This answer explains how to substitute a derivative in a symbolic expression. However, this approach fails when why function is "chained twice", e.g. sin(psi(phi(t))). See the example
clear;
syms t phi(t) psi(phi)
psi = psi(phi)
dphidt = diff(phi, t)
dpsidphi = diff(psi, phi)
x = sin(psi)
dxdt = diff(x, t)
syms phid psip % Substitution variables for derivatives
test1 = subs(dxdt,dphidt,phid) % This works
test2 = subs(dxdt,dpsidphi,psip) % This doesn't work
% Ideal result, without any dependencies
res = psip*phid*cos(psi)
Substituting phid for diff(phi(t), t) works, but substituting psip for D(psi)(phi(t)) does not work. What am I doing wrong? Do I have to specify the chained dependecies in a different way? I also don't quite understand the difference between D and diff.
Second question:
My actual terms are a lot more complicated and I have to take derivatives with respect to phid for example. Therefore, I have to get rid of all the dependencies because I want to generate a function using matlabFunction. I can remove the t dependency using formula, however I can't get rid of the phi(psi) dependency. Is there a command for this or do I have to make another substitution?
Related
I created a symbolic function in MATLAB R2021b using this script with the goal of solving an ODE.
syms phi(x) lambda L
eqn_x = diff(phi,x,2) == -lambda*phi;
dphi = diff(phi,x);
cond = [phi(0)==0, dphi(1)==0]; % this is the line where the problem starts
disp(cond)
phi = dsolve(eqn_x,x,cond);
This script runs without any errors, but I want to evaluate dphi(L)==0 instead of at x=1. When I try to do this instead, I get the following error.
Obviously, the way MATLAB is interpreting dphi(L)==0 is not correct since it is replacing the dependent parameter x with L which should be an independent parameter. Can someone recommend a fix or workaround?
I've also tried this dphi(x==L)==0 and I got the same error.
Addition following kikon's comment
I tried both options suggested based on the tutorial. These are the results.
dphi | x = L creates a parse error in MATLAB
dphi | x == L gives this result. This seems to indicate x=l OR dphi/dx which is not a condition. I'm tempted to try & in place of | to get an and, but this may not be sufficient.
subs(dphi, x = L) gives the same condition when I use dphi(L)==0 which is incorrect.
I am writing a code that generates a function f in a loop. This function f changes in every loop, for example from f = x + 2x to f = 3x^2 + 1 (randomly), and I want to evaluate f at different points in every loop. I have tried using subs, eval, matlabFunction etc but it is still running slowly. How would you tackle a problem like this in the most efficient way?
This is as fast as I have been able to do it. ****matlabFunction and subs go slower than this.
The code below is my solution and it is one loop. In my larger code the function f and point x0 change in every loop so you can imagine why I want this to go as fast as possible. I would greatly appreciate it if someone could go through this, and give me any pointers. If my coding is crap feel free to tell me :D
x = sym('x',[2,1]);
f = [x(1)-x(1)cos(x(2)), x(2)-3x(2)^2*cos(x(1))];
J = jacobian(f,x);
x0 = [2,1];
N=length(x0); % Number of equations
%% Transform into string
fstr = map2mat(char(f));
Jstr = map2mat(char(J));
% replace every occurence of 'xi' with 'x(i)'
Jstr = addPar(Jstr,N);
fstr = addPar(fstr,N);
x = x0;
phi0 = eval(fstr)
J = eval(Jstr)
function str = addPar(str,N)
% pstr = addPar(str,N)
% Transforms every occurence of xi in str into x(i)
% N is the maximum value of i
% replace every occurence of xi with x(i)
% note that we do this backwards to avoid x10 being
% replaced with x(1)0
for i=N:-1:1
is = num2str(i);
xis = ['x' is];
xpis = ['x(' is ')'];
str = strrep(str,xis,xpis);
end
function r = map2mat(r)
% MAP2MAT Maple to MATLAB string conversion.
% Lifted from the symbolic toolbox source code
% MAP2MAT(r) converts the Maple string r containing
% matrix, vector, or array to a valid MATLAB string.
