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Math Help - Transfer function by differential equation

  1. #1
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    Transfer function by differential equation

    How can I obtain the transfer function of system Y(s)/ X(s), which has the differential equation below ?


    d^3/dt^3 + 3(d^2*y/dt^2) +5(dy/dt) + y = d^3*x/dt^3 + 4(d^2* x/dt^2) + 6(dx/dt) + 8x


    If possible give me a tip to insert special characters here, as pi and division sign with a number over the other


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  2. #2
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    Quote Originally Posted by moisesbr View Post
    How can I obtain the transfer function of system Y(s)/ X(s), which has the differential equation below ?


    d^3/dt^3 + 3(d^2*y/dt^2) +5(dy/dt) + y = d^3*x/dt^3 + 4(d^2* x/dt^2) + 6(dx/dt) + 8x

    If possible give me a tip to insert special characters here, as pi and division sign with a number over the other

    For mathematical type setting we use LaTeX, the tutorial is here

    CB
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by moisesbr View Post
    How can I obtain the transfer function of system Y(s)/ X(s), which has the differential equation below ?


    d^3/dt^3 + 3(d^2*y/dt^2) +5(dy/dt) + y = d^3*x/dt^3 + 4(d^2* x/dt^2) + 6(dx/dt) + 8x

    If possible give me a tip to insert special characters here, as pi and division sign with a number over the other

    Write this as:

    D^3y+3D^2y+5Dy+y=D^3x+4D^2x+6Dx+8x

    Then if we assume that at time t=0 that the input and out put and all relevant derivatives are zero, and we take the Laplace transform of the above equation we get:

     <br />
s^3Y(s)+3s^2Y(s)+5sY(s)+Y(s)=s^3X(s)+4s^2X(s)+6sX(  s)+8X(s)<br />

    then the transfer function is:

     <br />
T(s)=\frac{Y(s)}{X(s)}=\frac{s^3+4s^2+6s+8}{s^3+3s  ^2+5s+1}<br />

    CB
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  4. #4
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    How did you pass from first to second line below ?

    Sorry for the silly question but I've been a long time out of maths
    and I having a hard time to remind all in a new course I am starting

    Thanks for your patience

    d^3/dt^3 + 3(d^2*y/dt^2) +5(dy/dt) + y = d^3*x/dt^3 + 4(d^2* x/dt^2) + 6(dx/dt) + 8x


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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by moisesbr View Post
    How did you pass from first to second line below ?

    Sorry for the silly question but I've been a long time out of maths
    and I having a hard time to remind all in a new course I am starting

    Thanks for your patience

    d^3/dt^3 + 3(d^2*y/dt^2) +5(dy/dt) + y = d^3*x/dt^3 + 4(d^2* x/dt^2) + 6(dx/dt) + 8x


    It is just another notation for the derivatives:

    D^n \equiv \frac{d^n}{dt^n}

    CB
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