The dependency graph in MATLAB - function

I've got the following function:
phi0 = [0; 0]; %initial values
[T,PHI] = ode45(#eqn,[0, 10], phi0,odeset('RelTol',2e-13,'AbsTol',1e-100));
plot(T, PHI(:,1),'-b',T, PHI(:,2),'-g');
title('\it d = 0.1')
w1 = PHI(end,1)/(10000*2*pi) %the limit frequency for phi1
w2 = PHI(end,2)/(10000*2*pi) %the limit frequency for phi2
delta_w = w2 - w1
phi1_at_t_10k = PHI(end,1) %the value phi1(t=10000)
phi2_at_t_10k = PHI(end,2)
function dy_dt = eqn(t,phi)
d = 0.1; %synchronization parameter
n = 3;
g = [ 1.01; 1.02];
f = g-sin(phi/n);
exch = [d;-d]*sin(phi(2)-phi(1));
dy_dt = f+exch;
end
The w is calculated by the formula: w_i = (1/2pi)(lim((phi(t)-phi(0))/t) where t->infinity (here it's equal to 10000).
The question is how to plot the dependence of delta_w on different values of d (from d=0 to d=5 with step = 0.1)?

To collect summarize my comments:
First make the parameter d explicit in the ODE function
function dy_dt = eqn(t,phi,d)
n = 3;
g = [ 1.01; 1.02];
f = g-sin(phi/n);
exch = [d;-d]*sin(phi(2)-phi(1));
dy_dt = f+exch;
end
Then put the ODE integration and evaluation of the result in its own procedure
function delta_w = f(d)
phi0 = [0; 0]; %initial values
opts = odeset('RelTol',2e-13,'AbsTol',1e-100);
[T,PHI] = ode45(#(t,y)eqn(t,y,d), [0, 10], phi0, opts);
w1 = PHI(end,1)/(10000*2*pi); %the limit frequency for phi1
w2 = PHI(end,2)/(10000*2*pi); %the limit frequency for phi2
delta_w = w2 - w1;
end
And finally evaluate for the list of d values under consideration
d = [0:0.1:5];
delta_w = arrayfun(#(x)f(x),d);
plot(d,delta_w);
This should give a result. If it is not the expected one, further research into assumptions, equations and code is necessary.

