diff options
| -rw-r--r-- | Functions/Heun/Heun.m | 44 | ||||
| -rw-r--r-- | Functions/days/README.md | 11 | ||||
| -rw-r--r-- | Functions/days/days.m | 22 | ||||
| -rw-r--r-- | Functions/falsePosition/falsePosition.m | 45 | ||||
| -rw-r--r-- | Functions/luFactor/luFactor.m | 87 | ||||
| -rw-r--r-- | Functions/simpson/README.md | 0 | ||||
| -rw-r--r-- | Functions/simpson/simpson.m | 39 | ||||
| -rw-r--r-- | Functions/specialMatrix/specialMatrix.m | 51 | ||||
| -rw-r--r-- | Scripts/README.md | 17 | ||||
| -rw-r--r-- | Scripts/hw02.m (renamed from Scripts/hw02/hw2.m) | 0 | ||||
| -rw-r--r-- | Scripts/hw03.m (renamed from Scripts/hw03/hw3.m) | 0 | ||||
| -rw-r--r-- | Scripts/hw11.m (renamed from Scripts/hw11/hw11.m) | 0 | ||||
| -rw-r--r-- | Scripts/hw2/hw2.m | 60 | ||||
| -rw-r--r-- | Scripts/hw3/hw3.m | 37 |
14 files changed, 17 insertions, 396 deletions
diff --git a/Functions/Heun/Heun.m b/Functions/Heun/Heun.m deleted file mode 100644 index 5812dfe..0000000 --- a/Functions/Heun/Heun.m +++ /dev/null @@ -1,44 +0,0 @@ -function [t,y] = Heun(dydt,tspan,y0,h,es,maxit) -% [t,y] = Heun(dydt,tspan,y0,h): uses the heun method to integrate an ODE -% inputs: -% dydt = the differential equation of interest (must be anonymous -% function) -% tspan = [ti,tf] the initial and final values of the independent -% variable -% y0 = the initial value of the dependent variable -% h = step size -% es = stopping criterion (%), optional (default = 0.001) -% maxit = maximum iterations of corrector, optional (default = 50) -% outputs: -% t = vector of independent variable values -% y = vector of solution for dependent variable - -%Error checking -if nargin<4, error('At least 4 aruments required.');end -if nargin<5|isempty(es), es = 0.001;end -if nargin<6|isempty(maxit), maxit = 50;end -if length(tspan)~=2, error('Argument tspan must have a length of 2');end - -%Trims tspan according to step size (h) -t=tspan(1):h:tspan(2); -%if rem(tspan(2),h)~=0 -if t(length(t))~=tspan(2) - t=[t,tspan(end)]; -end - -y=ones(1,length(t))*y0; -h=diff(t); - -for i=1:length(t)-1 - S=dydt(t(i),y(i)); - ynew=y(i)+S*h; - for j=1:maxit - yold=ynew; - ynew=y(i)+h(i)/2*(S+dydt(t(i+1),yold)); - ea = abs((ynew-yold)/ynew)*100; - if ea <= es, break, end - end - y(i+1)=ynew; -end -plot(t,y) -end
\ No newline at end of file diff --git a/Functions/days/README.md b/Functions/days/README.md deleted file mode 100644 index 8a11db2..0000000 --- a/Functions/days/README.md +++ /dev/null @@ -1,11 +0,0 @@ -#days.m -Function to count the total days elapsed in a year according to a given date. -The syntax of the function is `days(<months>, <days>, <leap>)` where `months` is a integer (1-12). `days` is an integer (1-31) and `leap` - -## Input -months - month number (1-12). Example: `8` represents August. -days - day number of the month. -leap - indicates if the year is a leapyear or a regular year. `0` for regular and `1` for leap year. - -## Output -`nd` - number of days elapsed in the year.
