I am trying to find the displacement x(t) as shown but i keep on getting the error
Error using +
Matrix dimensions must agree.
My code is as shown below and its for an over damped vibration system
for i = 1 : 100;
t(i)= i/40;
x(i) = (C1*exp(-s+(((s^2)-1)^.5)*Wn*t)) + (C2*exp(-s-(((s^2)-1)^.5)*Wn*t));
end
I looked up this problem earlier and I saw a similar problem where a (.) was missing in front of the operator and tried this on my code but still couldn't get it to work. I don't know if I placed them in the wrong places or not but I am still stuck.
Could anyone please show me where I went wrong?
This is very likely because you use t inside the expression for x, and not t(i). (Assuming all other variables are scalars.
Try:
for ii = 1 : 100;
t(ii)= ii/40
x(ii) = (C1*exp(-s+(((s^2)-1)^0.5)*Wn*t(ii)))+ (C2*exp(-s-(((s^2)-1)^0.5)*Wn*t(ii)))
end
A better solution would be to vectorize this:
t = (1:100)./40;
x = (C1.*exp(-s+(((s^2)-1)^.5)*Wn.*t))+ (C2.*exp(-s-(((s^2)-1)^.5)*Wn.*t))
Related
I am trying to code something in Matlab and it involves a lot of accessing elements in vectors. Below is a snippet of code that I am working on:
x(1)=1;
for i=2:18
x(i)=0;
end
for i=1:18
y(i)=1;
end
for i = 0:262124
x(i+18+1) = x(i+7+1) + mod(x(i+1),2);
y(i+18+1) = y(i+10+1) + y(i+7+1) + y(i+5+1) + mod(y(i+1), 2);
end
% n can be = 0, 1, 2,..., 262142
n = 2;
for i = 0: 262142
z(i+1) = x(mod(i+n+1, 262143)); %error: Subscript indices must either be real positive integers or logicals.
end
In the last "for" loop where I am initialising vector z(), I get an error saying: "Subscript indices must either be real positive integers or logicals." However, when I do not suppres z(i+1) by ommiting the semi colon, the program is able to run, and I can see the values of z in the workspace. Why is this?
The code I am writing in Matlab is based upon the series of instructions shown in the image below. However, I can't seem to track down my error which leads to me not being able to access the elements of x() (without not suppressing the output of z()).
I appreciate any ideas :-) Thank you!
The code breaks at that loop last iteration because , for i=262140 you get
(mod(i+n+1, 262143)) = 0
so you cant access x(0) in matlab. the first elements of any variable is x(1).
In addition, and not related to your question, this code doesn't use the advantages matlab has, instead of
for i=2:18
x(i)=0;
end
you can just write:
x(2:18)=0;
etc
I am trying to find the first term 'p' of a geometric series with common ratio 1.05 in MATLAB as follows. However the solve function is giving the error as below (posted right after the code). I can't seem to figure out the reason for this error, because when I display the expression for 'sum', it is correctly showing an expression in terms of'p', but the problem arises when I try to equate that to a value, and solve for 'p'. Any insights would be appreciated! Thanks.
clear all;
clc;
t=20; %no. of terms in geometric series
sum =0;
jackpot = 1000; %sum of geometric series
%p is first term
syms p
for x=1:t
sum = sum + p*((1.05)^(x-1));
end
disp(sum);
eqn1 = sum == jackpot;
solve(eqn1,p);
Output:
(18614477322052275759*p)/562949953421312000
??? Error using ==> char
Conversion to char from logical is not possible.
Error in ==> solve>getEqns at 169
vc = char(v);
Error in ==> solve at 67
[eqns,vars] = getEqns(varargin{:});
Error in ==> geometric_trial at 13
solve(eqn1,p);
So, I got the answer for this from a user Walter Roberson on another forum. Posting here, from his answer.
I was trying this on a version of MATLAB which is really old i.e. R2010a. In this, using 'symbolic_expression == symbolic_expression' does not set up an equation for later solving, but instead compares the two expressions for literal equality and returns a logical value immediately.
In versions that old, the easiest fix is to change
eqn1 = sum == jackpot
to
eqn1 = (sum) - (jackpot)
and let solve() deal with the implicit equality to 0.
Can anybody help me with this assignment please?
I am new to matlab, and passing this year depends on this assignment, i don't have much time to explore matlab and i already wasted alot of time trying to do this assignment in my way.
I have already wrote the equations on the paper, but transfering the equations into matlab codes is really hard for me.
All i have for now is:
syms h
l = (0.75-h.^2)/(3*sqrt((5*h.^2)/4)); %h is h_max
V_default = (h.^2/2)*l;
dv = diff(V_default); %it's max. when the derivative is max.
h1 = solve( dv ==0);
h_max = (h1>0);
l_max = (0.75-h_max.^2)/(3*sqrt((h_max/2).^2+(h_max.^2)));
V_max = ((h_max.^2)./(2.*l_max));
but it keep give me error "Error using ./
Matrix dimensions must agree.
Error in triangle (line 9)
V_max = ((h_max.^2)./(2.*l_max)); "
Not really helping with the assignment here, but with the Matlab syntax. In the following line:
l_max = (0.75-h_max.^2)/(3*sqrt((h_max/2).^2+(h_max.^2)));
you're using / that is a matrix divide. You might want to use ./ which will divide the terms element by element. If I do this
l_max = (0.75-h_max.^2) ./ (3*sqrt((h_max/2).^2+(h_max.^2)));
then your code doesn't return any error. But I have no idea if it's the correct solution of your assignment, I'll leave that to you!
