In this question we will determine the potential distribution of the boundary value problem (2) numerically using finite difference approximation of the derivatives. This

is very similar to the Euler formula most of you have learned in ECE 201. First we

discretize the independent variable using N interior points z = ∆z, 2∆z, . . . , N∆z and

supplement it with z0 = 0 and zN+1 = h. We write this as zn = n∆z, n = 1, 2, . . . , N,

z0 = 0, zN+1 = h. The potential distribution is then calculated at the N interior

points V (zn) =: Vn, n = 1, 2, . . . , N given the boundary data V (z0) =: V0 = 0 and

V (zN+1) =: VN+1 = Vb. In finite-differences, the derivatives are approximated as

dV

dz (zn) ≈

[Vn+1 − Vn−1]

2∆z

(3)

d

2V

dz2

(zn) ≈

[Vn+1 − 2Vn + Vn−1]

(∆z)

2

. (4)

We now define the unknown voltage array v = [V1, V2, . . . , Vn, . . . , VN−1, VN ]

T

, where

the superscript T stands for matrix transpose and the boundary array of known voltages

b = [−V0, 0, . . . , 0, . . . , 0, −Vb]

T and generate a linear system of equations of the form





a11, a12, . . . , a1n, . . . , a1N

a21, a22, . . . , a2n, . . . , a2N

.

.

.

aN1, aN2, . . . , aNn, . . . , aNN





v =:

|{z}

A

N × N

v

|{z}

N × 1

= b

|{z}

N × 1

. (5)

By making use of the boundary value equation (2) determine the entries of the matrix

A. To facilitate this, it is best to write the equations for n = 1, n = 2, n = N −1, n = N

and observe the trend. The matrix A will consist only of numerical values. Then obtain

the solution by matrix inversion v = A−1b. Using h = 1, ∆z = h/101, Vb = 1 [V] and

MATLAB, determine the unknown vector v. Plot v versus z = [z1, z2, . . . , z100]

T

.

Superimpose on the same plot the analytical solution you determine from Q2(i) and

compare the two solutions. Use the values given here for h and Vb in the analytical

solution.