University of Waterloo
Faculty of Engineering
Department of Electrical and Computer Engineering
ECE 204A
Pre-Laboratory 11
Prepared by
Surname/Last Name, Legal Given/First Name(s)
UW Student ID Number: 2NNNNNNN
UW User ID: uwuserid @uwaterloo.ca
2A Electrical/Computer Engineering
8 February 2016
11.1a Definitions:
1. An ordinary-differential equation (ODE) is any equation involving expressions
of a variable x, an unknown function u(x), and at least one of its derivatives:
u(1)(x), u(2)(x), u(3)(x), u(4)(x), ... .
2. A 2nd-order ordinary-differential equation (2nd-order ODE) is any equation
involving expressions of a variable x, the 2nd-derivative of an unknown function
u(2)(x) and possibly the function itself or its first derivatives, u(x) and u(1)(x).
3. A 2nd-order linear ODE (2nd-order LODE) is any equation of the form
p(x) u(2)(x) + q(x) u(1)(x) + r(x) u(x) = g(x)
where p(x), q(x), r(x), and g(x) are functions of x with p(x) ≠ 0. The function g(x)
is said to be the forcing function.
4. A 2nd-order homogenous LODE (2nd-order hom. LODE) is any equation of the
form
p(x) u(2)(x) + q(x) u(1)(x) + r(x) u(x) = 0
where p(x), q(x), and r(x) are functions of x with p(x) ≠ 0.
5. A 2nd-order LODE with constant coefficients (2nd-order LODE w/cc) is any
equation of the form
a u(2)(x) + b u(1)(x) + c u(x) = g(x)
where a, b, and c are constants with respect to x with a ≠ 0 and g(x) is a function
of x.
For each of the following, indicate with an x is the most specific description of the given
equation:
ODE
2nd-order
ODE
2nd-order
LODE
2nd-order
hom.
LODE
2nd-order
LODE
w/cc
2nd-order
hom.
LODE
w/cc
u(2)(x) = 0
u(2)(x) + 2u(1)(x) + 5u(x) = 4
u(2)(x) + u(1)(x) u(x) = 0
u(2)(x) + sin(x) 2u(1)(x) + u(x) = cos(x)
u(2)(x) + 2u(1)(x + 1) = 0
u(2)(x) + 2x u(x) = 0
u(3)(x) + 3u(2)(x) + u(x) = sin(x)
x2 + 4x – 2 = u(x)
11.1bA quick review: Given the 2nd-order LODE
u(2)(x) + 4u(1)(x) – 7u(x) = g(x),
we can write this as a system of two 1st-order LODEs:
1. Define w1 ≡ u and w2 ≡ u(1).
That is, w1(x) = u(x) and w2(x) = u(1)(x) for all values of x.
2. Notice now that if w1 and u are equal, then w1(1)(x) = u(1)(x); however, we also
defined w2(x) to be equal to u(1)(x), so we can write w1(1)(x) = w2(x).
3. Notice also that if w2(x) = u(1)(x) then u(2)(x) = w2(1)(x).
Therefore, given the differential equation above, we can write it as a system of linear
differential equations
w1(1)(x) = w2(x)
w2(1)(x) + 4w2(x) – 7w1(x) = g(x)
or
w1(1)(x) = w2(x)
w2(1)(x) = –4w2(x) + 7w1(x) + g(x).
The 2nd-order ODE above could be implemented as follows:
function [ddu] = f( x, u, du )
ddu = -4*du + 7*u + g(x);
end
2
None
of
These
where we would, of course, have to know what g(x) is defined as. Instead, rewrite this
function so that it takes a vector of two values w and returns a vector of two values, the
derivatives of the arguments.
function [dw] = f( x, w )
ddu = [...
...];
end
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11.2a The following code approximate the derivative f(1)(x) by using a formula found by
two steps of Richardson extrapolation:
function [dy] = diffr2( f, x, h )
dy = (f(x+h) - 40*f(x+h/2) + 256*f(x+h/4) ...
- 256*f(x-h/4) + 40*f(x-h/2) - f(x-h))/(90*h);
end
Suppose the function f11a is a vector-valued function, for example,
 sin  x1   x2  0.2 