%
% Examples: map2mat(matrix([[a,b], [c,d]]) returns
% [a,b;c,d]
% map2mat(array([[a,b], [c,d]]) returns
% [a,b;c,d]
% map2mat(vector([[a,b,c,d]]) returns
% [a,b,c,d]
% Deblank.
r(findstr(r,' ')) = [];
% Special case of the empty matrix or vector
if strcmp(r,'vector([])') | strcmp(r,'matrix([])') | ...
strcmp(r,'array([])')
r = [];
else
% Remove matrix, vector, or array from the string.
r = strrep(r,'matrix([[','['); r = strrep(r,'array([[','[');
r = strrep(r,'vector([','['); r = strrep(r,'],[',';');
r = strrep(r,']])',']'); r = strrep(r,'])',']');
end
There are several ways to get huge boosts in speed for this sort of problem:
The java GUI front end slows everything down. Go back to version 2010a or earlier. Go back to when it was based on C or fortran. The MATLAB script runs as fast as if you had put it into the MATLAB "compiler".
If you have MatLab compiler (or builder, I forget which) but not the coder, then you can process your code and have it run a few times faster without modifying the code.
write it to a file, then call it as a function. I have done this for changing finite-element expressions, so large ugly math that makes $y = 3x^2 +1$ look simple. In that it gave me solid speed increase.
vectorize, vectorize, vectorize. It used to reliably give 10x to 12x speed increase. Pull it out of loops. The java, I think, obscures this some by making everything slower.
have you "profiled" your function to make sure that "eval" or such are the problem? If you fix "eval" and your bottleneck is elsewhere then you will have problems.
If you have the choice between eval and subs, stick with eval. subs gives you a symbolic solution, not a numeric one.
If there is a clean way to have multiple instances of MatLab running, especially if you have a decently core-rich cpu that MatLab does not fully utilize, then get several of them going. If you are at an educational institution you might try running several different versions (2010a, 2010b, 2009a,...) on the same system. I (fuzzily) recall they didn't collide when I did it. Running more than about 8 started slowing things down more than it improved them. Make sure they aren't colliding on file access if you are using files to share control.
You could write your program in LabVIEW (not MathScript, not MatLab) and because it is a compiled language, there are times that code can run 1000x faster.
You could go all numeric and make it a matrix activity. This depends on your code, but if you could randomly populate the columns in the matrix then matrix multiply it to a matrix $ \left[ 1, x, x^{2}, ...\right] $, that would likely be several hundreds or thousands of times faster than your current level of equation handling and still in MatLab.
About your coding:
don't redeclare "x" as a symbol every loop, that is expensive.
what is this "map2mat" then "addPar" stuff?
the string handling functions are horrible for runtime. Stick to one language. The symbolic toolbox IS maple, and you don't have to get goofy hand-made parsing to make it work with the rest of MatLab.
I have the following code:
syms t x;
e=symfun(x-t,[x,t]);
In the problem I want to solve x is a function of t but I only know its value at the given t,so I modeled it here as a variable.I want to differentiate e with respect to time without "losing" x,so that I can then substitute it with x'(t) which is known to me.
In another question of mine here,someone suggested that I write the following:
e=symfun(exp(t)-t,[t]);
and after the differentiation check if I can substitute exp(t) with the value of x'(t).
Is this possible?Is there any other neater way?
I'm really not sure I understand what you're asking (and I didn't understand your other question either), but here's an attempt.
Since, x is a function of time, let's make that explicit by making it what the help and documentation for symfun calls an "abstract" or "arbitrary" symbolic function, i.e., one without a definition. In Matlab R2014b:
syms t x(t);
e = symfun(x-t,t)
which returns
e(t) =
x(t) - t
Taking the derivative of the symfun function e with respect to time:
edot = diff(e,t)
returns
edot(t) =
D(x)(t) - 1
the expression for edot(t) is a function of the derivative of x with respect to time:
xdot = diff(x,t)
which is the abstract symfun:
xdot(t) =
D(x)(t)
Now, I think you want to be able to substitute a specific value for xdot (xdot_given) into e(t) for t at t_given. You should be able to do this just using subs, e.g., something like this:
sums t_given xdot_given;
edot_t_given = subs(edot,{t,xdot},{t_given, xdot_given});
You may not need to substitute t if the only parts of edot that are a function of time are the xdot parts.
I have the following script that defines a function and sets up and equation:
H = #(f) sum(log(f));
f = rand(1, 1);
syms a
H(f)-H(f-a)
I want to solve H(f)-H(f-a)=0 for a. I tried using fzero in the following manner, fzero('H(f)-H(f-a)', 0), but this doesn't yield me anything useful.
well the clear answer for this would be that a=0. This is of course the obvious answer, and may be why you are not getting "anything useful", especially when only considering one value for f.
However, you can actually get useful information when you add more values and use the solve function.
