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# 04_linear_algebra | |
Linear Algebra HW 4 | |
## Problem 2 | |
### Part A | |
``` matlab | |
H4 = [1,1/2,1/3,1/4;1/2,1/3,1/4,1/5;1/3,1/4,1/5,1/6;1/4,1/5,1/6,1/7]; | |
norm(H4,2) | |
norm(H4,'fro') | |
norm(H4,1) | |
norm(H4,inf) | |
H5 = [1,1/2,1/3,1/4,1/5;1/2,1/3,1/4,1/5,1/6;1/3,1/4,1/5,1/6,1/7;1/4,1/5,1/6,1/7,1/8;1/5,1/6,1/7,1/8,1/9] | |
norm(H5,2) | |
norm(H5,'fro') | |
norm(H5,1) | |
norm(H5,inf) | |
``` | |
Output (4x4): | |
- 2-norm: 1.5002 | |
- frobenius-norm: 1.5097 | |
- 1-norm: 2.0833 | |
- inf-norm: 2.0833 | |
Output (5x5): | |
- 2-norm: 1.5671 | |
- frobenius-norm: 1.5809 | |
- 1-norm: 2.2833 | |
- inf-norm: 2.2833 | |
### Part B | |
```matlab | |
H4 = [1,1/2,1/3,1/4;1/2,1/3,1/4,1/5;1/3,1/4,1/5,1/6;1/4,1/5,1/6,1/7]; | |
iH4 = inv(H4) | |
norm(iH4,2) | |
norm(iH4,'fro') | |
norm(iH4,1) | |
norm(iH4,inf) | |
H5 = [1,1/2,1/3,1/4,1/5;1/2,1/3,1/4,1/5,1/6;1/3,1/4,1/5,1/6,1/7;1/4,1/5,1/6,1/7,1/8;1/5,1/6,1/7,1/8,1/9]; | |
iH5 = inv(H5) | |
norm(iH5,2) | |
norm(iH5,'fro') | |
norm(iH5,1) | |
norm(iH5,inf) | |
``` | |
Output (4x4): | |
- 2-norm: 1.0341e+04 | |
- frobenius-norm: 1.0342e+04 | |
- 1-norm: 1.3620e+04 | |
- inf-norm: 1.3620e+04 | |
Output (5x5): | |
- 2-norm: 3.0414e+05 | |
- frobenius-norm: 3.0416e+05 | |
- 1-norm: 4.1328e+05 | |
- inf-norm: 4.1328e+05 | |
### Part C | |
```matlab | |
H4 = [1,1/2,1/3,1/4;1/2,1/3,1/4,1/5;1/3,1/4,1/5,1/6;1/4,1/5,1/6,1/7]; | |
cond(H4,2) | |
cond(H4,'fro') | |
cond(H4,1) | |
cond(H4,inf) | |
H5 = [1,1/2,1/3,1/4,1/5;1/2,1/3,1/4,1/5,1/6;1/3,1/4,1/5,1/6,1/7;1/4,1/5,1/6,1/7,1/8;1/5,1/6,1/7,1/8,1/9]; | |
cond(H5,2) | |
cond(H5,'fro') | |
cond(H5,1) | |
cond(H5,inf) | |
``` | |
Output (4x4): | |
- 2-norm: 1.5514e+04 | |
- frobenius-norm: 1.5614e+04 | |
- 1-norm: 2.8375e+04 | |
- inf-norm: 2.8375e+04 | |
Output (5x5): | |
- 2-norm: 4.7661e+05 | |
- frobenius-norm: 4.8085e+05 | |
- 1-norm: 9.4366e+05 | |
- inf-norm: 9.4366e+05 | |
## Problem 3 | |
```matlab | |
function [d,u]=chol_tridiag(e,f); | |
% chol_tridiag calculates the components for Cholesky factorization of a symmetric | |
% positive definite tridiagonal matrix | |
% given its off-diagonal vector, e | |
% and diagonal vector f | |
% the output is | |
% the diagonal of the Upper matrix, d | |
% the off-diagonal of the Upper matrix, u | |
d = zeros(length(f),1); | |
u = zeros(length(f)-1,1); | |
d(1) = sqrt(f(1)); | |
u(1) = e(1)/d(1); | |
k = 2; | |
while k <= length(f)-1 | |
d(k) = sqrt(f(k)- (u(k-1))^2); | |
u(k) = e(k)/d(k); | |
k = k+1; | |
end | |
d(4) = sqrt(f(4)-(u(3))^2); | |
end | |
``` | |
## Problem 4 | |
```matlab | |
function [x] = solve_tridiag(u,d,b) | |
% Function that provides the solution of Ax=b | |
% given: | |
% diagonal of upper matrix of cholesky factorization, d | |
% off-diagonal of upper matrix of cholesky factorization, u | |
% the vector, b | |
x = zeros(1,length(b)); | |
y = zeros(1,length(b)); | |
% forward substitution | |
y(1) = b(1)/d(1); | |
for i=2:length(b) | |
y(i)=(b(i)-u(i-1)*y(i-1))/d(i); | |
end | |
% back substitution | |
x(length(b)) = y(length(b))/d(length(b)); | |
for i = length(b)-1:-1:1 | |
x(i) = ((y(i)-u(i))*x(i+1))/d(i); | |
end | |
end | |
``` | |
## Problem 5 | |
### Part A | |
```matlab | |
k = 1000; | |
k1 = k; | |
k2 = k1; | |
k3 = k2; | |
k4 = k3; | |
K = [2*k,-k,0,0;-k,2*k,-k,0;0,-k,2*k,-k;0,0,-k,k]; | |
