How would you plot these in SciLab or MatLab? I am new to these and have no idea how the software works. Please help.
$Plot following functions with different colors in Scilab or MatLab
– f2(x) = logn
– f3(x) = n
– f4(x) = nlogn
– f5(x) = n2
– f6(x) = nj (j > 2)
– f7(x) = cn (c > 1)
– f8(x) = n!
where x = linspace(1, 50, 50).
Well, a lot of these are built-in functions. For example
>> x = linspace(1,50,50);
>> plot(x,log(x))
>> plot(x,x)
>> plot(x,x.*log(x))
>> plot(x,x.^2)
I don't know what nj (j > 2) and cn (c > 1) are supposed to mean.
For the last one, you should look at the function factorial.
It's not clear from the context whether you're supposed to plot them on different graphs or all on the same graph. If all on the same graph, then you can use
>> hold on;
to freeze the current axes - that means that any new lines will get drawn on top of the old ones, instead of being drawn on a fresh set of axes.
In Matlab (and probably in Scilab) you can supply a "line spec" argument to the plot function, which tells it what color and style to draw the line in. For example,
>> figure
>> hold on
>> plot(x,log(x),'b')
>> plot(x,x/10,'r')
>> plot(x,x.^2/1000,'g')
Tells Matlab to plot the function f(x)=log(x) in blue, f(x)=x/10 in red and f(x)=x^2/1000 in green, which results in this plot:
I can't comment or upvote yet but I'd add to Chris Taylor's answer that in Scilab the hold on and hold off convention isn't used. All plot commands output to the current axes, which are 'held on' all the time. If you want to generate a new figure or change the current axes you can use figure(n), where n can be any (nonconsecutive) positive integer - just a label really.
See also clf(n), gcf() and gca() - Scilab's figure handling differs quite a bit from Matlab's, though the matplotlib ATOMS module goes some way towards making Scilab look and behave more like Matlab.
In Scilab, it will be
x = 1:50;
clf
plot("ll", x,log, x,x, x,x.*log(x), x,x.^2)
gca().sub_ticks(2) = 8;
xgrid(color("grey"))
legend("$"+["ln(x)", "x", "x.ln(x)", "x^2"]+"$", "in_upper_left")
Related
In Julia 1.0.5, I have a function f(x::Vector{<:Real}), defined as
f(x::Vector{<:Real}) = (x[1] - 2)^2 + ( x[2] - 1 )^2
The signature is like this, because I would like to use it with the ForwardDiff package, and it works with it just fine. I give the function to ForwardDiff.gradient, and everything works like a charm.
However, I would also like to do some visualizations with PyPlot, using this same function f. Namely, I would like to draw its contour with contourf. For this purpose, I have constructed two vectors X::Vector{<:Real} and Y::Vector{<:Real}, and would like to call the same function f with them to produce the contour.
However, making the call f.([X, Y]) is not broadcasting the vectors as I would like, as I get the error
LoadError: MethodError: no method matching (::getfield(Main, Symbol("#f#1044")))(::Int64)
Closest candidates are:
f(!Matched::Array{#s25,1} where #s25<:Real)
This of course prevents me from using the contourf function, as it needs the values of f on a 2D-grid.
Do I need to define an entirely different f(x::Vector{<:Real}, y::Vector{<:Real}) to be able to plot the contour as I would like, or is there an alternative where I can avoid this?
this problem can be resolved by the power of multiple dispatch:
f(x::Vector{<:Real}) = (x[1] - 2)^2 + ( x[2] - 1 )^2
f(x::Real,y::Real) = f([x,y])
nx = 10
ny = 20
X = rand(nx) #mesh of x points
Y = rand(ny) #mesh of y points
Z = f.(transpose(X),Y) #nx x ny matrix
for the two argument gradient:
two_point_gradient(f,x,y) = ForwardDiff.gradient(f,[x,y])
G = two_point_gradient.(f,transpose(X),Y) #returns a vector of gradients, where G[i..] = gradient(f,X[i..],Y[i...])
I need to solve this differential equation using Runge-Kytta 4(5) on Scilab:
The initial conditions are above. The interval and the h-step are:
I don't need to implement Runge-Kutta. I just need to solve this and plot the result on the plane:
I tried to follow these instructions on the official "Scilab Help":
https://x-engineer.org/graduate-engineering/programming-languages/scilab/solve-second-order-ordinary-differential-equation-ode-scilab/
The suggested code is:
// Import the diagram and set the ending time
loadScicos();
loadXcosLibs();
importXcosDiagram("SCI/modules/xcos/examples/solvers/ODE_Example.zcos");
scs_m.props.tf = 5000;
// Select the solver Runge-Kutta and set the precision
scs_m.props.tol(6) = 6;
scs_m.props.tol(7) = 10^-2;
// Start the timer, launch the simulation and display time
tic();
try xcos_simulate(scs_m, 4); catch disp(lasterror()); end
t = toc();
disp(t, "Time for Runge-Kutta:");
However, it is not clear for me how I can change this for the specific differential equation that I showed above. I have a very basic knowledge of Scilab.