\ No newline at end of file diff --git a/Functions/days/days.m b/Functions/days/days.m deleted file mode 100644 index 87163b7..0000000 --- a/Functions/days/days.m +++ /dev/null @@ -1,22 +0,0 @@ -function nd = days(mo, da, leap) -%days - Counts days elapsed in the year. -% -% days(<months>, <days>, <leap>) determines the days elapsed in a year -% based on the current date. This will include the current day. -% Months (1-12), -% Example: August 28, no leap year -% days(8,28,0) - -daysPeM=[0 31 59 90 120 151 181 212 243 273 304 334]; % cummulative days prior each month. - -nd = daysPeM(mo) + da; - -if leap == 1 & mo >= 3 - nd = nd + 1; -end -end - - -% MECH 105: Homework 4 - Part 1 -% Author: Christian Kolset -% Date: 5.21.21
\ No newline at end of file diff --git a/Functions/falsePosition/falsePosition.m b/Functions/falsePosition/falsePosition.m deleted file mode 100644 index d2cb776..0000000 --- a/Functions/falsePosition/falsePosition.m +++ /dev/null @@ -1,45 +0,0 @@ -function [root, fx, ea, iter] = falsePosition(func, xl, xu, es, maxit, varargin) -%falsePosition finds the root of a function using false position method -% Syntax: -% [root, fx, ea, iter] = falsePosition(func, xl, xu, es, maxit, varargin) -% -% Input: -% func = function being evaluated -% xl = lower guess -% xu = upper guess -% es = desired relative error (default 0.0001%) -% maxit = maximum number of iterations (default 200) -% varargin, ...= any additional parameters used by the function -% Output: -% root = estimated root location -% fx = function evaluated at root location -% ea = approximated relative error (%) -% iter = number of iterations performed - -if nargin < 3, error('Too few arguments: at least 3 needed'), end -if nargin < 4 | isempty(es), es = 0.0001; end -if nargin < 5 | isempty(maxit), maxit = 200; end - -iter=0; -ea=100; -xr= xl; - -while(1) - xrold= xr; - iter = iter + 1; - xr=xu-func(xu)*(xl-xu)/(func(xl)-func(xu)); - if xr~=0 - ea = abs((xr-xrold)/xr)*100; - end - if func(xl)*func(xr)<0 - xl=xr; - else - xl=xr; - end - if iter == maxit | ea <= es, break, end -end -root = xr; -fx=func(xr,varargin{:}); -ea=ea; -iter=iter; -end
\ No newline at end of file diff --git a/Functions/luFactor/luFactor.m b/Functions/luFactor/luFactor.m deleted file mode 100644 index 4c853dd..0000000 --- a/Functions/luFactor/luFactor.m +++ /dev/null @@ -1,87 +0,0 @@ -function [L, U, P] = luFactor(A) -% luFactor(A) -% LU decomposition with pivoting -% inputs: -% A = coefficient matrix -% outputs: -% L = lower triangular matrix -% U = upper triangular matrix -% P = the permutation matrix - - -[m,n] = size(A); -if m~=n, error('Enter a square matrix.'); end -nb = n + 1; -L=zeros(m,m); -U=zeros(m,m); -P=eye(m); - -%Partial pivoting -for j = 1:n - 1 - for p = j+1:m - if (abs(A(j,j)) < abs(A(p,j))) - A([j p],:) = A([p j],:); - P([j p],:) = P([p j],:); - end - end -end -% [big,i] = max(abs(A(j:n,j))); -% ipr = i + j - 1; -% if ipr ~= j -% A([j,ipr],:) = A([ipr,j],:); -% end -%end - -for i=1:m - % Computing L - for k=1:i-1 - L(i,k)=A(i,k); - for j=1:k-1 - L(i,k)= L(i,k)-L(i,j)*U(j,k); - end - L(i,k) = L(i,k)/U(k,k); - end - - % Computing U - for k=i:m - U(i,k) = A(i,k); - for j=1:i-1 - U(i,k)= U(i,k)-L(i,j)*U(j,k); - end - end -end - -for i=1:m - L(i,i)=1; -end - -%% Checks -if P*A ~= L*U, error('LU factorization failed to be computed correctly.'); end - -% Checks output size -if size(L) ~= size(A) | size(U) ~= size(A) | size(P) ~= size(A) - error('Incorrect output size') -end - -% Checks if matrices are triangular -%for i=1:m -% for j=1:n -% error('Incorrect output: LU decomposition are not triangular') -%end - -% Checks the permutation matrix P is valid -%P_str = string(P) -%for k = m -% num1=count(P(:,1:n),1) -% if num1 ~=1 -% error('Incorrect P matrix output. Too many 1's on the same line') -% end -% num0=count(P(:,1:n),0) -% if num0 ~= n-1 -% error('Incorrect P matric output. Wrong number of 0's') -% end -%end - -%} - -end
