Question

Sommation problem

Answer 1

Could someone explain step 1 to step 2?

Answer 2

I hope there's a "clever" way to do this. I'll be doing a bit of internet diving to find it. XD

Answer 3

haha... it a big question.. i know.. but i dont know how they get from the first step to the second one. I know you can write the sinus in euler form.. but then?

Answer 4

First I am only going to solve the \(\sin\) function.
\[\left.\begin{array}{rr}\sin\frac{2\pi(n-2)}{6}\\at \:\:\:\: n = 0, -\sin\frac{2\pi}{3}\\at \:\:\:\:n=1, \sin \frac{-\pi}{6}\\at \:\:\:\: n=2, \sin 0 \\at\:\:\:\: n=3, \sin\frac{\pi}{3}\\at \:\:\:\:n=4, \sin\frac{2\pi}{3} \\ at \:\:\:\:n=5, sin\pi \end{array}\right.\]

\[\sin \frac{2\pi}{3} = \frac{\sqrt{3}}{2}, \sin\frac{\pi}{3} =\frac{\sqrt{3}}{2}, \sin \pi = \sin = 0 = 0 \]

\[\frac{4}{6} \left(-\frac{\sqrt{3}}{2}e^{-j\pi k*0}{3} - e^{-j\pi k*1}{3}\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}e^{-j\pi n k*3}{3}+ \frac{\sqrt{3}}{2}e^{-j\pi n k*4}{3}\right)\]


\[\large \frac{1}{\sqrt{3}} \left(-e^{-j\pi k*0/3} - e^{-j\pi k*1/3}+ e^{-j\pi n k}+ e^{-j\pi n k*4/ 3}\right)\]

Answer 5

I am not getting the exact answer, but I think the method is right.

Answer 6

owhhh :O... now i understand how they get the \[1/\sqrt{3}\] i didnt knew sin(2pi/3) was \[\sqrt{3}/2\]

Answer 7

thank you so much lshaan:)

Answer 8

No problem, I hope you can do it now. :-)

Answer 9

for future reference, its "SUMMATION" not "SOMMATION"

hahaha grammar police strikes again >:)

Answer 10

i hope so too... i just didnt knew the sin(2pi/3) was like square(3)/2 how do you know that? is that something standard?

Answer 11

hhaha... owhh.. typefault.. sorry officer!

Answer 12

Nope, not standard enough to memorize. sin(2pi/3) = sin(120) = sin(90+30) = cos30.

Answer 13

phoee i wont forget it :)