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.
Related
I want to solve a system of linear equalities of the type Ax = b+u, where A and b are known. I used a function in MATLAB like this:
x = #(u) gmres(A,b+u);
Then I used fmincon, where a value for u is given to this expression and x is computed. For example
J = #(u) (x(u)' * x(u) - x^*)^2
and
[J^*,u] = fmincon(J,...);
withe the dots as matrices and vectors for the equalities and inequalities.
My problem is, that MATLAB delivers always an output with information about the command gmres. But I have no idea, how I can stop this (it makes the Program much slower).
I hope you know an answer.
Patsch
It's a little hidden in the documentation, but it does say
No messages are displayed if the flag output is specified.
So you need to call gmres with at least two outputs. You can do this by making a wrapper function
function x = gmresnomsg(varargin)
[x,~] = gmres(varargin{:});
end
and use that for your handle creation
x = #(u) gmresnomsg(A,b+u);
I am trying to find the roots of a nonlinear function that depends on several symbolic variables and a symbolic function. To guarantee the existence of a solution I need to set some assumptions on the symbolic function b(x).
By using the Matlab function assume I can set simple assumptions like
syms b(x) d
assume(b(0)==0 & b(x)>=0);
Now I am looking for a way to include this assumption: (symbolic function b(x), symbolic variable d)
There is a value y such that b(x) < d*x for x < y and b(x) > d*x for x > y.
To make it easier we can fix for example d=0.1 and b(10)=1 such that y=10. However I would prefer leaving d and y symbolic.
I tried to write a function that checks the assumption but unsurprisingly Matlab does not seem to accept a symbolic function as an input variable.
Do you know a solution to this problem or have an idea?
Update 2017-01-03:
While having fixed d=0.1,b(10)=1 and y=10 I am trying to check the above assumption by using
syms b(x)
d=0.1;
assume(x, 'real');
assume(x>=0);
assume(b(0)==0 & b(x)>=0 & b(10)==1);
checkAssumption(x)= mySign(x)*(d*x-b(x));
assume(checkAssumption(x)>=0);
where
function sign=mySign(x)
if x<10
sign=-1;
else if x==10
sign=0;
else if x>10
sign=-1;
end
end
end
Then I get the error "Conversion to logical from sym is not possible." in mySign. Using double(x) as the input variable of mySign leads to the error "DOUBLE cannot convert the input expression into a double array".
Do you know a way to get mySign and checkAssumptions evaluated or have an other idea how to assume the above assumption?
I first defined functions for dy/dt=y and dy/dt=t:
function dy=d(y):
dy=y
end
function ddy=dd(t):
ddy=t
end
And then I used ode45, respectively:
[t,y]=ode45('d',[1 10],1)
[t,y]=ode45('dd',[1 10],1)
which returns the following error: Error using d
Too many input arguments.
My question is:
Where did I go wrong?
How does Matlab know whether y or t is the independent variable? When I define the first function, it could be reasonably interpreted as dt/dy=y instead of dy/dt=y. Is there a built-in convention for defining functions?
First things first: the docs on ode45 are on the mathworks website, or you can get them from the console by entering help ode45.
The function you pass in needs to take two variables, y then t. As you noticed, with just one it would be impossible to distinguish a function of only y from a function of only t. The first argument has to be the independent, the second is the dependent.
Try defining your function as dy = d(t, y) and ddy = dd(t, y) with the same bodies.
one other note, while using a string representing the function name should work, you can use #d and #dd to reference the functions directly.
I'm trying to use the MATLAB function fzero properly but my program keeps returning an error message. This is my code (made up of two m-files):
friction_zero.m
function fric_zero = friction_zero(reynolds)
fric_zero = 0.25*power(log10(5.74/(power(reynolds,0.9))),-2);
flow.m
function f = flow(fric)
f = 1/(sqrt(fric))-1.873*log10(reynolds*sqrt(fric))-233/((reynolds*sqrt(fric))^0.9)-0.2361;
f_initial = friction_zero(power(10,4));
z = fzero(#flow,f_initial)
The goal is to return z as the root for the equation specified by f when flow.m is run.
