Related
The MATLAB program below is a function that references specific input values for S, E, r, sigma, and tau.
function [C, Cdelta, P, Pdelta] = ch08(S,E,r,sigma,tau)
% Input arguments: S = asset price at time t
% E = Exercise price
% r = interest rate
% sigma = volatility
% tau = time to expiry (T-t)
%
% Output arguments: C = call value, Cdelta = delta value of call
% P = Put value, Pdelta = delta value of put
%
% function [C, Cdelta, P, Pdelta] = ch08(S,E,r,sigma,tau)
if tau > 0
d1 = (log(S/E) + (r + 0.5*sigma^2)*(tau))/(sigma*sqrt(tau));
d2 = d1 - sigma*sqrt(tau);
N1 = 0.5*(1+erf(d1/sqrt(2)));
N2 = 0.5*(1+erf(d2/sqrt(2)));
C = S*N1-E*exp(-r*(tau))*N2;
Cdelta = N1;
P = C + E*exp(-r*tau) - S;
Pdelta = Cdelta - 1;
title('Graph of Call Value vs. Time to Expiry')
xlabel('Time to expiry')
ylabel('Call Value')
plot (tau,C)
else
C = max(S-E,0);
Cdelta = 0.5*(sign(S-E) + 1);
P = max(E-S,0);
Pdelta = Cdelta - 1;
title('Graph of Call Value vs. Time to Expiry')
xlabel('Time to expiry')
ylabel('Call Value')
plot (tau,C)
end
After running
ch08(3,2.5,0.03,0.25,1)
The following output is produced
After running the function again,
ch08(3,2.5,0.03,0.25,1)
hold on
ch08(3,2.5,0.03,0.25,0.9)
Two data points are produced
After manually typing decreasing tau values,
ch08(3,2.5,0.03,0.25,1)
hold on
ch08(3,2.5,0.03,0.25,0.9)
hold on
ch08(3,2.5,0.03,0.25,0.8)
hold on
ch08(3,2.5,0.03,0.25,0.7)
hold on
ch08(3,2.5,0.03,0.25,0.6)
hold on
ch08(3,2.5,0.03,0.25,0.5)
hold on
ch08(3,2.5,0.03,0.25,0.4)
hold on
ch08(3,2.5,0.03,0.25,0.3)
hold on
ch08(3,2.5,0.03,0.25,0.2)
hold on
ch08(3,2.5,0.03,0.25,0.1)
The graph will produce a bunch of data points,
Is there a way to automate the tau values entered in ch08(S,E,r,sigma,tau) so that the user doesn't have to type all of the input in?
As I suggested in comments, you need to use a for loop. You can create an array with values of tau that you want to use, and call your function with a different element from that array in each loop iteration:
figure, hold on
tau = 10.^(0:-1:-6);
for ii = 1:length(tau)
ch08(3,2.5,0.03,0.25,tau(ii))
end
However, a better solution would be to not plot within your ch08 function, and return the value C as you did in your first version of your question. Then you can do this:
tau = 10.^(0:-1:-6);
C = zeros(size(tau));
for ii = 1:length(tau)
C(ii) = ch08(3,2.5,0.03,0.25,tau(ii));
end
plot(tau,C,'.');
This would allow you to change your plot as you please, for example drawing a line through your points.
PS: Note that you only need to give hold on once. It sets a flag in the figure window that is not cleared until you do hold off or clf.
I have a MATLAB function, when I try to run it, it does not store the output variables in the workspace. Please let me know the proper solution such that the variables in the function are stored in the workspace.
I have a following code in which I have to store values of variables T and Y in workspace.