\ No newline at end of file diff --git a/Functions/simpson/README.md b/Functions/simpson/README.md deleted file mode 100644 index e69de29..0000000 --- a/Functions/simpson/README.md +++ /dev/null diff --git a/Functions/simpson/simpson.m b/Functions/simpson/simpson.m deleted file mode 100644 index acd4a76..0000000 --- a/Functions/simpson/simpson.m +++ /dev/null @@ -1,39 +0,0 @@ -function [I] = Simpson(x, y) -% Numerical evaluation of integral by Simpson's 1/3 Rule -% Inputs -% x = the vector of equally spaced independent variable -% y = the vector of function values with respect to x -% Outputs: -% I = the numerical integral calculated - -%Error cheching -if length(x)~=length(y), error('The vectors need to be of equal lengths'),end -if length(x)==1, error('Vectors must be at least 2 data points long'),end -if range(x(2:end)-x(1:end-1))~=0, error('The x vector needs to be equally spaced'), end -%Error Errorchecking END - -lower_bound=min(x); -interval=length(x)-1; -if rem(interval,2)==0 - trap_rule=0; - upper_bound=max(x); -else - warning('Odd number of intervals. Applying Trapezoidal rule on last interval') - trap_rule=1; - upper_bound=x(end-1); -end - -if length(x)==2 - I=(upper_bound-lower_bound)*(y(1)+y(2))/2; -else - h=(upper_bound-lower_bound)/interval; - SumOdd=sum(y(1:2:end))-y(end-1); - SumEven=sum(y(2:2:end))-y(end-2); - I=(upper_bound-lower_bound)*(y(1)+4*SumOdd+2*SumEven+y(end))/(3*interval); -end - -if trap_rule==1 & length(x)~=2 - I_trap=(max(x)-upper_bound)*(y(upper_bound)+y(max(x)))/2; - I=I+I_trap; -end -end
\ No newline at end of file diff --git a/Functions/specialMatrix/specialMatrix.m b/Functions/specialMatrix/specialMatrix.m deleted file mode 100644 index d6c31f8..0000000 --- a/Functions/specialMatrix/specialMatrix.m +++ /dev/null @@ -1,51 +0,0 @@ -function [A] = specialMatrix(n,m) -% This function should return a matrix A as described in the problem statement -% Inputs n is the number of rows, and m the number of columns -% It is recomended to first create the matrxix A of the correct size, filling it with zeros to start with is not a bad choice - -%-------------------------------------------- - -if nargin ~= 2 - error('Error: Please enter two arguments.') -end -if (n|m) <= 0 - error('Error: Index error. Arguments must be positive integers or logical values.') -end - -A = zeros(n,m); %Creates a "blank" n x m matrix full of zeros - -% Now the real challenge is to fill in the correct values of A - -A(1,:) = 1:m; %Lables the first row with column numbers -A(:,1) = 1:n; %Lables the first column with row numbers - -for j = 2:n - for k = 2:m - A(j,k)=A(j-1,k)+A(j,k-1); % fills each element of the matrix with the sum of the left and above element. - end -end -end - -%--------------------------------------------- - -%{ -if nargin(specialMatrix) ~= 2 - error('Error: Please enter two arguments.') -elseif nargin(specialMatrix) <= 0 - error('Error: Index errer. Argument must be positive integers or logical values.') -end - -if n<2 - A = [1:m]; %If matrix is 1 x m then make a matrix with 1 row. -else - A = [1:m;1:m]; %Create row 1 -end - -for j = 3:n %Loop to make - A(j,:) = sum(A); % -end % -A(:,1)=(1:n); - -end -% Things beyond here are outside of your functions -%}
\ No newline at end of file diff --git a/Scripts/README.md b/Scripts/README.md new file mode 100644 index 0000000..2e8e64f --- /dev/null +++ b/Scripts/README.md @@ -0,0 +1,17 @@ +# [Scripts](Scripts) +This directory contains the scripts to solve the homeworks assignments below: + +## hw02.m +Both part 1 and part 2 on the excercise are writen in the script. When the script is run part 2 will be skipped. To run part 2, put part 1 in comments and uncomment part 2. To uncomment remove the `%{` and `%}` (line 43 and 70) to prevent the MATLAB interpreter to skip the code. + +### Part 1 +Creates a supplots of two graphs. The first plot shows the charge on a capacitor in an electrical circuit. First graph ranges from t = 0 to 0.8 witj, q0=10, R=60, L=9, and