Related

Simulating waves in disc or circle form

I am getting an error when I run this code while selecting disc view or circle view option for wave simulation. The code and error are attached. I think there is some problem in this part of code typically in fzero function. Any help would be great.
Code:
function z = bjzeros(n,k)
% BJZEROS Zeros of the Bessel function.
% z = bjzeros(n,k) is the first k zeros of besselj(n,x)
% delta must be chosen so that the linear search can take
% steps as large as possible
delta = .99*pi;
Jsubn = inline('besselj(n,x)''x','n');
a = n+1;
fa = besselj(n,a);
z = zeros(1,k);
j = 0;
while j < k
b = a + delta;
fb = besselj(n,b);
if sign(fb) ~= sign(fa)
j = j+1;
z(j) = fzerotx(Jsubn,[a b],n);
end
a = b;
fa = fb;
end
Error:
Undefined function 'fzerotx' for input arguments of type 'inline'.
Error in waves>bjzeros (line 292)
z(j) = fzerotx(Jsubn,[a b],n);
Error in waves (line 137)
mu = [bjzeros(0,2) bjzeros(1,2)];
Function Declarations and Syntax
The fzerotx() function may not be declared. You can follow the file structure below to create the required M-files/functions in the same directory. Another small error may be caused by a missing comma, I got rid of the error by changing the line:
Jsubn = inline('besselj(n,x)''x','n');
to
Jsubn = inline('besselj(n,x)','x','n');
File 1: Main File/Function Call → [main.m]
mu = [bjzeros(0,2) bjzeros(1,2)];
File 2: bjzeros() Function → [bjzeros.m]
function z = bjzeros(n,k)
% BJZEROS Zeros of the Bessel function.
% z = bjzeros(n,k) is the first k zeros of besselj(n,x)
% delta must be chosen so that the linear search can take
% steps as large as possible
delta = .99*pi;
Jsubn = inline('besselj(n,x)','x','n');
a = n+1;
fa = besselj(n,a);
z = zeros(1,k);
j = 0;
while j < k
b = a + delta;
fb = besselj(n,b);
if sign(fb) ~= sign(fa)
j = j+1;
z(j) = fzerotx(Jsubn,[a b],n);
end
a = b;
fa = fb;
end
end
File 3: fzerotx() Function → [fzerotx.m]
Function Reference: MATLAB: Textbook version of FZERO
function b = fzerotx(F,ab,varargin)
%FZEROTX Textbook version of FZERO.
% x = fzerotx(F,[a,b]) tries to find a zero of F(x) between a and b.
% F(a) and F(b) must have opposite signs. fzerotx returns one
% end point of a small subinterval of [a,b] where F changes sign.
% Arguments beyond the first two, fzerotx(F,[a,b],p1,p2,...),
% are passed on, F(x,p1,p2,..).
%
% Examples:
% fzerotx(#sin,[1,4])
% F = #(x) sin(x); fzerotx(F,[1,4])
% Copyright 2014 Cleve Moler
% Copyright 2014 The MathWorks, Inc.
% Initialize.
a = ab(1);
b = ab(2);
fa = F(a,varargin{:});
fb = F(b,varargin{:});
if sign(fa) == sign(fb)
error('Function must change sign on the interval')
end
c = a;
fc = fa;
d = b - c;
e = d;
% Main loop, exit from middle of the loop
while fb ~= 0
% The three current points, a, b, and c, satisfy:
% f(x) changes sign between a and b.
% abs(f(b)) <= abs(f(a)).
% c = previous b, so c might = a.
% The next point is chosen from
% Bisection point, (a+b)/2.
% Secant point determined by b and c.
% Inverse quadratic interpolation point determined
% by a, b, and c if they are distinct.
if sign(fa) == sign(fb)
a = c; fa = fc;
d = b - c; e = d;
end
if abs(fa) < abs(fb)
c = b; b = a; a = c;
fc = fb; fb = fa; fa = fc;
end
% Convergence test and possible exit
m = 0.5*(a - b);
tol = 2.0*eps*max(abs(b),1.0);
if (abs(m) <= tol) | (fb == 0.0)
break
end
% Choose bisection or interpolation
if (abs(e) < tol) | (abs(fc) <= abs(fb))
% Bisection
d = m;
e = m;
else
% Interpolation
s = fb/fc;
if (a == c)
% Linear interpolation (secant)
p = 2.0*m*s;
q = 1.0 - s;
else
% Inverse quadratic interpolation
q = fc/fa;
r = fb/fa;
p = s*(2.0*m*q*(q - r) - (b - c)*(r - 1.0));
q = (q - 1.0)*(r - 1.0)*(s - 1.0);
end;
if p > 0, q = -q; else p = -p; end;
% Is interpolated point acceptable
if (2.0*p < 3.0*m*q - abs(tol*q)) & (p < abs(0.5*e*q))
e = d;
d = p/q;
else
d = m;
e = m;
end;
end
% Next point
c = b;
fc = fb;
if abs(d) > tol
b = b + d;
else
b = b - sign(b-a)*tol;
end
fb = F(b,varargin{:});
end
Ran using MATLAB R2019b

