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Math Help - Differential Equation system Problem

  1. #1
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    Differential Equation system Problem

    Can anybody help?

    A system is modelled by Tx' + x = Ky

    given the following results deduce the parameters T and K

    . When the input y(t) of 5 units was applied the output x(t) ultimately settled with a value of 20 units

    . when a sinusoidal input was applied at an input frequency of 10 rad/s the output lagged behind the input by exactly -45 degrees.

    ii) if y(t) was a step change drive, at what time would x(t) reach half of its final steady state value ( T & K deduced previously)
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  2. #2
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    Quote Originally Posted by al2308 View Post
    Can anybody help?

    A system is modelled by Tx' + x = Ky

    given the following results deduce the parameters T and K

    . When the input y(t) of 5 units was applied the output x(t) ultimately settled with a value of 20 units

    . when a sinusoidal input was applied at an input frequency of 10 rad/s the output lagged behind the input by exactly -45 degrees.

    ii) if y(t) was a step change drive, at what time would x(t) reach half of its final steady state value ( T & K deduced previously)
    T \frac{dx}{dt} + x = 5K has the solution x = 20.

    T \frac{dx}{dt} + x = K \sin (10t) has the solution x = A \sin \left( 10\left( t - \frac{\pi}{4} \right) \right).

    Use the above to get the value of T and K.
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  3. #3
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    I have no idea on where to even start this one. Has anybody got any pointers? I assume you end up with a simultaneous equation of some form. But i'm not sure how to get there.
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  4. #4
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    Quote Originally Posted by al2308 View Post
    I have no idea on where to even start this one. Has anybody got any pointers? I assume you end up with a simultaneous equation of some form. But i'm not sure how to get there.
    You substitute the solutions given into the DE's and solve for what you don't know:

    Sub x = 20 into the first:

    20 = 5K => K = 4.

    Sub K = 4 into the second:

    T \frac{dx}{dt} + x = 4 \sin (10t)

    Now sub x = A \sin \left( 10\left( t - \frac{\pi}{4} \right) \right) = (expand using compound angle formula) into the DE. Equate coefficients of sin and cos to get two simultaneous equations in A and T. Solve those equations to get the value of A and T.
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    My knowledge of the compound angle formula is a bit rusty. So i've cheated and used my calculator. This has returned a very long expansion.

    A(sqrt2/2 sin(t) - sqrt2/2 Cos(t)) x {512(sqrt2/2 cos(t) + sqrt2/2 sin(t))^9 - 1024(sqrt2/2 Cos(t) + sqrt2/2 sin(t))^7 + 672(sqrt2/2 Cos(t) + sqrt2/2 sin(t))^5 - 160(sqrt2/2 Cos(t) + sqrt2/2 sin(t))^3 + 10(sqrt2/2 Cos(t) + sqrt2/2 sin(t))}

    this is not the smallest expresion to put into a DE. is my calculator right? i suspect it is! as i've tested it out on some smaller equations, or is there an easier method of solving the original problem.
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    Quote Originally Posted by al2308 View Post
    My knowledge of the compound angle formula is a bit rusty. So i've cheated and used my calculator. This has returned a very long expansion.

    A(sqrt2/2 sin(t) - sqrt2/2 Cos(t)) x {512(sqrt2/2 cos(t) + sqrt2/2 sin(t))^9 - 1024(sqrt2/2 Cos(t) + sqrt2/2 sin(t))^7 + 672(sqrt2/2 Cos(t) + sqrt2/2 sin(t))^5 - 160(sqrt2/2 Cos(t) + sqrt2/2 sin(t))^3 + 10(sqrt2/2 Cos(t) + sqrt2/2 sin(t))}

    this is not the smallest expresion to put into a DE. is my calculator right? i suspect it is! as i've tested it out on some smaller equations, or is there an easier method of solving the original problem.
    Actually I was anticipating you'd do something like the following:

    \sin \left( 10\left( t - \frac{\pi}{4} \right) \right)

    = \sin \left( 10 t - \frac{10 \pi}{4} \right)

    = \sin \left( 10 t - \frac{5 \pi}{2} \right)

    = \sin \left( 10 t - \frac{\pi}{2} \right)

    = - \cos (10 t).

    You could also use symmetry and the complementary angle formula to get the last line.

    Substituting a solution where the argument of the trig function is 10t is pretty essential since the term on the right hand side is a a trig with an argument of 10t. Hard to justify equating coefficients of sin and cos on each side if they are sining and cosing different arguments ......
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    I wasn't far off then!!!

    ok so X=-Acos(10t)

    X'=10Asin(10t) (hopefully)

    putting these into the DE

    T(10ASin(10t) + Acos(10t) = 4Sin(10t)

    I'm not confident this is correct or where i'm headed as trig tends to do stange things!
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  8. #8
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    Quote Originally Posted by al2308 View Post
    I wasn't far off then!!!

    ok so X=-Acos(10t)

    X'=10Asin(10t) (hopefully)

    putting these into the DE

    T(10ASin(10t) + Acos(10t) = 4Sin(10t)

    I'm not confident this is correct or where i'm headed as trig tends to do stange things!
    Well, there's a problem now because you require:

    10AT = 4 AND A = 0 .......

    I'm not sure how you'll resolve this - perhaps get clarification on the original question .....?
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