function him1
k0 = ones(1,5);
exp=[0.2;0.12;0.24;0.2]; %//experimental data
time=[0;1;2;3]; %//time span
%// initial values of state variables
x01=1;
x02=1;
x03=1;
x04=1;
x0 = [x01,x02,x03,x04];
tspan = [min(time),max(time)];
k_opt = fminsearch(#minimize, k0)
function e = minimize(k0)
[~,y_hat] = ode45(#unit2, tspan, x0,[], k0);
% size(y_hat)
% y_hat = deval(sol, time(end)); % // evaluate solution at given times
e = sum((y_hat(end,:)' - exp).^2) % // compute squarederror '
end
% // plot with optimal parameter
[T,Y] = ode45(#unit2, tspan, [x01,x02,x03,x04], [], k_opt);
figure
subplot(1,2,1)
plot(time(end), exp, '*', 'markersize',15)
hold on
plot(T,Y, 'linewidth', 2)
end
function dx = unit2(t, x, k)
dx = zeros(4,1);
dx(1)=-k(1)*x(1)*7 + k(2)*x(2);
dx(2)=k(1)*x(1)*7 - k(2)*x(2) -k(3)*x(2)*x(2) + k(4)*x(3);
dx(3)=k(3)*x(2)*x(2) - k(4)*x(3)-k(5)*x(4);
dx(4)=k(5)*x(4);
end
It is a characateristic of functions, that only variables which are defined in the declatation are handed over to the workspace. (If you ignore more complex concepts like global variables and assignin)
If your main target is to get the calculation done and the variables in the workspace, than I suggest to convert it to an m-script instead of a function.
See: http://de.mathworks.com/help/matlab/matlab_prog/scripts-and-functions.html
You can try to identify your variables as global.
I have a simple function as below:
function dz = statespace(t,z)
dz = A*z + B*k*z
end
My main script is :
clear all;close all;format short;clc;
% step 1 -- input system parameters
% structure data
M1 = 1; M2= 1; M3= 1; %Lumped Mass(Tons)
M= diag([M1,M2,M3]);
k(1)= 980; k(2)= 980; k(3)= 980; %Stiffness Coefficient(KN/M)
K = zeros(3,3);
for i=1:2
K(i,i)=k(i)+k(i+1);
end
K(3,3)=k(3);
for i=1:2
K(i,i+1)=-k(i+1);
end
for i=2:3
K(i,i-1)=-k(i);
end %Stiffness Matrix(KN/M)
c(1)= 1.407; c(2)= 1.407; c(3)= 1.407; %Damping Coefficient(KN.sec/M)
C = zeros(3,3);
for i=1:2
C(i,i)=c(i)+c(i+1);
end
C(3,3)=c(3);
for i=1:2
C(i,i+1)=-c(i+1);
end
for i=2:3
C(i,i-1)=-c(i);
end %Damping Matrix(KN.sec/M)
A = [zeros(3) eye(3);-inv(M)*K -inv(M)*C]
H = [1;0;0]
B = [0;0;0;inv(M)*H]
k = [-1 -1 -1 -1 -1 -1]
t = 0:0.004:10;
[t,z] = ode45(statespace,t);
When I run my main script it comes with following error:
Undefined function or variable 'A'.
Error in statespace (line 2)
dz = A*z + B*k*z
As you can see I defined A in main script. Why this problem happening?
There multiple things wrong with your code. First, you need to supply the values of A and B to your function but as you are invoking it (incorrectly without the # and additional parameter y0 as I commented below) in the toolbox ode45, you have to keep just two parameters so you cannot supply A and B as additional parameters. If you define A and B within your function or share them via global variables you will get further. However, as noted below the definitions don't seem to be correct as A * z and B * k * z don't have the same dimensions. z is a scalar so B * k needs to be same size and shape as A which currently it is not.
Edit from:
As Jubobs suggested change your function's parameters to include A, B and k. Also you don't need t as it is never used in the function. So:
function dz = statespace(A, B, k, z)
dz = A*z + B*k*z
end
As others have pointed out, A, B and k are not defined in the function workspace, so you either need to define them again (ugly, not recommended), declare them as global variables (slightly better, but still not good practice), or pass them as arguments to your function (the better solution). However, because you then want to use the function with ode45, you need to be a bit careful with how you do it:
function dz = statespace(t,z,A,B,k)
dz = A*z + B*k*z
end
and then the call to ode45 would like this:
[t,z] = ode45(#(t,z)statespace(t,z,A,B,k),[0 Tf],z0); % where Tf is your final time and z0 your initial conditions
See Error using ode45 and deval for a similar problem.
this post is related to my previous question : image processing in matlab
as I have uploaded my algorithme there.
the think is that I am trying to change the filtering part of the code.
in matlab filter.m function can accpet filter(B, A, my pixels evolutions in Time) and it return me the filtered values.
but at the moment I have to pass the the whole time series together.
but the problem is that now I want to change the code in a way instead of passing the whole timeseries into the filter, I want to pass one value at a time, but I want filter function treat the value like the nth value not the first valu.
