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Control Engineering Basics

With all initial values zero, which of the following is the correct pair of operations that represent differentiation and integration of a time function in the s domain after the Laplace transform?

  1. ADifferentiation is multiplying by s, and integration is dividing by s
  2. BDifferentiation is dividing by s, and integration is multiplying by s
  3. CDifferentiation is adding s, and integration is subtracting s
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Answer: A. Differentiation is multiplying by s, and integration is dividing by s

Under the Laplace transform, the derivative of f(t) corresponds to sF(s) − f(0) and the integral to F(s)/s. With zero initial values, differentiation becomes multiplication by s and integration becomes division by s, so a differential equation turns into calculations with polynomials in s. For example, the RC-circuit equation RC·dv/dt + v = vin becomes (RCs + 1)V(s) = Vin(s), which gives the transfer function 1/(1 + RCs). The unit step function transforms to 1/s, and e^(−at) to 1/(s + a).

  • A. Correct. This is why an integrator is written 1/s and a differentiator s.
  • B. It is the other way round. Multiplying by s is differentiation, and dividing by s is integration.
  • C. They become multiplication and division, not addition and subtraction.
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