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UnkleRhaukus
laplace transform of a periodic function/
im not very pleased with the final form of my answer,
mabye this is better \[F(p)=\frac{2}{(1-e^{-4p})p^2}-\frac{2e^{-2p}}{(1-e^{-4p})p^2}-\frac{4e^{-2p}}{(1-e^{-4p})p}\]
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
also how did go in this one
|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
|dw:1354804514349:dw|
|dw:1354804529087:dw|
|dw:1354804592320:dw|
yes! your answer isn't correct because you had some mistake to solving integrals.
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
|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
\[F(s)=\int\limits_{-\infty}^{\infty}f(t)e^-st*dt\]
\[\delta=\sigma\]?
@UnkleRhaukus am i explain it more if u want?
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?