User:Arctanx.tk

Maximum Power Transfer

in the notes, he says $P={\textrm {Re}}\left\{{\frac {V_{s}^{2}\left(R_{L}+jX_{L}\right)}{(R_{S}+R_{L})^{2}+(X_{S}+X_{L})^{2}}}\right\}$ but in the derivation the denominator is $\left[(R_{S}+R_{L})+j(X_{S}+X_{L})\right]^{2}$ which doesn't expand out. I'm wondering if you could have a look at your notes to see if they match

Question 5a

Evaulate $\int _{0}^{2}(x^{2}+1)dx$ by calculating $\lim _{||P||\rightarrow 0}\sum _{i=1}^{n}f(x_{i}^{*})\Delta x_{i}$, where $P$ is the partition of $[0,2]$ into $n$ equal intervals and $x_{i}^{*}$ is the right end of the $i$th. subinterval.

(Hint: $\sum _{i=1}^{n}i^{2}={\frac {n(n+1)(2n+1)}{6}}$)

The Sum Thing

$\sum _{x=1}^{n}f(x)$ means that you take $f(1)$, $f(2)$, $f(3)$, etc. up to $f(n)$, and add them all together.

For example,

$\sum _{x=1}^{5}2x=2(1)+2(2)+2(3)+2(4)+2(5)=30$

Note that:

$\sum _{x=1}^{5}2x=2\sum _{x=1}^{5}x$

and that a rule exists that

$\sum _{x=1}^{n}x={\frac {n(n+1)}{2}}$

so

$\sum _{x=1}^{5}2x=2\sum _{x=1}^{5}x=2{\frac {5(5+1)}{2}}=2{\frac {30}{2}}=30$