I am doing two updates for h and x given in this [paper] http://paris.cs.illinois.edu/pubs/nasser-icassp2015.pdf (You don't have to read the paper, just look for the equations 4,10 and 14 for updating h and x given on page 2 and 3).
This is the code snippet that I have tried so far. Can you tell me if its correct? Also, is there any way to optimize these for loops?
In some cases (t-tau) term was negative and MATLAB was giving an error. So, I put a condition that only implement if (t-tau)>0. Doing this is correct or is there any other way to take care of the negative indices?
%updates
Lh=10;
lambda = 0.1*(sum(reverberatedspeechspec(:))/(size(Y,1)*size(Y,2)));
S=reverberatedspeechspec;
W=basis_mel_act; W=gather(W); W = double(W); %W is the dictionary
%---initialization for H(RIR)----
H=rand(size(Y,1),Lh);
nmfIter = 50;
%---initialization for X(Activations)-----
W_trans=W';
X=W_trans*S; X=double(X);
Y = zeros(size(S,1));
for idx = 1 : size(Y,1)
Y(idx,:) = filter(S(idx,:),1,H(idx,:));
end
Stilde = zeros(size(S));
Ytilde = zeros(size(S));
for iter=1:nmfIter
% update for H
Stilde = W*X;
Ytilde = zeros(size(S));
for j=1:size(Stilde,1)
Ytilde (j,:) = filter(Stilde(j,:),1,H(j,:));
end
ratio = Y./Ytilde;
numerator = zeros(size(H));
denominator = numerator;
for k = 1 :size(Y,1)
for tau = 1:Lh
for t= 1:size(Y,2)
if gt (t-tau , 0)
numerator (k,tau) = numerator(k,tau) + ratio(k,t) * Stilde(k,t-tau);
denominator(k,tau) = denominator (k,tau) + Stilde (k,t-tau);
end
end
end
end
H = H .* numerator ./denominator ;
%updating Ytilde after getting a new value for H
for j=1:size(Stilde,1)
Ytilde (j,:) = filter(Stilde(j,:),1,H(j,:));
end
%update for X
ratio = Y./Ytilde;
ratio = [ratio zeros(size(Y,1),Lh)]; %zero padding Y and Ytilde for (t+tau) term in update of X.
Product = H.' * W; % Product of H_transpose and W in th update which is equivalent to the term ∑H(k,tau)W(k,r)
numerator = zeros(size(H));
denominator = numerator;
for r = 1:size(W,2)
for t = 1:size(Y,2)
for k = 1:size(Y,1)
for tau = 1:Lh
numerator(r,t) = numerator(r,t) + ratio(k,t+tau) * Product(tau,r);
denominator(r,t) = denominator(r,t) + Product(tau,r) + lambda;
end
end
end
end
X = X .* numerator ./denominator;
end
Related
I want to determine the Steepest descent of the Rosenbruck function using Armijo steplength where x = [-1.2, 1]' (the initial column vector).
The problem is, that the code has been running for a long time. I think there will be an infinite loop created here. But I could not understand where the problem was.
Could anyone help me?
n=input('enter the number of variables n ');
% Armijo stepsize rule parameters
x = [-1.2 1]';
s = 10;
m = 0;
sigma = .1;
beta = .5;
obj=func(x);
g=grad(x);
k_max = 10^5;
k=0; % k = # iterations
nf=1; % nf = # function eval.
