Given that:
Y = 0.299R + 0.587G + 0.114B
What values do we put in for R,G, and B? I’m assuming 0-255. For arguments sake, if R, G, B are each 50, then does it mean Y=0.299(50) + 0.587(500) + 0.11(50)?
The next two are also confusing. How can B - Y even be possible if Y contains Blue then isn’t B - Y just taking away itself?
Cb = 0.564( B − Y )
Cr =0.713(R−Y)
It's just simple (confusing) math ...
Remark: There are few standards of YCbCr following formula applies BT.601, with "full range":
Equation (1): Y = 0.299R + 0.587G + 0.114B
The common definition of YCbCr assumes that R, G, and B are 8 bits unsigned integers in range [0, 255].
There are cases where R, G, B are floating point values in range [0, 1] (normalized values).
There are also HDR cases where range is [0, 1023] for example.
In case R=50, G=50, B=50, you just need to assign the values:
Y = 0.299*50 + 0.587*50 + 0.114*50
Result: Y = 50.
Since Y represents the Luma (line luminescence), and RGB=(50,50,50), is a gray pixel, it does make sense that Y = 50.
The following equations Cb = 0.564(B - Y), Cr = 0.713(R - Y) are incorrect.
Instead of Cb, and Cr they should be named Pb and Pr.
Equation (2): Pb = 0.564(B - Y)
Equation (3): Pr = 0.713(R - Y)
The equations mean that you can compute Y first, and then use the result for computing Pb and Pr.
Remark: don't round the value of Y when you are using it for computing Pb and Pr.
You can also assign Equation (1) in (2) and (3):
Pb = 0.564(B - Y) = 0.564(B - (0.299R + 0.587G + 0.114B)) = 0.4997*B - 0.3311*G - 0.1686*R
Pr = 0.713(R - Y) = 0.713(R - (0.299R + 0.587G + 0.114B)) = 0.4998*R - 0.4185*G - 0.0813*B
There are some small inaccuracies.
Wikipedia is more accurate (but still just a result of mathematical assignments):
Y = 0.299*R + 0.587*G + 0.114*B
Pb = -0.168736*R - 0.331264*G + 0.5*B
Pr = 0.5*R - 0.418688*G - 0.081312*B
In the above formulas the range of Pb, Pr is [-127.5, 127.5].
In the "full range" formula of YCbCr (not YPbPr), an offset of 128 is added to Pb and Pr (so result is always positive).
In case of 8 bits, the final result is limited to range [0, 255] and rounded.
What you're referencing is the conversion of RGB to YPbPr.
Conversion to YCbCr is as follows:
Y = 0.299 * R + 0.587 * G + 0.114 * B
Cb = -0.167 * R - 0.3313 * G - 0.5 * B + 128
Cr = 0.5 * R - 0.4187 * G - 0.0813 * B + 128
Yours is YPbPr (which is better for JPEG Compression, see below):
delta = 0.5
Y = 0.299 * R + 0.587 * G + 0.114 * B (same as above)
Pb: (B - Y) * 0.564 + delta
Pr: (B - Y) * 0.713 + delta
The above answer did a better job of explaining this.
I've been looking into JPEG Compression for implementation in Pytorch and found this thread (https://photo.stackexchange.com/a/8357) useful to explain why we use YPbPr for compression over YCbCr.
Pb and Pr versions are better for image compression because the luminance information (which contains the most detail) is retained in only one (Y) channel, while Pb and Pr would contain the chrominance information. Thus when you're doing down-sampling later down the line, there's less loss of valuable info.
Related
I've created this code and it gives me this error message:
Error using *
Incorrect dimensions for matrix multiplication.
Error in poli3 = sin(pi*a) ...
