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I wanted to plot the load voltage across the resistor in series with a diode using matlab. This is a rather simple example involving piecewise functions, however I ran into an unexpected error.
t=[0:0.001:10]
vs=4*sin(pi * t)
for i =1:length(vs)
if(vs(i)<=0.7)
v(i)=0;
else
v(i)=vs(i)-0.7;
end
end
plot(t,v)
t is the time, vs is the source voltage, and v is the load voltage. Now, running this gave me an error saying "error, t and v are of different sizes..". Using length() I found out that while t and vs are of lengths 10001, v is somehow of length 1000001.
This is truly baffling to me. How can v and vs possibly differ in size? every element of vs was mapped to an element of v, and yet the size of v comes out to be about 100 times the size of vs.
Being new to matlab, I still am not very comfortable with not explicitly declaring the array v before using it in the for loop. However, I went through with it, because the example I worked on prior to this, used the same thing and it worked without any problems. We simply had a plot a piecewise function:
x=[-2 : 0.00001 : 20];
for i=1: length(x)
if(x(i)>=-2 && x(i)<0)
y(i)=sqrt(x(i)^2+1);
else if(x(i)>=0 && x(i)<10)
y(i)=3*x(i)+1;
else
y(i)=9*sin(5*x(i)-50);
end
end
end
plot(x,y)
This code worked flawlessly, and in my opinion the initial code is fundamentally doing the same thing, so again, I'm clueless as to why it failed.
The original code works if you initialise v to be of the same size as t (and therefore, vs), but still, I want to know why the code involving x,y worked (where y wasn't initialised) and the code involving (t,v) failed.
Also, my friend copy pasted the entire code into the command window of matlab 2016, and it worked. (and I'm using the 2021 version).
Its good practice to initialize variables before entering a loop. It will help avoid undefinied behaviour when you run the script multiple times. If you run the script with different lengths for t, it would fail the second run. One solution would be:
t=0:0.001:10;
vs=4*sin(pi * t);
v=nan(size(t));
for i =1:length(vs)
if(vs(i)<=0.7)
v(i)=0;
else
v(i)=vs(i)-0.7;
end
end
figure;
plot(t,v);
You could also avoid the for loop and use matrix operations instead:
t=0:0.001:10;
vs=4*sin(pi * t);
v=vs-0.7;
v(vs<=0.7)=0;
figure;
plot(t,v);
Dear community of stack overflow. Is it possible to improve the speed of this code in Matlab? Can I use vectorization? Note that I have to use in every loop the "vpasolve" or "fsolve" function.
Here is the code which calls the function with a double loop:
for i=1:1000
for j=1:1000
SilA(i,j)=SolW(i,j);
end
end
Here is the function which is called by the above code. Can I vectorize the input of w, xi and make the code run faster?
function [ffw] = SolW(w,xi)
format long e;
z=0;mm=0.46;sop=80;
epit=0.1;voP=80.;
rho=2.1;aposC=0.1;aposD=0.1;
parEh=0.2*10^6;parEv=0.2*10^6;parGv=0.074*10^6;
parpos=0.35;hp=0.2;Ep=30*10^6;
parposV=0.20;ll=0.15;dd=2*ll;
C11=(parEh*(parEv-parEh*parpos^2)/((1+parpos)*(parEv-parEv*parpos-2*parEh*parpos^2)))*(1+2*1i*aposD);
C13=(parEh*parEv*parpos/((parEv-parEv*parpos-2*parEh*parpos^2)))*(1+2*1i*aposD);
C33=((1-parpos)*parEv^2/(parEv-parEv*parpos-2*parEh*parpos^2))*(1+2*1i*aposD);
C44=parGv*(1+2*1i*aposD);
DD=(Ep*hp^3)/(12*(1-parposV^2));
a1=C44;
a2=C33;
c1=(C13+C44)*1i*xi;
c2=(C13+C44)*1i*xi;
b1=rho*w^2-C11*xi^2;
b2=rho*w^2-C44*xi^2;
syms xr
rsol=vpasolve((a1*xr+b1).*(a2*xr+b2)-c1*c2*xr==0,xr);
rsol=eval(rsol);
r=[-sqrt(rsol)];
r1=r(1,:);
r2=r(2,:);
Fdf1=sop*sqrt(2*pi/(1i*epit*xi))*exp(1i*(xi*voP+w)^2/(2*epit*xi));
BC11=C44*(r1-1i*xi*c2*r1/(a2*r1^2+b2));
BC21=(C13*1i*xi)-((C33*c2*r1^2)/(a2*r1^2+b2))+(DD*xi^4-mm*w^2+1i*aposC*w)*c2*r1/(a2*r1^2+b2);
BC12=C44*(r2-1i*xi*c2*r2/(a2*r2^2+b2));
BC22=(C13*1i*xi)-((C33*c2*r2^2)/(a2*r2^2+b2))+(DD*xi^4-mm*w^2+1i*aposC*w)*c2*r2/(a2*r2^2+b2);
syms As1 As2;
try
[Ass1,Ass2]=vpasolve(BC11*As1+BC12*As2==0,BC21*As1+BC22*As2+Fdf1==0,As1,As2);
A1=eval(Ass1);
A2=eval(Ass2);
catch
A1=0.0;
A2=0.0;
end
Bn1=-(c2*r1/(a2*r1^2+b2))*A1;
Bn2=-(c2*r2/(a2*r2^2+b2))*A2;
ffw=Bn1*exp(r1*z)+Bn2*exp(r2*z);
end
Everything in your function but the calls to vpasolve, and try.... can be vectorize.