x_new = zeros([],1) ; % empty vector which can be filled if length is not known ;
[X,Y]=meshgrid(-2:0.5:2);
fx = 100*(X.^2 - Y).^2 + (X-1).^2;
contour(X, Y, fx, 20)
while (norm(g)>10^(-3)) && (k<k_max)
d = -g./abs(g); % steepest descent direction
s = 1;
newobj = func(x + beta.^m*s*d);
m = m+1;
if obj > newobj - (sigma*beta.^m*s*g'*d)
t = beta^m *s;
x = x + t*d;
m_new = m;
newobj = func(x + t*d);
nf = nf+1;
else
m = m+1;
end
obj=newobj;
g=grad(x);
k = k + 1;
x_new = [x_new, x];
end
% Output x and k
x_new, k, nf
fprintf('Optimal Solution x = [%f, %f]\n', x(1), x(2))
plot(x_new)
function y = func(x)
y = 100*(x(1)^2 - x(2))^2 + (x(1)-1)^2;
end
function y = grad(x)
y(1) = 100*(2*(x(1)^2-x(2))*2*x(1)) + 2*(x(1)-1);
end
I'm having serious problems in a simple example of fem assembly.
I just want to assemble the Mass matrix without any coefficient. The geometry is simple:
conn=[1, 2, 3];
p = [0 0; 1 0; 0 1];
I made it like this so that the physical element will be equal to the reference one.
my basis functions:
phi_1 = #(eta) 1 - eta(1) - eta(2);
phi_2 = #(eta) eta(1);
phi_3 = #(eta) eta(2);
phi = {phi_1, phi_2, phi_3};
Jacobian matrix:
J = #(x,y) [x(2) - x(1), x(3) - x(1);
y(2) - y(1), y(3) - y(1)];
The rest of the code:
M = zeros(np,np);
for K = 1:size(conn,1)
l2g = conn(K,:); %local to global mapping
x = p(l2g,1); %node x-coordinate
y = p(l2g,2); %node y-coordinate
jac = J(x,y);
loc_M = localM(jac, phi);
M(l2g, l2g) = M(l2g, l2g) + loc_M; %add element masses to M
end
localM:
function loc_M = localM(J,phi)
d_J = det(J);
loc_M = zeros(3,3);
for i = 1:3
for j = 1:3
loc_M(i,j) = d_J * quadrature(phi{i}, phi{j});
end
end
end
quadrature:
function value = quadrature(phi_i, phi_j)
p = [1/3, 1/3;
0.6, 0.2;
0.2, 0.6;
0.2, 0.2];
w = [-27/96, 25/96, 25/96, 25/96];
res = 0;
for i = 1:size(p,1)
res = res + phi_i(p(i,:)) * phi_j(p(i,:)) * w(i);
end
value = res;
end
For the simple entry (1,1) I obtain 0.833, while computing the integral by hand or on wolfram alpha I get 0.166 (2 times the result of the quadrature).
I tried with different points and weights for quadrature, but really I do not know what I am doing wrong.
I'm trying to build make a code where an equation is not calculated for some certain values. I have a meshgrid with several values for x and y and I want to include a for loop that will calculate some values for most of the points in the meshgrid but I'm trying to include in that loop a condition that if the points have a specified index, the value will not be calculated. In my second group of for/if loops, I want to say that for all values of i and k (row and column), the value for z and phi are calculated with the exception of the specified i and k values (in the if loop). What I'm doing at the moment is not working...
The error I'm getting is:
The expression to the left of the equals sign is not a valid target for an assignment.
Here is my code at the moment. I'd really appreciate any advice on this! Thanks in advance
U_i = 20;
a = 4;
c = -a*5;
b = a*10;
d = -20;
e = 20;
n = a*10;
[x,y] = meshgrid([c:(b-c)/n:b],[d:(e-d)/n:e]');
for i = 1:length(x)
for k = 1:length(x)
% Zeroing values where cylinder is
if sqrt(x(i,k).^2 + y(i,k).^2) < a
x(i,k) = 0;
y(i,k) = 0;
end
end
end
r = sqrt(x.^2 + y.^2);
theta = atan2(y,x);
z = zeros(length(x));
phi = zeros(length(x));
for i = 1:length(x)
for k = 1:length(x)
if (i > 16 && i < 24 && k > 16 && k <= length(x))
z = 0;
phi = 0;
else
z = U_i.*r.*(1-a^2./r.^2).*sin(theta); % Stream function
phi = U_i*r.*(1+a^2./r.^2).*cos(theta); % Velocity potential
end
end
end
The original code in the question can be rewritten as seen below. Pay attention in the line with ind(17:24,:) since your edit now excludes 24 and you original question included 24.
