i'm new in matlab. I didn't understand how to derive a dirac delta function and then shift it using symbolic toolbox.
syms t
x = dirac(t)
why can't i see the dirac delta function using ezplot(x,[-10,10]) for example?
As others have noted, the Dirac delta function is not a true function, but a generalized function. The help for dirac indicates this:
dirac(X) is not a function in the strict sense, but rather a
distribution with int(dirac(x-a)*f(x),-inf,inf) = f(a) and
diff(heaviside(x),x) = dirac(x).
Strictly speaking, it's impossible for Matlab to plot the Dirac delta function in the normal way because part of it extends to infinity. However, there are numerous workarounds if you want a visualization. A simple one is to use the stem plot function and the > operator to convert the one Inf value to something finite. This produces a unit impulse function (or Kronecker delta):
t = -10:10;
x = dirac(t) > 0;
stem(t,x)
If t and x already exist as symbolic variables/expressions rather than numeric ones you can use subs:
syms t
x = dirac(t);
t2 = -10:10;
x2 = subs(x,t,t2)>0;
stem(t2, x2)
You can write your own plot routine if you want something that looks different. Using ezplot is not likely to work as it doesn't offer as much control.
First, I've not met ezplot before; I had to read up on it. For things that are functionals like your x, it's handy, but you still have to realize it's exactly giving you what it promises: A plot.
If you had the job of plotting the dirac delta function, how would you go about doing it correctly? You can't. You must find a convention of annotating your plot with the info that there is a single, isolated, infinite point in your plot.
Plotting something with a line plot hence is unsuitable for anything but smooth functions (that's a well-defined term). Dirac Delta definitely isn't amongst the class of functions that are smooth. You would typically use a vertical line or something to denote the point where your functional is not 0.
Related
I want to minimize a function like below:
Here, n can be 5,10,50 etc. I want to use Matlab and want to use Gradient Descent and Quasi-Newton Method with BFGS update to solve this problem along with backtracking line search. I am a novice in Matlab. Can anyone help, please? I can find a solution for a similar problem in that link: https://www.mathworks.com/help/optim/ug/unconstrained-nonlinear-optimization-algorithms.html .
But, I really don't know how to create a vector-valued function in Matlab (in my case input x can be an n-dimensional vector).
You will have to make quite a leap to get where you want to be -- may I suggest to go through some basic tutorial first in order to digest basic MATLAB syntax and concepts? Another useful read is the very basic example to unconstrained optimization in the documentation. However, the answer to your question touches only basic syntax, so we can go through it quickly nevertheless.
The absolute minimum to invoke the unconstraint nonlinear optimization algorithms of the Optimization Toolbox is the formulation of an objective function. That function is supposed to return the function value f of your function at any given point x, and in your case it reads
function f = objfun(x)
f = sum(100 * (x(2:end) - x(1:end-1).^2).^2 + (1 - x(1:end-1)).^2);
end
Notice that
we select the indiviual components of the x vector by matrix indexing, and that
the .^ notation effects that the operand is to be squared elementwise.
For simplicity, save this function to a file objfun.m in your current working directory, so that you have it available from the command window.
Now all you have to do is to call the appropriate optimization algorithm, say, the quasi Newton method, from the command window:
n = 10; % Use n variables
options = optimoptions(#fminunc,'Algorithm','quasi-newton'); % Use QM method
x0 = rand(n,1); % Random starting guess
[x,fval,exitflag] = fminunc(#objfun, x0, options); % Solve!
fprintf('Final objval=%.2e, exitflag=%d\n', fval, exitflag);
On my machine I see that the algorithm converges:
Local minimum found.
Optimization completed because the size of the gradient is less than
the default value of the optimality tolerance.
Final objval=5.57e-11, exitflag=1
Lets say, I have a function 'x' and a function '2sin(x)'
How do I output the intersects, i.e. the roots in MATLAB? I can easily plot the two functions and find them that way but surely there must exist an absolute way of doing this.
If you have two analytical (by which I mean symbolic) functions, you can define their difference and use fzero to find a zero, i.e. the root:
f = #(x) x; %defines a function f(x)
g = #(x) 2*sin(x); %defines a function g(x)
%solve f==g
xroot = fzero(#(x)f(x)-g(x),0.5); %starts search from x==0.5
For tricky functions you might have to set a good starting point, and it will only find one solution even if there are multiple ones.
The constructs seen above #(x) something-with-x are called anonymous functions, and they can be extended to multivariate cases as well, like #(x,y) 3*x.*y+c assuming that c is a variable that has been assigned a value earlier.
When writing the comments, I thought that
syms x; solve(x==2*sin(x))
would return the expected result. At least in Matlab 2013b solve fails to find a analytic solution for this problem, falling back to a numeric solver only returning one solution, 0.
An alternative is
s = feval(symengine,'numeric::solve',2*sin(x)==x,x,'AllRealRoots')
which is taken from this answer to a similar question. Besides using AllRealRoots you could use a numeric solver, manually setting starting points which roughly match the values you have read from the graph. This wa you get precise results:
[fzero(#(x)f(x)-g(x),-2),fzero(#(x)f(x)-g(x),0),fzero(#(x)f(x)-g(x),2)]
For a higher precision you could switch from fzero to vpasolve, but fzero is probably sufficient and faster.
I have been working on solving some equation in a more complicated context. However, I want to illustrate my question through the following simple example.
