Solutions of Exercise 2.0

Solutions of Exercise 2.0#

Consider a second-order system $G(s) = \frac{1}{s^2 + 3s + 2}$ and a proportional controller $F(s) = 4$.

Draw the block diagram of the closed loop.
Does this controller ensure closed-loop stability?
Compute the steady-state output $y(\infty)$ for step references with amplitude $1$.
Compute the steady-state input $u(\infty)$ for step references with amplitude $1$.

Solution#

Question 1#

Question 2#

To check for asymptotic stability, we first find the closed-loop transfer function:

\[ Y(s) = G(s) U(s) = G(s) U(s) (\underbrace{R(s) - Y(s)}_{E(s)}) \]

Therefore: $$ Y(s) = \underbrace{\frac{G(s) F(s)}{1 + G(s) F(s)}}_{G_c(s)} R(s) $$

Let’s replace $F(s) = 4$ and $G(s) = \frac{1}{s^2 + 3s + 2}$. By so doing, we get $$ G_c(s) = \frac{4}{s^2 + 3s + 6} $$

To assess whether the controller guarantees closed-loop stability, we compute the closed-loop poles, i.e., the roots of the denominator of $G_c(s)$.

\[ s^2 + 3s + 6 = 0 \quad \rightarrow \quad s = \frac{-3 \pm \sqrt{9-24}}{2} = -\frac{3}{2} \pm \frac{\sqrt{-15}}{2} = \boxed{-1.5 \pm i \frac{\sqrt{15}}{2}} \]

The closed-loop poles are complex-conjugate numbers. They are stable, because their real part is negative ($-1.5 < 0$).

Note

$\sqrt{-15}$ is the square root of a negative number, and therefore an imaginary number. This has be re-written as follows

\[ \sqrt{-15} = \sqrt{(-1) 15} = \sqrt{-1} \sqrt{15} = i \sqrt{15} \]

Question 3#

To compute $y(\infty)$ we use the Final Value Theorem which states that

\[ \lim_{t \to \infty} = \lim_{s \to 0} s \, Y(s) \]

Final Value Theorem

The Final Value Theorem can be applied only if the (closed-loop) system is stable! In this exercise, we have checked it in Question 2.

Remembering that $Y(s) = G_c(s) R(s)$, where the reference is $r(t) = \textrm{step}(t)$ and its Laplace transform is $R(s) = \frac{1}{s}$, we can compute $y(\infty)$ as follows:

\[ y(\infty) = \lim_{s \to 0} \cancel{s} \frac{4}{s^2 + 3s + 6} \frac{1}{\cancel{s}} = \frac{4}{s^2 + 3s + 6} = \frac{2}{3} \approx 0.67 \]

Question 4#

To compute $u(t)$, we first compute its Laplace function, $U(s)$.

\[ U(s) = F(s) E(s) = F(s) (R(s) - Y(s)) = F(s) (R(s) - G(s)U(s)) \]

Therefore, moving $-F(s)G(s) U(s)$ on the left-hand side and dividing by $1+F(s)G(s)$, we get: $$ U(s) = \frac{F(s)}{1 + F(s)G(s)} R(s) = \frac{4(s^2 + 3s + 2)}{s^2 + 3s + 6} R(s) $$

Note that the poles of this transfer function are the same of $G_c(s)$. Thus, it is stable. We can now apply the Final Value theorem:

\[ u(\infty) = \lim_{s \to 0} s U(s) = \lim_{s \to 0} \cancel{s} \frac{4(s^2 + 3s + 2)}{s^2 + 3s + 6} \frac{1}{\cancel{s}} = \frac{8}{6} \approx 1.33 \]