I believe I have the correct syntax as I have spent a couple of hours online looking at examples. What happens is that it returns the following error message:
"Undefined function or variable 'fric'."
(Of course it's undefined, it's the variable I'm trying to solve!)
Can someone point out to me what I've done wrong? Thanks
EDIT
Thanks to all who helped! You have assisted me to eventually figure out my problem.
I had to add another file. Here is a full summary of the completed code with output.
friction_zero.m
function fric_zero = friction_zero(re)
fric_zero = 0.25*power(log10(5.74/(power(re,0.9))),-2); %starting value for fric
flow.m
function z = flow(fric)
re = power(10,4);
z = 1/(sqrt(fric))-1.873*log10(re*sqrt(fric))-233/((re*sqrt(fric))^0.9)-0.2361;
flow2.m
f_initial = friction_zero(re); %arbitrary starting value (Reynolds)
x = #flow;
fric_root = fzero(x,f_initial)
This returns an output of:
fric_root = 0.0235
Which seems to be the correct answer (phew!)
I realised that (1) I didn't define reynolds (which is now just re) in the right place, and (2) I was trying to do too much and thus skipped out on the line x = #flow;, for some reason when I added the extra line in, MATLAB stopped complaining. Not sure why it wouldn't have just taken #flow straight into fzero().
Once again, thanks :)
You need to make sure that f is a function in your code. This is simply an expression with reynolds being a constant when it isn't defined. As such, wrap this as an anonymous function with fric as the input variable. Also, you need to make sure the output variable from your function is z, not f. Since you're solving for fric, you don't need to specify this as the input variable into flow. Also, you need to specify f as the input into fzero, not flow. flow is the name of your main function. In addition, reynolds in flow is not defined, so I'm going to assume that it's the same as what you specified to friction_zero. With these edits, try doing this:
function z = flow()
reynolds = power(10,4);
f = #(fric) 1/(sqrt(fric))-1.873*log10(reynolds*sqrt(fric))-233/((reynolds*sqrt(fric))^0.9)-0.2361;
f_initial = friction_zero(reynolds);
z = fzero(#f, f_initial); %// You're solving for `f`, not flow. flow is your function name
The reason that you have a problem is because flow is called without argument I think. You should read a little more about matlab functions. By the way, reynolds is not defined either.
I am afraid I cannot help you completely since I have not been doing fluid mechanics. However, I can tell you about functions.
A matlab function definition looks something like this:
function x0 = f(xGuess)
a = 2;
fcn =#(t) a*t.^3+t; % t must not be an input to f.
disp(fcn);
a = 3;
disp(fcn);
x0 = fsolve(fcn1,xGuess); % x0 is calculated here
The function can then ne called as myX0 = f(myGuess). When you define a matlab function with arguments and return values, you must tell matlab what to do with them. Matlab cannot guess that. In this function you tell matlab to use xGuess as an initial guess to fsolve, when solving the anonymous function fcn. Notice also that matlab does not assume that an undefined variable is an independent variable. You need to tell matlab that now I want to create an anonymous function fcn which have an independent variable t.
Observation 1: I use .^. This is since the function will take an argument an evaluate it and this argument can also be a vector. In this particulat case I want pointwise evaluation. This is not really necessary when using fsolve but it is good practice if f is not a matrix equation, since "vectorization" is often used in matlab.
Observation 2: notice that even if a changes its value the function does not change. This is since matlab passes the value of a variable when defining a function and not the variable itself. A c programmer would say that a variable is passed by its value and not by a pointer. This means that fcn is really defined as fcn = #(x) 2*t.^3+t;. Using the variable a is just a conveniance (constants can may also be complicated to find, but when found they are just a value).
Armed with this knowledge, you should be able to tackle the problem in front of you. Also, the recursive call to flow in your function will eventuallt cause a crash. When you write a function that calls itself like this you must have a stopping criterium, something to tell the program when to stop. As it is now, flow will call ifself in the last row, like z = fzero(#flow,f_initial) for 500 times and then crash. Alos it is possible as well to define functions with zero inputs:
function plancksConstant = h()
plancksConstant = 6.62606957eā34;
Where the call h or h() will return Plancks constant.
Good luck!
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 ā 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.