Second Order Circuit Analysis - Mesh - Issue with DE Setup

[ghoti]

Hi,

I am sure I have mistaken my DE setup or my initial conditions assumption.

Homework Statement

[URL]http://ivila.net/E8.png[/URL]
R1 will be the middle Resistor
R2 the top one.

Homework Equations

i1' + R1/L i1 = R1/L i2
i2(R1+R2)*1/(R1) + 1/(R1*C)*integral(i2)=i1

The Attempt at a Solution

My initial conditions are i1(0) = 1A
Vl = L di1/dt
at (0+) VL = 0 therefore L di1/dt = 0 therefore di1/dt = 0
i1'(0) = 0
Substitute for i1 I end up with sqrt(2)exp(-2)cos(t+45')

 

[vela]

Your first initial condition, i₁(0) = 1A, is fine, but your second one, v_L(0⁺)=0, is wrong. The current through an inductor must be continuous, but the voltage across it doesn't have to be.

The opposite is true for capacitors. The voltage must be continuous, but the current doesn't have to be. So the second initial condition would be v_C(0) = 1 V.

 

[ghoti]

Thanks vela, I have changed my working however I am still tragically stuck.

for the mesh, is = i1

initial conditions
i1(0) = 1A
Vc(0) = 1V

Mesh Equations
i_1 = (\frac{R_1+R_2}{R_1})i_2 + (\frac{1}{R_1C}) \int i_2
i_2 = (\frac{L}{R_1})\frac{di_1}{dt} + i_1

Sub (2) into (1)
i_1 = (\frac{R_1+R_2}{R_1})((\frac{L}{R_1})\frac{di_1}{dt} + i_1) + (\frac{1}{R_1C}) \int ((\frac{L}{R_1})\frac{di_1}{dt} + i_1)

Expand
i_1 = ((\frac{R_1+R_2}{R_1})(\frac{L}{R_1})\frac{di_1}{dt} + (\frac{R_1+R_2}{R_1})i_1) + (\frac{1}{R_1C})(\frac{L}{R_1}) (\int \frac{di_1}{dt} + \int i_1)

Simplify
i_1 = (\frac{L(R_1+R_2)}{R_1^2}\frac{di_1}{dt} + (\frac{R_1+R_2}{R_1})i_1) + (\frac{L}{R_1^2C})i_1 + (\frac{L}{R_1^2C})\int i_1

i_1 = \frac{L(R_1+R_2)}{R_1^2}\frac{di_1}{dt} + (\frac{(R_1^2C + R_1CR_2 + L)}{R_1^2C})i_1 + (\frac{L}{R_1^2C})\int i_1

\frac{L(R_1+R_2)}{R_1^2}\frac{d^2i_1}{dt} + (\frac{(R_1^2C + R_1CR_2 + L)}{R_1^2C} -1)\frac{di_1}{dt} + (\frac{L}{R_1^2C})i_1 = 0

Check if DE is dimensionally correct.
\frac{d^2i_1}{dt} + (\frac{R_1CR_2 + L}{CL(R_1+R_2)})\frac{di_1}{dt} + (\frac{1}{C(R_1+R_2)})i_1 = 0

I believe this to be incorrect due to the i1' term with an R/L coefficient, this is not dimensionally correct. CR^2/CLR gives a R/L term (no good) and L/CLR gives a 1/RC term, also no good. My understanding is that this co-ef should be some mixture of \tau

We will continue just to check

Put in a few numbers,
L = 0.5
C = 0.5
R1 = 1
R2 = 1

\frac{d^2i_1}{dt} - 0.5\frac{di_1}{dt} + i_1 = 0

i_1(0) = 1A
V_c(0) = 1V therefore V_r(0) = 1V
V_r(0) = V_l(0) = 1V
V_l = L\frac{di_1}{dt}

\frac{di_1}{dt} = 2

Putting the DE into MatLab (to save some time)
pretty(simplify(dsolve(' D2y-0.5*Dy+y=0' , ' Dy(0)=2' , ' y(0)=1 ' )))
results in,

1/15*exp(1/4*t)*(7*sin(1/4*15^(1/2)*t)*15^(1/2)+15*cos(1/4*15^(1/2)*t))

As predicted, miles off, I am expecting a pair of sine and cosine terms with a common coefficient of 1

 

[vela]

After you substitute (2) into (1), differentiate the result to get rid of the integral, and let \tau_C = R_1C and \tau_L = L/R_1, both of which have units of time. Then you'll have

i_1' = \frac{R_1+R_2}{R_1}(\tau_L i_1'' + i_1') + \frac{1}{\tau_C} (\tau_L i_1' + i_1)

Collect terms to get

\frac{R_1+R_2}{R_1}\tau_L i_1'' + \left(\frac{R_1+R_2}{R_1}+\frac{\tau_L}{\tau_C}-1\right)i_1' + \frac{1}{\tau_C} i_1 = 0

It's pretty easy to see that result is dimensionally correct.

 

[ghoti]

Working with \tau is a much simpler way to work the Algebra. I never thought of carrying around the time-constants are pairs but it makes complete sense. thankyou! should save me a lot of paper if nothing else.

I re-worked my solution from my initial mesh equations to your final result but I am confused about the final DE.

Does the \frac{\tau_L}{\tau_C} term not reduce to dimensionless?

 

[vela]

Yes, that combination is dimensionless as is (R1+R2)/R1 and 1.

You also made a mistake deriving di₁/dt. Check the signs in your equations.