Difference results when solving equation with Calculus and LAPLACE

[david753]

In RC circuit, if I use Calculus to solve the differential equation,
the output is Vo=Vi*(1-exp(-t/(R*C))),

But, If I use LAPLACE to solve the differential equation, the answer seems to be different. How to explain it? Or what wrong with me?

please refer to
**broken link removed**

 

[safwatonline]

Calculus or LAPLACE?

both answers are correct but what you are doing wrong is that you are comparing the Vo(t) with h(t) which are certainly not the same , to get the right answer try convoluting the h(t) with Vi(t)=u(t) to get the Vo(t) or try to calculate the inverse laplace of Vo(s)=Vi(s)*H(s)=(1/s)*H(s) u will get the correct answer
regards,
Safwat

 

[david753]

Re: Calculus or LAPLACE?

Dear friends,
Thanks for your reply.
But, your reply is not completed.
Could u please prove both ways that can get the same result in details?
Thanks again.

 

[safwatonline]

Calculus or LAPLACE?

Vo(s)=Vi(s)*H(s)
Vi(s)=P/s
H(s)=wo/(s+wo)
Vo(s)=(P*wo)(s/(s+wo))
given that inverse_laplace(s*F(s0))=integration(f(t)) from 0 to t
then vo(t)=inverse_laplace(Vo(s))=integration(f(t)) from 0 to t
where f(t)=inverse_laplace((P*w0)(1/(s+wo)))=
therefore f(t)=(P*wo)exp(-wo*t)
therefore vo(t)=integration(f(t) from (0,t)
therefore vo(t)=integration((P*wo)*exp(-wo*t)) from t=0 to t=t
therefore vo(t)=P*(1-exp(-wo*t))


note that the output you write in the question is not writen correctly as you wrote vo(t)=vi(t)*(1-exp(-wo*t))
while the right answer is vo(t)=P*(1-exp(-wo*t)) as written in the link you posted
regards,
safwat

 

[david753]

Re: Calculus or LAPLACE?

1.
Vo(s)=Vi(s)*H(s)
Vi(s)=P/s
H(s)=wo/(s+wo)
Vo(s)=(P*wo)(s/(s+wo))
It is wrong, I supposed.
The correct equation is
Vo(s)=(P*wo)(s*(s+wo))
Is it right?

2.
the right answer is vo(t)=P*(1-exp(-wo*t))
My question is
What does "P" mean?
Impluse or anything else?

 

[safwatonline]

Re: Calculus or LAPLACE?

Vo(s)=Vi(s)*H(s)
Vi(s)=P/s
H(s)=wo/(s+wo)
Vo(s)=(P*wo)(1/(s*(s+wo)))
given that inverse_laplace((1/s)*F(s))=integration(f(t)) from 0 to t
then vo(t)=inverse_laplace(Vo(s))=integration(f(t)) from 0 to t
where f(t)=inverse_laplace((P*w0)(1/(s+wo)))=
therefore f(t)=(P*wo)exp(-wo*t)
therefore vo(t)=integration(f(t) from (0,t)
therefore vo(t)=integration((P*wo)*exp(-wo*t)) from t=0 to t=t
therefore vo(t)=P*(1-exp(-wo*t))


note that the output you write in the question is not writen correctly as you wrote vo(t)=vi(t)*(1-exp(-wo*t))
while the right answer is vo(t)=P*(1-exp(-wo*t)) as written in the link you posted
regards,
safwat

i mistyped the "1/s" to be "s" so i corrected it above, but the flow + final answer is still correct.

P is the unit step amplitude u can put it as 1 (this was the Vi(t) used in the link u posted
regards,
Safwat