I use function with multiple outputs farina4 that computes coefficients a, b, e, f and a vector out_p5tads_final (1 x n array) through a minimization of a system of equations using the data input set p5tads (1 x n array):
function [a b e f fval out_p5tads_final] = farina4(p5tads)
f = #(coeff)calculs_farina4(coeff,p5tads);
[ans,fval] = fminsearchcon(f,coeff0,[0 0 0 0],[1 1 1 1]);% fminsearch with constrains
a = ans(1);
b = ans(2);
e = ans(3);
f = ans(4);
out_p5tads_final = p5tads_farina4(a,b,e,f);
function out_coeff = calculs_farina4(coeff0,p5tads)
%bla-bla-bla
end
function out_p5tads = p5tads_farina4(a,b,e,f)
%bla-bla-bla
end
end
After calculating a, b, e, f and out_p5tads_final I need to calculate/minimize the RMS function with respect to out_p5tads_f4.
RMS = sqrt(mean((p5tads(:) - out_p5tads_f4(:)).^2))*100
and to repeat function farina4 in order to find the optimal set of the parameters a, b, e, f and out_p5tads_final.
I am trying to build up an algorithm of such optimization and do not see a way so far.
For instance, it seems to be not possible to introduce a function with multiple output inside the above RMS equation unless there is a way to index somehow the output of this function farina4.
If there can be an alternative optimization algorithm for RMS without fminsearch (or similar) ?
a b e and f are values between 0 and 1
out_p5tads_final is an (1 x 10) array
%
function out_coeff = calculs_farina4(coeff0,p5tads)
%
mmmm = p5tads(1);
mmmr = p5tads(2);
rmmr = p5tads(3);
mmrm = p5tads(4);
mmrr = p5tads(5);
rmrm = p5tads(6);
rmrr = p5tads(7);
mrrm = p5tads(8);
mrrr = p5tads(9);
rrrr = p5tads(10);
%
a = coeff0(1);
b = coeff0(2);
e = coeff0(3);
f = coeff0(4);
%
f_mmmm = mmmm - ((a^2*b^2*(a + b) + e^2*f^2*(e + f))/2);
f_mmmr = mmmr - (a^2*b^2*(e + f) + e^2*f^2*(a + b));
f_rmmr = rmmr - ((a^2*f^2*(b + e) + b^2*e^2*(a + f))/2);
f_mmrm = mmrm - 2*a*b*e*f;
f_mmrr = mmrr - b*f*(a^3 + e^3) + a*e*(a^3 + f^3);
f_rmrm = rmrm - 2*a*b*e*f;
f_rmrr = rmrr - 2*a*b*e*f;
f_mrrm = mrrm - ((a^2*b^2*(e + f) + e^2*f^2*(a + b))/2);
f_mrrr = mrrr - (a^2*f^2*(b + e) + b^2*e^2*(a + f));
f_rrrr = rrrr - ((a^2*f^2*(a + f) + b^2*e^2*(b + e))/2);
%
out_coeff = f_mmmm^2 + f_mmmr^2 + f_rmmr^2 + f_mmrm^2 + f_mmrr^2 + f_rmrm^2 + f_rmrr^2 + f_mrrm^2 + f_mrrr^2 + f_rrrr^2;
end
%
function out_p5tads = p5tads_farina4(a,b,e,f)
%
p_mmmm = ((a^2*b^2*(a + b) + e^2*f^2*(e + f))/2);
p_mmmr = (a^2*b^2*(e + f) + e^2*f^2*(a + b));
p_rmmr = ((a^2*f^2*(b + e) + b^2*e^2*(a + f))/2);
p_mmrm = 2*a*b*e*f;
p_mmrr = b*f*(a^3 + e^3) + a*e*(a^3 + f^3);
p_rmrm = 2*a*b*e*f;
p_rmrr = 2*a*b*e*f;
p_mrrm = ((a^2*b^2*(e + f) + e^2*f^2*(a + b))/2);
p_mrrr = (a^2*f^2*(b + e) + b^2*e^2*(a + f));
p_rrrr = ((a^2*f^2*(a + f) + b^2*e^2*(b + e))/2);
%
out_p5tads = [p_mmmm,p_mmmr,p_rmmr,p_mmrm,p_mmrr,p_rmrm,p_rmrr,p_mrrm,p_mrrr,p_rrrr];
end
end
Thanks much in advance !
19/08/2014 3:35 pm
I need to get an optimal set of coefficients a b e f that the RMS value , which is calculated from
RMS = sqrt(mean((p5tads(:) - out_p5tads_f4(:)).^2))*100
is minimal. Here, the vector p5tads is used to calculate/optimize the set of a b e f coefficients, which are in turn used to calculate the vector out_p5tads_f4. The code should run a desired number of optimizations cycles (e.g. by default 100) and then select the series of a b e f and out_p5tads_f4 afforded the minimal RMS error value (with respect to out_p5tads_f4).
Related
I've created a code for the Simpson's rule but I think I got the function wrong. I have no other sources to refer to (or they're too difficult to be understood). Here is my code:
function s = simpson(f_str, a, b, h)
f = inline(f_str);
n = (b-a)/h;
x = a + [1:n-1]*h;
xi = a + [1:n]*h;
s = h/3 * (f(a) + f(b) + 2*sum(f(x)) + 4*sum(f(xi)));
end
Can anybody help see where is the wrong part?