The final plot should be something like the picture bellow, an ellipse:
Just to provide some mathematical context, this is the differential equation that describes the pendulum problem.
Could someone help me, please?
=========
UPDATE
Based on #luizpauloml comments, I am updating this post.
I need to convert the second-order ODE into a system of first-order ODEs and then I need to write a function to represent such system.
So, I know how to do this on pen and paper. Hence, using z as a variable:
OK, but how do I write a normal script?
The Xcos is quite disposable. I only kept it because I was trying to mimic the example on the official Scilab page.
To solve this, you need to use ode(), which can employ many methods, Runge-Kutta included. First, you need to define a function to represent the system of ODEs, and Step 1 in the link you provided shows you what to do:
function z = f(t,y)
//f(t,z) represents the sysmte of ODEs:
// -the first argument should always be the independe variable
// -the second argument should always be the dependent variables
// -it may have more than two arguments
// -y is a vector 2x1: y(1) = theta, y(2) = theta'
// -z is a vector 2x1: z(1) = z , z(2) = z'
z(1) = y(2) //first equation: z = theta'
z(2) = 10*sin(y(1)) //second equation: z' = 10*sin(theta)
endfunction
Notice that even if t (the independent variable) does not explicitly appear in your system of ODEs, it still needs to be an argument of f(). Now you just use ode(), setting the flag 'rk' or 'rkf' to use either one of the available Runge-Kutta methods:
ts = linspace(0,3,200);
theta0 = %pi/4;
dtheta0 = 0;
y0 = [theta0; dtheta0];
t0 = 0;
thetas = ode('rk',y0, t0, ts, f); //the output have the same order
//as the argument `y` of f()
scf(1); clf();
plot2d(thetas(2,:),thetas(1,:),-5);
xtitle('Phase portrait', 'theta''(t)','theta(t)');
xgrid();
The output:
Unlike Matlab, Octave Symbolic has no piecewise function. Is there a work around? I would like to do something like this:
syms x
y = piecewise(x0, 1)
Relatedly, how does one get pieces of a piecewise function? I ran the following:
>> int (exp(-a*x), x, 0, t)
And got the following correct answer displayed and stored in a variable:
t for a = 0
-a*t
1 e
- - ----- otherwise
a a
But now I would like to access the "otherwise" part of the answer so I can factor it. How do I do that?
(Yes, I can factor it in my head, but I am practicing for when more complicated expressions come along. I am also only really looking for an approach using symbolic expressions -- even though in any single case numerics may work fine, I want to understand the symbolic approach.)
Thanks!
Matlab's piecewise function seems to be fairly new (introduced in 2016b), but it basically just looks like a glorified ternary operator. Unfortunately I don't have 2016 to check if it performs any checks on the inputs or not, but in general you can recreate a 'ternary' operator in octave by indexing into a cell using logical indexing. E.g.
{#() return_A(), #() return_B(), #() return_default()}([test1, test2, true]){1}()
Explanation:
Step 1: You put all the values of interest in a cell array. Wrap them in function handles if you want to prevent them being evaluated at the time of parsing (e.g. if you wanted the output of the ternary operator to be to produce an error)
Step 2: Index this cell array using logical indexing, where at each index you perform a logical test
Step 3: If you need a 'default' case, use a 'true' test for the last element.
Step 4: From the cell (sub)array that results from above, select the first element and 'run' the resulting function handle. Selecting the first element has the effect that if more than one tests succeed, you only pick the first result; given the 'default' test will always succeed, this also makes sure that this is not picked unless it's the first and only test that succeeds (which it does so by default).
Here are the above steps implemented into a function (appropriate sanity checks omitted here for brevity), following the same syntax as matlab's piecewise:
function Out = piecewise (varargin)
Conditions = varargin(1:2:end); % Select all 'odd' inputs
Values = varargin(2:2:end); % Select all 'even' inputs
N = length (Conditions);
if length (Values) ~= N % 'default' case has been provided
Values{end+1} = Conditions{end}; % move default return-value to 'Values'
Conditions{end} = true; % replace final (ie. default) test with true
end
% Wrap return-values into function-handles
ValFuncs = cell (1, N);
for n = 1 : N; ValFuncs{n} = #() Values{n}; end
% Grab funhandle for first successful test and call it to return its value
Out = ValFuncs([Conditions{:}]){1}();
end
Example use:
>> syms x t;
>> F = #(a) piecewise(a == 0, t, (1/a)*exp(-a*t)/a);
>> F(0)
ans = (sym) t
>> F(3)
ans = (sym)
-3⋅t
ℯ
─────
9
I was having fun with image processing and hough transforms on Octave but the results are not the expected ones.
