Problem Statement

Evaluate the remaining coefficient of Matrix by using degrees of Freedom

Solution

We have

${\displaystyle Z(s)=\sum _{i=0}^{3}c_{i}s^{i}=\sum _{i=0}^{3}N_{i}(s)d_{i}}$

such that

${\displaystyle d_{1}=Z_{i}}$

${\displaystyle d_{2}={\dot {Z_{i}}}}$

${\displaystyle d_{3}=Z_{i+1}}$

${\displaystyle d_{4}={{\dot {Z}}_{i+1}}}$

where

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

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

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

${\displaystyle {\begin{bmatrix}1&0&0&0\\0&1&0&0\end{bmatrix}}}$

Using the equations above we have

${\displaystyle d_{3}=Z_{i+1}=Z(s=1)=c_{0}+c_{1}+c_{2}+c_{3}}$

${\displaystyle d_{4}={\dot {Z}}_{i+1}={\frac {1}{h}}Z'(s=1)=c_{1}+2c_{2}+3c_{3}}$

Putting the results in matrix form we obtain

${\displaystyle {\begin{bmatrix}1&0&0&0\\0&1&0&0\\1&1&1&1\\0&1&2&3\end{bmatrix}}{\begin{Bmatrix}c_{0}\\c_{1}\\c_{2}\\c_{3}\end{Bmatrix}}={\begin{Bmatrix}Z_{i}\\Z'_{i}\\Z_{i+1}\\Z'_{i+1}\end{Bmatrix}}}$

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

Problem Statement

Identify the basis functions

${\displaystyle N_{i}(s)}$

where ${\displaystyle i=1,2,3,4}$

Solution

We have

${\displaystyle z(s)=\sum _{i=0}^{3}c^{i}s^{i}=\sum _{i=1}^{4}N_{i}(s)d_{i}}$

Expanding above we obtain

${\displaystyle N_{1}d_{1}+N_{2}d_{2}+N_{3}d_{3}+N_{4}d_{4}=C_{0}s^{0}+C_{1}s^{1}+C_{2}s^{2}+C_{3}s^{3}}$

${\displaystyle =C_{0}+C_{1}s^{1}+C_{2}s^{2}+C_{3}s^{3}}$

${\displaystyle d_{1}=z_{i}=z(s=0)=C_{0}}$

${\displaystyle d_{2}={\dot {z}}_{i}={\dot {z}}_{i}(s=0)=C_{1}}$

${\displaystyle d_{3}=z_{i+1}=z_{i+1}(s=1)=C_{0}+C_{1}+C_{2}+C_{3}}$

${\displaystyle d_{4}={\dot {z}}_{i+1}=C_{1}+2C_{2}+3C_{3}}$

Inserting above values in first eq we obtain

${\displaystyle (N_{1}+N_{3})C_{0}+(N_{2}+N_{3}+N_{4})C_{1}+(N_{3}+2N_{4})C_{2}+(N_{3}+3N_{4})C_{3}=C_{0}+C_{1}s+C_{2}s^{2}+C_{3}s^{3}}$

Comparing both LHS and RHS we obtain

${\displaystyle N_{1}+N_{3}=1}$

${\displaystyle N_{2}+N_{3}+N_{4}=s}$

${\displaystyle N_{3}+2N_{4}=s^{2}}$

${\displaystyle N_{3}+3N_{4}=s^{3}}$

Solving above we obtain basis functions

${\displaystyle N_{1}=1-3s^{2}+2s^{3}}$

${\displaystyle N_{2}=s^{3}-2s^{2}+s}$

${\displaystyle N_{3}=3s^{2}-2s^{3}}$

Below is the plot of above basis functions

${\displaystyle N_{4}=s^{3}-s^{2}}$

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))

${\displaystyle t(s)=(1-s)t_{i}+st_{i+1}}$

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

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

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

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

${\displaystyle t(s)=t_{i}+hs}$ ${\displaystyle hs=t-t_{i}}$ ${\displaystyle s={\frac {(t-t_{i})}{h}}}$ ${\displaystyle s=s(t)}$

Hence Proved