Scilab plot shows incorrect ode values

I'm trying to create a model of an asynchronous electrical motor in scilab, and display graphs of how the rpm, currents and torque change over time. It looks quite long but you don't need to read it all.
fHz = 50;
Um = 230;
p = 3;
we = 2*%pi*fHz/p;
wb = 2*%pi*50;
Rs = 0.435;
Rr = 0.64;
Ls = 0.0477;
Xls = wb*Ls; // [Ohm]
Lr = 0.0577;
Xlr = wb*Lr; // [Ohm]
Lm = 0.012;
Xm = wb*Lm; // [Ohm]
Xml = 1/(1/Xls + 1/Xm + 1/Xlr) // [Ohm];
D = 0.0002;
J = 0.28;
Mt = 0.0;
function [xdot]=AszinkronGep(t, x, Um, fHz)
xdot = zeros(12, 1);
Fsq = x(1);
Fsd = x(2);
Frq = x(3);
Frd = x(4);
wr = x(5);
isabc(1) = x(6);
isabc(2) = x(7);
isabc(3) = x(8);
irabc(1) = x(9);
irabc(2) = x(10);
irabc(3) = x(11);
Ua = Um*sin(2*%pi*fHz*t);
Ub = Um*sin(2*%pi*fHz*t - 2*%pi/3);
Uc = Um*sin(2*%pi*fHz*t + 2*%pi/3);
Uab = 2/3*[1, -0.5, -0.5; 0, sqrt(3)/2, -sqrt(3)/2]*[Ua;Ub;Uc];
phi = 2*%pi*fHz*t;
Udq = [cos(phi), sin(phi); -sin(phi), cos(phi)]*Uab;
Usd = Udq(1);
Usq = Udq(2);
Urd = 0;
Urq = 0;
isd = ( Fsd-Xml*(Fsd/Xls + Frd/Xlr) )/Xls;
isq = ( Fsq-Xml*(Fsq/Xls + Frq/Xlr) )/Xls;
ird = ( Frd-Xml*(Fsd/Xls + Frd/Xlr) )/Xlr;
irq = ( Frq-Xml*(Fsq/Xls + Frq/Xlr) )/Xlr;
isdq = [isd; isq];
isalphabeta = [cos(phi), -sin(phi); sin(phi), cos(phi)]*isdq;
isabc = [1, 0; -0.5, sqrt(3)/2; -0.5, -sqrt(3)/2]*isalphabeta;
irdq = [ird; irq];
iralphabeta = [cos(phi), -sin(phi); sin(phi), cos(phi)]*irdq;
irabc = [1, 0; -0.5, sqrt(3)/2; -0.5, -sqrt(3)/2]*iralphabeta;
//TORQUE
Me = (3/2)*p*(Fsd*isq - Fsq*isd)/wb
Fmq = Xml*( Fsq/Xls + Frq /Xlr );
Fmd = Xml*( Fsd/Xls + Frd /Xlr );
//Differential equations
xdot(1) = wb*( Usq - we/wb*Fsd + Rs/Xls*(Fmq - Fsq) );
xdot(2) = wb*( Usd + we/wb*Fsq + Rs/Xls*(Fmd - Fsd) );
xdot(3) = wb*( Urq - (we - wr)/wb*Frd + Rr/Xlr *(Fmq - Frq) );
xdot(4) = wb*( Urd + (we - wr)/wb*Frq + Rr/Xlr *(Fmd - Frd ) );
xdot(5) = p*(Me - D*wr - Mt)/J;
xdot(6) = isabc(1);
xdot(7) = isabc(2);
xdot(8) = isabc(3);
xdot(9) = irabc(1);
xdot(10) = irabc(2);
xdot(11) = irabc(3);
xdot(12) = Me;
if t <= 5 then
disp(Me);
end
endfunction
//Simulation parameter
t = 0:0.001:5;
t0 = 0;
//Starting parameters
y0 = [0;0;0;0;0;0;0;0;0;0;0;0]
y = ode(y0,t0,t,list(AszinkronGep,Um,fHz));
//Graphs
figure(1)
plot(t,y(5,:), "linewidth", 3);
xlabel("time [s]", "fontsize", 3, "color", "blue");
ylabel("rpm [rpm]", "fontsize", 3, "color", "blue");
figure(4)
plot(t,y(12,:), "linewidth", 3);
xlabel("time [s]", "fontsize", 3, "color", "blue");
ylabel("torque [Nm]", "fontsize", 3, "color", "blue");
I want a graph that shows 'Me' as a function of time. So I write: xdot(12) = Me, then plot that, but it doesn't looks like how it should at all. Just to check, I added 'disp(Me)' at the end of the function, to see if the calculations are correct at all. And yes, those are the right values. Why does it give me different values when I plot it?
As noted in the comments, y(12) is the integral of Me over t.
If you want Me, you just need to differentiate it:
//after you run ode() and have y values
h = t(2) - t(1);
Me = diff(y(12,:)) ./ h;
//plotting
scf(); clf();
subplot(2,1,1);
xtitle("y(12,:)");
plot2d(t,y(12,:));
subplot(2,1,2);
xtitle("Me");
plot2d(t(1:$-1),Me);
Here is the output:

Using a metric predictor when modelling ordinal predicted variable in PyMC3

I am trying to implement the ordered probit regression model from chapter 23.4 of Doing Bayesian Data Analysis (Kruschke) in PyMC3. After sampling, the posterior distribution for the intercept and slope are not really comparable to the results from the book. I think there is some fundamental issue with the model definition, but I fail to see it.
Data:
X is the metric predictor (standardized to zX), Y are ordinal outcomes (1-7).
nYlevels3 = df3.Y.nunique()
# Setting the thresholds for the ordinal outcomes. The outer sides are
# fixed, while the others are estimated.
thresh3 = [k + .5 for k in range(1, nYlevels3)]
thresh_obs3 = np.ma.asarray(thresh3)
thresh_obs3[1:-1] = np.ma.masked
#as_op(itypes=[tt.dvector, tt.dvector, tt.dscalar], otypes=[tt.dmatrix])
def outcome_probabilities(theta, mu, sigma):
out = np.empty((mu.size, nYlevels3))
n = norm(loc=mu, scale=sigma)
out[:,0] = n.cdf(theta[0])
out[:,1] = np.max([np.repeat(0,mu.size), n.cdf(theta[1]) - n.cdf(theta[0])])
out[:,2] = np.max([np.repeat(0,mu.size), n.cdf(theta[2]) - n.cdf(theta[1])])
out[:,3] = np.max([np.repeat(0,mu.size), n.cdf(theta[3]) - n.cdf(theta[2])])
out[:,4] = np.max([np.repeat(0,mu.size), n.cdf(theta[4]) - n.cdf(theta[3])])
out[:,5] = np.max([np.repeat(0,mu.size), n.cdf(theta[5]) - n.cdf(theta[4])])
out[:,6] = 1 - n.cdf(theta[5])
return out
with pm.Model() as ordinal_model_metric:
theta = pm.Normal('theta', mu=thresh3, tau=np.repeat(1/2**2, len(thresh3)),
shape=len(thresh3), observed=thresh_obs3, testval=thresh3[1:-1])
# Intercept
zbeta0 = pm.Normal('zbeta0', mu=(1+nYlevels3)/2, tau=1/nYlevels3**2)
# Slope
zbeta = pm.Normal('zbeta', mu=0.0, tau=1/nYlevels3**2)
# Mean of the underlying metric distribution
mu = pm.Deterministic('mu', zbeta0 + zbeta*zX)
zsigma = pm.Uniform('zsigma', nYlevels3/1000.0, nYlevels3*10.0)
pr = outcome_probabilities(theta, mu, zsigma)
y = pm.Categorical('y', pr, observed=df3.Y.cat.codes)
http://nbviewer.jupyter.org/github/JWarmenhoven/DBDA-python/blob/master/Notebooks/Chapter%2023.ipynb
For reference, here is the JAGS model used by Kruschke on which I based my model:
Ntotal = length(y)
# Threshold 1 and nYlevels-1 are fixed; other thresholds are estimated.
# This allows all parameters to be interpretable on the response scale.
nYlevels = max(y)
thresh = rep(NA,nYlevels-1)
thresh[1] = 1 + 0.5
thresh[nYlevels-1] = nYlevels-1 + 0.5
modelString = "
model {
for ( i in 1:Ntotal ) {
y[i] ~ dcat( pr[i,1:nYlevels] )
pr[i,1] <- pnorm( thresh[1] , mu[x[i]] , 1/sigma[x[i]]^2 )
for ( k in 2:(nYlevels-1) ) {
pr[i,k] <- max( 0 , pnorm( thresh[ k ] , mu[x[i]] , 1/sigma[x[i]]^2 )
- pnorm( thresh[k-1] , mu[x[i]] , 1/sigma[x[i]]^2 ) )
}
pr[i,nYlevels] <- 1 - pnorm( thresh[nYlevels-1] , mu[x[i]] , 1/sigma[x[i]]^2 )
}
for ( j in 1:2 ) { # 2 groups
mu[j] ~ dnorm( (1+nYlevels)/2 , 1/(nYlevels)^2 )
sigma[j] ~ dunif( nYlevels/1000 , nYlevels*10 )
}
for ( k in 2:(nYlevels-2) ) { # 1 and nYlevels-1 are fixed, not stochastic
thresh[k] ~ dnorm( k+0.5 , 1/2^2 )
}
}
It was not a fundamental issue after all: I forgot to indicate the axis for np.max() in the function below.
#as_op(itypes=[tt.dvector, tt.dvector, tt.dscalar], otypes=[tt.dmatrix])
def outcome_probabilities(theta, mu, sigma):
out = np.empty((mu.size, nYlevels3))
n = norm(loc=mu, scale=sigma)
out[:,0] = n.cdf(theta[0])
out[:,1] = np.max([np.repeat(0,mu.size), n.cdf(theta[1]) - n.cdf(theta[0])], axis=0)
out[:,2] = np.max([np.repeat(0,mu.size), n.cdf(theta[2]) - n.cdf(theta[1])], axis=0)
out[:,3] = np.max([np.repeat(0,mu.size), n.cdf(theta[3]) - n.cdf(theta[2])], axis=0)
out[:,4] = np.max([np.repeat(0,mu.size), n.cdf(theta[4]) - n.cdf(theta[3])], axis=0)
out[:,5] = np.max([np.repeat(0,mu.size), n.cdf(theta[5]) - n.cdf(theta[4])], axis=0)
out[:,6] = 1 - n.cdf(theta[5])
return out