H = #(f) sum(log(f));
f = rand(5, 1);
syms a
temp = H(f)-H(f-a);
out = double(solve(temp == 0,a));
This will output a vector with all values of a that are solutions to your function. Just note that this may not result in the best answers in general, numerical solving of high order functions is rather difficult and unreliable. Because of this, it may take a long time to run as well.
I'm trying to model the effect of different filter "building blocks" on a system which is a construct based on these filters.
I would like the basic filters to be "modular", i.e. they should be "replaceable", without rewriting the construct which is based upon the basic filters.
For example, I have a system of filters G_0, G_1, which is defined in terms of some basic filters called H_0 and H_1.
I'm trying to do the following:
syms z
syms H_0(z) H_1(z)
G_0(z)=H_0(z^(4))*H_0(z^(2))*H_0(z)
G_1(z)=H_1(z^(4))*H_0(z^(2))*H_0(z)
This declares the z-domain I'd like to work in, and a construct of two filters G_0,G_1, based on the basic filters H_0,H_1.
Now, I'm trying to evaluate the construct in terms of some basic filters:
H_1(z) = 1+z^-1
H_0(z) = 1+0*z^-1
What I would like to get at this point is an expanded polynomial of z.
E.g. for the declarations above, I'd like to see that G_0(z)=1, and that G_1(z)=1+z^(-4).
I've tried stuff like "subs(G_0(z))", "formula(G_0(z))", "formula(subs(subs(G_0(z))))", but I keep getting result in terms of H_0 and H_1.
Any advice? Many thanks in advance.
Edit - some clarifications:
In reality, I have 10-20 transfer functions like G_0 and G_1, so I'm trying to avoid re-declaring all of them every time I change the basic blocks H_0 and H_1. The basic blocks H_0 and H_1 would actually be of a much higher degree than they are in the example here.
G_0 and G_1 will not change after being declared, only H_0 and H_1 will.
H_0(z^2) means using z^2 as an argument for H_0(z). So wherever z appears in the declaration of H_0, z^2 should be plugged in
The desired output is a function in terms of z, not H_0 and H_1.
A workable hack is having an m-File containing the declarations of the construct (G_0 and G_1 in this example), which is run every time H_0 and H_1 are redefined. I was wondering if there's a more elegant way of doing it, along the lines of the (non-working) code shown above.
This seems to work quite nicely, and is very easily extendable. I redefined H_0 to H_1 as an example only.
syms z
H_1(z) = 1+z^-1;
H_0(z) = 1+0*z^-1;
G_0=#(Ha,z) Ha(z^(4))*Ha(z^(2))*Ha(z);
G_1=#(Ha,Hb,z) Hb(z^(4))*Ha(z^(2))*Ha(z);
G_0(H_0,z)
G_1(H_0,H_1,z)
H_0=#(z) H_1(z);
G_0(H_0,z)
G_1(H_0,H_1,z)
This seems to be a namespace issue. You can't define a symbolic expression or function in terms of arbitrary/abstract symfuns and then later on define these symfuns explicitly and be able to use them to obtain an exploit form of the original symbolic expression or function (at least not easily). Here's an example of how a symbolic function can be replaced by name:
syms z y(z)
x(z) = y(z);
y(z) = z^2; % Redefines y(z)
subs(x,'y(z)',y)
Unfortunately, this method depends on specifying the function(s) to be substituted exactly – because strings are used, Matlab sees arbitrary/abstract symfuns with different arguments as different functions. So the following example does not work as it returns y(z^2):
syms z y(z)
x(z) = y(z^2); % Function of z^2 instead
y(z) = z^2;
subs(x,'y(z)',y)
But if the last line was changed to subs(x,'y(z^2)',y) it would work.
So one option might be to form strings for case, but that seems overly complex and inelegant. I think that it would make more sense to simply not explicitly (re)define your arbitrary/abstract H_0, H_1, etc. functions and instead use other variables. In terms of the simple example:
syms z y(z)
x(z) = y(z^2);
y_(z) = z^2; % Create new explicit symfun
subs(x,y,y_)
which returns z^4. For your code:
syms z H_0(z) H_1(z)
G_0(z) = H_0(z^4)*H_0(z^2)*H_0(z);
G_1(z) = H_1(z^4)*H_0(z^2)*H_0(z);
H_0_(z) = 1+0*z^-1;
H_1_(z) = 1+z^-1;
subs(G_0, {H_0, H_1}, {H_0_, H_1_})
subs(G_1, {H_0, H_1}, {H_0_, H_1_})
which returns
ans(z) =
1
ans(z) =
1/z^4 + 1
You can then change H_0_ and H_1_, etc. at will and use subs to evaluateG_1andG_2` again.