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UnkleRhaukus

  • 2 years ago

laplace transform of a periodic function/

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  1. UnkleRhaukus
    • 2 years ago
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  2. UnkleRhaukus
    • 2 years ago
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    im not very pleased with the final form of my answer,

  3. UnkleRhaukus
    • 2 years ago
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    mabye this is better \[F(p)=\frac{2}{(1-e^{-4p})p^2}-\frac{2e^{-2p}}{(1-e^{-4p})p^2}-\frac{4e^{-2p}}{(1-e^{-4p})p}\]

  4. lonliness
    • 2 years ago
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    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

  5. UnkleRhaukus
    • 2 years ago
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    @lonliness , what region should i integrate for figure 2

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  6. UnkleRhaukus
    • 2 years ago
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    also how did go in this one

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  7. mahmit2012
    • 2 years ago
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    |dw:1354803234598:dw|

  8. mahmit2012
    • 2 years ago
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    |dw:1354803478460:dw|

  9. mahmit2012
    • 2 years ago
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    |dw:1354803608097:dw|

  10. UnkleRhaukus
    • 2 years ago
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    but if i go from 0 to 2 pi ( in fig2) the the integral will be messy

  11. mahmit2012
    • 2 years ago
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    wait !

  12. UnkleRhaukus
    • 2 years ago
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    ok

  13. mahmit2012
    • 2 years ago
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    |dw:1354804514349:dw|

  14. mahmit2012
    • 2 years ago
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    |dw:1354804529087:dw|

  15. mahmit2012
    • 2 years ago
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    |dw:1354804592320:dw|

  16. mahmit2012
    • 2 years ago
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    yes! your answer isn't correct because you had some mistake to solving integrals.

  17. UnkleRhaukus
    • 2 years ago
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    i see that now,

  18. mahmit2012
    • 2 years ago
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    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

  19. mahmit2012
    • 2 years ago
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    |dw:1354805030916:dw|

  20. UnkleRhaukus
    • 2 years ago
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    |dw:1354861124769:dw|

  21. mahmit2012
    • 2 years ago
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    |dw:1354861368611:dw|

  22. UnkleRhaukus
    • 2 years ago
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    the slope is 1 and then -1

  23. UnkleRhaukus
    • 2 years ago
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    yes?

  24. UnkleRhaukus
    • 2 years ago
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    |dw:1354862069169:dw|

  25. UnkleRhaukus
    • 2 years ago
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    @mahmit2012

  26. UnkleRhaukus
    • 2 years ago
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    @TuringTest

  27. ali110
    • 2 years ago
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    @mahmit2012 s=ROC(region of convergence)+jw(frequency responce) so where the ROC of this function?

  28. UnkleRhaukus
    • 2 years ago
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    im not sure , i havent herd that term before, is it a restriction on p?

  29. ali110
    • 2 years ago
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    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

  30. ali110
    • 2 years ago
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    |dw:1354866466108:dw|

  31. ali110
    • 2 years ago
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    \[F(s)=\int\limits_{-\infty}^{\infty}f(t)e^-st*dt\]

  32. UnkleRhaukus
    • 2 years ago
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    \[\delta=\sigma\]?

  33. ali110
    • 2 years ago
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    @UnkleRhaukus am i explain it more if u want?

  34. UnkleRhaukus
    • 2 years ago
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    yes please

  35. ali110
    • 2 years ago
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    oh yes \[s=\sigma+jw\] actually i have not seen it on table before

  36. UnkleRhaukus
    • 2 years ago
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    for some reason my text book has been using \(p\) for \(s\) \(x\) for \(t\) \(n\) fr \(\omega\) \(i\) for \(j\)

  37. ali110
    • 2 years ago
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    so for moving function(signal) |dw:1354867100786:dw| now for left sided signal |dw:1354867294896:dw| |dw:1354867371794:dw|

  38. ali110
    • 2 years ago
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    i am using book of ALAN.V.OPPENHEIM and also book of SAMARJIT.GHOSH

  39. ali110
    • 2 years ago
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    Have u know before why we use laplace transform?

  40. UnkleRhaukus
    • 2 years ago
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    do we use laplace transform to understand the frequencies of the system?

  41. UnkleRhaukus
    • 2 years ago
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    (also to solve Initial value problems )

  42. ali110
    • 2 years ago
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    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!!

  43. UnkleRhaukus
    • 2 years ago
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    yes i would be interested to try out your matlab code , also what is a modulation problem ?

  44. ali110
    • 2 years ago
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    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

  45. ali110
    • 2 years ago
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    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

  46. UnkleRhaukus
    • 2 years ago
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    ok, can you tell me more about the region of convergence for my triangles signal

  47. ali110
    • 2 years ago
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    i ask it for mohan gholami(@mahmit2012)!! he will tell u!!i think he know it!!!

  48. ali110
    • 2 years ago
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    btw fren which book u use for signals and systems can u tell me its authur name?

  49. UnkleRhaukus
    • 2 years ago
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  50. UnkleRhaukus
    • 2 years ago
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    \[ \begin{align*} \mathcal L\big\{f(x)\big\}&=\int\limits_0^\infty f(x)e^{-px}\cdot\text dx\\ &=\tfrac1{1-e^{-4p}}\int\limits_0^{4}f(x)e^{-px}\cdot\text dx\\ &=\tfrac1{1-e^{-4p}}\left[\int\limits_0^{2}xe^{-px}\cdot\text dx+\int\limits_2^4(4-x)e^{-px}\cdot\text dx\right]\\ &=\tfrac1{1-e^{-4p}}\left[\frac{xe^{-px}}{-p}\Big|_0^2-\int\limits_0^{2}\frac{e^{-px}}{-p}\cdot\text dx+4\int\limits_2^4e^{-px}\cdot\text dx-\int\limits_2^4xe^{-px}\cdot\text dx\right]\\ &=\tfrac1{1-e^{-4p}}\left[\frac{2e^{-2p}}{-p}-\frac{e^{-px}}{p^2}\Big|_0^2+\frac{4e^{-px}}{-p}\Big|_2^4-\frac{xe^{-px}}{-p}\Big|_2^4+\int\limits_2^{4}\frac{e^{-px}}{-p}\cdot\text dx\right]\\ &=\tfrac1{1-e^{-4p}}\left[\frac{-2e^{-2p}}{p}-\frac{e^{-2p}-1}{p^2}+\frac{4e^{-4p}-4e^{-2p}}{-p}-\tfrac{4e^{-4p}-2e^{-2p}}{-p}+\frac{e^{-px}}{p^2}\Big|_2^4\right]\\ &=\tfrac1{1-e^{-4p}}\left[\frac{-2e^{-2p}}{p}+\frac{1-e^{-2p}}{p^2}+\frac{2e^{-2p}}{p}+\frac{e^{-4p}-e^{-2p}}{p^2}\right]\\ &=\frac1{1-e^{-4p}}\left[\frac{1-2e^{-2p}+4e^{-4p}}{p^2}\right]\\ &=\frac{1-2e^{-2p}+4e^{-4p}}{(1-e^{-4p})p^2}\\ \end{align*}\]

  51. UnkleRhaukus
    • 2 years ago
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    is this right for the wave of triangles?

  52. UnkleRhaukus
    • 2 years ago
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    is there some way to check?

  53. UnkleRhaukus
    • 2 years ago
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    @sirm3d

  54. UnkleRhaukus
    • 2 years ago
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