C=0.00005. A second plot shows a capacitor with 10 times greater charge (C=0.0005) + +### Part 2 +Creates a plot displaying the measured data (red diamond shapes) and the expected function (green dashed line). Ranges from 0 to 70 minutes using a a step size of 30 seconds. + +# hw03.m +This script solves the volume in a tank with the shape of a cylinder with a tuncated cone on top. This script should be turned into a function to solve the volume of a tank with any given the height of cylinder and truncated cone, and diameter of both the cylinder and top of the truncated cone. + +## hw11.m +Script to solve the angle of which to pull a 25 kg block with 150 N using the bisection method of finding roots. diff --git a/Scripts/hw02/hw2.m b/Scripts/hw02.m index 6736240..6736240 100644 --- a/Scripts/hw02/hw2.m +++ b/Scripts/hw02.m diff --git a/Scripts/hw03/hw3.m b/Scripts/hw03.m index 7caf82c..7caf82c 100644 --- a/Scripts/hw03/hw3.m +++ b/Scripts/hw03.m diff --git a/Scripts/hw11/hw11.m b/Scripts/hw11.m index 312e816..312e816 100644 --- a/Scripts/hw11/hw11.m +++ b/Scripts/hw11.m diff --git a/Scripts/hw2/hw2.m b/Scripts/hw2/hw2.m deleted file mode 100644 index 6736240..0000000 --- a/Scripts/hw2/hw2.m +++ /dev/null @@ -1,60 +0,0 @@ -% Homework:2 -% Author: Christian Kolset -clear - -%% Part 1 - -% Function parameters -q0 = 10; -R = 60; -L = 9; -C = 0.00005; - -% Use linspace to create an array of 100 points between 0 and 0.8 -t = linspace(0,0.8); - -% Calculate the values of q -q = q0.*2.718.^(-R.*t/(2*L)).*cos(sqrt((1/(L*C))-(R/(2*L))^2).*t); - -% Plot q vs t -hold on -subplot(2,1,1) -plot (t,q,'b--*') -title('Capacity vs Time Graph') -xlabel('Time') -ylabel('Charge') -%legend('Charge','Time') -hold off - -% Make the capacitor 10x bigger -q2 = q0.*2.718.^(-R.*t/(2.*L)).*cos(sqrt((1/(L.*10.*C))-(R/(2.*L))^2).*t); - -% Plot q2 vs t -hold on -subplot(2,1,2) -plot(t,q2,'rs:') -title('10x Capacity vs Time Graph') -xlabel('Time') -ylabel('Charge') -%legend('Charge','Time') -hold off - -%% Part 2 -%{ -% Given experimental data -t_exp = 10:10:60; -c_exp = [3.4 2.6 1.6 1.3 1.0 0.5]; - -% Expected function -t_func = 0:0.5:70; -c_func = 4.84*2.718.^(-0.034*t_func); - -% Plot -hold on -plot(t_exp,c_exp,'rd:') -plot(t_func,c_func,'g--') -xlabel('Time [minutes]') -ylabel('Concentration [ppm]') -legend('Experimental','Predicted') -hold off -%}
\ No newline at end of file diff --git a/Scripts/hw3/hw3.m b/Scripts/hw3/hw3.m deleted file mode 100644 index 7caf82c..0000000 --- a/Scripts/hw3/hw3.m +++ /dev/null @@ -1,37 +0,0 @@ -% HW3 -% Author: Christian Kolset -% Date: 20.5.21 - -%% HW3 -% Specify the variables needed to solve this problem (ie. height of each section, diameter, radiaus, ...) -% It is alwasy easier to work with variables (diameter_cyl = 25) than to use numbers everywhere, since a -% diameter indicates something specific but the number 25 could mean anything -diameter_Bot = 25; -r_bot = diameter_Bot/2; -h_cone = h-19; - -% Specify the height of the water -h = 20 -% You can comment / uncomment lines below for testing. This will overwrite the previous line for h = 20. -% For submission, make sure all of the following lines are commented out and h = 20! (OR IT IS MARKED AS WRONG) -%h = 5 -%h = 19 -%h = 47 -%h = -1 - -% Now compute the volume. Using conditional statments you will want to first check the height makes sense, -% and then solve the volume depending on what portion of the tank has been filled. -% Make sure that your volume is stored in the variable v! (OR IT WILL BE MARKED AS WRONG) -% You may find it more convenient to move v around in you code, it is only given here to indicate what variable to use. -%v = - -v_cyl = @(h,r_bot) pi*r_bot^2*h; -v_truncone = @(h_cone,r_bot) 1/3*pi*h_cone*(r_bot^2+r_bot*((h_cone+16.625)/1.33)+((h_cone+16.625)/1.33)^2); - -if h > 19 - v = v_cyl(19,r_bot)+v_truncone(h_cone,r_bot); -else - v = v_cyl(h,r_bot); -end - -fprintf(1,'Volume of Tank: %6.2f\n',v)
\ No newline at end of file |