laplace transform of a periodic function/

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

laplace transform of a periodic function/

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

1 Attachment
im not very pleased with the final form of my answer,
mabye this is better \[F(p)=\frac{2}{(1-e^{-4p})p^2}-\frac{2e^{-2p}}{(1-e^{-4p})p^2}-\frac{4e^{-2p}}{(1-e^{-4p})p}\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

yes this one is better but the one that you have solved is much better and simplified , so i dont think you need to do this
@lonliness , what region should i integrate for figure 2
1 Attachment
also how did go in this one
1 Attachment
|dw:1354803234598:dw|
|dw:1354803478460:dw|
|dw:1354803608097:dw|
but if i go from 0 to 2 pi ( in fig2) the the integral will be messy
wait !
ok
|dw:1354804514349:dw|
|dw:1354804529087:dw|
|dw:1354804592320:dw|
yes! your answer isn't correct because you had some mistake to solving integrals.
i see that now,
Do you know how I understood my mistake? Because I knew that the bounded signals should have Laplace with bounded limit in s=p=0
|dw:1354805030916:dw|
|dw:1354861124769:dw|
|dw:1354861368611:dw|
the slope is 1 and then -1
yes?
|dw:1354862069169:dw|
@mahmit2012 s=ROC(region of convergence)+jw(frequency responce) so where the ROC of this function?
im not sure , i havent herd that term before, is it a restriction on p?
actually we moved from fourier transform to laplace transform because fourier transform tells only frequency responce of a signal/function but laplace transform can explain stability/unstability of function including with its frequency responce so its easy to design system by using laplace transform
|dw:1354866466108:dw|
\[F(s)=\int\limits_{-\infty}^{\infty}f(t)e^-st*dt\]
\[\delta=\sigma\]?
@UnkleRhaukus am i explain it more if u want?
yes please
oh yes \[s=\sigma+jw\] actually i have not seen it on table before
for some reason my text book has been using \(p\) for \(s\) \(x\) for \(t\) \(n\) fr \(\omega\) \(i\) for \(j\)
so for moving function(signal) |dw:1354867100786:dw| now for left sided signal |dw:1354867294896:dw| |dw:1354867371794:dw|
i am using book of ALAN.V.OPPENHEIM and also book of SAMARJIT.GHOSH
Have u know before why we use laplace transform?
do we use laplace transform to understand the frequencies of the system?
(also to solve Initial value problems )
yes we use laplace transform to find both frequency responce(jw) as well as its stabality and unstabaliy(sigma) of function s=sigma+jw as see above yes also for initial value problem+modulation problem+final value problem i can give u simple function matlab code if u want!!
yes i would be interested to try out your matlab code , also what is a modulation problem ?
modulation theorem!!!! application of signal and system in communication it is a technique that is use to convey the info over a long distance i.e for voice signal as voice signal dont have high range for long distance actual signal+carrier signal 3400hz+high freq signal(1 Mhz) voice data may die over a long distance if we not use carrier signal signal send for long distance through amplitude modulation as well as frequency modulation we can fix the band width of AM,short band width require if we want to send signal over a lond distance for FM,long band width is needed if we want to send signal for long distance but here there is some cost problem!!i will discuss if u want
but frequency modulation is best as noise signals in atmosphere donot affect on it like mobile phone signals we use frequency modulation technique!!! but still its so costly to use Freq modulation
ok, can you tell me more about the region of convergence for my triangles signal
i ask it for mohan gholami(@mahmit2012)!! he will tell u!!i think he know it!!!
btw fren which book u use for signals and systems can u tell me its authur name?
\[ \begin{align*} \mathcal L\big\{f(x)\big\}&=\int\limits_0^\infty f(x)e^{-px}\cdot\text dx\\ &=\tfrac1{1-e^{-4p}}\int\limits_0^{4}f(x)e^{-px}\cdot\text dx\\ &=\tfrac1{1-e^{-4p}}\left[\int\limits_0^{2}xe^{-px}\cdot\text dx+\int\limits_2^4(4-x)e^{-px}\cdot\text dx\right]\\ &=\tfrac1{1-e^{-4p}}\left[\frac{xe^{-px}}{-p}\Big|_0^2-\int\limits_0^{2}\frac{e^{-px}}{-p}\cdot\text dx+4\int\limits_2^4e^{-px}\cdot\text dx-\int\limits_2^4xe^{-px}\cdot\text dx\right]\\ &=\tfrac1{1-e^{-4p}}\left[\frac{2e^{-2p}}{-p}-\frac{e^{-px}}{p^2}\Big|_0^2+\frac{4e^{-px}}{-p}\Big|_2^4-\frac{xe^{-px}}{-p}\Big|_2^4+\int\limits_2^{4}\frac{e^{-px}}{-p}\cdot\text dx\right]\\ &=\tfrac1{1-e^{-4p}}\left[\frac{-2e^{-2p}}{p}-\frac{e^{-2p}-1}{p^2}+\frac{4e^{-4p}-4e^{-2p}}{-p}-\tfrac{4e^{-4p}-2e^{-2p}}{-p}+\frac{e^{-px}}{p^2}\Big|_2^4\right]\\ &=\tfrac1{1-e^{-4p}}\left[\frac{-2e^{-2p}}{p}+\frac{1-e^{-2p}}{p^2}+\frac{2e^{-2p}}{p}+\frac{e^{-4p}-e^{-2p}}{p^2}\right]\\ &=\frac1{1-e^{-4p}}\left[\frac{1-2e^{-2p}+4e^{-4p}}{p^2}\right]\\ &=\frac{1-2e^{-2p}+4e^{-4p}}{(1-e^{-4p})p^2}\\ \end{align*}\]
is this right for the wave of triangles?
is there some way to check?
http://www.wolframalpha.com/input/?i=inverse+laplace+transform+of+%281-2e%5E%28-2s%29%2B4e%5E%28-4s%29%29%2F%28%281-e%5E%28-4s%29%29s%5E2%29 .. damn it
1 Attachment

Not the answer you are looking for?

Search for more explanations.

Ask your own question