University of Florida/Egm6341/s10.team3.aks/HW6

(3) Evaluate the rest of the coefficient of matrix

Ref Lecture Notes p.35-3

Problem Statement

Evaluate the remaining coefficient of Matrix by using degrees of Freedom

Solution

We have

$Z(s)=\sum _{i=0}^3{c}_i{s}^i=\sum _{i=0}^3{N}_i(s)d_i$

such that

$d_1=Z_i$

$d_2=\dot{Z_i}$

$d_3=Z_{i+1}$

$d_4={\dot{Z}}_{i+1}$

where

$\dot{Z}=\frac{dZ}{dt}=\frac{dZ}{ds}\frac{ds}{dt}$

such that $\frac{ds}{dt}=\frac{1}{h}$

We know the coefficient of matrix for first two rows from lecture notes p.35-3

$\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\end{array}\right]$

Using the equations above we have

$d_3=Z_{i+1}=Z(s=1)=c_0+c_1+c_2+c_3$

$d_4={\dot{Z}}_{i+1}=\frac{1}{h}{Z}^{\prime }(s=1)=c_1+2c_2+3c_3$

Putting the results in matrix form we obtain

$\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 1& 1& 1& 1\\ 0& 1& 2& 3\end{array}\right]\left\{\begin{array}{c}{c}_0\\ c_1\\ c_2\\ c_3\end{array}\right\}=\left\{\begin{array}{c}{Z}_i\\ Z{\text{'}}_i\\ Z_{i+1}\\ Z{\text{'}}_{i+1}\end{array}\right\}$

(4) Verify the inverse of matrix using Matlab

Ref Lecture Notes p.35-4

Problem Statement

Find the inverse of given Matrix

A =

    1     0     0     0
    0     1     0     0
    1     1     1     1
    0     1     2     3

Solution

 

%Defining Matrix in Matlab%

>> A = [1 0 0 0 ; 0 1 0 0 ; 1 1 1 1 ; 0 1 2 3]

A =

     1     0     0     0
     0     1     0     0
     1     1     1     1
     0     1     2     3

%Taking Inverse of matrix%
>> B = inv (A)

B =

     1     0     0     0
     0     1     0     0
    -3    -2     3    -1
     2     1    -2     1

which is same as the one given on p.35-4

Hence Verified

(5) Identify basis functions and plot them

Ref Lecture Notes p.35-4

Problem Statement

Identify the basis functions

$N_i(s)$

where $i=1,2,3,4$

Solution

We have

$z(s)=\sum _{i=0}^3{c}^i{s}^i=\sum _{i=1}^4{N}_i(s)d_i$

Expanding above we obtain

$N_1{d}_1+N_2{d}_2+N_3{d}_3+N_4{d}_4=C_0{s}^0+C_1{s}^1+C_2{s}^2+C_3{s}^3$

$=C_0+C_1{s}^1+C_2{s}^2+C_3{s}^3$

$d_1=z_i=z(s=0)=C_0$

$d_2={\dot{z}}_i={\dot{z}}_i(s=0)=C_1$

$d_3=z_{i+1}=z_{i+1}(s=1)=C_0+C_1+C_2+C_3$

$d_4={\dot{z}}_{i+1}=C_1+2C_2+3C_3$

Inserting above values in first eq we obtain

$(N_1+N_3)C_0+(N_2+N_3+N_4)C_1+(N_3+2N_4)C_2+(N_3+3N_4)C_3=C_0+C_{1}s+C_2{s}^2+C_3{s}^3$

Comparing both LHS and RHS we obtain

$N_1+N_3=1$

$N_2+N_3+N_4=s$

$N_3+2N_4=s^2$

$N_3+3N_4=s^3$

Solving above we obtain basis functions

$N_1=1-3s^2+2s^3$

$N_2=s^3-2s^2+s$

$N_3=3s^2-2s^3$

Below is the plot of above basis functions

File:Basis function plot.jpg

$N_4=s^3-s^2$

(6) Show that s = s(t)

Ref Lecture Notes p.36-1

Problem Statement

We have to show that s is the function of t (s = s(t) )

Solution

We have (from p.35-1 eq (1))

$t(s)=(1-s)t_i+s{t}_{i+1}$

$=t_i+s(t_{i+1}-t_i)$

$but\frac{ds}{dt}=\frac{1}{h}$

$or\frac{dt}{ds}=h$

so $t_{i+1}-t_i=\frac{t_{i+1}-t_i}{1-0}=h$

$t(s)=t_i+hs$ $hs=t-t_i$ $s=\frac{(t-t_i)}{h}$ $s=s(t)$

Hence Proved