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UnkleRhaukus

laplace transform of a periodic function/

  • one year ago
  • one year ago

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

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

    • one year ago
  4. lonliness
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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

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

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

    • one year ago
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  7. mahmit2012
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    |dw:1354803234598:dw|

    • one year ago
  8. mahmit2012
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    |dw:1354803478460:dw|

    • one year ago
  9. mahmit2012
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    |dw:1354803608097:dw|

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

    • one year ago
  11. mahmit2012
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    wait !

    • one year ago
  12. UnkleRhaukus
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    ok

    • one year ago
  13. mahmit2012
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    |dw:1354804514349:dw|

    • one year ago
  14. mahmit2012
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    |dw:1354804529087:dw|

    • one year ago
  15. mahmit2012
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    |dw:1354804592320:dw|

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

    • one year ago
  17. UnkleRhaukus
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    i see that now,

    • one year ago
  18. mahmit2012
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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

    • one year ago
  19. mahmit2012
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    |dw:1354805030916:dw|

    • one year ago
  20. UnkleRhaukus
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    |dw:1354861124769:dw|

    • one year ago
  21. mahmit2012
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    |dw:1354861368611:dw|

    • one year ago
  22. UnkleRhaukus
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    the slope is 1 and then -1

    • one year ago
  23. UnkleRhaukus
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    yes?

    • one year ago
  24. UnkleRhaukus
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    |dw:1354862069169:dw|

    • one year ago
  25. UnkleRhaukus
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    @mahmit2012

    • one year ago
  26. UnkleRhaukus
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    @TuringTest

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

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

    • one year ago
  29. ali110
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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

    • one year ago
  30. ali110
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    |dw:1354866466108:dw|

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

    • one year ago
  32. UnkleRhaukus
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    \[\delta=\sigma\]?

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

    • one year ago
  34. UnkleRhaukus
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    yes please

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

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

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

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

    • one year ago
  39. ali110
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    Have u know before why we use laplace transform?

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

    • one year ago
  41. UnkleRhaukus
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    (also to solve Initial value problems )

    • one year ago
  42. ali110
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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!!

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

    • one year ago
  44. ali110
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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

    • one year ago
  45. ali110
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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

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

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

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

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

    • one year ago
  51. UnkleRhaukus
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    is this right for the wave of triangles?

    • one year ago
  52. UnkleRhaukus
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    is there some way to check?

    • one year ago
  53. UnkleRhaukus
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    @sirm3d

    • one year ago
  54. UnkleRhaukus
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    • one year ago
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