I am new to matlab and am trying to loop through an equation from 1 to 1000 and plot the result, but every value is turning out the same and the plot is incrorrect.
Here is my current attempt, but the variable Rm is giving only one results (instead of 1000 separate ones):
Ds = 1.04e-10;
Gamma = 1.9;
Omega = 1.09e-29;
Deltas = 2.5e-10;
Boltzmann = 1.3806e-23;
T = 1123.15;
Beta = 0.83;
Zo = 6.7;
EtaNi = 0.39;
EtaYSZ = 0.61;
Rm0 = 0.29;
RmYSZ0 = 0.265;
lambda = 4.70e-3;
C = Ds * ((Gamma * Omega * Deltas) / (2 * Boltzmann * T)) * (Beta / ((1 - Beta^2) * ((1 + Beta^2)^0.5) * ((1 + Beta)^3))) * Zo * (EtaNi/((EtaNi/Rm0) + (EtaYSZ/RmYSZ0)));
tinit=1; % Initial time value (h)
tend=1000; % End time value (h)
tinc=1; % Increment in time value (h)
t= [tinit:tinc:tend]; % Time vector
n = 1000;
for i=1:n
Rm(i) = (((5*C)/lambda) * (1 - exp(-lambda*i)) + (Rm0)^5)^(1/5);
end
plot(t,Rm);
Expected result is an exponential curve, any help would be appreciated
Your term before R0 is an exponential that ranges from 0 to 4.5e-27. R0^5 is 0.0021. Floating point precision is not enough to preserve the first term when it is added to the second. So (5C/L*(...) + Rm0^5) == Rm0^5, so it is constant.
Related
I'm curious to know if there is a better way to express the following loop:
To = 1;
fileName = "fourier/signal.txt";
spectrum_left(abs(spectrum_left) < 1e-3) = 0+0i;
Ts = 1 / Fs;
t = 0:Ts:To-Ts;
signal = load(fileName, "-ascii");
Ns = numel(signal);
Fs = Ns / To;
fr = f(abs(spectrum_left) > 0)
a = abs(spectrum_left(abs(spectrum_left) > 0))
p = angle(spectrum_left(abs(spectrum_left) > 0))
signal_synth = 0;
len = length(a);
for i=1:len
a_i = a(i);
p_i = p(i);
f_i = fr(i);
s_i = a_i * cos(2*pi*f_i*t + p_i)
signal_synth = signal_synth + s_i;
end
Any suggestions are greatly appreciated. And Happy New Year!
In MATLAB, element-wise operators, that preceded by the period character (.), can be applied on arrays. I used (:) operator to convert the vectors to column vectors. Note that t is a row vector, when it is combined with column vectors the result will be a matrix that is formed by implicit expansion. In order to sum up the values the function sum is used that by default sums along the first dimension of the matrix and the result will be a row vector with the same size as t.
signal_synth = sum(a(:) .* cos(2 * pi * f(:) .* t + p(:)));
A more compact form using matrix multiplication:
signal_synth = a(:).' * cos(2 * pi * f(:) .* t + p(:));
For the purpose of generalization, I hope Matlab can automatically compute the 1st & 2nd derivatives of the associated function f(x). (in case I change f(x) = sin(6x) to f(x) = sin(8x))
I know there exists built-in commands called diff() and syms, but I cannot figure out how to deal with them with the index i in the for-loop. This is the key problem I am struggling with.
How do I make changes to the following set of codes? I am using MATLAB R2019b.
n = 10;
h = (2.0 * pi) / (n - 1);
for i = 1 : n
x(i) = 0.0 + (i - 1) * h;
f(i) = sin(6 * x(i));
dfe(i) = 6 * cos(6 * x(i)); % first derivative
ddfe(i) = -36 * sin(6 * x(i)); % second derivative
end
You can simply use subs and double to do that. For your case:
% x is given here
n = 10;
h = (2.0 * pi) / (n - 1);
syms 'y';
g = sin(6 * y);
for i = 1 : n
x(i) = 0.0 + (i - 1) * h;
f(i) = double(subs(g,y,x(i)));
dfe(i) = double(subs(diff(g),y,x(i))); % first derivative
ddfe(i) = double(subs(diff(g,2),y,x(i))); % second derivative
end
By #Daivd comment, you can vectorize the loop as well:
% x is given here
n = 10;
h = (2.0 * pi) / (n - 1);
syms 'y';
g = sin(6 * y);
x = 0.0 + ((1:n) - 1) * h;
f = double(subs(g,y,x));
dfe = double(subs(diff(g),y,x)); % first derivative
ddfe = double(subs(diff(g,2),y,x)); % second derivative
Using Matlab I am trying to construct a neural network that can classify handwritten digits that are 30x30 pixels. I use backpropagation to find the correct weights and biases. The network starts with 900 inputs, then has 2 hidden layers with 16 neurons and it ends with 10 outputs. Each output neuron has a value between 0 and 1 that represents the belief that the input should be classified as a certain digit. The problem is that after training, the output becomes almost indifferent to the input and it goes towards a uniform belief of 0.1 for each output.
