Question

periodic rectangle wave , Help :D

Answer 1

Help with?

Answer 2

first of all u can say that the series will not contain any cosine terms so average value will be zero so
a0=0 and
an=0
so its a odd function

Answer 3

ahaa okkk

Answer 4

but why /.

Answer 5

i got that a_0=0
but why this will not contain any cosine terms

Answer 6

in any odd symmetry like when
f(t)=-f(-t)
a0 and an becomes zero
u can check it it will produce zero

Answer 7

aha continue plz :D

Answer 8

an=\[an=\frac{ 1 }{ \pi}\int\limits_{0}^{2\pi}k*\cos(nwt)d(wt)\]

Answer 9

now break the limits 0 to (pi) and
(pi) to 2*(pi)
then integrate it will produce zero

Answer 10

are u getting it ?

Answer 11

\[an=\frac{ 1 }{ \pi }[\int\limits_{0}^{\pi}kcos(nwt)d(wt)+\int\limits_{\pi}^{2\pi}(-k)\cos(nwt)d(wt)\]
okay with u?

Answer 12

yess im oki with it , also i can see from the graph without integration
area under the curve of f(x) btw -pi , pi =0
okkk continue :D

Answer 13

haha :D

Answer 14

now then only have to find bn

Answer 15

so
\(\large b_n=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin \text {nx dx} \)

Answer 16

\[bn=\frac{ 1 }{ \pi}[\int\limits_{0}^{\pi}k*\sin(nwt)d(wt)+\int\limits_{\pi}^{2\pi}(-k)\sin(nwt)d(wt)]\]

Answer 17

ahha how did u change second term to( pi ,2 pi ?)

Answer 18

just breaking the limit 0 to 2pi into 0 to pi and pi to 2pi

Answer 19

mmm limit from
-pi to pi
we break it like this right ?
\(\large b_n=\frac{1}{\pi}\int_{-\pi}^{0 } f(x) \sin \text {nx dx} +\int_{0}^{\pi } f(x) \sin \text {nx dx} \)

Answer 20

why -pi to pi??

Answer 21

so it would be like this
\(\large b_n=\frac{1}{\pi}\int_{-\pi}^{0 } (-k) \sin \text {nx dx} +\int_{0}^{\pi } (k) \sin \text {nx dx} \)

Answer 22

u have to integrate the function over its period so it will be 0 to 2pi

Answer 23

well , u can see the range of the wave
its oki if it from 0 to 2 pi
or -pi to pi
the same length after all , but i thought it would be more easy
since f(x) change according to this , what do u think ?

Answer 24

its the same
f(0)=f(2pi)

Answer 25

no in 0 to 2pi
0 to pi gives u k and
pi to 2pi gives u (-k)
is'nt it
that is why i'm breaking the limits

Answer 26

oki :P
but its the same hehe
oki continue plz

Answer 27

what happen is this