I have created a sudo code, as I am injecting one picture into the code, but I dont know how can change the filtering part., any body has any idea??
clear all
j=1;
for i=0:3000
str = num2str(i);
str1 = strcat(str,'.mat');
load(str1);
D{j}=A(20:200,130:230);
j=j+1;
end
N=5;
w = [0.00000002 0.05;0.05 0.1;0.1 0.15;0.15 0.20;
0.20 0.25;0.25 0.30;0.30 0.35;0.35 0.40;
0.40 0.45;0.45 0.50;0.50 0.55;0.55 0.60;
0.60 0.65;0.65 0.70;0.70 0.75;0.75 0.80;
0.80 0.85;0.85 0.90;0.90 0.95;0.95 0.99999999];
for ind=1:20
wn = w(ind,:);
[b,a] = butter(N,wn);
bCoeff(ind,:)=b;
aCoeff(ind,:)=a;
end
ts=[];
sumLastImages=[];
for k=1:10 %number of images
for bands=1:20 %number of bands
for i=1:10 %image h
for j=1:10 %image w
pixelValue = D{k}(i,j);
% reflectivity elimination
% for the current pixel, have the summation of the same position from before
% images and create a mean value base on the temporal values
sumLastImages(i,j)=pixelValue+sumLastImages(i,j);
meanValue = sumLastImages(i,j)/k;
if(meanValue==0)
filteredimages{bands}(i,j)=0;
continue;
else
pixel_calculated_meanvalue = pixelValue/meanValue;
end
% filter part that need to be changed, and because we are adding
% one value then the reutrn of the filter is one too
ts_f = filter(bCoeff(bands,:), aCoeff(bands,:), ...
pixel_calculated_meanvalue);
filteredimages{bands}(i,j)=ts_f;
end
end
finalImagesSummation{bands}(:,:) = ...
(filteredimages{bands}(:,:)^2)+finalImagesSummation{bands}(:,:);
finalImages{bands}(:,:)=finalImagesSummation/k;
end
end
EDIT
I managed to change the code like this, which now I load the fist 200 images, and after that I am able to inject the images one by one into the filter, but now the problem is that I am getting "Out of memory. Type HELP MEMORY for your options." error for big
images.
here is my code any idea to efficent the code :
%%
cd('D:\MatlabTest\06-06-Lentils');
clear all
%%
N=5;
W = [0.0 0.10;0.10 0.20;0.20 0.30;0.30 0.40;
0.40 0.50;0.50 0.60 ;0.60 0.70;0.70 0.80 ;
0.80 0.90;0.90 1.0];
[bCoeff{1},aCoeff{1}] = butter(N,0.1,'Low');
for ind=2:9
wn = W(ind,:);
[b,a] = butter(N,wn);
bCoeff{ind}=b;
aCoeff{ind}=a;
end
[bCoeff{10},aCoeff{10}] = butter(N,0.9,'high');
%%
j=1;
D = zeros(200,380,320);
T = 200;
K = 0:399;
l = T+1;
Yout = cell(1,10);
figure;
for i = 1:length(K)-200
disp(i)
if i == 1
for j = 1:T
str = int2str(K(1)+j-1);
str1 = strcat(str,'.mat');
load(str1);
D(j,1:380,1:320) = A;
end
else
str = int2str(l);
str1 = strcat(str,'.mat');
load(str1);
temp = D(2:200,1:380,1:320) ;
temp(200,1:380,1:320) = A;
D = temp;
clear temp
l = l +1;
end
for p = 1:380
for q = 1:320
x = D(:,p,q) - mean(D(:,p,q));
for k = 1:10
temp = filter(bCoeff{k},aCoeff{k},x);
if i == 1
Yout{k}(p,q) = mean(temp);
else
Yout{k}(p,q) = (Yout{k}(p,q) + mean(temp))/2;
end
end
end
end
for k = 1:10
subplot(5,2,k)
subimage(Yout{k}*1000,[0 255]);
color bar
colormap jet
end
pause(0.01);
end
disp('Done Loading...')
No need to rewrite the filter function, there is a simple solution!