In line 5, the result h1 is a vector of two values but the variable itself remains symbolic, from the Symbolic Math Toolbox. MATLAB treats such variables slightly different. For that reason, the line h_max = (h1>0) doesn't really do what you expect. As I think from this point, you are interested in one value h_max, I would convert h1 to a regular MATLAB variable and change your code to the following:
h1 = double(solve( dv ==0)); % converts symbolic to regular vectors
h_max = h1(h1>0); % filters out all negative and zero values
l_max = (0.75-h_max.^2)/(3*sqrt((h_max/2).^2+(h_max.^2)));
V_max = ((h_max.^2)./(2.*l_max));
EDIT.
If you still have error, it means solve( ...) returns more than 1 positive values. In this case, as suggested, use dotted operations, such as ./ but the results in l_max and V_max will not be a single value but vectors of the same size as h_max. Which means you don't have one max Volume.
I'm simply trying to find the exact minimum of a simple function in MATLAB. I've been experimenting with the use of built-in functions such as "fminbnd" and inline function definition, but I don't think I quite know what I'm doing.
My code is below. I want to find the x and y of Error's minimum.
clear all
A = 5;
tau = linspace(1,4,500); %Array of many tau values between 1 and 4
E1 = qfunc(((-tau) + 5) /(sqrt(2.5)));
E0 = qfunc((tau)/(sqrt(2.5)));
Error = 0.5*E0 + 0.5*E1;
figure
subplot (311), plot(tau, E0);
xlabel('Threshold (Tau)'), ylabel('E0')
title('Error vs. Threshold (E0, 1 <= T <= 4)')
subplot (312), plot(tau, E1);
xlabel('Threshold (Tau)'), ylabel('E1')
title('Error vs. Threshold (E1, 1 <= T <= 4)')
subplot (313), plot(tau, Error);
xlabel('Threshold (Tau)'), ylabel('Pr[Error]');
title('Error vs. Threshold (Pr[Error], 1 <= T <= 4)')
I mean, I can use the cursor when the function is graphed to get close (though not right at the point where it occurs (Threshold = 2.5), but there must be a method just to print the number to the window. So far I have tried:
fminbnd('Error', 'E0', 'E1')
And many other variants. Also tried using anonymous and inline function definitions with no luck.
Can anyone point me in the right direction? Feel foolish for being stuck with this simple problem... Any help greatly appreciated!
See fminbnd
You should try something like this:
Error =#(tau) 0.5*qfunc(((-tau) + 5) /(sqrt(2.5))) + 0.5*qfunc((tau)/(sqrt(2.5)));
x = fminbnd(Error,0,10)
The first argument of fminbnd(f,x1,x2) is the function and the other arguments are the bounds. I did f=Error, x1=0 and x2=10.
Output:
x=2.5000
Another way is to save your error function in .m file. See the webpage above.
I don't understand why you're using E0 and E1 as the limits of the range where the minimum should be found. Or am I misunderstanding something in your code?
Maybe if you have your function as a discrete collection of samples (as seems implied from your way of constructing it, error is going to be a matrix, I think), you could use the "min" command: http://www.mathworks.es/es/help/matlab/ref/min.html
Hope this helped!
having a problem with my "new love", matlab: I wrote a function to calculate an integral using the trapz-method: `
function [L]=bogenlaenge_innen(schwingungen)
R = 1500; %Ablegeradius
OA = 1; %Amplitude
S = schwingungen; %Schwingungszahl
B = 3.175; %Tapebreite
phi = 0:2.*pi./10000:2.*pi;
BL = sqrt((R-B).^2+2.*(R-B).*OA.*sin(S.*phi)+OA.^2.*(sin(S.*phi)).^2+OA.^2.*S.^2.*(cos(S.*phi)).^2);
L = trapz(phi,BL)`
this works fine when i start it with one specific number out of the command-window. Now I want to plot the values of "L" for several S.
I did the following in a new *.m-file:
W = (0:1:1500);
T = bogenlaenge_innen(W);
plot(W,T)
And there it is:
Error using .*
Matrix dimensions must agree.
What is wrong? Is it just a dot somewhere? I am using matlab for the second day now, so please be patient.... ;) Thank you so much in advance!
PS: just ignore the german part of the code, it does not really matter :)
In your code, the arrays S and phi in the expression sin(S.*phi) should have same size or one of them should be a constant in order the code works
The error is most likely because you have made it so that the number of elements in schwingungen, i.e. W in your code, must be equal to the number of elements in phi. Since size(W) gives you a different result from size(0:2.*pi./10000:2.*pi), you get the error.
The way .* works is that is multiplies each corresponding elements of two matrices provided that they either have the same dimensions or that one of them is a scalar. So your code will work when schwingungen is a scalar, but not when it's a vector as chances are it has a different number of elements from the way you hard coded phi.
The simplest course of action (not necessarily the most Matlabesque though) for you is to loop through the different values of S:
W = (0:1:1500);
T = zeros(size(W); %Preallocate for speed)
for ii = 1:length(W)
T(ii) = bogenlaenge_innen(W(ii));
end
plot(W,T)
In your function you define phi as a vector of 10001 elements.
In this same function you do S.*phi, so if S is not the same length as phi, you will get the "dimensions must agree" error.
In your call to the function you are doing it with a vector of length 1501, so there is your error.
Regards