I want to use ezplot in MATLAB, and because the function I want to plot consists of a large number of terms I may split it into smaller functions. Let me give an example of a small number of terms and it can be generalized to a large number of terms. To plot the function:
y2+xy+xy3+x+1=0
I let y1=x+1 and I write the following in MATLAB:
x=[0:1:5]
y1=x+1
ezplot('y.^2+x*y+x*y.^3+y1')
But there is an error. Please tell me how can I correct the error. Is it possible to use this feature (splitting the equation or function into a number of terms)?
Your error is caused by trying to replace x+1 with y1. ezplot requires that symbolic expressions are functions of only 2 symbolic variables. However, there are 3 symbolic variables (x, y, and y1) in your call to ezplot:
ezplot('y^2+x*y+x*y^3+y1');
If you use your original equation, everything should work fine:
ezplot('y^2+x*y+x*y^3+x+1');
EDIT: In case you were curious...
If you want to plot an equation with 3 variables, you will first need to solve the equation for one of them and then use the function ezsurf (this is illustrated in this answer I gave to another SO question). Technically, y1 is a dependent variable the way you have defined it (since it depends on the variable x). However, for the sake of the following example, let's assume it's an independent variable. The equation:
y^2 + x*y + x*y^3 + y1 = 0
would be solved for y1 to get the following:
y1 = -y^2 - x*y - x*y^3
and y1 would be plotted in the following way:
ezsurf('-y^2-x*y-x*y^3');
Related
I have to fit the dots, results of measurements, by an exponential function on Matlab. My profesor asked me to use only
fminsearch
polyval
polyfit
One of them or both. I have to find the parameters a and b (the value) which are fitting it.
There is the lines I wrote :
x=[1:10:70]
y=[0:10:70]
x=[12.5,11.8,10.8,10.9,6.5,6.2,6.1,5.423,4.625]
y=[0,0.61,1.3,1.4,14.9,18.5,20.1,29.7,58.2]
xlabel('Conductivité')
ylabel('Inductance')
The function has the form a*e^(-b*x) +c
Well polyfit and polyval are only usefull for working with polynomials. So you would have to write a minimization problem of the form min(f(x)).
functionToMinimize = #(pars, x, y)(norm(pars(1).*exp(-pars(2).*x) - y));
targetFunctionForFminseardch = #(pars)(functionToMinimize(pars, x, y));
minPars = fminsearch(targetFunctionForFminseardch, [0, 1])
Read up on anonymous functions and the use of vector norms if you have questions how to construct such a minimization problem.
Your code also has some flaws. Why are you defining x and y twice when you only want to use the actual measured data?
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.
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.
My project require me to use Matlab to create a symbolic equation with square wave inside.
I tried to write it like this but to no avail:
syms t;
a=square(t);
Input arguments must be 'double'.
What can i do to solve this problem? Thanks in advance for the helps offered.
here are a couple of general options using floor and sign functions:
f=#(A,T,x0,x) A*sign(sin((2*pi*(x-x0))/T));
f=#(A,T,x0,x) A*(-1).^(floor(2*(x-x0)/T));
So for example using the floor function:
syms x
sqr=2*floor(x)-floor(2*x)+1;
ezplot(sqr, [-2, 2])
Here is something to get you started. Recall that we can express a square wave as a Fourier Series expansion. I won't bother you with the details, but you can represent any periodic function as a summation of cosines and sines (à la #RTL). Without going into the derivation, this is the closed-form equation for a square wave of frequency f, with a peak-to-peak amplitude of 2 (i.e. it goes from -1 to 1). Recall that the frequency is the amount of cycles per seconds. Therefore, f = 1 means that we repeat our square wave every second.
Basically, what you have to do is code up the first line of the equation... but how in the world would you do that? Welcome to the world of the Symbolic Math Toolbox. What we will need to do before hand is declare what our frequency is. Let's assume f = 1 for now. With the Symbolic Math Toolbox, you can define what are considered as mathematics variables within MATLAB. After, MATLAB has a whole suite of tools that you can use to evaluate functions that rely on these variables. A good example would be if you want to use this to define a closed-form solution of a function f(x). You can then use diff to differentiate and see what the derivative is. Try it yourself:
syms x;
f = x^4;
df = diff(f);
syms denotes that you are declaring anything coming after the statement to be a mathematical variable. In this case, x is just that. df should now give you 4x^3. Cool eh? In any case, let's get back to our problem at hand. We see that there are in fact two variables in the periodic square function that need to be defined: t and k. Once we do this, we need to create our function that is inside the summation first. We can do this by:
syms t k;
f = 1; %//Define frequency here
funcSum = (sin(2*pi*(2*k - 1)*f*t) / (2*k - 1));
That settles that problem... now how do we encapsulate this into an infinite sum!? The sum command in MATLAB assumes that we have a finite array to sum over. If you want to symbolically sum over a function, we must use the symsum function. We usually call it like this:
funcOut = symsum(func, v, start, finish);
func is the function we wish to sum over. v is the summation variable that we wish to use to index in the sum. In our case, that's k. start is the beginning of the sum, which is 1 in our case, and finish is where we wish to finish up our summation. In our case, that's infinity, and so MATLAB has a special keyword called Inf to denote that. Therefore:
xsquare = (4/pi) * symsum(funcSum, k, 1, Inf);
xquare now contains your representation of a square wave defined in terms of the Symbolic Math Toolbox. Now, if you want to plot your square wave and see if we have this right. We can do the following. Let's go between -3 <= t <= 3. As such, you would do something like this:
tVector = -3 : 0.01 : 3; %// Choose a step size of 0.01
yout = subs(xsquare, t, tVector);
You will notice though that there will be some values that are NaN. The reason why is because right at a multiple of the period (T = 1, 2, 3, ...), the behaviour is undefined as the derivative right at these points is undefined. As such, we can fill this in using either 1 or -1. Let's just choose 1 for now. Also, because the Fourier Series is generally a complex-valued function, and the square-wave is purely real, the output of this function will actually give you a complex-valued vector. As such, simply chop off the complex parts to get the real parts only:
yout = real(double(yout)); %// To cast back to double.
yout(isnan(yout)) = 1;
plot(tVector, yout);
You'll get something like:
You could also do this the ezplot way by doing: ezplot(xsquare). However, you'll see that at the points where the wave repeats itself, we get NaN values and so there is a disconnect between the high peak and low peak.
Note:
Natan's solution is much more elegant. I was still writing this post by the time he put something up. Either way, I wanted to give a more signal processing perspective to how to do this. Go Fourier!
A Fourier series for the square wave of unit amplitude is:
alpha + 2/Pi*sum(sin( n * Pi*alpha)/n*cos(n*theta),n=1..infinity)
Here is a handy trick:
cos(n*theta) = Re( exp( I * n * theta))
and
1/n*exp(I*n*theta) = I*anti-derivative(exp(I*n*theta),theta)
Put it all together: pull the anti-derivative ( or integral ) operator out of the sum, and you get a geometric series. Then integrate and finally take the real part.
Result:
squarewave=
alpha+ 1/Pi*Re(I*ln((1-exp(I*(theta+Pi*alpha)))/(1-exp(I*(theta-Pi*alpha)))))
I tried it in MAPLE and it works great! (probably not very practical though)