Solving a circuit with Laplace

[eehelp150]

Homework Statement

Assume zero initial conditions

Step 1. Write the nodal equations to find i(t) in the time domain.
Step 2.Solve the differential equation obtained in step 1 using laplace to obtain i(t).

Homework Equations

The Attempt at a Solution

Convert to sdomain
##7e^{-6t}## becomes ##\frac{7}{s+6}##
4H becomes 4s

##\frac{7}{s+6}+\frac{V_A}{5}+\frac{1}{4s}\int_{0}^{t}(V_A-V_B)=0##

##\frac{1}{4s}\int_{0}^{t}(V_B-V_A)+\frac{V_B}{3}=0##

Am I doing this right?

 

[gneill]

There should be no explicit integrals or derivatives required in the equations that you write. Laplace takes care of all that implicitly.

The Laplace "variable" s is more of an operator than a variable. When s multiplies a Laplace domain function (such as sVa(s)) it is equivalent to differentiating the corresponding time-domain function. When a Laplace domain function is divided by s, it is equivalent to integrating in the time domain. So (ignoring initial conditions):

##sF(s) ⇔ \frac{d}{dt}f(t)##

##\frac{1}{s} F(s) ⇔ \int_o^t f(\tau)~d \tau##

A nice thing about the Laplace domain is that reactive components have "impedances" that automatically take care of their own differentiation and integration. Thus Laplace domain "impedances" sL for inductors and 1/(sC) for capacitors carry with them their own operators that automatically handle the calculus. All you have to do is write your circuit equations with these "impedances", manipulating them with standard algebra, and it's as though all the calculus were being done for you behind the scenes.

 

[eehelp150]

There should be no explicit integrals or derivatives required in the equations that you write. Laplace takes care of all that implicitly.

The Laplace "variable" s is more of an operator than a variable. When s multiplies a Laplace domain function (such as sVa(s)) it is equivalent to differentiating the corresponding time-domain function. When a Laplace domain function is divided by s, it is equivalent to integrating in the time domain. So (ignoring initial conditions):

##sF(s) ⇔ \frac{d}{dt}f(t)##

##\frac{1}{s} F(s) ⇔ \int_o^t f(\tau)~d \tau##

A nice thing about the Laplace domain is that reactive components have "impedances" that automatically take care of their own differentiation and integration. Thus Laplace domain "impedances" sL for inductors and 1/(sC) for capacitors carry with them their own operators that automatically handle the calculus. All you have to do is write your circuit equations with these "impedances", manipulating them with standard algebra, and it's as though all the calculus were being done for you behind the scenes.

##\frac{7}{s+6}+\frac{V_A}{5}+\frac{V_A-V_B}{4s}=0##
##\frac{V_B-V_A}{4s}+\frac{V_B}{3}=0##
Does this look better?

 

[gneill]

Yes, that looks better. Just check the sign on the first term of the first equation. As written you have the source current flowing out of node A (judging by how you've written the other terms).

 

[eehelp150]

Yes, that looks better. Just check the sign on the first term of the first equation. As written you have the source current flowing out of node A (judging by how you've written the other terms).

##-\frac{7}{s+6}+\frac{V_A}{5}+\frac{V_A-V_B}{4s}=0##
##\frac{V_B-V_A}{4s}+\frac{V_B}{3}=0##

##\frac{-28s}{s+6}+\frac{4sV_A}{5}+V_A-V_B=0##
##\frac{-28s}{s+6}+\frac{4sV_A}{5}+V_A=V_B##

##\frac{\frac{-28s}{s+6}+\frac{4sV_A}{5}+V_A-V_A}{4s}+\frac{\frac{-28s}{s+6}+\frac{4sV_A}{5}+V_A}{3}=0##

##\frac{-7s}{s+6}+\frac{V_A}{5}+\frac{-28s}{3s+18}+\frac{4sV_A}{15}+\frac{V_A}{3}=0####V_A(\frac{1}{5}+\frac{1}{3}+\frac{4s}{15})=\frac{49s}{3s+18}##
##V_A(\frac{8}{15}+\frac{4s}{15})=\frac{49s}{3s+18}##
##V_A=\frac{49s}{3s+18}*\frac{15}{8+4s}##

Am I on the right track?