Why is this Elgamal Digital Signature Algorithm implemented in MATLAB not working properly?

%%%CODE:
clc;
disp('Implementation of ELGAMAL Digital Signature');
clear all;
close all;
%%Hardcoded values (vpi stands for VariablePrecisionIntegers, and are used to store large values)
g = vpi(5) %alpha in example
k = vpi(9) %Random number 1<k<p-1 and gcd(k, p − 1) = 1.
p = vpi(23) %Prime Number
x = vpi(3) %Secret Key 1 < x < p − 1
m = vpi(7) %Message
y = vpi(2)
r = vpi(2)
s = vpi(2)
%%Key Generation
y = powermod(g,x,p) %y = g^x mod p
%%Signature Generation
r = powermod(g,k,p) %r = g^k mod p
multinver = mulinv(9,23) %Generates multiplicative inverse k^-1 mod p
s = mod(((multinver)*(m-x*r)),p-1) %s = (k^-1)*(m-x*r) mod p-1
%%Verification
zvg = vpi(2);
zvg = powermod (g,m,p) %zvg = g^m mod p
zvyr = vpi(2);
zvyr = mod(((y^r)*(r^s)),p) %zvyr = y^r * r^s mod p
Output:
Implementation of ELGAMAL Digital Signature
g = 5
k = 9
p = 23
x = 3
m = 7
y = 2
r = 2
s = 2
y = 10
r = 11
multinver = 18
s = 16
zvg = 17
zvyr = 5
zvg should be equal to zvyr, but it's coming out wrong.
You are computing multinver wrong
It should be
multinver = mulinv(k,p-1) %Generates multiplicative inverse k^-1 mod (p-1)
Notice that we are computing the inverse mod (p-1) not mod (p).

Fourier coefficients using matlab numerical integration

I have been trying to display the an and bn fourier coefficients in matlab but no success, I was able to display the a0 because that is not part of the iteration.
I will highly appreciate your help, below is my code
syms an;
syms n;
syms t;
y = sym(0);
L = 0.0005;
inc = 0.00001; % equally sample space of 100 points
an = int(3*t^2*cos(n*pi*t/L),t,-L,L)*(1/L);
bn = int(3*t^2*sin(n*pi*t/L),t,-L,L)*(1/L);
a0 = int(3*t^2,t,-L,L)*(1/L);
a0 = .5 *a0;
a0=a0
for i=1:5
y = subs(an, n, i)*cos(i*pi*t/0.0005)
z = subs(bn, n, i)*sin(i*pi*t/0.0005)
end
If everything you stated within your question is correct, I would tend to solve it this way:
clc, clear all,close all
L = 0.0005;
n = 5;
an = zeros(1,n);
bn = zeros(1,n);
for i = 1:5
f1 = #(t) 3.*(t.^2).*cos(i.*pi.*t./L);
f2 = #(t) 3.*(t.^2).*sin(i.*pi.*t./L);
an(i) = quad(f1,-L,L).*(1./L);
bn(i) = quad(f2,-L,L).*(1./L);
a0 = .5.*quad(#(t) 3.*t.^2,-L,L).*(1./L);
end
I hope this helps.