First, all this does not depend on w or xi, so could be computed once only:
format long e;
z=0;mm=0.46;sop=80;
epit=0.1;voP=80.;
rho=2.1;aposC=0.1;aposD=0.1;
parEh=0.2*10^6;parEv=0.2*10^6;parGv=0.074*10^6;
parpos=0.35;hp=0.2;Ep=30*10^6;
parposV=0.20;ll=0.15;dd=2*ll;
C11=(parEh*(parEv-parEh*parpos^2)/((1+parpos)*(parEv-parEv*parpos-2*parEh*parpos^2)))*(1+2*1i*aposD);
C13=(parEh*parEv*parpos/((parEv-parEv*parpos-2*parEh*parpos^2)))*(1+2*1i*aposD);
C33=((1-parpos)*parEv^2/(parEv-parEv*parpos-2*parEh*parpos^2))*(1+2*1i*aposD);
C44=parGv*(1+2*1i*aposD);
DD=(Ep*hp^3)/(12*(1-parposV^2));
a1=C44;
a2=C33;
From what I know, eval is slow, and I'm pretty sure that you don't need it:
rsol=eval(rsol);
Here is an example of vectorization. You should first generate all indices combination using meshgrid, and then use the . to noticed matlab to use element wise operations:
[I, J] = meshgrid(1:1000, 1:1000)
c1=(C13+C44)*1i.*xi;
c2=(C13+C44)*1i.*xi;
b1=rho.*w.^2 - C11.*xi.^2;
b2=rho.*w.^2-C44.*xi.^2;
But you won't be able to vectorize vpasolve, and try.... litteraly, without changing it to something else. vpasolve is probably the bottleneck of you computation (verify this using matlab profiler), so optimizing as proposed above will probably not reduce your computation time much.
Then you have several solutions:
use parfor if you have access to it to parallelize your computations, which depending on your architecture, could give you a 4x speedup or so.
it may be possible to linearize your equations and solve them all in one operation. Anyway, using a linear solver will be probably much faster than using vpasolve. This will probably give you the fastest speedup (guessing factor 10 -100 ?)
because you have continuous data, you could reduce the number of steps, if you dare loosing precision.
Hope this helps
In the above program everything can be vectorized as #beesleep said above.