Below I show one function used in the code. I don't know if the problem comes from the value given by derivadan or what.
x = -1:0.01:1; % Intervalo en el que se evaluará el polinomio de Taylor
y = sin(pi*x); % Función
a = 0;
derivada3 = derivadan(0.01, 3, a);
derivada7 = derivadan(0.01, 7, a);
derivada3_vec = repmat(derivada3, size(x - a));
derivada7_vec = repmat(derivada7, size(x - a));
poli3 = sin(pi*a) + derivada3_vec*(x - a) + (derivada3_vec*(x - a).^2)/factorial(2) + (derivada3_vec*(x - a).^3)/factorial(3);
poli7 = sin(pi*a) + derivada7_vec*(x - a) + (derivada7_vec*(x - a).^2)/factorial(2) + (derivada7_vec*(x - a).^3)/factorial(3) + (derivada7_vec*(x - a).^4)/factorial(4) + (derivada7_vec*(x - a).^5)/factorial(5) + (derivada7_vec*(x - a).^6)/factorial(6) + (derivada7_vec*(x - a).^7)/factorial(7);
figure
plot(x, poli3, 'r', x, poli7, 'b')
legend('Taylor grau 3', 'Taylor grau 7')
title('Grafica Taylor 3 grau vs Grafica Taylor 7 grau')
function Yd = derivadan(h, grado, vecX)
Yd = zeros(size(vecX));
for i = 1:grado
Yd = (vecX(2:end) - vecX(1:end-1)) / h;
vecX = Yd;
end
end
In MATLAB one can go 2 ways when developing taylor series; the hard way and the easy way.
As asked, first the HARD way :
close all;clear all;clc
dx=.01
x = -1+dx:dx:1-dx;
y = sin(pi*x);
a =0;
[va,na]=find(x==a)
n1=3
D3y=zeros(n1,numel(x));
for k=1:1:n1
D3y(k,1:end-k)=dn(dx,k,y);
end
T3=y(na)+sum(1./factorial([1:n1])'.*D3y(:,na).*((x(1:end-n1)-a)'.^[1:n1])',1);
n2=7
D7y=zeros(n2,numel(x));
for k=1:1:n2
D7y(k,1:end-k)=dn(dx,k,y);
end
T7=y(na)+sum([1./factorial([1:n2])]'.*D7y(:,na).*((x(1:end-n2)-a)'.^[1:n2])',1);
figure(1);ax=gca
plot(ax,x(1:numel(T7)),T3(1:numel(T7)),'r')
grid on;hold on
xlabel('x')
plot(ax,x(1:numel(T7)),T7(1:numel(T7)),'b--')
plot(ax,x(1:numel(T7)),y(1:numel(T7)),'g')
axis(ax,[-1 1 -1.2 1.2])
legend('T3', 'T7','sin(pi*x)','Location','northeastoutside')
the support function being
function Yd = dn(h, n, vecX)
Yd = zeros(size(vecX));
for i = 1:n
Yd = (vecX(2:end) - vecX(1:end-1))/h;
vecX = Yd;
end
end
Explanation
1.- The custom function derivadan that I call dn shortens one sample for every unit up in grado where grado is the derivative order.
For instance, the 3rd order derivative is going to be 3 samples shorter than the input function.
This causes product mismatch and when later on attempting plot it's going to cause plot error.
2.- To avoid such mismatchs ALL vectors shortened to the size of the shortest one.
x(1:end-a)
is a samples shorter than x and y and can be used as reference vector in plot.
3.- Call function derivadan (that I call dn) correctly
dn expects as 3rd input (3rd from left) a vector, the function values to differentiate, yet you are calling derivadan with a in 3rd input field. a is scalar and you have set it null. Fixed it.
derivada3 = derivadan(0.01, 3, a);
should be called
derivada3 = derivadan(0.01, 3, y);
same for derivada7
4.- So
error using * ...
error in poly3=sin(pi*a) ...
MATLAB doesn't particularly mean that there's an error right on sin(pi*a) , that could be, but it's not the case here >
MATLAB is saying : THERE'S AN ERROR IN LINE STARTING WITH
poly3=sin(pi*a) ..
MATLAB aborts there.
Same error is found in next line starting with
poly7 ..
Since sin(pi*a)=0 because a=0 yet all other terms in sum for poly3 are repmat outcomes with different sizes all sizes being different and >1 hence attempting product of different sizes.
Operator * requires all terms have same size.
5.- Syntax Error
derivada3_vec = repmat(derivada3, size(x - a))
is built is not correct
this line repeats size(x) times the nth order derivative !
it's a really long sequence.
