Solutions of Exercise 2.5

Solutions of Exercise 2.5#

We are trying to design the controller for the system:

\[ G(s) = \frac{5}{s^3 + 8s^2 + 20s + 25} \]

Assume we go for a family of PI controllers described by the following transfer function, where \(K > 0\) is a parameter to be designed.

\[ F(s) = \frac{K(s+1)}{s} \]

Compute the closed-loop transfer function \(G_c(s)\).
Apply the Routh-Hurwitz criterion to find for what values of \(K\) the system is stable.

Hint

Question 2 can be solved using the Routh-Hurwitz criterion.

The characteristic polynomial \(\Phi(s) = a^0 s^n + b_0 s^{n-1} + a_1 s^{n-2} + b_1 s^{n-3} + ...\) is the denominator of the system we are assessing the stability of. In this case, it is the denominator of \(G_c(s)\).
Note that the coefficients \(a_i\) and \(b_i\) are defined starting from the highest degree, \(s^n\), and they are “alternating”. \(b_n\) might be zero if \(n\) is an even number.

Solution#

Question 1#

The closed-loop transfer function is given by:

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

First, let’s compute the loop transfer function \(G_o(s) = F(s)G(s)\):

\[ G_o(s) = \frac{K(s+1)}{s} \cdot \frac{5}{s^3 + 8s^2 + 20s + 25} = \frac{5K(s+1)}{s^4 + 8s^3 + 20s^2 + 25s} \]

Now we can form \(G_c(s)\):

\[ G_c(s) = \frac{\frac{5K(s+1)}{s^4 + 8s^3 + 20s^2 + 25s}}{1 + \frac{5K(s+1)}{s^4 + 8s^3 + 20s^2 + 25s}} = \frac{5K(s+1)}{s^4 + 8s^3 + 20s^2 + 25s + 5K(s+1)} \]

Simplifying the denominator:

\[ G_c(s) = \frac{5K(s+1)}{s^4 + 8s^3 + 20s^2 + (25 + 5K)s + 5K} \]

Question 2#

Computing the Routh table#

Warning

Note that we could not solve this exercise computing the poles explicitly as in Solutions of Exercise 2.2. An alternative solution is based on the root locus.

The characteristic polynomial of the closed-loop system is the denominator of \(G_c(s)\):

\[ \Phi(s) = s^4 + 8s^3 + 20s^2 + (25 + 5K)s + 5K \]

We construct the Routh table, where \(a_0 = 1\), \(a_1 = 20\), \(a_2 = 5K\), and \(b_0 = 8\), \(b_1 = 25 + 5K\), \(b_2 = 0\).

\( \begin{array}{ccc} 1 & 20 & 5K \\ 8 & 25 + 5K & 0 \\ \hline c_1 & c_2 & 0 \\ d_1 & d_2 & 0 \\ e_1 & e_2 & 0 \end{array} \)

We remind that the coefficient \(c_k\) is computed as

(4)#\[\begin{split} c_{k} = \frac{1}{b_0} \textrm{det} \begin{bmatrix} b_0 & b_{k+1} \\ a_0 & a_{k+1} \end{bmatrix} = \frac{b_0 \cdot a_{k+1} - b_{k+1} \cdot a_0}{b_0} \end{split}\]

We now compute the coefficients \(c_0\), \(c_1\), and \(c_2\).

\[ c_0 = \frac{b_0 a_1 - b_1 a_0}{b_0} = \frac{160-25-5K}{8} = \frac{135-5K}{8} \]

\[ c_1 = \frac{b_0 a_2 - b_2 a_0}{b_0} = \frac{40K-0}{8} = 5K \]

\[ c_2 = \frac{b_0 a_3 - b_3 a_0}{b_0} = \frac{8 \cdot 0 - 0 \cdot 1}{8} = 0 \]

To compute the coefficient \(d_k\), we just shift (4):

\[\begin{split} d_{k} = \frac{1}{c_0} \textrm{det} \begin{bmatrix} c_0 & c_{k+1} \\ b_0 & b_{k+1} \end{bmatrix} = \frac{c_0 \cdot b_{k+1} - c_{k+1} \cdot b_0}{c_0} \end{split}\]

Then:

\[ d_0 = \frac{c_0 b_1 - c_1 b_0}{c_0} = \frac{\frac{135-5K}{8} \cdot (25 + 5K) - 5K \cdot 8}{\frac{135-5K}{8}} = \frac{(135-5K) (25 + 5K) - 5K \cdot 8 \cdot 8}{135 - 5K} = \frac{ 3375 + 230K - 25K^2}{135 - 5K} \]

\[ d_1 = \frac{c_0 b_2 - c_2 b_0}{c_0} = \frac{c_0 \cdot 0 - 0 \cdot b_0}{c_0} = 0 \]

\[ d_2 = 0 \]

The \(e_k\) coefficients can be computed similarly:

\[ e_0 = \frac{d_0 c_1 - c_0 d_1}{d_0} \]

Because \(d_1 =0\), \(e_0\) reads:

\[ e_0 = \frac{\cancel{d_0} c_1}{\cancel{d_0}} = c_1 = 5K \]

The remaining coefficients \(e_1 = e_2 = 0\).

Assessing the stability with the Routh criterion#

According to the Routh criterion, the system is stable if the element in the first column are all strictly positive.

\( \begin{dcases} 1 > 0 & (A) \\ 8 > 0 & (B) \\ \frac{135-5K}{8} > 0 \qquad & (C) \\ \frac{ 3375 + 230K - 25K^2}{135 - 5K} \qquad & (D)\\ 5K > 0 & (E) \end{dcases} \)

We now solve the inequalities one by one, and later reconstruct the system.

Inequality (A)

Satisfied \(\forall K\)

Inequality (B)

Satisfied \(\forall K\)

Inequality (C)

Because \(8\) is known and positive, we can get rid of it:

\[ 135 - 5K > 0 \quad \rightarrow \boxed{K < 27} \]

Inequality (D)

This is a fractional inequality. It is solved by computing the intervals where the numerator is positive, and the intervals where the numerator is negative. Then, we apply the rule of sign.

Numerator > 0

\[ 3375 + 230K - 25K^2 > 0 \]

We change the sign and flip the inequality:

\[ 25 K^2 - 230K - 3375 < 0 \]

Then, we compute the roots, that specify the boundary of the interval

\[ 25K^2 - 230 - 3375 = 0 \]

The discriminant is \(\Delta = 230^2 + 4 \cdot 25 \cdot 3375 = 390400 > 0\). Because the discriminant is positive, the system admits two real solutions:

\[ K_1 = \frac{230 - \sqrt{390400}}{50} = - 7.9 \text{ and } K_2 = \frac{230 + \sqrt{390400}}{50} = 17.1 \]

Since the inequality had a \(<\), we conclude that the numerator is positive when \(-7.9 < K < 17.1\).

Denominator > 0

\[ 135 - 5K > 0 \]

This inequality is satisfied for \(K < 27\).

Rule of signs

Inequality (D) was a \(>0\) inequality, therefore the solution is

\[ -7.9 < K < 17.1 \cup K > 27 \]

Inequality (E)

Satisfied for \(K>0\).

Computing the interval#

We can finally reconstruct the system of inequalities (A)-(E)

\( \begin{dcases} \forall K \\ \forall K \\ K < 27 \\ -7.9 < K < 17.1 \cup K > 27 \\ K > 0 \end{dcases} \)

To find the largest interval that satisies all the inequalities at the same, we can sketch the intervals

The interval of \(K\) for which the controller achieves closed-loop stability is \(\boxed{0 < K < 17.1 }\).