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Engineering, 23.02.2022 15:10 chelsey54
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.
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