x = cond(K) | |
c = log10(x); | |
t_k = 0; % accuracy of coefficients of K: 10^-t (accuracy is 1 so t = 0) | |
e = 10^(c-t_k) | |
``` | |
Condition: | |
- 29.2841 | |
Error: | |
- 29.2841 | |
### Part B | |
```matlab | |
k = 1000; | |
k1 = k; | |
k2 = 1000*10^12; | |
k3 = k1; | |
k4 = k3; | |
K = [k1+k2,-k2,0,0;-k2,k2+k3,-k3,0;0,-k3,k3+k4,-k4;0,0,-k4,k4]; | |
x = cond(K) | |
c = log10(x); | |
t_k = 9; % accuracy of coefficients of K: 10^-t | |
e = 10^(c-t_k) | |
``` | |
Condition: | |
- 1.1291e13 | |
Error: | |
- 1.1291e22 | |
### Part C | |
```matlab | |
k = 1000; | |
k1 = k; | |
k2 = 1000*10^-12; | |
k3 = k1; | |
k4 = k3; | |
K = [k1+k2,-k2,0,0;-k2,k2+k3,-k3,0;0,-k3,k3+k4,-k4;0,0,-k4,k4]; | |
x = cond(K) | |
c = log10(x); | |
t_k = 9; % accuracy of coefficients of K: 10^-t | |
e = 10^(c-t_k) | |
``` | |
Condition: | |
- 9.001e12 | |
Error: | |
- 9.001e3 | |
## Problem 6 | |
### Part A | |
``` matlab | |
k = 1000; | |
k1 = k; | |
k2 = k1; | |
k3 = k2; | |
k4 = k3; | |
e = [-k2,-k3,-k4]; | |
f = [k1+k2,k2+k3,k3+k4,k4]; | |
[d,u] = chol_tridiag(e,f) | |
b = [1;1;1;1]; | |
[x] = solve_tridiag(u,d,b) | |
``` | |
Displacements: | |
- mass 1: 0.0025 m | |
- mass 2: 0.0050 m | |
- mass 3: 0.0075 m | |
- mass 4: 0.0100 m | |
### Part B | |
```matlab | |
k = 1000; | |
k1 = k; | |
k2 = 1000e12; | |
k3 = k1; | |
k4 = k3; | |
e = [-k2,-k3,-k4]; | |
f = [k1+k2,k2+k3,k3+k4,k4]; | |
[d,u] = chol_tridiag(e,f) | |
b = [1;1;1;1]; | |
[x] = solve_tridiag(u,d,b) | |
``` | |
Displacements: | |
- mass 1: 0.0023 m | |
- mass 2: 0.0023 m | |
- mass 3: 0.0047 m | |
- mass 4: 0.0070 m | |
### Part C | |
``` matlab | |
k = 1000; | |
k1 = k; | |
k2 = 1000e-12; | |
k3 = k1; | |
k4 = k3; | |
e = [-k2,-k3,-k4]; | |
f = [k1+k2,k2+k3,k3+k4,k4]; | |
[d,u] = chol_tridiag(e,f) | |
b = [1;1;1;1]; | |
[x] = solve_tridiag(u,d,b) | |
``` | |
Displacements: | |
- mass 1: 3008695.77 m | |
- mass 2: 3008695769.04 m | |
- mass 3: 3005690078.97 m | |
- mass 4: 2999690697.57 m | |
## Problem 7 | |
```matlab | |
function [v,w] = mass_spring_vibrate(k1,k2,k3,k4) | |
% Function that finds the vibration modes (v) and natural frequencies (w) of a | |
% spring - mass system given four parameters k1,k2,k3,k4 | |
m1 = 1; %kg | |
m2 = 2;%kg | |
m3 = 4; %kg | |
% x1 = A1*sin(w*t); | |
% x2 = A2*sin(w*t); | |
% x3 = A3*sin(w*t); | |
%((k2+k1)/m1-w^2)*x1 - (k2/m1)*x2 = 0; differential equation 1 | |
%((k2+k3)/m2)*x2 - (k2/m2)*x1 - (k3/m2)*x3 = 0; differential equation 2 | |
%((k3+k4)/m3)*x3 - (k4/m3)*x2 = 0; differential equation 3 | |
K = [(k2+k1)/m1, -k2/m1, 0; -k2/m2,(k2+k3)/m2, -k3/m2; 0, -k4/m3, (k2+k1)/m3]; | |
[v,lambda] = eig(K); | |
w = sqrt(lambda); | |
end | |
``` | |
Outputs: | |
- Mode 1: Natural Frequency = 6.345 rad/sec | |
- Mass 1 and 3 in phase with each other, mass 2 out of phase | |
- Mode 2: Natural Frequency = 3.594 rad/sec | |
- Mass 1 and 2 in phase with each other, mass 3 out of phase | |
- Mode 3: Natural Frequency = 2.08 rad/sec | |
- Mass 1,2 and 3 in phase with each other | |
## Problem 8 | |
| # of segments | largest load (N) | smallest load (N) | # of eigenvalues | | |
| --- | --- | --- | --- | | |
| 5 | 10998.82 | 1161.18 | 4 | | |
| 6 | 16337.43 | 1172.97 | 5 | | |
| 10 | 47449.69 | 1190.31 | 9 | | |
As the number of segments approaches zero, the eigenvalues will decrease until there is only one |