U_i = 20;
a = 4;
c = -a*5;
b = a*10;
d = -20;
e = 20;
n = a*10;
[x,y] = meshgrid([c:(b-c)/n:b],[d:(e-d)/n:e]');
ind = find(sqrt(x.^2 + y.^2) < a);
x(ind) = 0;
y(ind) = 0;
r = sqrt(x.^2 + y.^2);
theta = atan2(y,x);
ind = true(size(x));
ind(17:24,17:length(x)) = false;
z = zeros(size(x));
phi = zeros(size(x));
z(ind) = U_i.*r(ind).*(1-a^2./r(ind).^2).*sin(theta(ind)); % Stream function
phi(ind) = U_i.*r(ind).*(1+a^2./r(ind).^2).*cos(theta(ind)); % Velocity potential
I'm writing a script for an aerodynamics class and I'm getting the following error:
Undefined function or variable 'dCt_dx'.
Error in Project2_Iteration (line 81)
Ct = trapz(x,dCt_dx)
I'm not sure what the cause is. It's something to do with my if statement. My script is below:
clear all
clc
global dr a n Vinf Vr w rho k x c cl dr B R beta t
%Environmental Parameters
n = 2400; %rpm
Vinf = 154; %KTAS
rho = 0.07647 * (.7429/.9450); %from mattingly for 8kft
a = 1084; %speed of sound, ft/s, 8000 ft
n = n/60; %convert to rps
w = 2*pi*n;
Vinf = (Vinf*6076.12)/3600; %convert from KTAS to ft/s
k = length(c);
dr = R/k; %length of each blade element
for i = 1:k
r(i) = i*dr - (.5*dr); %radius at center of blade element
dA = 2*pi*r*dr; %Planform area of blade element
x(i) = r(i)/R;
if x(i) > .15 && x(i-1) < .15
i_15 = i;
end
if x(i) > .75 && x(i-1) < .75
i_75h = i;
i_75l = i-1;
end
Vr(i) = w*r(i) + Vinf;
%Aerodynamic Parameters
M = Vr(i)/a;
if M > 0.9
M = 0.9;
end
m0 = 0.9*(2*pi/(1-M^2)^0.5); %lift-curve slope (2pi/rad)
%1: Calculate phi
phi = atan(Vinf/(2*pi*n*r(i)));
%2: Choose Vo
Vo = .00175*Vinf;
%3: Calculate Theta
theta = atan((Vinf + Vo)/(2*pi*n*r(i)))-phi;
%4:
if option == 1
%calculate cl(i) from c(i)
sigma = (B*c(i))/(pi*R);
if sigma > 0
cl(i) = (8*x(i)*theta*cos(phi)*tan(phi+theta))/sigma;
else
cl(i) = 0;
end
else %option == 2
%calculate c(i) from cl(i)
if cl(i) ~= 0
sigma = (8*x(i)*theta*cos(phi)*tan(phi+theta))/cl(i);
else
sigma = 0;
end
c(i) = (sigma*pi*R)/B;
if c(i) < 0
c(i) = 0;
end
end
%5: Calculate cd
cd(i) = 0.0090 + 0.0055*(cl(i)-0.1)^2;
%6: calculate alpha
alpha = cl(i)/m0;
%7: calculate beta
beta(i) = phi + alpha + theta;
%8: calculate dCt/dx and dCq/dx
phi0 = phi+theta;
lambda_t = (1/(cos(phi)^2))*(cl(i)*cos(phi0) - cd(i)*sin(phi0));
lambda_q = (1/(cos(phi)^2))*(cl(i)*sin(phi0) + cd(i)*cos(phi0));
if x(i) >= 0.15
dCt_dx(i) = ((pi^3)*(x(i)^2)*sigma*lambda_t)/8; %Roskam eq. 7.47, pg. 280
dCq_dx(i) = ((pi^3)*(x(i)^3)*sigma*lambda_q)/16; %Roskam eq. 7.48, pg 280
else
dCt_dx(i) = 0;
dCq_dx(i) = 0;
end
%calculate Mdd
t(i) = (0.04/(x(i)^1.2))*c(i);
Mdd(i) = 0.94 - (t(i)/c(i)) - cl(i)/10;
end
%9: calculate Ct, Cq, Cd
Ct = trapz(x,dCt_dx)
Cq = trapz(x,dCq_dx)
D = 2*R;
Q=(rho*(n^2)*(D^5)*Cq)
T=(rho*(n^2)*(D^4)*Ct)
When I step through your script, I see that the the entire for i = 1:k loop is skipped because k=0. You set k = length(c), but c was never initialized to a value, so it has length zero.
