Feedback Control Closed-Loop Transfer Function

Figure source:

B. Bequette, Process control: modeling, design. and simulation, Prentice Hall Press, Upper Saddle River, NJ 2002

Block diagram (Laplace domain) variables:

c(s) = controller output

e(s) = error

r(s) = setpoint (reference signal)

u(s) = manipulated variable

l(s) = load disturbance

y(s) = process output

ym(s) = measured output

Transfer functions:

g_c(s) = controller

g_v(s) = valve

g_p(s) = process

g_m(s) = measurement

g_CL(s) = closed-loop

g_d(s) = disturbance

Calculations:

y(s) = l(s)g_d(s) + u(s)g_p(s)

u(s) = c(s)g_v(s)

c(s) = e(s)g_c(s)

e(s) = r(s) – y_m(s)

y_m(s) = y(s)g_m(s)

Lump all expressions together, we can get the following:

y(s) = l(s)g_d(s) + c(s)g_v(s)g_p(s)

y(s) = l(s)g_d(s) + e(s)g_c(s)g_v(s)g_p(s)

y(s) = l(s)g_d(s) + (r(s) – y_m(s))g_c(s)g_v(s)g_p(s)

y(s) = l(s)g_d(s) + (r(s) – y(s)g_m(s))g_c(s)g_v(s)g_p(s)

then:

Equation (1)

If there is no disturbance:

Equation (2)

Equation (3)

L'Hospital's Rule

I found the L'Hospital's rule extremely useful when I was doing process control homework tonight. So if you are not familiar with it, or can't recall it clearly, here it is.

In words, if taking limit directly doesn't work out, one thing you can do is to take derivative of the numerator and the denominator at the same time. Then take limit. Remember, after this step, if the limit still doesn't work, you can apply the L'Hospital's rule again and again until you find a way that the limit can be taken.

This can be simply applied to the initial value theorem and final value theorem in process control.

                    Initial Value Theorem

                    Final Value Theorem

                                                   
                    where F(s) is the Laplace transformation of f(t).

Hope this will help someone who are looking for a way to solve the limit problems.