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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