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Math Help - Unit step function

  1. #1
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    Unit step function

    Q1)f(t) = { 10e^-t 0<t<1 }
    -10^e-t t>1

    Express the function in terms of the Heaviside unit step function and hence obtain its Laplace transform.


    Q2) f(t) = 2 [u(t) u(t 4)]+ t u(t 4)
    = 2u(t) 2 u(t 4)+ t u(t 4)
    = 2u(t)+ ( t 2 ) u(t 4) <----- how to derive to this step, it was addition and now ts multiplication. explaination and guide pls.

    Also does anyone know any good sites for Laplace transform read-up(not wiki) and exercises to practise. Unit step fuction sites too.. my notes explaination is not clear enough for me or i am just stupid.
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  2. #2
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    Quote Originally Posted by mathsnerd View Post
    Q1)f(t) = { 10e^-t 0<t<1 }
    -10^e-t t>1

    Express the function in terms of the Heaviside unit step function and hence obtain its Laplace transform.
    f(t) = 10 e^{-t} 2[1/2 - u(t-1)], t>0,

    so:

    Lf = L[10e^{-t}] - 20 L[u(t-1)e^{-t}]

    Lf(s) = 10/(s+1) - 20 e^{-s}/(s+1)

    Q2) f(t) = 2 [u(t) u(t 4)]+ t u(t 4)
    = 2u(t) 2 u(t 4)+ t u(t 4)
    = 2u(t)+ ( t 2 ) u(t 4) <----- how to derive to this step, it was addition and now ts multiplication. explaination and guide pls.
    You have:

    f(t) = 2u(t) 2 u(t 4)+ t u(t 4)

    now you take out the common factor u(t-4) from the last two terms on the right:

    f(t) = 2u(t) + u(t 4)(-2+ t)

    rearrange and you have your mysterious line.

    RonL
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  3. #3
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    f(t) = 10 e^{-t} 2[1/2 - u(t-1)], t>0,

    so:

    Lf = L[10e^{-t}] - 20 L[u(t-1)e^{-t}]

    Lf(s) = 10/(s+1) - 20 e^{-s}/(s+1)

    thanks for the help...i apply what u taught me in Q2 and i got this

    f(t) = 10 e^{-t} - (10e^{-t} - 10e^{-t}) u(t-1)

    how do i get your workings in bold....care to help me explain further... thanks Ron.

    btw...the answer is 10/(s+1) - 20 e^-{s+1}/(s+1) .... you got it almost right or my book has the wrong answer.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by mathsnerd View Post
    f(t) = 10 e^{-t} 2[1/2 - u(t-1)], t>0,
    u(t-1) is zero up to t=1 and 1 from there on, but we want a step that is
    1 up to t=1 and -1 from there on so that f(t) = 10 e^{-t} up to t=0
    and is -10 e^{-t} from there onwards.

    So we wamt the step height to be -2 ( +1 to -1) and we need it to be offset
    by 1, so we use:

    s(t-1) = 2[1/2 - u(t-1)]

    which is a step function, which is 1 up to t=1 and -1 from there on.

    so:

    Lf = L[10e^{-t}] - 20 L[u(t-1)e^{-t}]
    We have:

    f(t) = 10 e^{-t} 2[1/2 - u(t-1)], t>0,

    so multiplying this out:

    f(t) = 10 e^{-t} - 20 e^{-t} u(t-1), t>0,

    Now tale LT:

    Lf = L[10 e^{-t} - 20 e^{-t} u(t-1)],

    but the LT is a linear transform, so:

    Lf = 10 L[e^{-t}] - 20 L[e^{-t} u(t-1)],

    Lf(s) = 10/(s+1) - 20 e^{-s}/(s+1)

    thanks for the help...i apply what u taught me in Q2 and i got this

    f(t) = 10 e^{-t} - (10e^{-t} - 10e^{-t}) u(t-1)

    how do i get your workings in bold....care to help me explain further... thanks Ron.

    btw...the answer is 10/(s+1) - 20 e^-{s+1}/(s+1) .... you got it almost right or my book has the wrong answer.
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