For example:
[I, J] = meshgrid(1:1000, 1:1000)
c1=(C13+C44)*1i.*xi;
c2=(C13+C44)*1i.*xi;
b1=rho.*w.^2 - C11.*xi.^2;
b2=rho.*w.^2-C44.*xi.^2;
The vpasolve part ,i.e.,
syms xr
rsol=vpasolve((a1*xr+b1).*(a2*xr+b2)-c1*c2*xr==0,xr);
rsol=eval(rsol);
r=[-sqrt(rsol)];
r1=r(1,:);
r2=r(2,:);
can also be vectorized by using fsolve as it is shown here:
fun=#(Xr) (a1.*Xr+b1).*(a2.*Xr+b2)-c1.*c2.*Xr
x01=-ones(2*Nxi);
x02=ones(2*Nxi);
options.Algorithm= 'trust-region-reflective'
options.JacobPattern=speye(4*Nxi^2);
options.PrecondBandWidth=0;
[rsol1]=fsolve(fun,x01,options);
[rsol2]=fsolve(fun,x02,options);
The other part, i.e,
syms As1 As2;
try
[Ass1,Ass2]=vpasolve(BC11*As1+BC12*As2==0,BC21*As1+BC22*As2+Fdf1==0,As1,As2);
A1=eval(Ass1);
A2=eval(Ass2);
catch
A1=0.0;
A2=0.0;
end
since contains linear equations and in each row has two dependent equations can be solved as it is shown here:
funAB1=#(As1) BC11.*As1+BC12.*((-Fdf2-BC21.*As1)./BC22);
x0AB=ones(2*Nxi)+1i*ones(2*Nxi);
options.Algorithm= 'trust-region-reflective';
options.JacobPattern=speye(4*Nxi^2);
options.PrecondBandWidth=0;
[A1]=fsolve(funAB1,x0AB,options);
A2=((-Fdf2-BC21.*A1)./BC22);
However, both can also be solved analytically, i.e.,
AAr=a1.*a2;
BBr=a1.*b2+b1.*a2-c1.*c2;
CCr=b1.*b2;
Xr1=(-BBr+sqrt(BBr^.2-4.*AAr.*CCr))./(2.*AAr);
Xr2=(-BBr-sqrt(BBr^.2-4.*AAr.*CCr))./(2.*AAr);
r1=-sqrt(Xr1(:,:));
r2=-sqrt(Xr2(:,:));
Maple generates a strange solution form for this PDE. I am having hard time plotting the solution.
The solution is in terms of infinite series. I set the number of terms to say 20, and then set the time to plot the solution at t=2 seconds. Then want to plot the solution now for x=0..1. But the plot comes out empty.
When I sample the solution, and use listplot, I get correct solution plot.
Here is MWE
restart;
pde:=diff(u(x,t),t)=diff(u(x,t),x$2)+x;
bc:=u(0,t)=0,u(1,t)=0;
ic:=u(x,0)=x*(1-x);
sol:=pdsolve({pde,ic,bc},u(x,t)):
sol:=value(sol);
Now set the number of terms to 20 and set t=2
sol2:=subs(t=2,sol):
sol2:=subs(infinity=20,sol2);
The above is what I want to plot.
plot(rhs(sol2),x=0..1);
I get empty plot
So had to manually sample it and use listplot
f:=x->rhs(sol2);
data:=[seq([x,f(x)],x=0..1,.01)]:
plots:-listplot(data);
Solution looks correct, when I compare it to Mathematica's result. But Mathematica result is simpler as it does not have those integrals in the sum.
pde=D[u[x,t],t]==D[u[x,t],{x,2}]+x;
bc={u[0,t]==0,u[1,t]==0};
ic=u[x,0]==x(1-x);
DSolve[{pde,ic,bc},u[x,t],x,t];
%/.K[1]->n;
%/.Infinity->20;
%/.t->2;
And the plot is
Question is: How to plot Maple solution without manually sampling it?
Short answer seems to be that it is a regression in Maple 2017.3.
For me, your code works directly in Maple 2017.2 and Maple 2016.2 (without any unevaluated integrals). I will submit a bug report against the regression.
[edited] Let me know if any of these four ways work for your version (presumably Maple 2017.3).
restart;
pde:=diff(u(x,t),t)=diff(u(x,t),x$2)+x;
bc:=u(0,t)=0,u(1,t)=0;
ic:=u(x,0)=x*(1-x);
sol:=pdsolve({pde,ic,bc},u(x,t)):
sol:=value(sol);
sol5:=value(combine(subs([sum=Sum,t=2,infinity=20],sol))):
plot(rhs(sol5),x=0..1);
sol4:=combine(subs([sum=Sum,t=2,infinity=20],sol)):
(UseHardwareFloats,oldUHF):=false,UseHardwareFloats:
plot(rhs(sol4),x=0..1);
UseHardwareFloats:=oldUHF: # re-instate
sol2:=subs([sum=Sum,int=Int,t=2],sol):
# Switch integration and summation in second summand of rhs(sol).
sol3:=subsop(2=Sum(int(op([2,1,1],rhs(sol2)),op([2,2],rhs(sol2))),
op([2,1,2],rhs(sol2))),rhs(sol2)):
# Rename dummy index and combine summations.
sol3:=Sum(subs(n1=n,op([1,1],sol3))+op([2,1],sol3),
subs(n1=n,op([1,2],sol3))):
# Curtail to first 20 terms.
sol3:=lhs(sol2)=subs(infinity=20,simplify(sol3));
plot(rhs(sol3),x=0..1);
F:=unapply(subs([Sum='add'],rhs(sol3)),x):
plot(F,0..1);
[edited] Here is yet another way, working for me in Maple 2017.3 on 64bit Linux.