Now the EASY way
6.- MATLAB already has command taylor
syms x;T3=taylor(sin(pi*x),x,0)
T3 = (pi^5*x^5)/120 - (pi^3*x^3)/6 + pi*x
syms x;T3=taylor(sin(pi*x),x,0,'Order',3)
T3 = pi*x
syms x;T3=taylor(sin(pi*x),x,0,'Order',7)
T7 = (pi^5*x^5)/120 - (pi^3*x^3)/6 + pi*x
T9=taylor(sin(pi*x),x,0,'Order',9)
T9 =- (pi^7*x^7)/5040 + (pi^5*x^5)/120 - (pi^3*x^3)/6 + pi*x
It really simplfies taylor series development because it readily generates all that is needed to such endeavour :
syms f(x)
f(x) = sin(pi*x);
a=0
T_default = taylor(f, x,'ExpansionPoint',a);
T8 = taylor(f, x, 'Order', 8,'ExpansionPoint',a);
T10 = taylor(f, x, 'Order', 10,'ExpansionPoint',a);
figure(2)
fplot([T_default T8 T10 f])
axis([-2 3 -1.2 1.2])
hold on
plot(a,f(a),'r*')
grid on;xlabel('x')
title(['Taylor Series Expansion x =' num2str(a)])
a=.5
T_default = taylor(f, x,'ExpansionPoint',a);
T8 = taylor(f, x, 'Order', 8,'ExpansionPoint',a);
T10 = taylor(f, x, 'Order', 10,'ExpansionPoint',a);
figure(3)
fplot([T_default T8 T10 f])
axis([-2 3 -1.2 1.2])
hold on
plot(a,f(a),'r*')
grid on;xlabel('x')
title(['Taylor Series Expansion x =' num2str(a)])
a=1
T_default = taylor(f, x,'ExpansionPoint',a);
T8 = taylor(f, x, 'Order', 8,'ExpansionPoint',a);
T10 = taylor(f, x, 'Order', 10,'ExpansionPoint',a);
figure(4)
fplot([T_default T8 T10 f])
axis([-2 3 -1.2 1.2])
hold on
plot(a,f(a),'r*')
grid on;xlabel('x')
title(['Taylor Series Expansion x =' num2str(a)])
thanks for reading
I checked the line that you were having issues with; it seems that the error is in the derivada3_vec*(x - a) (as well as in the other terms that use derivada3_vec).
Looking at the variable itself: derivada3_vec is an empty vector. Going back further, the derivada3 variable is also an empty vector.
Your issue is in your function derivadan. You're inputting a 1x1 vector (a = [0]), but the function assumes that a is at least 1x2 (or 2x1).
I suspect there are other issues, but this is the cause of your error message.
I would like to create a realistic pool game and to implement at least some basic ball effects. I started from scratch with SceneKit and at this point I'm just studying the proper technology to go with it.SceneKit would be the ideal.
I managed to achieve an acceptable ball effect for sidespin and some sort of forward spin. The one I'm struggle with is backspin. I'm playing around with the position parameter of applyForce method, but it seems that alone will not give me the result I'm looking for. Either I'm missing something (I've got limited knowledge of physics) or SceneKit's physics simulation is just not enough for what I want. Basically I have a sphere of 1.5 radius and I went from -1.5 to 1.5 on Y component for the position vector and the result is either the white ball or the ball I'm hitting jumps when collision occurs.
The first screenshot shows the moment of impact whilst the latter shows after the collision and how it jumps.
The two spheres are configured like this
let sphereGeometry = SCNSphere(radius: 1.5)
sphere1 = SCNNode(geometry: sphereGeometry)
sphere1.position = SCNVector3(x: -15, y: 0, z: 0)
sphere2 = SCNNode(geometry: sphereGeometry)
sphere2.position = SCNVector3(x: 15, y: 0, z: 0)
And the code that gives me that effect is the following:
sphere1.physicsBody?.applyForce(SCNVector3Make(350, 0, 0), atPosition:SCNVector3Make(1.5, -0.25, 0), impulse: true)
What I'm trying to do in that code is to hit the ball roughly a bit below the center. How I got to -0.25 was to get an angle of 10 degrees and calculate its sin function. Then I multiplied it by sphere radius so I can get a point that lies right on the sphere's surface.
