You want to analyse for the case where v_s is a DC source (constant voltage), R

Question

You want to analyse for the case where v_s is a DC source (constant voltage), R = 20000 ohm, L = 0.01 H, C = 5 times 10^-5 F. a) Show that the above circuit can be modelled as the ODE d^2i/dt^2 + R di/L dt + 1/LC I = 1/L d V_s/dt, b)Convert the 2nd order ODE to a system of two first order ODEs (in matrix form). c) Solve the eigenvalues of the matrix A. d) Solve the corresponding eigenvectors for the matrix A. e) Verify your answers for b) and c) using the eig function in MATLAB. f) Solve the system of first order ODEs, given that i (0) = 10 A and i^'(0) = 0.

Explanation / Answer

Given

Vs = constant

R = 20,000 ohm

L = 0.01 H

C = 5*10^-5 F

a. So from kirchoff's voltage law

Vs = iR + Ldi/dt + q/C [ where i is current in circuit and q is charge on capacitor]

now dq/dt = i

so, differentiating the whole equation

d(Vs)/dt = Rdi/dt + Ld^2i/dt^2 + i/c

d^2i/dt + (R/L)di/dt + i/LC = (1/L) d(Vs)/dt

b. Now, d(Vs)/dt = 0 [ because Vs is a constant]

so, the equation becomes

d^2i/dt + (R/L)di/dt + i/LC = 0

let i = m

di/dt = n

then differentiating both equations

di/dt = dm/dt = n

d^2i/dt^2 = dn/dt

so, the equations become

dn/dt + (R/L)n + m/LC = 0

dm/dt - n = 0

dn/dt = -(R/L)n - m/LC

dm/dt = n

so the matrices become

[n'] = [-R/L -1/LC][n]

[m'] = [1 0 ][m]

c. Matrix A = [-R/L -1/LC]

[1 0]

substituting vaslues

Matrix A = [-2,000,000 -2,000,000]

[1 0]

to solve for eigenvalues

assume |(A - lambda*I)| = 0

so, [-2,000,000 - lambda -2,000,000] = 0

[1 -lambda ]

-(2,000,000 + lambda)(-lambda) - (-2,000,000)(1) = 0

2,000,000*lambda + lambda^2 + 2,000,000 = 0

solving for lambda = -1.00000005 , -1,999,998.999

d. so let the eigenvectors be v1 and v2

[-2,000,000 - lambda1 -2,000,000][v11] = 0

[1 -lambda1 ][v12]

(-2,000,000 - lambda1)V11 - 2,000,000V12 = 0

V11 - lambda1*v12 = 0

-2000001.00000005v11 - 2,000,000v12 = 0

v11 - 1.00000005v12 = 0

solving we get

-2000001.00000005(1.00000005v12 ) - 2,000,000v12 = 0

v12 = 0, v11 = 0

so v1 = [0 0]

similiarly

[-2,000,000 - lambda2 -2,000,000][v21] = 0

[1 -lambda2 ][v22]

(-2,000,000 - lambda2)V11 - 2,000,000V22 = 0

V11 - lambda2*v22 = 0

v22 = 0, v21 = 0

so v22 = [0 0]

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