If you want to feed filter with one sample at a time, you need to pass the state parameters as well so that each input sample is processed depending on its predecessor. After filtering, the new state is actually returned as a second parameter, so that most of the work is already done by MATLAB for you. This is good news!
For the sake of readability, allow me to temporarily replace your variable names with simple letters:
a = aCoeff(bands, :);
b = bCoeff(bands, :);
x = pixel_calculated_meanvalue;
ts_f is represented by y.
And so, this:
y = filter(b, a, x);
is actually equivalent to this:
N = numel(x);
y = zeros(N, 1); %# Preallocate memory for output
z = zeros(max(length(a), length(b)) - 1, 1); %# This is the initial filter state
for i = 1:N
[y(i), z] = filter(b, a, x(i), z);
end
You can check for yourself that the result is the same!
For your example, the code would be:
N = numel(pixel_calculated_meanvalue);
ts_f = zeros(N, 1);
z = zeros(max(length(aCoeff(bands, :)), length(bCoeff(bands, :))) - 1, 1);
for i = 1:N
[ts_f(i), z] = filter(bCoeff(bands, :), aCoeff(bands, :), ...
pixel_calculated_meanvalue(i), z);
end
With this method you can process one input sample at a time, just make sure you store the last filter state after every filter call. If you plan on using multiple filters, you'll have to store a state vector per filter!
Overview
If all you want to have is an IIR filter, which you can feed incrementally, i.e. where you do not have to supply the full vector at once, you can simply implement your own function and use either persistent variables.
Simple approach
The code snippet below defines a function myFilter, which makes use of persistent
variables, which you can control with the following commands:
init: set up the IIR coefficients
getA: returns the A coefficients, i.e. the outputweights
getB: returns the B coefficients, i.e. the input weights
getX: returns the stored input data x[0], x[1], ... x[M]
getY: returns the output data y[0], y[1], ... y[N]
getCurrentY: returns the last output data y[N]
Here is the function:
function result = myFilter(varargin)
% myFilter A simple IIR filter which can be fed incrementally.
%
% The filter is controlled with the following commands:
% myFilter('init', B, A)
% Initializes the coefficients B and A. B are the weights for the
% input and A for the output.
% myFilter('reset')
% Resets the input and output buffers to zero.
% A = myFilter('getA')
% B = myFilter('getB')
% Returns the filter coefficients A and B, respectively.
% x = myFilter('getX')
% y = myFilter('getY')
% Returns the buffered input and output vectors.
% y = myFilter('getCurrentY')
% Returns the current output value.
% myFilter(x)
% Adds the new value x as input to the filter and updates the
% output.
persistent Bcoeff
persistent Acoeff
persistent x
persistent y
if ischar(varargin{1})
% This is a command.
switch varargin{1}
case 'init'
Bcoeff = varargin{2};
Acoeff = varargin{3};
Bcoeff = Bcoeff / Acoeff(1);
Acoeff = Acoeff / Acoeff(1);
x = zeros(size(Bcoeff));
y = zeros(1, length(Acoeff) - 1);
return
case 'reset'
x = zeros(size(Bcoeff));
y = zeros(1, length(Acoeff) - 1);
return
case 'getA'
result = Acoeff;
return
case 'getB'
result = Bcoeff;
return
case 'getX'
result = x;
return
case 'getY'
result = y;
return
case 'getCurrentY'
result = y(1);
return
otherwise
error('Unknown command');
end
end
% A new value has to be filtered.
xnew = varargin{1};
x = [xnew, x(1:end-1)];
ynew = sum(x .* Bcoeff) - sum(y .* Acoeff(2:end));
y = [ynew, y(1:end-1)];
end
And a usage example:
% Define the filter coefficients. Bcoeff acts on the input, Acoeff on
% the output.
Bcoeff = [4, 5];
Acoeff = [1, 2, 3];
% Initialize the filter.
myFilter('init', Bcoeff, Acoeff);
% Add a value to be filtered.
myFilter(10)
myFilter('getCurrentY')
ans =
40
% Add another one.
myFilter(20)
myFilter('getCurrentY')
ans =
50
% And a third one.
myFilter(30)
myFilter('getCurrentY')
ans =
0
% Compare with the Matlab filter function.
filter(Bcoeff, Acoeff, [10 20 30])
ans =
40 50 0
The drawback of this approach is that it is only possible to have one active filter
simultaneously. This is problematic e.g. in your question, where you have different
filters which are updated in an alternating fashion.