 

[gneill]

##-\frac{7}{s+6}+\frac{V_A}{5}+\frac{V_A-V_B}{4s}=0##
##\frac{V_B-V_A}{4s}+\frac{V_B}{3}=0##

##\frac{-28s}{s+6}+\frac{4sV_A}{5}+V_A-V_B=0##
##\frac{-28s}{s+6}+\frac{4sV_A}{5}+V_A=V_B##

##\frac{\frac{-28s}{s+6}+\frac{4sV_A}{5}+V_A-V_A}{4s}+\frac{\frac{-28s}{s+6}+\frac{4sV_A}{5}+V_A}{3}=0##

##\frac{-7s}{s+6}+\frac{V_A}{5}+\frac{-28s}{3s+18}+\frac{4sV_A}{15}+\frac{V_A}{3}=0##

Check the first term on the last line

 

[eehelp150]

Check the first term on the last line

##\frac{-7}{s+6}+\frac{V_A}{5}+\frac{-28s}{3s+18}+\frac{4sV_A}{15}+\frac{V_A}{3}=0##
##V_A(\frac{1}{5}+\frac{4s}{15}+\frac{1}{3})+\frac{-28s-21}{3s+18}=0##
##V_A(\frac{4s+8}{15})=\frac{28s+21}{3s+18}##
##V_A=\frac{28s+21}{3s+18}*\frac{15}{4s+8}##

 

[eehelp150]

Check the first term on the last line

Trying to get Vb instead of Va. Could you check my work? i(t) = VB/3, correct?
##\frac{V_B-V_C}{4s}+\frac{V_B}{3}=0##
##V_B-V_A+\frac{4sV_B}{3}##
##V_A=V_B+\frac{4sV_B}{3}##
sub into eqA
##-7e^{-6t}+\frac{V_B(1+\frac{4s}{3})}{5}+\frac{V_B(1+\frac{4s}{3})}{4s}##
simplify
##-7e^{-6t}+\frac{V_B}{5}+\frac{4sV_B}{15}+\frac{V_B}{4s}+\frac{V_B}{3}=0##
##-7e^{-6t}+\frac{8V_B+V_B4s}{15}+\frac{V_B}{4s}=0##
##\frac{8V_B+V_B4s}{15}+\frac{V_B}{4s}=7e^{-6t}##
mult everything by 60s
##32sV_B+16s^2V_B+15V_B=420se^{-6t}##
##e^{-6t} = \frac{1}{s+6}##
##32sV_B+16s^2V_B+15V_B=\frac{420s}{s+6}##

##V_B=\frac{420s}{(4s+3)(4s+5)(s+6)}##

Partial fraction decomp:

Inverse Laplace:
VB=

i(t)=VB/3 =

is this correct?

 

[gneill]

Trying to get Vb instead of Va. Could you check my work? i(t) = VB/3, correct?

Yes, that's correct.

##\frac{V_B-V_C}{4s}+\frac{V_B}{3}=0##
##V_B-V_A+\frac{4sV_B}{3}##
##V_A=V_B+\frac{4sV_B}{3}##
sub into eqA
##-7e^{-6t}+\frac{V_B(1+\frac{4s}{3})}{5}+\frac{V_B(1+\frac{4s}{3})}{4s}##

Problem: You can't mix time-domain and Laplace domain functions. ##7e^{-6t}## doesn't belong here.

 

[eehelp150]

Yes, that's correct.

Problem: You can't mix time-domain and Laplace domain functions. ##7e^{-6t}## doesn't belong here.

I ended up getting VB=

i(s) = VB(s)/3 =

i(t)=

However, the answer should be:

Any idea what I'm doing wrong?

EDIT: I think I see my mistake. I will rework and post results

 

[gneill]

All I can tell is that your Laplace expression for VB has not turned out correctly. There's a lot of tedious algebra involved so you'll need to be very careful at each step.

I 'll admit that reviewing every line of algebra is not my favorite thing to do, so instead I'm going to give you a goal to work towards. The solution for VB(s) should be able to be expressed in the form:

##VB = A \frac{1}{(s+6)(s+2)}##, where A is a numeric constant.