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UnkleRhaukus

  • 3 years ago

Use transform definitions, and the evaluation of a suitable double integral, to calculate the Laplace transform of the following integral (ii)\[\mathcal L\left\{\int\limits_0^tf(t-u)g(u)\cdot\text du\right\}\]

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  1. UnkleRhaukus
    • 3 years ago
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    \[\begin{align*}\mathcal L\left\{\int\limits_0^tf(t-u)g(u)\cdot\text du\right\}&=\int\limits_0^\infty\int\limits_0^tf(t-u)g(u)\cdot\text du\cdot e^{-pt}\text dt\\ &=\int\limits_0^\infty\int\limits_0^\infty f(t-u)g(u)H(t-u)\cdot\text du\cdot e^{-pt}\cdot\text dt\\ &=\int\limits_0^\infty\int\limits_0^\infty f(t-u)g(u)H(t-u)e^{-pt}\cdot\text dt\cdot\text du\\ &=\\ \end{align*} \]

  2. gohangoku58
    • 3 years ago
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    \(\begin{align*}\mathcal L\left\{\int\limits_0^tf(t-u)g(u)\cdot\text du\right\}&=\int\limits_0^\infty\int\limits_0^tf(t-u)g(u)\cdot\text du\cdot e^{-pt}\text dt\\ &=\int\limits_0^\infty e^{-pt}[\int \limits_0^tf(t-u)g(u)du]dt \\ &=\int\limits_0^\infty g(u)[\int \limits_u^\infty e^{-pt}f(t-u)dt]du \\ by \:changing \:the\:order\:of \:integration \end{align*}\) the region from u=0 to u=t is same as region from t=u to t=\(\infty\) now, substitute t-u = x in inner integral. then you will be able to split the integrals , one only containing t, and other only containing x....

  3. gohangoku58
    • 3 years ago
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    @UnkleRhaukus are u trying that ^ ?

  4. UnkleRhaukus
    • 3 years ago
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    yeah im working on it

  5. gohangoku58
    • 3 years ago
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    :)

  6. gohangoku58
    • 3 years ago
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    numbers ? which numbers ?

  7. UnkleRhaukus
    • 3 years ago
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    \[\begin{align*}\mathcal L\left\{\int\limits_0^tf(t-u)g(u)\cdot\text du\right\}&=\int\limits_0^\infty\int\limits_0^tf(t-u)g(u)\cdot\text du\cdot e^{-pt}\text dt\\ &=\int\limits_0^\infty\int\limits_0^\infty f(t-u)g(u)H(t-u)\cdot\text du\cdot e^{-pt}\cdot\text dt\\ &=\int\limits_0^\infty\int\limits_0^\infty f(t-u)g(u)H(t-u)e^{-pt}\cdot\text dt\cdot\text du\\ &=\int\limits_0^\infty g(u)\int\limits_0^\infty f(t-u)H(t-u)e^{-pt}\cdot\text dt\cdot\text du\\ \text{let }v=t-u&\\ \text dv =\text dt\\ t=0\rightarrow v=-u\\ t=\infty \rightarrow v=\infty\\ &=\int\limits_0^\infty g(u)\int\limits_{-u}^\infty f(v)H(v)e^{-p(v+u)}\cdot\text dv\cdot\text du\\ &=\int\limits_0^\infty g(u)\int\limits_{-u}^\infty f(v)e^{-p(v+u)}\cdot\text dv\cdot\text du\\ \end{align*}\]

  8. gohangoku58
    • 3 years ago
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    didn't change the order of integration.....? plus you H is throwing me of...why(and how) its used there ... ?

  9. UnkleRhaukus
    • 3 years ago
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    im not sure wether or not the unsure the unit heaviside step function

  10. gohangoku58
    • 3 years ago
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    if u follow (and understand) what i am suggesting, u don't need H

  11. UnkleRhaukus
    • 3 years ago
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    \[\begin{align*}\mathcal L\left\{\int\limits_0^tf(t-u)g(u)\cdot\text du\right\} &=\int\limits_0^\infty\int\limits_0^t f(t-u)g(u)\cdot\text du\cdot e^{-pt}\cdot\text dt\\ &=\int\limits_0^\infty e^{-pt}\int\limits_0^t f(t-u)g(u)\cdot\text du\cdot\text dt\\ &=\int\limits_0^\infty e^{-pt}\int\limits_0^t f(t-u)g(u)\cdot\text dt\cdot\text du\\ \text{let }v=t-u&\\ \text dv =\text dt\\ t=0\rightarrow v=-u\\ t=\infty \rightarrow v=\infty\\ &=\int\limits_0^\infty g(u)\int\limits_{u}^\infty e^{-p(v+u)}f(v)\cdot\text dv\cdot\text du\\ &=\\ \end{align*}\]

  12. gohangoku58
    • 3 years ago
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    still, didn't change the order of integration.......