Because of this, dCt_dx is never given a value--and more importantly the majority of your script is never run.
If you're going to be using MATLAB in the future, I really suggest learning how to do this. It makes it a lot easier to find bugs. Try looking at this video.
I've written some code to implement an algorithm that takes as input a vector q of real numbers, and returns as an output a complex matrix R. The Matlab code below produces a plot showing the input vector q and the output matrix R.
Given only the complex matrix output R, I would like to obtain the input vector q. Can I do this using least-squares optimization? Since there is a recursive running sum in the code (rs_r and rs_i), the calculation for a column of the output matrix is dependent on the calculation of the previous column.
Perhaps a non-linear optimization can be set up to recompose the input vector q from the output matrix R?
Looking at this in another way, I've used an algorithm to compute a matrix R. I want to run the algorithm "in reverse," to get the input vector q from the output matrix R.
If there is no way to recompose the starting values from the output, thereby treating the problem as a "black box," then perhaps the mathematics of the model itself can be used in the optimization? The program evaluates the following equation:
The Utilde(tau, omega) is the output matrix R. The tau (time) variable comprises the columns of the response matrix R, whereas the omega (frequency) variable comprises the rows of the response matrix R. The integration is performed as a recursive running sum from tau = 0 up to the current tau timestep.
Here are the plots created by the program posted below:
Here is the full program code:
N = 1001;
q = zeros(N, 1); % here is the input
q(1:200) = 55;
q(201:300) = 120;
q(301:400) = 70;
q(401:600) = 40;
q(601:800) = 100;
q(801:1001) = 70;
dt = 0.0042;
fs = 1 / dt;
wSize = 101;
Glim = 20;
ginv = 0;
R = get_response(N, q, dt, wSize, Glim, ginv); % R is output matrix
rows = wSize;
cols = N;
figure; plot(q); title('q value input as vector');
ylim([0 200]); xlim([0 1001])
figure; imagesc(abs(R)); title('Matrix output of algorithm')
colorbar
Here is the function that performs the calculation:
function response = get_response(N, Q, dt, wSize, Glim, ginv)
fs = 1 / dt;
Npad = wSize - 1;
N1 = wSize + Npad;
N2 = floor(N1 / 2 + 1);
f = (fs/2)*linspace(0,1,N2);
omega = 2 * pi .* f';
omegah = 2 * pi * f(end);
sigma2 = exp(-(0.23*Glim + 1.63));
sign = 1;
if(ginv == 1)
sign = -1;
end
ratio = omega ./ omegah;
rs_r = zeros(N2, 1);
rs_i = zeros(N2, 1);
termr = zeros(N2, 1);
termi = zeros(N2, 1);
termr_sub1 = zeros(N2, 1);
termi_sub1 = zeros(N2, 1);
response = zeros(N2, N);
% cycle over cols of matrix
for ti = 1:N
term0 = omega ./ (2 .* Q(ti));
gamma = 1 / (pi * Q(ti));
% calculate for the real part
