Discussion
The unit step response of a system is given.
\(Y(t) = 1 - (7/3)e^{-t} + (3/2)e^{-2t} - (1/6)e^{-4t}\)
What is its transfer function?
Answer: A
Take the Laplace transform of the step response, then divide by the transform of the unit step, \(1/s\).
\[ \begin{aligned} Y(s) \\ &= \frac{1}{s} - \frac{7/3}{s+1} + \frac{3/2}{s+2} - \frac{1/6}{s+4} \end{aligned} \]
\[ \begin{aligned} H(s) = sY(s) \\ &= 1 - \frac{7s}{3(s+1)} + \frac{3s}{2(s+2)} - \frac{s}{6(s+4)} \end{aligned} \]
Putting this over a common denominator gives \(H(s) = \dfrac{s + 8}{(s+1)(s+2)(s+4)}\). A check at \(s = 0\): the dc gain is \(8/8 = 1\), which matches the final value of the step response, \(y(\infty) = 1\). Also the residue of \(Y(s)\) at \(s = -1\) is \(H(-1)/(-1) = -7/3\), which agrees with the coefficient of \(e^{-t}\). The answer is option A.
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