Question

The open loop transfer function of a first order lag process has a dead-time delay of 2 seconds and is controlled by a proportional controller that is given by: GH(s)=40[1/1+100s]e⁻²ˢ

Determine the transfer function in frequency domain.

Answer 1

The transfer function in the frequency domain for the given first-order lag process with a dead-time delay can be represented as GH(jω) = 40 / (1 + j100ω) e^-2jω.

Step-by-step explanation:

The transfer function of a first-order lag process with a dead-time delay can be represented as:

GH(s) = K / (1 + Ts) e^-Td s

where K is the gain, T is the time constant, and Td is the dead-time delay.

In the given problem, the transfer function GH(s) is given as:

GH(s) = 40 / (1 + 100s) e^-2s

This can be represented in the frequency domain as:

GH(jω) = 40 / (1 + j100ω) e^-2jω

where j is the imaginary unit and ω is the frequency.