So I've been reading several papers/chapters about pool physics and I think I found something that at least proves me I can do it with SceneKit. So what I was missing was i. right formulae ii. angular velocity. The physics still need a lot of polish but at least it seems to get roughly the trajectory one would expect when applying these effects. Here's the code in case anyone's interested in:
//Cue strength
let strength : Float = 1000
//Cue mass expressed in terms of ball's mass
let cueMass : Float = self.balls[0].mass * 1.25
//White ball
let whiteBall = self.balls[0]
//The ball we are trying to hit
let targetBall = self.balls[1]
//White ball radius
let ballRadius = whiteBall.radius
//This should be in the range of {-R, R} where R is the ball radius. It determines how much off the center we would like to hit the ball along the z-axis. Produces left/right spin
let a : Float = 0
//This should be in the range of {-R, R} where R is the ball radius. It determines how much off the center we would like to hit the ball along the y-axis. Produces top/back spin
let b : Float = -ballRadius * 0.7
//This is calculated based off a and b and it is the position that we will be hitting the ball along the x-axis.
let c : Float = sqrt(ballRadius * ballRadius - a * a - b * b)
//This is the angle of the cue expressed in degrees. Values greater than zero will produce jump shots
let cueAngle : Float = 0
//Cue angle in radians for math functions
let cueAngleInRadians : Float = (cueAngle * 3.14) / 180
let cosAngle = cos(cueAngleInRadians)
let sinAngle = sin(cueAngleInRadians)
//Values to calculate the magnitude to be applied given the above variables
let m0 = a * a
let m1 = b * b * cosAngle * cosAngle
let m2 = c * c * sinAngle * sinAngle
let m3 = 2 * b * c * cosAngle * sinAngle
let w = (5 / (2 * ballRadius * ballRadius)) * (m0 + m1 + m2 + m3)
let n = 2 * whiteBall.mass * strength
let magnitude = n / (1 + whiteBall.mass / cueMass + w)
//We would like to point to the target ball
let targetVector = targetBall.position
//Get the unit vector of our target
var target = (targetVector - whiteBall.position).normal
//Multiply our direction by the force's magnitude. Y-axis component reflects the angle of the cue
target.x *= magnitude
target.y = (magnitude / whiteBall.mass) * sinAngle
target.z *= magnitude
//Apply the impulse at the given position by c, b, a
whiteBall.physicsBody?.applyForce(target, atPosition: SCNVector3Make(c, b, a), impulse: true)
//Values to calculate angular force
let i = ((2 / 5) * whiteBall.mass * ballRadius * ballRadius)
let wx = a * magnitude * sinAngle
let wy = -a * magnitude * cosAngle
let wz = -c * magnitude * sinAngle + b * magnitude * cosAngle
let wv = SCNVector3Make(wx, wy, wz) * (1 / i)
//Apply a torque
whiteBall.physicsBody?.applyTorque(SCNVector4Make(wv.x, wv.y, wv.z, 0.4), impulse: true)
Note that values of a, b, c should take into account the target vector's direction.
This question already has answers here:
Numerical derivative of a vector
(3 answers)
Closed 7 years ago.
I have two arrays: x and y. In practice, y is dependent on x, but both arrays are measured values. I would like to obtain the derivative of y with respect to x. If x were uniformly spaced (i.e. x=[1 2 3 4 5]), I could do something with diff like this:
h = 0.001;
x = -pi:h:pi;
f = sin(X);
y = diff(f)/h;
However, x is not uniformly spaced (i.e. x=[1 1.9 2.8 4.1]). How can I obtain the partial derivative of this data set?