Advanced approach
In order to operate multiple filters simultatenously, you need some way to identify
the filter. The solution I present here works with handles. A handle is simple an
integer. To be more precise, it is actually an index into a persistent array, which
itself holds the filter state, i.e. the filter coefficients and the buffers for the
input and the output.
The calling syntax is a bit more complicated, because you have to pass a handle. However,
it is possible that multiple filters are active simultaneously.
function result = myFilterH(varargin)
% myFilterH A simple IIR filter which can be fed incrementally.
% Operates on a filter handle.
%
% The filter is controlled with the following commands:
% handle = myFilterH('create')
% Creates a new filter handle.
% myFilterH(handle, 'init', B, A)
% Initializes the coefficients B and A. B are the weights for the
% input and A for the output. handle identifies the filter.
% myFilterH(handle, 'reset')
% Resets the input and output buffers to zero.
% A = myFilterH(handle, 'getA')
% B = myFilterH(handle 'getB')
% Returns the filter coefficients A and B, respectively.
% x = myFilterH(handle, 'getX')
% y = myFilterH(handle, 'getY')
% Returns the buffered input and output vectors.
% y = myFilterH(handle, 'getCurrentY')
% Returns the current output value.
% myFilterH(handle, x)
% Adds the new value x as input to the filter and updates the
% output.
persistent handles
if ischar(varargin{1})
if strcmp(varargin{1}, 'create')
if isempty(handles)
handles = struct('x', [], 'y', [], 'A', [], 'B', []);
result = 1;
else
result = length(handles) + 1;
handles(result).x = [];
end
return
else
error('First argument must be a filter handle or ''create''');
end
end
% The first input should be the handles(hIdx).
hIdx = varargin{1};
if hIdx < 0 || hIdx > length(handles)
error('Invalid filter handle')
end
if ischar(varargin{2})
% This is a command.
switch varargin{2}
case 'init'
handles(hIdx).B = varargin{3};
handles(hIdx).A = varargin{4};
handles(hIdx).B = handles(hIdx).B / handles(hIdx).A(1);
handles(hIdx).A = handles(hIdx).A / handles(hIdx).A(1);
handles(hIdx).x = zeros(size(handles(hIdx).B));
handles(hIdx).y = zeros(1, length(handles(hIdx).A) - 1);
return
case 'reset'
handles(hIdx).x = zeros(size(handles(hIdx).B));
handles(hIdx).y = zeros(1, length(handles(hIdx).A) - 1);
return
case 'getA'
result = handles(hIdx).A;
return
case 'getB'
result = handles(hIdx).B;
return
case 'getX'
result = handles(hIdx).x;
return
case 'getY'
result = handles(hIdx).y;
return
case 'getCurrentY'
result = handles(hIdx).y(1);
return
otherwise
error('Unknown command');
end
end
% A new value has to be filtered.
xnew = varargin{2};
handles(hIdx).x = [xnew, handles(hIdx).x(1:end-1)];
ynew = sum(handles(hIdx).x .* handles(hIdx).B) ...
- sum(handles(hIdx).y .* handles(hIdx).A(2:end));
handles(hIdx).y = [ynew, handles(hIdx).y(1:end-1)];
end
And the example:
% Define the filter coefficients.
Bcoeff = [4, 5];
Acoeff = [1, 2, 3];
% Create new filter handles.
fh1 = myFilterH('create');
fh2 = myFilterH('create');
% Initialize the filter handle. Note that reversing Acoeff and Bcoeff creates
% two totally different filters.
myFilterH(fh1, 'init', Bcoeff, Acoeff);
myFilterH(fh2, 'init', Acoeff, Bcoeff);
% Add a value to be filtered.
myFilterH(fh1, 10);
myFilterH(fh2, 10);
[myFilterH(fh1, 'getCurrentY'), myFilterH(fh2, 'getCurrentY')]
ans =
40.0000 2.5000
% Add another one.
myFilterH(fh1, 20);
myFilterH(fh2, 20);
[myFilterH(fh1, 'getCurrentY'), myFilterH(fh2, 'getCurrentY')]
ans =
50.0000 6.8750
% And a third one.