My approach is to take each image with 30x30 pixels and reshape it to be vectors of 900x1 (note that 'Images_vector' is already in the vector format when it is loaded). The weights and biases are initiated with random values between 0 and 1. I am using stochastic gradiënt descent to update the weights and biases with 10 randomly selected samples per batch. The equations are as described by Nielsen.
The script is as follows.
%% Inputs
numberofbatches = 1000;
batchsize = 10;
alpha = 1;
cutoff = 8000;
layers = [900 16 16 10];
%% Initialization
rng(0);
load('Images_vector')
Images_vector = reshape(Images_vector', 1, 10000);
labels = [ones(1,1000) 2*ones(1,1000) 3*ones(1,1000) 4*ones(1,1000) 5*ones(1,1000) 6*ones(1,1000) 7*ones(1,1000) 8*ones(1,1000) 9*ones(1,1000) 10*ones(1,1000)];
newOrder = randperm(10000);
Images_vector = Images_vector(newOrder);
labels = labels(newOrder);
images_training = Images_vector(1:cutoff);
images_testing = Images_vector(cutoff + 1:10000);
w = cell(1,length(layers) - 1);
b = cell(1,length(layers));
dCdw = cell(1,length(layers) - 1);
dCdb = cell(1,length(layers));
for i = 1:length(layers) - 1
w{i} = rand(layers(i+1),layers(i));
b{i+1} = rand(layers(i+1),1);
end
%% Learning process
batches = randi([1 cutoff - batchsize],1,numberofbatches);
cost = zeros(numberofbatches,1);
c = 1;
for batch = batches
for i = 1:length(layers) - 1
dCdw{i} = zeros(layers(i+1),layers(i));
dCdb{i+1} = zeros(layers(i+1),1);
end
for n = batch:batch+batchsize
y = zeros(10,1);
disp(labels(n))
y(labels(n)) = 1;
% Network
a{1} = images_training{n};
z{2} = w{1} * a{1} + b{2};
a{2} = sigmoid(0, z{2});
z{3} = w{2} * a{2} + b{3};
a{3} = sigmoid(0, z{3});
z{4} = w{3} * a{3} + b{4};
a{4} = sigmoid(0, z{4});
% Cost
cost(c) = sum((a{4} - y).^2) / 2;
% Gradient
d{4} = (a{4} - y) .* sigmoid(1, z{4});
d{3} = (w{3}' * d{4}) .* sigmoid(1, z{3});
d{2} = (w{2}' * d{3}) .* sigmoid(1, z{2});
dCdb{4} = dCdb{4} + d{4} / 10;
dCdb{3} = dCdb{3} + d{3} / 10;
dCdb{2} = dCdb{2} + d{2} / 10;
dCdw{3} = dCdw{3} + (a{3} * d{4}')' / 10;
dCdw{2} = dCdw{2} + (a{2} * d{3}')' / 10;
dCdw{1} = dCdw{1} + (a{1} * d{2}')' / 10;
c = c + 1;
end
% Adjustment
b{4} = b{4} - dCdb{4} * alpha;
b{3} = b{3} - dCdb{3} * alpha;
b{2} = b{2} - dCdb{2} * alpha;
w{3} = w{3} - dCdw{3} * alpha;
w{2} = w{2} - dCdw{2} * alpha;
w{1} = w{1} - dCdw{1} * alpha;
end
figure
plot(cost)
ylabel 'Cost'
xlabel 'Batches trained on'
With the sigmoid function being the following.
function y = sigmoid(derivative, x)
if derivative == 0
y = 1 ./ (1 + exp(-x));
else
y = sigmoid(0, x) .* (1 - sigmoid(0, x));
end
end
Other than this I have also tried to have 1 of each digit in each batch, but this gave the same result. Also I have tried varying the batch size, the number of batches and alpha, but with no success.