A good way to do it is gradient,
dydx = gradient(y, x);
I like it because it returns a vector which is the same length as x and y. The downside though, is it's first order accurate. This can sometimes be a problem, a fix could be to write your own,
x = unique([linspace(0, 2*pi, 50), logspace(0, log10(2*pi), 50)]);
y = cos(x) ;
subplot(2,1,1) ;
plot(x, Gradient(y, x), x, gradient(y,x), x, -sin(x));
legend('2^{nd} order', '1^{st} order', 'exact') ;
subplot(2,1,2) ;
plot(x, Gradient(y, x) + sin(x), x, gradient(y,x) + sin(x));
legend('2^{nd} order - exact', '1^{st} order - exact')
With Gradient being
function dydx = Gradient(y,x)
y = y(:);
p = x(3:end) - x(2:end-1);
p = p(:);
m = x(2:end-1) - x(1:end-2);
m = m(:);
p1 = x(2) - x(1);
p2 = x(3) - x(1);
m1 = x(end) - x(end-1);
m2 = x(end) - x(end-2);
dydx = reshape([ ((-p1^2 + p2^2)*y(1) - p2^2*y(2) + p1^2*y(3))/(p1*(p1 - p2)*p2);
((-m.^2 + p.^2).*y(2:end-1) - p.^2.*y(1:end-2) + m.^2.*y(3:end))./(m.*p.*(m + p));
((m1^2 - m2^2)*y(end) + m2^2*y(end-1) - m1^2*y(end-2))/(m1^2*m2 - m1*m2^2) ...
], size(x));
end
Edit:
Improved it for multidimensional array and constant spacing support
function dydx = Gradient(y,x)
if length(y) < 3
dydx = gradient(y,x);
return
end
[~, n] = max(size(y));
N = ndims(y);
i = repmat({':'},1,N-1);
y = permute(y, [n, 1:n-1, n+1:N]);
if isscalar(x)
%"x" is actually a spacing value
p = x;
m = x;
p1 = x;
p2 = x;
m1 = x;
m2 = x;
else
if isvector(x)
x = repmat(x(:), size(y(1, i{:})));
else
x = permute(x, [n, 1:n-1, n+1:N]);
end
if all(size(x) ~= size(y))
error('Sizes of arrays must be the same.')
end
p = x(3:end, i{:}) - x(2:end-1, i{:});
m = x(2:end-1, i{:}) - x(1:end-2, i{:});
p1 = x(2, i{:}) - x(1, i{:});
p2 = x(3, i{:}) - x(1, i{:});
m1 = x(end, i{:}) - x(end-1, i{:});
m2 = x(end, i{:}) - x(end-2, i{:});
end
dydx = ipermute([ ((-p1.^2 + p2.^2).*y(1,i{:}) - p2.^2.*y(2,i{:}) + p1.^2.*y(3,i{:}))./(p1.*(p1 - p2).*p2);
((-m.^2 + p.^2).*y(2:end-1,i{:}) - p.^2.*y(1:end-2,i{:}) + m.^2.*y(3:end,i{:}))./(m.*p.*(m + p));
((m1.^2 - m2.^2).*y(end,i{:}) + m2.^2.*y(end-1,i{:}) - m1.^2.*y(end-2,i{:}))./(m1.^2.*m2 - m1.*m2.^2) ...
], [n, 1:n-1, n+1:N]);
end
I have the antenna array factor expression here:
I have coded the array factor expression as given below:
lambda = 1;
M = 100;N = 200; %an M x N array
dx = 0.3*lambda; %inter-element spacing in x direction
m = 1:M;
xm = (m - 0.5*(M+1))*dx; %element positions in x direction
dy = 0.4*lambda;
n = 1:N;
yn = (n - 0.5*(N+1))*dy;
thetaCount = 360; % no of theta values
thetaRes = 2*pi/thetaCount; % theta resolution
thetas = 0:thetaRes:2*pi-thetaRes; % theta values
phiCount = 180;
phiRes = pi/phiCount;
phis = -pi/2:phiRes:pi/2-phiRes;
cmpWeights = rand(N,M); %complex Weights
AF = zeros(phiCount,thetaCount); %Array factor
tic
for i = 1:phiCount
for j = 1:thetaCount
for p = 1:M
for q = 1:N
AF(i,j) = AF(i,j) + cmpWeights(q,p)*exp((2*pi*1j/lambda)*(xm(p)*sin(thetas(j))*cos(phis(i)) + yn(q)*sin(thetas(j))*sin(phis(i))));
end
end
end
end
How can I vectorize the code for calculating the Array Factor (AF).
I want the line:
AF(i,j) = AF(i,j) + cmpWeights(q,p)*exp((2*pi*1j/lambda)*(xm(p)*sin(thetas(j))*cos(phis(i)) + yn(q)*sin(thetas(j))*sin(phis(i))));
to be written in vectorized form (by modifying the for loop).