myFilterH(fh1, 30);
myFilterH(fh2, 30);
[myFilterH(fh1, 'getCurrentY'), myFilterH(fh2, 'getCurrentY')]
ans =
0 16.4063
% Compare with the Matlab filter function.
filter(Bcoeff, Acoeff, [10 20 30])
ans =
40 50 0
filter(Acoeff, Bcoeff, [10 20 30])
ans =
2.5000 6.8750 16.4063
Using it for your example
To use the advanced approach in your example, you could first create an array of filters:
fh = [];
for ind = 1:20
wn = w(ind,:);
[b,a] = butter(N,wn);
fh(ind) = myFilterH('create');
myFilterH(fh(ind), 'init', b, a);
end
Later on in the filter part simply add your value to all of the filters. However, for this
you also need to reverse the loops because right now you would need one filter per
band per pixel. If you loop over pixels first and then over bands, you can reuse the filters:
for height = 1:10
for width = 1:10
for bands = 1:20
myFilterH(fh(bands), 'reset');
for k = 1:10
[...]
ts_f = myFilterH(fh(bands), ...
pixel_calculated_meanvalue);
filteredimages{bands}(i,j) = myFilterH(fh(bands), 'getCurrentY');
end
end
end
end
Hope that helps!
If I understand the question correctly, the crux is "How do I return multiple things from a Matlab function?"
You can return multiple things like this:
function [a, b, np, x, y] = filter(ord, a, b, np, x, y)
%code of function here
%make some changes to a, b, np, x, and y
end
If you want to call the filter from another function and catch its return values, you can do this:
function [] = main()
%do some stuff, presumably generate initial values for ord, a, b, np, x, y
[new_a, new_b, new_np, new_x, new_y] = filter(ord, a, b, np, x, y)
end
In short, if you do function [x, y] = myfunc(a, b), then myfunc returns both x and y.
Currently I'm doing a code on the Secant Method, and so far the code runs ok.
However, I still have to update my "count.funcCount" by counting the number of function calls that I used in the "secant" function. How should I modify my code ?
This is what I have so far for the code:
function [ root, fcneval, count ] = secant(f, x0, x1, my_options)
my_options = optimset('MaxIter', 50, 'TolFun', 1.0e-5);
count = struct('iterations',0,'funcCount',0,'message',{'empty'});
xp = x0; %# Initialize xp and xc to match with
xc = x1; %# x0 and x1, respectively
MaxIter = 100;
TolFun = 1.0e-5;
%# Secant Method Loop
for i = 2:MaxIter
fd = f(xc) - f(xp); %# Together, the ratio of d and fd yields
d = xc - xp; %# the slope of the secant line going through xc and xp
xn = ((xc * fd) - (f(xc) * d)) / fd; %# Secant Method step
if (abs(xc - xn) < TolFun) && (abs(f(xn)) < TolFun) %# Stopping condition
break;
elseif i > MaxIter %# If still can't find a root after maximum
%# 100 iterations, consider not converge
count.message = sprintf('Do not converge');
end
xp = xc; % Update variables xc and xp
xc = xn;
end
%# Get outputs:
root = xn;
fcneval = f(root);
count.iterations = i;
end %# end function
---------------------------------------
function [f] = fun(x)
f = cos(x) + x;
end
Please help me, thank you in advance
Although I did not understand your question but still I think you can use a counter. Initialize it with zero and increment it every time your function is called. You are using an inbuilt function, just make necessary changes in its source code( FYI :you can do this in matlab) then save it with a new name and then use it in your main code.
You could create a nested function which have access to the variables of its parent function (including the function f and the counter count.funcCount). This function would call the actual method which does the computation, and then increment the counter.
Here is a rather silly example to illustrate the concept:
function [output,count] = myAlgorithm(f, x0)
count = 0;
output = x0;
while rand()<0.99 %# simulate a long loop
output = output + fcn(x0);
end
%# nested function with closure
function y = fcn(x)
%# access `f` and `count` inside the parent function
y = f(x); %# call the function
count = count + 1; %# increment counter
end
end
Now you call it as:
[output,count] = myAlgorithm(#(x)cos(x)+x, 1)