Consider the following two functions:
function y=f1(x)
y=1-x;
end
function y=f2(x)
if x<0
y=0;
else
y=x;
end
end
I want to solve the following equation: f1(x)=f2(x). The code I used is:
syms x;
x=solve(f1(x)-f2(x));
And I got the following error:
??? Error using ==> sym.sym>notimplemented at 2621
Function 'lt' is not implemented for MuPAD symbolic objects.
Error in ==> sym.sym>sym.lt at 812
notimplemented('lt');
Error in ==> f2 at 3
if x<0
I know the error is because x is a symbolic variable and therefore I could not compare x with 0 in the piecewise function f2(x).
Is there a way to fix this and solve the equation?
First, make sure symbolic math is even the appropriate solution method for your problem. In many cases it isn't. Look at fzero and fsolve amongst many others. A symbolic method is only needed if, for example, you want a formula or if you need to ensure precision.
In such an old version of Matlab, you may want to break up your piecewise function into separate continuous functions and solve them separately:
syms x;
s1 = solve(1-x^2,x) % For x >= 0
s2 = solve(1-x,x) % For x < 0
Then you can either manually examine or numerically compare the outputs to determine if any or all of the solutions are valid for the chosen regime – something like this:
s = [s1(double(s1) >= 0);s2(double(s2) < 0)]
You can also take advantage of the heaviside function, which is available in much older versions.
syms x;
f1 = 1-x;
f2 = x*heaviside(x);
s = solve(f1-f2,x)
Yes, the Heaviside function is 0.5 at zero – this gives it the appropriate mathematical properties. You can shift it to compare values other than zero. This is a standard technique.
In Matlab R2012a+, you can take advantage of assumptions in addition to the normal relational operators. To add to #AlexB's comment, you should convert the output of any logical comparison to symbolic before using isAlways:
isAlways(sym(x<0))
In your case, x is obviously not "always" on one side or the other of zero, but you may still find this useful in other cases.
If you want to get deep into Matlab's symbolic math, you can create piecewise functions using MuPAD, which are accessible from Matlab – e.g., see my example here.
I'm trying to find the power in an aperture from a Gaussian beam, where the aperture is offset from the beam center. The solution is the following equation (reference) (sorry, no LaTeX here):
Wz is a constant, along with a and r. I'm not sure how I can do something like this with MATLAB. Does anyone have a suggestion? I know there's a dblquad() function, but it assumes that the limits of integration are fixed, and not dependent on each other.
Using a bit of mathematical footwork, you could reduce the double integral to a single one (albeit containing the error function) which should be easier to calculate numerically in MATLAB:
(With reservation for errors; check the calculations yourself if possible.)
It turns out that more recent versions of MATLAB now have a quad2d() function, which does a 2d integral over a surface. Example 2 on the reference page details an example of doing this type of integration.
My code ended up looking something like this:
powerIntegral = #(x,y) 2/(pi*W^2)*exp(-2*((x - offsetDist).^2 + y.^2)/(W^2));
ymin = #(x) -sqrt(radius.^2 - x.^2);
ymax = #(x) sqrt(radius.^2 - x.^2);
powerRatioGaussian = quad2d(powerIntegral,-radius,radius,ymin,ymax);
Pretty nifty. Thanks for the help.
I am not sure, but I think that the symbolic toolbox can help you here. It is suited for this kind of problems. You can define your variables as symbolic vars using the syms command, and compute the integral symbolically. Then, you can assign the variables values and find the actual value.
Disclaimer : I have never actually used it myself.
Generally speaking, for numeric integration, you can transform an integral with dependent boundary conditions to one with independent boundaries by multiplying by a function that is 1 if you are inside the original boundary and 0 if you are outside. Then take your limits to be a square that contains your original conditions. In other words here you would multiply by
g(x,y) = ((x^2 + y^2) < a^2)
and your limits would be -a
You have to be a little careful about the continuity assumptions in your integration method, but you should be OK unless something is very weird. You can always check by changing your cell size and making sure the computed integral value doesn't change.
In this particular case, you could also make the transformation from cartesian to polar coordinates;
x = rcos(t)
y = rsin(t)
dxdy = rdrdt
Then your limits of integration would be r from 0 to a and t from 0 to 2*pi
If I have an RC circuit with transfer function 1/(1+sRC) how do I draw the transfer function using MATLAB?
Num2=[1];
Den2=[R*C 1];
RCcirc=tf(Num2,Den2);
How do I declare the R and the C so that there are no errors?
tf is the wrong tool for plotting the transfer function. Try these instead:
Use linspace to generate a range of values for s. Give R and C reasonable values of your choice.
Read up on arithmetic operations in MATLAB, especially ./
Look at how to use plot and familiarize yourself with the command using some simple examples from the docs.
With these you should be able to plot the transfer function in MATLAB :)
First of all you need to understand what transfer function you want. Without defined values of R and C you won't get any transfer function. Compare it to this, you want to plot a sine wave: x = sin(w*t), I hope you can agree with me that you cannot plot such a function (including axes) unless I specifically say e.g. t is the time, ranging from 0 seconds to 10 seconds and w is a pulsation of 1 rad/s. It's exactly the same with your RC network: without any values, it is impossible for numerical software such as MATLAB to come up with a plot.
If you fill in those values, you can use th tf function to display the transfer function in whatever way you like (e.g. a bode plot).
On the other hand, if you just want the expression 1/(1+s*R*C), take a look at the symbolic toolbox, you can do such things there. But to make a plot, you will still have to fill in the R and C value (and even a value for your Laplace variable in this case).