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UnkleRhaukus
 4 years ago
laplace transform of a periodic function/
UnkleRhaukus
 4 years ago
laplace transform of a periodic function/

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UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0im not very pleased with the final form of my answer,

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0mabye this is better \[F(p)=\frac{2}{(1e^{4p})p^2}\frac{2e^{2p}}{(1e^{4p})p^2}\frac{4e^{2p}}{(1e^{4p})p}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes 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

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0@lonliness , what region should i integrate for figure 2

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0also how did go in this one

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354803234598:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354803478460:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354803608097:dw

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0but if i go from 0 to 2 pi ( in fig2) the the integral will be messy

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354804514349:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354804529087:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354804592320:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes! your answer isn't correct because you had some mistake to solving integrals.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Do you know how I understood my mistake? Because I knew that the bounded signals should have Laplace with bounded limit in s=p=0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354805030916:dw

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354861124769:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354861368611:dw

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0the slope is 1 and then 1

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354862069169:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@mahmit2012 s=ROC(region of convergence)+jw(frequency responce) so where the ROC of this function?

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0im not sure , i havent herd that term before, is it a restriction on p?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0actually 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354866466108:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[F(s)=\int\limits_{\infty}^{\infty}f(t)e^st*dt\]

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0\[\delta=\sigma\]?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@UnkleRhaukus am i explain it more if u want?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh yes \[s=\sigma+jw\] actually i have not seen it on table before

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0for some reason my text book has been using \(p\) for \(s\) \(x\) for \(t\) \(n\) fr \(\omega\) \(i\) for \(j\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so for moving function(signal) dw:1354867100786:dw now for left sided signal dw:1354867294896:dw dw:1354867371794:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i am using book of ALAN.V.OPPENHEIM and also book of SAMARJIT.GHOSH

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Have u know before why we use laplace transform?

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0do we use laplace transform to understand the frequencies of the system?

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0(also to solve Initial value problems )

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes 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!!

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0yes i would be interested to try out your matlab code , also what is a modulation problem ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0modulation 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but 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

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0ok, can you tell me more about the region of convergence for my triangles signal

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i ask it for mohan gholami(@mahmit2012)!! he will tell u!!i think he know it!!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0btw fren which book u use for signals and systems can u tell me its authur name?

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0\[ \begin{align*} \mathcal L\big\{f(x)\big\}&=\int\limits_0^\infty f(x)e^{px}\cdot\text dx\\ &=\tfrac1{1e^{4p}}\int\limits_0^{4}f(x)e^{px}\cdot\text dx\\ &=\tfrac1{1e^{4p}}\left[\int\limits_0^{2}xe^{px}\cdot\text dx+\int\limits_2^4(4x)e^{px}\cdot\text dx\right]\\ &=\tfrac1{1e^{4p}}\left[\frac{xe^{px}}{p}\Big_0^2\int\limits_0^{2}\frac{e^{px}}{p}\cdot\text dx+4\int\limits_2^4e^{px}\cdot\text dx\int\limits_2^4xe^{px}\cdot\text dx\right]\\ &=\tfrac1{1e^{4p}}\left[\frac{2e^{2p}}{p}\frac{e^{px}}{p^2}\Big_0^2+\frac{4e^{px}}{p}\Big_2^4\frac{xe^{px}}{p}\Big_2^4+\int\limits_2^{4}\frac{e^{px}}{p}\cdot\text dx\right]\\ &=\tfrac1{1e^{4p}}\left[\frac{2e^{2p}}{p}\frac{e^{2p}1}{p^2}+\frac{4e^{4p}4e^{2p}}{p}\tfrac{4e^{4p}2e^{2p}}{p}+\frac{e^{px}}{p^2}\Big_2^4\right]\\ &=\tfrac1{1e^{4p}}\left[\frac{2e^{2p}}{p}+\frac{1e^{2p}}{p^2}+\frac{2e^{2p}}{p}+\frac{e^{4p}e^{2p}}{p^2}\right]\\ &=\frac1{1e^{4p}}\left[\frac{12e^{2p}+4e^{4p}}{p^2}\right]\\ &=\frac{12e^{2p}+4e^{4p}}{(1e^{4p})p^2}\\ \end{align*}\]

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0is this right for the wave of triangles?

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0is there some way to check?
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