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# c14 - 18.03 Class 14 Exponential signals higher order...

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18.03 Class 14, March 10, 2006 Exponential signals, higher order equations, operators [1] Exponential signals x" + bx' + kx = A e^{rt} (*) We want to find some solution. Try for a solution of the form xp = B e^{rt} : k] xp = B e^{rt} b] xp' = B r e^{rt} xp" = B r^2 e^{rt} _____________________ A e^{rt} = B ( r^2 + br + k ) e^{rt} or xp = ( A / p(r)) e^{rt} where p(s) = s^2 + bs + k is the characteristic polynomial. Eg x" + 2x' + 2x = 4 e^{3t} p(3) = 3^2 + 2(3) + 2 = 17 so xp = ( 4 / 17 ) e^{3t} . The general solution is given by x = (4/17) e^{3t} + e^{-t} ( a cos(t) + b sin(t) ) Of course this will let us solve (*) for sinusoidal signals as well: Eg y" + 2y' + 2y = sin(3t) This is the imaginary part of z" + 2z' + 2z = e^{3it} p(3i) = (3i)^2 + 2(3i) + 2 = -7 + 6i so zp = ( 1 / (-7 + 6i) ) e^{it} We want the imaginary part. Lets do it by writing out real and imaginary parts: zp = ( (-7 - 6i) / 85 ) ( cos(3t) + i sin(3t) ) and find the imagnary part of the product: yp = - (7/85) cos(3t) + (6/85) sin(3t) The general solution is y = - (7/85) cos(3t) + (6/85) sin(3t) + e^{-t} ( a cos t + b sin t ).

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The work shows: The Exponential Response Formula: a solution to x" + bx' + kx = A e^ {rt} is given by xp = ( A / p(r) ) e^{rt} provided p(r) is not zero.
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c14 - 18.03 Class 14 Exponential signals higher order...

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