  13. gohangoku58
    • 3 years ago
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    u=0 to u=t cahnges to t=u to t=infinity

  14. gohangoku58
    • 3 years ago
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    then put t-u=v

  15. UnkleRhaukus
    • 3 years ago
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    this question is confusing me

  16. gohangoku58
    • 3 years ago
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    whats the confusion about ? changing order of integration ?

  17. UnkleRhaukus
    • 3 years ago
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    yes

  18. gohangoku58
    • 3 years ago
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    |dw:1353245519310:dw|

  19. UnkleRhaukus
    • 3 years ago
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    \[\begin{align*}\mathcal L\left\{\int\limits_0^tf(t-u)g(u)\cdot\text du\right\} &=\int\limits_0^\infty\int\limits_0^t f(t-u)g(u)\cdot\text du\cdot e^{-pt}\cdot\text dt\\ &=\int\limits_0^\infty e^{-pt}\int\limits_0^t f(t-u)g(u)\cdot\text du\cdot\text dt\\ u=0\rightarrow t=u\\ u=t \rightarrow t=\infty\\ &=\int\limits_0^\infty g(u)\int\limits_u^\infty e^{-pt}f(t-u)\cdot\text dt\cdot\text du\\ \text{let }v=t-u&\\ \text dv =\text dt\\ t=0\rightarrow v=-u\\ t=\infty \rightarrow v=\infty\\ &=\int\limits_0^\infty g(u)\int\limits_{u}^\infty e^{-p(v+u)}f(v)\cdot\text dv\cdot\text du\\ \end{align*}\]

  20. gohangoku58
    • 3 years ago
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    yes, now u see, how they can be separated into 2 different integrals, both going from 0 to infinity..

  21. gohangoku58
    • 3 years ago
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    and both integrals are definition of LT

  22. UnkleRhaukus
    • 3 years ago
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    \[=\int\limits_0^\infty g(u)\int\limits_{u}^\infty e^{-p(v+u)}f(v)\cdot\text dv\cdot\text du\] \[=\int\limits_0^\infty g(u)e^{-pu}\cdot\text du\quad\times\quad \int\limits_{u}^\infty e^{-pv}f(v)\cdot\text dv\quad?\]

  23. gohangoku58
    • 3 years ago
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    =L(g(x))L(f(x)) is the most simplified form...

  24. gohangoku58
    • 3 years ago
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    when u put v= t-u the limits change to 0 to infinity...

  25. UnkleRhaukus
    • 3 years ago
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    i dont understand that bit

  26. gohangoku58
    • 3 years ago
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    v=t-u when t=u v= 0 when t=infty v=infty

  27. gohangoku58
    • 3 years ago
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    everything else is correct.. just limits will be 0 to infinity

  28. UnkleRhaukus
    • 3 years ago
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    that is starting to make sense now

  29. gohangoku58
    • 3 years ago
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    \(=\int\limits_0^\infty g(u)e^{-pu}\cdot\text du\quad\times\quad \int\limits_{0}^\infty e^{-pv}f(v)\cdot\text dv\quad =L[g(t)]L[f(t)]\)

  30. gohangoku58
    • 3 years ago
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    ok ?

  31. UnkleRhaukus
    • 3 years ago
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    is this all right now? \[\begin{align*}\mathcal L\left\{\int\limits_0^tf(t-u)g(u)\cdot\text du\right\} &=\int\limits_0^\infty\int\limits_0^t f(t-u)g(u)\cdot\text du\cdot e^{-pt}\cdot\text dt\\ &=\int\limits_0^\infty e^{-pt}\int\limits_0^t f(t-u)g(u)\cdot\text du\cdot\text dt\\ u=0\rightarrow t=u\\ u=t \rightarrow t=\infty\\ &=\int\limits_0^\infty g(u)\int\limits_u^\infty e^{-pt}f(t-u)\cdot\text dt\cdot\text du\\ \text{let }v=t-u\\ \text dv =\text dt\\ t=u\rightarrow v=0\\ t=\infty \rightarrow v=\infty\\ &=\int\limits_0^\infty g(u)\int\limits_{0}^\infty e^{-p(v+u)}f(v)\cdot\text dv\cdot\text du\\ &=\int\limits_0^\infty g(u)e^{-up}\cdot\text du\times\int\limits_{0}^\infty f(v)e^{-pv}\cdot\text dv\\ \\&=G(p)F(p) \end{align*}\]

  32. gohangoku58
    • 3 years ago
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    yes

  33. gohangoku58
    • 3 years ago
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    finally!

  34. UnkleRhaukus
    • 3 years ago
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    Thank you so much!

  35. UnkleRhaukus
    • 3 years ago
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    what a convoluted process

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