Question

(b) A certain linear, time-invariant dynamic system is described by the following differential equation: d^2y(t)/dt^2 - dy(t)/dt - 6y(t) = x(t) where x(t) is the input function. (i) Find the Transfer Function of the system. Is the system stable? Justify your answer. (ii) If x(t) = 2u(t), where u(t) is the unit step function, and the initial conditions are y(0) = 1 and dy(0)/dt = 0, solve for y(t) when t > 0. 

          (b) A certain linear, time-invariant dynamic system is described by the following differential equation:

d^2y(t)/dt^2 - dy(t)/dt - 6y(t) = x(t)

where x(t) is the input function.

(i) Find the Transfer Function of the system. Is the system stable? Justify your answer.

(ii) If x(t) = 2u(t), where u(t) is the unit step function, and the initial conditions are y(0) = 1 and dy(0)/dt = 0, solve for y(t) when t > 0.
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(b) A certain linear, time-invariant dynamic system is described by the following differential equation: d^2y(t)/dt^2 - dy(t)/dt - 6y(t) = x(t) where x(t) is the input function. (i) Find the Transfer Function of the system. Is the system stable? Justify your answer. (ii) If x(t) = 2u(t), where u(t) is the unit step function, and the initial conditions are y(0) = 1 and dy(0)/dt = 0, solve for y(t) when t > 0.

Added by Rocio M.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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A certain linear, time-invariant dynamic system is described by the following differential equation: dy(t)/dt + 6y(t) = x(t)/dt^2 where x(t) is the input function. Find the Transfer Function of the system. Is the system stable? Justify your answer. (ii) If x(t) = 2u(t), where u(t) is the unit step function, and the initial conditions are y(0) = 1 and dy(0)/dt = 0, solve for y(t) when t > 0.
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Transcript

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00:01 Hello students given the certain linear time variant dynamic system that is d square y of t divided by d t squared minus d t by d t minus 6y of t is equal to x of t let be equation one where x of t is the input function.
00:23 First we need to find transfer function and stability convert the above equation.
00:30 To laplace by applying laplace transformation.
00:33 Applying laplace transformation we have s square y of s minus s into y of s minus 6 into y of s minus 6 into y of s is equal to x of s.
00:48 This implies we have s square minus yes minus 6 into y of s is equal to x of s.
00:58 This implies we have y of s by x of s is equal to 1 by s square minus s minus 6...
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