It produces the plot quickly, and doesn't involve curtailing any sum at 20 terms. Note that it does not do your earlier step of sol:=value(sol); since it does active int rather than Int before hitting any Sum with value. It also uses an assumption on x corresponding to the plotting range.
restart;
pde:=diff(u(x,t),t)=diff(u(x,t),x$2)+x:
bc:=u(0,t)=0,u(1,t)=0:
ic:=u(x,0)=x*(1-x):
sol:=pdsolve({pde,ic,bc},u(x,t)):
solA:=subs(sum=Sum,value(eval(eval(sol,t=2),Int=int))) assuming x>0, x<1;
plot(rhs(solA),x=0..1) assuming x>0, x<1;
I was trying to create an image resize code using C to do exactly what imresize.m in Matlab does. I stuck at the line calling imresizemex in imresize.m. It seems that imresizemex is a compiled machine code that can only be called in Matlab (I found it as imresizemex.mexw64 file in a Matlab private folder, no source code available). I also tried to call it in C, but failed. Does anyone know where to find the source code for imresizemex? Thanks a lot!
I think I figured it out ^ ^. It does the weighted multiplication and sum part of the cubic convolution interpolation. Here's my Matlab code replacing imresizemex. Although almost 6 seconds slower, it produces exactly the same result.
function outimg=reducesize(inimg, weights,indices,dim)
% reduce first dimension
reduce1=zeros(dim(1),size(inimg,2));
weight1=weights{1};
index1=indices{1};
for i=1:size(inimg,2)
for j=1:dim(1)
w11=weight1(j,:);
ind11=index1(j,:);
B=double(inimg(ind11,i));
v=w11.*B';
reduce1(j,i)=sum(v);
end
end
% reduce second dimension
reduce2=zeros(dim(1),dim(2));
weight2=weights{2};
index2=indices{2};
for i=1:dim(1)
for j=1:dim(2)
w22=weight2(j,:);
ind22=index2(j,:);
B=reduce1(i,ind22);
v=w22.*B;
reduce2(i,j)=sum(v);
end
end
outimg=round(reduce2);
I am trying to solve a very large non-linear problem using the MatCont package. Due to the large number of dimensions, and the non-linear nature, I believe that supplying the Jacobian for my system to the MatCont algorithm will speed things up immensely. However, I cannot get it to recognise that it has a Jacobian to use!
As a minimal working example, I have modified the circle finder from the help documentation to include a Jacobian:
function out = curve()
out{1}=#curvefunc;
out{2}=#defaultprocessor;
out{3}=#options;
out{4}=#jacobian;
out{13}=#adapt;
end
function f=curvefunc(x)
f=x(1)^2+x(2)^2-1;
end
function J=jacobian(x)
disp('USE JACOBIAN')
J=[2*x(1) , 2*x(2)];
end
function varargout=defaultprocessor(varargin)
if nargin>2
varargout{3}=varargin{3};
end
varargout{2}=[0];
varargout{1}=0;
end
function option=options()
option=contset;
end
function [res,x,v]=adapt(x,v)
res=[];
end
I then try to run this program from the command line using
[x,v,s,h,f] = cont(#curve,[1;0]);
However, the response is
first point found
tangent vector to first point found
elapsed time = 0.2 secs
npoints curve = 300
Since I told it to output 'USE JACOBIAN' every time the Jacobian function was called, it is clear that MatCont is not using it.
How do I get the Jacobian to be used?
I solved my own problem! Seems I was quite close to getting it working. The below is a bit of a botch, so if anyone knows how to do it with options please post an answer also.
Firstly I edit the options settings so that when it performs the continuation it only locates the first point:
function option=options()
option = contset;
option = contset(option,'MaxNumPoints',1);
end
its fine for it to do this using numerical Jacobians, the first point is known very well in most problems. This is then called from a script or a function using the following:
[x,v,s,h,f] = cont(#curve,[1;0]);
global cds
cds.options.MaxNumPoints=[];
cds.symjac=1;
[x,v,s,h,f] = cont(x,v,s,h,f,cds);
The first line finds the initial point using numerical Jacobians, as it was set up to do. The continuer is then manually adjusted to firstly have no limit on the maximal number of points (this can be set to any appropriate number) and then the use of the user provided Jacobian is set to 1 (true). The continuer is then resumed with the new settings and uses the Jacobain properly.