10/01 - Vectors & Functions

Vectors & Functions

• How to related the vector v to the sampling?

We could sample a continuous function every T seconds, creating a "bar graph".

$f(t)=\sum _{i=0}^{N-1}\underset{coefficients}{\underbrace{f(iT)}}\cdot \underset{basisfunctions}{\underbrace{p(t-iT)}}$

• Where $p(t)$ is a rectangle 1 unit high and T units wide

In an effort to make this more exact, will will continue to shrink the rectangle down to the Dirac Delta function, $\delta$

• $\delta (x)=\{\begin{array}{ll}+\mathrm{\infty },& x=0\\ 0,& x\ne 0\end{array}$
• $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta (x)dx=1.$

By using the Dirac Delta function the summation becomes an integral

$f(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(u)\cdot \delta (t-u)du$

Changing from one orthogonal basis set to another

We have a vector $\hat{v}=\sum _{j=1}^3{a}_j{\hat{a}}_j$ and wish to change it to $\hat{v}=\sum _{j=1}^3{b}_j{\hat{b}}_j$. We know each basis set, and their relationship to each other. We are trying to find the coefficients, (the $b_j$) that go with the new basis set.

• Working from the $\hat{a}$ basis set:

$\hat{v}\cdot {\hat{b}}_m=\sum _{j=1}^3{v}_j{\hat{a}}_j\cdot {\hat{b}}_m=\sum _{j=1}^3{v}_j\underset{projof{\hat{a}}_{j}on{\hat{b}}_m}{\underbrace{({\hat{a}}_j\cdot {\hat{b}}_m)}}$

• Working from the $\hat{b}$ basis set:

$\hat{v}\cdot {\hat{b}}_m=\sum _{j=1}^3{b}_j{\hat{b}}_j\cdot {\hat{b}}_m=\sum _{j=1}^3{b}_j\underset{projof{\hat{b}}_{j}on{\hat{b}}_m}{\underbrace{({\hat{b}}_j\cdot {\hat{b}}_m)}}=\sum _{j=1}^3{b}_j{k}_m\delta mj=k_m\sum _{j=1}^3{b}_j\delta mj=b_m{k}_m\sum _{j=1}^3=b_m{k}_m$

• Now taking the $\hat{v}\cdot {\hat{b}}_m$ that was derived from both basis sets and equating them:

$b_m{k}_m=\sum _{j=1}^3{v}_j{\hat{a}}_j\cdot {\hat{b}}_m⟹b_m=\frac{1}{k_m}\sum _{j=1}^3{v}_j({\hat{a}}_j\cdot {\hat{b}}_m)$

Defining $k_m$

Taking $k_m$ from the previous section:

${\hat{b}}_j\cdot {\hat{b}}_m=k_m\delta mj⟹{\hat{b}}_m\cdot {\hat{b}}_m=k_m⟹{\left|{\hat{b}}_m\right|}^2=k_m$

Thus $k_m$ is the length of ${\hat{b}}_m$ squared

Questions

• What does the ${\hat{b}}_m$ represent, say compared to ${\hat{b}}_j$?
  • ${\hat{b}}_m$ is a unit vector in the direction we're interested in finding the new coefficient for the new basis set -- rewrite this
  • ${\hat{b}}_j$ is
• When you do the dot product of say A \cdot B, is it always the projection of A onto B and not the opposite way around?
• Why did you decide to make it $k_m$ instead of $k_j$?
  • Completely arbitrary, the end result is the same either way
• $\hat{x}$ is a unit vector and $\overrightarrow{x}$ is a vector