f11a  x    cos  x2   x3  0.3 


e x3  x1  1


which may be implemented using
function [y] = f11a( x )
y = [sin( x(1)) - x(2) + 0.2;
cos( x(2)) + x(3) - 0.3;
exp(-x(3)) - x(1) - 1];
end
In this case, root-finding means that we wish to find to find a value of the vector x such
0
 
that f11a (x)   0  . This is quite common in electrical engineering: suppose you have a
0
 
non-linear circuit and you apply KCL at three nodes. At each node, the sum of the
voltages must be zero and if the circuit elements are non-linear, you may have to solve a
system similar to the one shown above. We will use something similar to Newton’s
method in one dimension where we solve
xk 1  xk 
f  xk 
f 1  xk 
.
In higher dimensions, however, what is the derivative of an n-dimensional function? The
solution appears by rewriting this equation as
f 1  xk  xk 1  xk    f  xk  .
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If we write this as f    xk  xk 1   f  xk  , then we solve for xk 1 and set
1
xk 1  xk  xk 1 . The question is what is the derivative in higher dimensions? It turns
out to be a matrix, the Jacobian (written as J(f)(x)) which can be evaluated as
 f1

 x1
 f
J  f  x    2
 x1
 f
 3
 x1
f1
x2
f 2
x2
f3
x2
f1 

x3 
f 2 
.
x3 
f3 

x3 
To do create this matrix, we must calculate the partial derivatives of the function f(x)
with respect to each of the three variables. The modification
function [dy] = diffr2( f, x, h, n )
dh = zeros( size( x ) );
dh(n) = h;
dy = ( f(x+dh) - 40*f(x+dh/2) + 256*f(x+dh/4) ...
- f(x-dh) + 40*f(x-dh/2) - 256*f(x-dh/4) )/(90*h);
end
 f1 


 xn 
 f 
approximates  2  evaluated at the point x. By evaluating this function at n = 1, 2, and
 xn 
 f 
 3 
 xn 
3, we can create three vectors and can create the Jacobian matrix. Once we have this
matrix, we can then solve J  f  xk  xk 1  f  xk  for xk 1 by solving the system of
linear equations and then just setting xk 1  xk  xk 1 . Implement the function
newton_nd which takes the same arguments as newton but where f is a function as
described above and x0 is an n-dimensional column vector. Note that the dimension n is
different from the maximum number of iterations N.
5
function [x] = newton_nd( f, x0, h, eps_step, eps_abs, N )
% REPLACE THIS WITH YOUR IMPLEMENTATION
% get the dimension n of the vector x0
x = x0;
for k = 1:N
x_old = x;
% Create a matrix J of zeros
for j = 1:n
% Set the jth column of J to the vector of
% partials w.r.t. x(j) evaluated at x
end
% Solve the system described above for the unknown dx
x = x + dx;
if norm( x - x_old ) < eps_step && norm( f(x) ) < eps_abs
return;
end
end
throw( MException( 'MATLAB:non_convergence', ...
'Newton''s method didn''t converge.' ) );
end
Test your code by running
x = newton_nd( @f11a, [1 1 -1]', 0.01, 1e-10, 1e-10, 10 )
% Copy and paste your output here
You have a value which is returned from newton_nd. How can you test the output to
determine whether or not you have a reasonable solution?
Your answer here.
6
11.2b Find a root of the following system of non-linear equations:
e x1  2 x2  x3  1
x1  e0.1x2  3x3  2
 x1  x2  e0.3x3  1
by first converting the three equations into expressions where the right-hand side is zero.
Then, define a vector-valued function f which returns the function which evaluates the
non-zero left-hand side. Enter the function here and make sure that you can find a root
with your choice of initial conditions.
function [y] = f11b(x)
% Your implementation here
end
x = newton_nd( @f11b, [? ? ?]', 0.01, 1e-10, 1e-10, 10 )
% Your output here
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