Approach #1: Full-throttle
The innermost nested loop generates this every iteration - cmpWeights(q,p)*exp((2*pi*1j/lambda)*(xm(p)*sin(thetas(j))*cos(phis(i)) + yn(q)*sin(thetas(j))*sin(phis(i)))), which are to summed up iteratively to give us the final output in AF.
Let's call the exp(.... part as B. Now, B basically has two parts, one is the scalar (2*pi*1j/lambda) and the other part
(xm(p)*sin(thetas(j))*cos(phis(i)) + yn(q)*sin(thetas(j))*sin(phis(i))) that is formed from the variables that are dependent on
the four iterators used in the original loopy versions - i,j,p,q. Let's call this other part as C for easy reference later on.
Let's put all that into perspective:
Loopy version had AF(i,j) = AF(i,j) + cmpWeights(q,p)*exp((2*pi*1j/lambda)*(xm(p)*sin(thetas(j))*cos(phis(i)) + yn(q)*sin(thetas(j))*sin(phis(i)))), which is now equivalent to AF(i,j) = AF(i,j) + cmpWeights(q,p)*B, where B = exp((2*pi*1j/lambda)*(xm(p)*sin(thetas(j))*cos(phis(i)) + yn(q)*sin(thetas(j))*sin(phis(i)))).
B could be simplified to B = exp((2*pi*1j/lambda)* C), where C = (xm(p)*sin(thetas(j))*cos(phis(i)) + yn(q)*sin(thetas(j))*sin(phis(i))).
C would depend on the iterators - i,j,p,q.
So, after porting onto a vectorized way, it would end up as this -
%// 1) Define vectors corresponding to iterators used in the loopy version
I = 1:phiCount;
J = 1:thetaCount;
P = 1:M;
Q = 1:N;
%// 2) Create vectorized version of C using all four vector iterators
mult1 = bsxfun(#times,sin(thetas(J)),cos(phis(I)).'); %//'
mult2 = bsxfun(#times,sin(thetas(J)),sin(phis(I)).'); %//'
mult1_xm = bsxfun(#times,mult1(:),permute(xm,[1 3 2]));
mult2_yn = bsxfun(#times,mult2(:),yn);
C_vect = bsxfun(#plus,mult1_xm,mult2_yn);
%// 3) Create vectorized version of B using vectorized C
B_vect = reshape(exp((2*pi*1j/lambda)*C_vect),phiCount*thetaCount,[]);
%// 4) Final output as matrix multiplication between vectorized versions of B and C
AF_vect = reshape(B_vect*cmpWeights(:),phiCount,thetaCount);
Approach #2: Less-memory intensive
This second approach would reduce the memory traffic and it uses the distributive property of exponential - exp(A+B) = exp(A)*exp(B).
Now, the original loopy version was this -
AF(i,j) = AF(i,j) + cmpWeights(q,p)*exp((2*pi*1j/lambda)*...
(xm(p)*sin(thetas(j))*cos(phis(i)) + yn(q)*sin(thetas(j))*sin(phis(i))))
So, after using the distributive property, we would endup with something like this -
K = (2*pi*1j/lambda)
part1 = K*xm(p)*sin(thetas(j))*cos(phis(i));
part2 = K*yn(q)*sin(thetas(j))*sin(phis(i));
AF(i,j) = AF(i,j) + cmpWeights(q,p)*exp(part1)*exp(part2);
Thus, the relevant vectorized approach would become something like this -
%// 1) Define vectors corresponding to iterators used in the loopy version
I = 1:phiCount;
J = 1:thetaCount;
P = 1:M;
Q = 1:N;
%// 2) Define the constant used at the start of EXP() call
K = (2*pi*1j/lambda);
%// 3) Perform the sine-cosine operations part1 & part2 in vectorized manners
mult1 = K*bsxfun(#times,sin(thetas(J)),cos(phis(I)).'); %//'
mult2 = K*bsxfun(#times,sin(thetas(J)),sin(phis(I)).'); %//'
%// Perform exp(part1) & exp(part2) in vectorized manners
part1_vect = exp(bsxfun(#times,mult1(:),xm));
part2_vect = exp(bsxfun(#times,mult2(:),yn));
%// Perform multiplications with cmpWeights for final output
AF = reshape(sum((part1_vect*cmpWeights.').*part2_vect,2),phiCount,[])
Quick Benchmarking
Here are the runtimes with the input data listed in the question for the original loopy approach and proposed approach #2 -
---------------------------- With Original Approach
Elapsed time is 358.081507 seconds.