Here is my edges image:
and here is my hough accumulator (x-axis is angle in deg, y-axis is radius):
I feel like I am missing the horizontal streaks but there is no local maximum in the accumulator for the 0/180 angle values.
Also, for the vertical streaks, the value of the radius should be equal to the x value of the edge's image, but instead the values of r are very high:
exp: the first vertical line on the left of the image has an equation of x=20(approx) -> r.r = x.x + y.y -> r=x -> r=20
The overall resulting lines detected do not match the edges at all:
Acculmulator with detected maxima:
Resulting lines:
As you can see the maximas of the accumulator are satisfyingly detected but the resulting lines' radius values are too high and theta values are missing.
It almost looks like the hough transform accumulator does not correspond to the image...
Can someone help me figure out why and how to correct it?
Here is my code:
function [r, theta] = findScratches (img, edge)
hough = houghtf(edge,"line", pi*[0:360]/180);
threshHough = hough>.5*max(hough(:));
[r, theta] = find(threshHough>0);
%deg to rad for the trig functions
theta = theta/180*pi;
%according to octave doc r range is 2*diagonal
%-> bring it down to 1*diagonal or all lines are out of the picture
r = r/2;
%coefficients of the line y=ax+b
a = -cos(theta)./sin(theta);
b = r./sin(theta);
x = 1:size(img,2);
y = a * x + b;
figure(1)
imagesc(edge);
colormap gray;
hold on;
for i=1:size(y,1)
axis ij;
plot(y(i,:),x,'r','linewidth',1);
end
hold off;
endfunction
Thank you in advance.
You're definitely on the right track. Blurring the accumulator image would help before looking for the hotspots. Also, why not do a quick erode and dilate before doing the hough transform?
I had the same issue - detected lines had the correct slope but were shifted. The problem is that the r returned by the find(threshHough>0) function call is in the interval of [0,2*diag] while the Hough transform operates with values of r from the interval of [-diag,diag]. Therefore if you change the line
r=r/2
to
r=r-size(hough,1)/2
you will get the correct offset.
Lets define a vector of angles (in radians):
angles=pi*[0:360]/180
You should not take this operation: theta = theta/180*pi.
Replace it by: theta = angles(theta), where theta are indices
Some one commented above suggesting adjusting r to -diag to +diag range by
r=r-size(hough,1)/2
This worked well for me. However another difference was that I used the default angle to compute Hough Transform with angles -90 to +90. The theta range in the vector is +1 to +181. So It needs to be adjusted by -91, then convert to radian.
theta = (theta-91)*pi/180;
With above 2 changes, rest of the code works ok.
I am trying to plot the function
f(x, y) = (x – 3).^2 – (y – 2).^2.
x is a vector from 2 to 4, and y is a vector from 1 to 3, both with increments of 0.2. However, I am getting the error:
"Subscript indices must either be real positive integers or logicals".
What do I do to fix this error?
I (think) I see what you are trying to achieve. You are writing your syntax like a mathematical function definition. Matlab is interpreting f as a 2-dimensional data type and trying to assign the value of the expression to data indexed at x,y. The values of x and y are not integers, so Matlab complains.
If you want to plot the output of the function (we'll call it z) as a function of x and y, you need to define the function quite differently . . .
f = #(x,y)(x-3).^2 - (y-2).^2;
x=2:.2:4;
y=1:.2:3;
z = f( repmat(x(:)',numel(y),1) , repmat(y(:),1,numel(x) ) );
surf(x,y,z);
xlabel('X'); ylabel('Y'); zlabel('Z');
This will give you an output like this . . .
The f = #(x,y) part of the first line states you want to define a function called f taking variables x and y. The rest of the line is the definition of that function.
If you want to plot z as a function of both x and y, then you need to supply all possible combinations in your range. This is what the line containing the repmat commands is for.
EDIT
There is a neat Matlab function meshgrid that can replace the repmat version of the script as suggested by #bas (welcome bas, please scroll to bas' answer and +1 it!) ...
f = #(x,y)(x-3).^2 - (y-2).^2;
x=2:.2:4;
y=1:.2:3;
[X,Y] = meshgrid(x,y);
surf(x,y,f(X,Y));
xlabel('x'); ylabel('y'); zlabel('z');
I typically use the MESHGRID function. Like so:
x = 2:0.2:4;
y = 1:0.2:3;
[X,Y] = meshgrid(x,y);
F = (X-3).^2-(Y-2).^2;
surf(x,y,F);
xlabel('x');ylabel('y');zlabel('f')
This is identical to the answer by #learnvst. it just does the repmat-ing for you.
Your problem is that the function you are using uses integers, and you are trying to assign a double to it. Integers cannot have decimal places. To fix this, you can make it to where it increases in increments of 1, instead of 0.2