Assuming that the h in your function is the step size:
function s = simpson(f_str, a, b, h)
% The sample vector will be
xi = a:h:b;
f = inline(f_str);
% the function at the endpoints
fa = f(xi(1));
fb = f(xi(end));
% the even terms.. i.e. f(x2), f(x4), ...
feven = f(xi(3:2:end-2));
% similarly the odd terms.. i.e. f(x1), f(x3), ...
fodd = f(xi(2:2:end));
% Bringing everything together
s = h / 3 * (fa + 2 * sum(feven) + 4 * sum(fodd) + fb);
end
Source:
https://en.wikipedia.org/wiki/Simpson%27s_rule
I have forward transformations given as below:
Tx: x' = a1x + a2y + a3xy + a4.
Tx: y' = b1x + b2y + b3xy + b4.
function y = transformationTest (I,a1,a2,a3,a4,b1,b2,b3,b4)
[M N] = size(I);
for i = 1: M
for j = 1: N
x1 = a1*i + a2*j + a3*i*j + a4;
y1 = b1*i + b2*j + b3*i*j + b4;
y(round(x1+1),round(y1+1)) = I(i,j);
end
end
end
Is my implementation wrong or what because I am getting to large images and meaningless.
I have to restrict it with the folowings:
P(-1) = f(-1), P(0)=f(0), P(1)=f(1), P'(1)=f'(1)
Let the polynomial be
ax³ + bx² + cx + d
By the given equations,
- a + b - c + d = f(-1)
d = f(0)
a + b + c + d = f(1)
3a +2b + c = f'(1)
You should be able to solve.
Here is a transfer function:
S = [tf([10 2 4],[1 95 2000 3450])];
How can I get real(S) and Imag(S)?
It sounds like you want the Fourier form of your transfer function. As far as I know, there's no builtin function for this so you'll need to use symbolic math:
num = [10 2 4];
den = [1 95 2000 3450];
syms s;
syms omega real; % Define as real-valued
f1 = poly2sym(num,s)/poly2sym(den,s)
f2 = subs(f1,s,1i*omega)
f2_real = simplify(real(f2))
f2_imag = simplify(imag(f2))
which returns
f1 =
(10*s^2 + 2*s + 4)/(s^3 + 95*s^2 + 2000*s + 3450)
f2 =
(- 10*omega^2 + omega*2i + 4)/(- omega^3*1i - 95*omega^2 + omega*2000i + 3450)
f2_real =
(4*(237*omega^4 - 7720*omega^2 + 3450))/(omega^6 + 5025*omega^4 + 3344500*omega^2 + 11902500)
f2_imag =
-(2*omega*(5*omega^4 - 9907*omega^2 + 550))/(omega^6 + 5025*omega^4 + 3344500*omega^2 + 11902500)
You can then use subs and vpa/double to evaluate these for a particular value of omega.
Using DeMorgan's theorem show that:
a. (A + B)'(A' +B)' = 0
b. A + A'B + A'B' = 1
Boolean expression F = x'y + xyz':
Derive an algebraic expression for the complement F'
Show that F·F' = 0
Show that F + F' = 1
Please Help me
Assuming you know how DeMorgan's law works and you understand the basics of AND, OR, NOT operations:
1.a) (A + B)'(A' + B)' = A'B'(A')'B' = A'B'AB' = A'AB'B' = A'AB' = 0 B' = 0.
I used two facts here that hold for any boolean variable A:
AA' = 0 (when A = 0, A' = 1 and when A = 1, A' = 0 so (AA') has to be 0)
0A = 0 (0 AND anything has to be 0)
1.b) A + A'B + A'B' = A + A'(B + B') = A + A' = 1.
I used the following two facts that hold for any boolean variables A, B and C:
AB + AC = A(B + C) - just like you would do with numeric variables and multiplication and addition. Only here we work with boolean variables and AND (multiplication) and OR (addition) operations.
A + A' = 0 (when A = 0, A' = 0 and when A = 1, A' = 0 so (A + A') has to be 1)
2.a) Let's first derive the complement of F:
F' = (x'y + xyz')' = (x'y)'(xyz')' = (x + y')((xy)' + z) = (x + y')(x' + y' + z) = xx' + xy' + xz + x'y' + y'y' + y'z = 0 + xy' + xz + x'y' + y' + y'z = xy' + xz + y'(x + 1) + y'z = xy' + xz + y' + y'z = xy' + xz + y'(z + 1) = xy' + y' + xz = y'(x + 1) = xz + y'.
There is only one additional fact that I used here, that for any boolean variables A and B following holds:
A + AB = A(B + 1) = A - logically, variable A completely determines the output of such an expression and part AB cannot change the output of entire expression (you can check this one with truth tables if it's not clear, but boolean algebra should be enough to understand). And of course, for any boolean variable A + 1 = A.
2.b) FF' = (x'y + xyz')(xz + y') = x'yxz + x'yy' + xyz'xz + xyz'y'.
x'yxz = (xx')yz = 0xz = 0
xyy'= x0 = 0
xyz'xz = xxy(zz') = xy0 = 0
xyz'y' = xz'(yy') = xz'0 = 0
Therefore, FF' = 0.
2.c) F + F' = x'y + xyz' + xz + y'
This one is not so obvious. Let's start with two middle components and see what we can work out:
xyz' + xz = x(yz' + z) = x(yz' + z(y + y')) = x(yz' + yz + y'z) = x(y(z + z') + y'z) = x(y + y'z) = xy + xy'z.
I used the fact that we can write any boolean variable A in the following way:
A = A(B + B') = AB + AB' as (B + B') evaluates to 1 for any boolean variable B, so initial expression is not changed by AND-ing it together with such an expression.
Plugging this back in F + F' expression yields:
x'y + xy + xy'z + y' = y(x + x') + y'(xz + 1) = y + y' = 1.