if(ti == 1)
Lambda = ones(N2, 1);
termr_sub1(1) = 0;
termr_sub1(2:end) = term0(2:end) .* (ratio(2:end).^-gamma);
else
termr(1) = 0;
termr(2:end) = term0(2:end) .* (ratio(2:end).^-gamma);
rs_r = rs_r - dt.*(termr + termr_sub1);
termr_sub1 = termr;
Beta = exp( -1 .* -0.5 .* rs_r );
Lambda = (Beta + sigma2) ./ (Beta.^2 + sigma2); % vector
end
% calculate for the complex part
if(ginv == 1)
termi(1) = 0;
termi(2:end) = (ratio(2:end).^(sign .* gamma) - 1) .* omega(2:end);
else
termi = (ratio.^(sign .* gamma) - 1) .* omega;
end
rs_i = rs_i - dt.*(termi + termi_sub1);
termi_sub1 = termi;
integrand = exp( 1i .* -0.5 .* rs_i );
if(ginv == 1)
response(:,ti) = Lambda .* integrand;
else
response(:,ti) = (1 ./ Lambda) .* integrand;
end
end % ti loop
No, you cannot do so unless you know the "model" itself for this process. If you intend to treat the process as a complete black box, then it is impossible in general, although in any specific instance, anything can happen.
Even if you DO know the underlying process, then it may still not work, as any least squares estimator is dependent on the starting values, so if you do not have a good guess there, it may converge to the wrong set of parameters.
It turns out that by using the mathematics of the model, the input can be estimated. This is not true in general, but for my problem it seems to work. The cumulative integral is eliminated by a partial derivative.
N = 1001;
q = zeros(N, 1);
q(1:200) = 55;
q(201:300) = 120;
q(301:400) = 70;
q(401:600) = 40;
q(601:800) = 100;
q(801:1001) = 70;
dt = 0.0042;
fs = 1 / dt;
wSize = 101;
Glim = 20;
ginv = 0;
R = get_response(N, q, dt, wSize, Glim, ginv);
rows = wSize;
cols = N;
cut_val = 200;
imagLogR = imag(log(R));
Mderiv = zeros(rows, cols-2);
for k = 1:rows
val = deriv_3pt(imagLogR(k,:), dt);
val(val > cut_val) = 0;
Mderiv(k,:) = val(1:end-1);
end
disp('Running iteration');
q0 = 10;
q1 = 500;
NN = cols - 2;
qout = zeros(NN, 1);
for k = 1:NN
data = Mderiv(:,k);
qout(k) = fminbnd(#(q) curve_fit_to_get_q(q, dt, rows, data),q0,q1);
end
figure; plot(q); title('q value input as vector');
ylim([0 200]); xlim([0 1001])
figure;
plot(qout); title('Reconstructed q')
ylim([0 200]); xlim([0 1001])
Here are the supporting functions:
function output = deriv_3pt(x, dt)
% Function to compute dx/dt using the 3pt symmetrical rule
% dt is the timestep
N = length(x);
N0 = N - 1;
output = zeros(N0, 1);
denom = 2 * dt;
for k = 2:N0
output(k - 1) = (x(k+1) - x(k-1)) / denom;
end
function sse = curve_fit_to_get_q(q, dt, rows, data)
fs = 1 / dt;
N2 = rows;
f = (fs/2)*linspace(0,1,N2); % vector for frequency along cols
omega = 2 * pi .* f';
omegah = 2 * pi * f(end);
ratio = omega ./ omegah;
gamma = 1 / (pi * q);
termi = ((ratio.^(gamma)) - 1) .* omega;
Error_Vector = termi - data;
sse = sum(Error_Vector.^2);