---------------------------- With Proposed Approach #2
Elapsed time is 0.405038 seconds.
The runtimes suggests a crazy performance improvement with Approach #2!
The basic trick is to figure out what things are constant, and what things depend on the subscript term - and therefore are matrix terms.
Within the sum:
C(n,m) is a matrix
2π/λ is a constant
sin(θ)cos(φ) is a constant
x(m) and y(n) are vectors
So the two things I would do are:
Expand the xm and ym into matrices using meshgrid()
Take all the constant term stuff outside the loop.
Like this:
...
piFactor = 2 * pi * 1j / lambda;
[xgrid, ygrid] = meshgrid(xm, ym); % xgrid and ygrid will be size (N, M)
for i = 1:phiCount
for j = 1:thetaCount
xFactor = sin(thetas(j)) * cos(phis(i));
yFactor = sin(thetas(j)) * sin(phis(i));
expFactor = exp(piFactor * (xgrid * xFactor + ygrid * yFactor)); % expFactor is size (N, M)
elements = cmpWeights .* expFactor; % elements of sum, size (N, M)
AF(i, j) = AF(i, j) + sum(elements(:)); % sum and then integrate.
end
end
You could probably figure out how to vectorise the outer loop too, but hopefully that gives you a starting point.
I am to implement the following formula in Delphi:
Understanding the formula:
Y_k is a floating point number which we'll call just Y.
w[i][j] is an array containing floating point numbers as well (1<=i<=43 and 1<=j<=30).
According to my source paper (p. 12) nabla(w)*Y_k is the partial derivative of (column) vector w with respect to the value Y". Is this correct?
Coding in Delphi (implementation):
So how do I implement this in Delphi?
Thanks a lot in advance!
I will assume that the problem is: "How can I compute a gradient of a scalar function in Delphi?" [I still think that the symbol given above looks more like a connection/covariant derivative as known from differential geometry/tensor calculus!]
You need to specify the input you got. The simplest case is when you got the function f whose gradient you wish to compute. Per definition, if f is a function of the k variables x1, x2, ..., xk, that is, if you'd write
f(x1, x2, ..., xk)
which is a scalar field in ℝk then the gradient is
∇f = (∂f/∂x1, ∂f/∂x2, ..., ∂f/∂xk)
that is, a vector field in ℝk (at each point in ℝk you get a k-dimensional vector).
This is rather easily implemented in Delphi. The following is an example for the case where k = 3.
type
TVector = record
x, y, z: real;
constructor Create(ax, ay, az: real);
end;
TRealValuedFunction = function(v: TVector): real;
function gradient(f: TRealValuedFunction; v: TVector): TVector;
const
h = 0.001;
begin
result.x := (f(TVector.Create(v.x + h, v.y, v.z)) - f(TVector.Create(v.x - h, v.y, v.z))) / (2*h);
result.y := (f(TVector.Create(v.x, v.y + h, v.z)) - f(TVector.Create(v.x, v.y - h, v.z))) / (2*h);
result.z := (f(TVector.Create(v.x, v.y, v.z + h)) - f(TVector.Create(v.x, v.y, v.z - h))) / (2*h);
end;
where, of course,
constructor TVector.Create(ax, ay, az: real);
begin
x := ax;
y := ay;
z := az;
end;
Sample usage:
function SampleFunction(v: TVector): real;
begin
result := 5*v.x + 7*v.y;
end;
procedure TForm4.FormCreate(Sender: TObject);
begin
with gradient(SampleFunction, TVector.Create(2, 6, 3)) do
ShowMessage(FloatToStr(x) + ', ' + FloatToStr(y) + ', ' + FloatToStr(z))
end;
The result is 5.00000000000256, 7.000000000005, 0 which is a very good approximation to the gradient of SampleFunction at v (in this case the gradient is constant in space, that is, it doesn't depend on which point v in space you select).
Of course, if you are writing anything serious, you will probably use your own vector algebra library.
Also, h = 0.001 might not be a great value in your case.