Solutions of Exercise 4.3

Fig. 28 Block diagram of a system affected by a disturbance \(v\)

A system \(G(s) = \frac{5}{s+2}\) is affected by a disturbance \(v\) via a transfer function \(H(s) = \frac{s+10}{s+2}\).

The system is controlled with a proportional controller \(F(s) = K\), with \(K > 0\), as shown in Fig. 28.

1. For what values of \(K\) the closed-loop stability is guaranteed?

2. For what values of \(K\) the steady-state effect of a disturbance \(v = \text{step}(t)\) on the output is less than \(5\%\) in closed-loop?

3. Show how the control scheme can be changed to explicitly compensate the disturbance.

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Solution

Question 1

The closed-loop transfer function is still

\[ G_c(s) = \frac{F(s)G(s)}{1 + F(s)G(s)} = \frac{5K}{s + 2 +5 K} \]

The closed-loop pole is in \(s = -2 - 5K\), which is negative for any \(K>0\).

Question 2

To appreciate the effect of \(v\) on the output, we compute the transfer function between \(v\) and \(y\). We can ignore the reference at this stage (\(r=0\)) because of the superimposition principle.

\[ Y(s) = G(s) U(s) + H(s) V(s) = G(s) F(s) (\cancel{R(s)} - Y(s)) + H(s) V(s) \]

The transfer function between \(V(s)\) and \(Y(s)\) is therefore

\[ \frac{Y(s)}{V(s)} = G_v(s) = \frac{H(s)}{1 + F(s)G(s)} = \frac{\frac{s+10}{s+2}}{1 + \frac{5K}{s+2}} = \frac{s+10}{s + 2 + 5K} \]

This transfer function is stable for \(K>0\) (it has the same poles of \(G_c(s)\)). Therefore, we can apply the Final Value Theorem to compute \(y(\infty)\) in response to \(v(t) = \text{step}(t)\), i.e., \(V(s) = \frac{1}{s}\).

\[ \lim_{t\to \infty} y(t) = \lim_{s \to 0} s Y(s) = \lim_{s \to 0} s G_v(s) V(s) = \lim_{s \to 0} \cancel{s} \frac{s+10}{s + 2 + 5K} \frac{1}{\cancel{s}} = \frac{10}{2+5K} \]

The effect is less than \(5\%\) if

\[ \left\lvert \frac{10}{2+5K} \right\lvert < 0.05 \]

This is equivalent to

\[ \lvert 2 + 5K \lvert > 200 = \frac{10}{0.05} \]

Mathematically, this is satisfied by \(2 + 5 K < - 200\) or \(2 + 5K > 200\), that is, \(K < - 40.4\) or \(K > 39.6\). The former solution must, however, be discarded because it would make the closed-loop unstable. Therefore, \(K > 39.6\).

Question 3

The disturbance \(v\) is measurable, therefore we can use a feed-forward compensator \(F_{ff}(s)\) as in the figure below.

Fig. 29 Control scheme with feed-forward disturbance compensator, \(F_{ff}(s)\).

In order to design this compensator \(F_{ff}(s)\), we can write down the closed-loop transfer function with this additional control action.

\[ Y(s) = G(s) \left[ F(s) (\cancel{R(s)} - Y(s)) + F_{ff}(s) V(s) \right] + H(s) V(s) \]

In this case, the transfer function between \(V(s)\) and \(Y(s)\) is

\[ \frac{Y(s)}{V(s)} = \frac{G(s) F_{ff}(s) + H(s)}{1 + F(s)G(s)} \]

An ideal compensator \(F_{ff}^{\star}(s)\) is one that makes the transfer function null, so that \(Y(s) = 0 V(s)\) for any possible \(V(s)\). This is achieved if

\[ G(s) F_{ff}^{\star}(s) + H(s) = 0 \]

Which means

\[ F_{ff}(s)^{\star} = - \frac{H(s)}{G(s)} = -\frac{\frac{s+10}{s+2}}{\frac{5}{s+2}} = - 0.2 (s + 10) \]

Warning

This feed-forward controller \(F^{\star}_{ff}(s)\) cannot be implemented, because its transfer function is non-proper (it has more zeros than poles).

In practice we would implement a static compensator

\[ F_{ff}(s) = F_{ff}^{\star}(0) = -2 \]