Does anyone know what I am doing wrong?
Correct me if I'm wrong: You have 10000 samples in you're data, which you divide into 1000 batches of 10 samples. Your training process consists of running over these 10000 samples once.
This might be too little, normally your training process consists of several epochs (one epoch = iterating over every sample once). You can try going over your batches multiple times.
Also for 900 inputs your network seems small. Try it with more neurons in the second layer. Hope it helps!
I have a problem in which I have to implement the following question in Matlab.
i(t) = A2 * sin(wr*t) * exp(-alpha*t); for t [0, 10] with step 0.5s
My approach is as follows
clc;
clear;
% Given Data
Vs = 220;
L = 5e-3;
C = 10e-6;
R = 22;
Vo = 50;
% a)
alpha = R / (2 * L);
omega_not = 1 / sqrt(L*C);
omega_r = sqrt( omega_not^2 - alpha^2 );
A2 = Vs / (omega_r * L);
t = 1:0.5:10;
i = A2 * sin( omega_r * t ) .* exp(-alpha * t);
% b)
t1 = pi / omega_r;
% c)
plot(t, i);
But it yields all the values of current equal to zero. Please help me solve the problem.
I think the problem is this part of the expression:
exp(-alpha * t)
When I run your code, -alpha equals -2200. The exponential for such a large negative number is so small that the code returns zero.
>> exp(-200)
ans =
1.3839e-87
>> exp(-1000)
ans =
0
I'm wondering if there is a way of finding the intersection point between a line and a circle written in polar coordinates.
% Line
x_line = 10 + r * cos(th);
y_line = 10 + r * sin(th);
%Circle
circle_x = circle_r * cos(alpha);
circle_y = circle_r * sin(alpha);
So far I've tried using the intersect(y_line, circle_y) function without any success. I'm relatively new to MATLAB so bear with me.
I have generalised the below so that other values than a=10 can be used...
a = 10; % constant line offset
th = 0; % constant angle of line
% rl = ?? - variable to find
% Coordinates of line:
% [xl, yl] = [ a + rl * cos(th), a + rl * sin(th) ];
rc = 1; % constant radius of circle
% alpha = ?? - variable to find
% Coordinates of circle:
% [xc, yc] = [ rc * cos(alpha), rc * sin(alpha) ];
We want the intersection, so xl = xc, yl = yc
% a + rl * cos(th) = rc * cos(alpha)
% a + rl * sin(th) = rc * sin(alpha)
Square both sides of both equations and sum them. Simplifying sin(a)^2 + cos(a)^2 = 1. Expanding brackets and simplifying further gives
% rl^2 + 2 * a * rl * (cos(th) + sin(th)) + 2 * a - rc^2 = 0
Now you can use the quadratic formula to get the value of rl.
Test discriminant:
dsc = (2 * a * (cos(th) + sin(th)) )^2 - 4 * (2 * a - rc^2);
rl = [];
if dsc < 0
% no intersection
elseif dsc == 0
% one intersection at
rl = - cos(th) - sin(th);
else
% two intersection points
rl = -cos(th) - sin(th) + [ sqrt(dsc)/2, -sqrt(dsc)/2];
end
% Get alpha from an earlier equation
alpha = acos( ( a + rl .* cos(th) ) ./ rc );
Now you have 0, 1 or 2 points of intersection of the line with the circle, from certain known and unknown values about each line. Essentially this is just simultaneous equations, see the start of this article for a basis of the maths
https://en.wikipedia.org/wiki/System_of_linear_equations
Do you need to do it numerically? This problem would have an easy analytical solution: The point (10 + r*cos(th),10 + r*sin(th)) is on a circle with radius R iff
(10+r*cos(th))^2 + (10+r*sin(th))^2 == R^2
<=> 200+r^2 + 2*r*(cos(th)+sin(th)) == R^2
<=> r^2 + 2*r*sqrt(2)*sin(th+pi/4) + 200 - R^2 = 0
which is a quadratic equation in r. If the discriminant is positive, there are two solutions (corresponding to two intersection points), otherwise, there are none.
If you work out the math, the condition for intersection is 100*(sin(2*th)-1)+circle_r^2 >= 0 and the roots are -10*sqrt(2)*sin(th+pi/4)*[1,1] + sqrt(100*(sin(2*th)-1)+circle_r^2)*[1,-1].
Here is a Matlab plot as an example for th = pi/3 and circle_r = 15. The magenta markers are calculated in closed-form using the equation shown above.