases.

I have to use pure D.E. - I cannot use laplace transform.

 

[goldsmith]

Dear Jonny
Hi
Ic= c(dv/dt) and you can get the voltage : ic*dt=c*dv ...> dvc=(ic/c*dt) ...> vc = 1/c integral (ic dt) .
And at square wave and DC , we can say : Vc(t) = vs-(vs-v0)e^(-t/RC)
With top considerations you can analyze your filter , simply.
Best Wishes
Goldsmith

---------- Post added at 17:21 ---------- Previous post was at 17:20 ----------

And you can find the transform function in laplace form simply.

 

[_Eduardo_]

C=22µF
R=1kΩ

Vout + VR= Vin
q/C + R dq/dt = Vin (1)
Vin=5e^(-100t)


(1) is the same of: Vout + RC dVout/dt = Vin

There are many methods to solve ODEs, perhaps in this case the simplest is write a general solution of unknowns coefficients.

That is: Vout = A e^(u t) + B e^(v t)

then dVout/dt = u·A·e^(u t) + v·B·e^(v t)

replacing in (1)

A e^(u t) + B e^(v t) + RC (u·A·e^(u t) + v·B·e^(v t)) = 5·e^(-100t)

(1+u·RC)·A·e^(u t) + (1+v·RC)·B·e^(v t) = 5·e^(-100t)

Then: u = -100 --> A = 5/(1-100RC)
And : v = -1/RC --> B = -A = 5/(1-100RC) because at t=0 must be Vout=0

Vout = ( e^(-100t) - e^(-t/RC) )·5/(1-100RC)



Note: Special case when RC = 1/100 .
Then evaluate the limit or use A·t·e^(-100t) as general equation

 

[jonnybgood]

I found the dq/dt in terms of time only with other constants worked out. I showed it to my teacher an he said it looks good. Now I have to find the minimum current.
Now the dq/dt that was found by differentiating q made subject of the formula. i = dq/dt. do I have to differentiate again and equate to zero to get the time for the minimum current?

 

[_Eduardo_]

...
Now the dq/dt that was found by differentiating q made subject of the formula. i = dq/dt. do I have to differentiate again and equate to zero to get the time for the minimum current?

Yes. But...

View attachment 68376

It's OK, but

View attachment 68378

No.

The expression of i is

i(t) = dq/dt = 11·e^(- 100·t)/1200 - e^(- t/rc)/240​

Then

i' = 6250000·e^(- 500·t/11)/33 - 2750000·e^(- 100·t)/3 = 0
==> t1 = 11·ln(11/5)/300 ( 28.91ms )​

and finally

i(t1) = -275^(1/3)/10648 (-0.6107mA )​

 

[jonnybgood]

Cheers. I managed to proof the minimum current at the particular time. Obviously I did not put the exact values here as I don't think it is ethical. Thanks for helping me out.