Question

A system is described by the following differential equation \[ 0.01 \frac{d^2y(t)}{dt^2} + 0.2 \frac{dy(t)}{dt} + y(t) = 6x(t), \] where time \((t)\) is in seconds. If \(x(t)\) is the unit step input applied at \(t=0\) s to this system, the magnitude of the output at \(t=1\) s is ________. (Round off the answer to two decimal places.)

Hint:

- Always normalize coefficients before forming transfer functions. - Use partial fractions carefully for repeated roots. - For unit step response, remember \(X(s)=1/s\). - Exponentials with large negative arguments (like \(e^{-10}\)) often approximate to nearly zero.

Answer 1

Correct Answer: 5.8

Step 1: Write transfer function. Taking Laplace transform with zero initial conditions: \[ 0.01 s^2 Y(s) + 0.2 s Y(s) + Y(s) = \frac{6}{s}. \] So the transfer function is \[ \frac{Y(s)}{X(s)} = \frac{6}{0.01s^2 + 0.2s + 1}. \] Step 2: Normalize coefficients. Divide numerator and denominator by 0.01: \[ \frac{Y(s)}{X(s)} = \frac{600}{s^2 + 20s + 100}. \] Step 3: Unit step input. For \(x(t) = u(t)\), \(X(s) = \tfrac{1}{s}\). Thus, \[ Y(s) = \frac{600}{s(s^2+20s+100)}. \] Step 4: Partial fraction decomposition. We factorize denominator: \[ s^2+20s+100 = (s+10)^2. \] So, \[ Y(s) = \frac{600}{s(s+10)^2}. \] Let \[ \frac{600}{s(s+10)^2} = \frac{A}{s} + \frac{B}{s+10} + \frac{C}{(s+10)^2}. \] Step 5: Solve for constants. Multiply both sides: \[ 600 = A(s+10)^2 + B\cdot s(s+10) + C\cdot s. \] - Put \(s=0\): \(600 = A(100)\) \(\Rightarrow A=6\). - Put \(s=-10\): \(600 = C(-10)\) \(\Rightarrow C=-60\). - Put \(s=1\): \(600 = 6(11^2) + B(11) + (-60)(1)\). \[ 600 = 726 + 11B - 60 \Rightarrow 600 = 666 + 11B \Rightarrow B = -6. \] So, \[ Y(s) = \frac{6}{s} - \frac{6}{s+10} - \frac{60}{(s+10)^2}. \] Step 6: Inverse Laplace. \[ y(t) = 6(1) - 6e^{-10t} - 60\cdot te^{-10t}, t \geq 0. \] Step 7: Evaluate at \(t=1\). \[ y(1) = 6 - 6e^{-10} - 60 e^{-10}. \] \[ = 6 - 66 e^{-10}. \] Since \(e^{-10} \approx 4.54 \times 10^{-5}\), \[ y(1) \approx 6 - 66(0.0000454) \approx 6 - 0.00299 = 5.997. \] Rounded to two decimal places: \(\mathbf{6.00}\). Final Answer: 6.00