Exercise 3.1

Exercise 3.1#

Consider the following systems:

  1. \(G(s) = \frac{2}{s + 1}\)

  2. \(G(s) = \frac{1}{s^3 + 1}\)

Let the input be \(u(t) = \sin(t)\). Compute, in the cases where it is meaningful, the output when the transients have disappeared.

Solution#

According to the Frequency Response Theorem, if \(G(s)\) is stable then any sinusoidal input \(u(t)=A \sin(\omega t)\) leads (asymptotically) to the sinusoidal output

\[ y(t) = A \lvert G(i \omega) \lvert \cdot \sin(\omega t + \angle G(i \omega)) \]

Note that

  • The sinusoidal output has the same frequency \(\omega\) as the input

  • The sinusoidal input gets amplified (or attenuated) by a factor equal to the magnitude of \(G(s)\) at frequency \(\omega\), i.e., \(\lvert G(i \omega) \lvert\)

  • The sinusoidal input gets shifted by the phase of \(G(s)\) at frequency \(\omega\), i.e., \(\angle G(i \omega)\)

System 1#

The system \(G(s) = \frac{2}{s+1}\) is stable (there is one pole in \(s = -1\)), therefore we can apply the Frequency Response Theorem.

\[ G(i \omega) = \frac{2}{i \omega + 1} = \frac{2}{i \omega + 1} \cdot \frac{-i \omega + 1}{-i \omega + 1} = \frac{-i 2 \omega + 2}{-i^2 \omega^2 + 1} = \frac{2}{1 + \omega^2} - i \frac{2 \omega}{1 + \omega^2} \]

The input has frequency \(\omega = 1\). We therefore first compute \(G(i 1)\), and then its modulus and phase.

\[ G(i 1) = 1 - i1 \]
\[ \lvert G(i1) \lvert = \sqrt{1^2 +(-1)^2} = \sqrt{2} \]
\[ \angle G(i1) = \textrm{atan2}(-1, 1) = - \frac{\pi}{4} \]

The asymptotic output signal is therefore

\[ y(t) = \sqrt{2} \sin \left(t - \frac{\pi}{4} \right) \]

System 2#

First, we have to compute the poles of \(G(s) = \frac{1}{s^3 + 1}\) to make sure that it is stable.

To compute the poles, we look for the roots of \(s^3 + 1 = 0\). Realizing that it’s a sum of cubes, we can factorize the left-hand side as

\[ (s + 1)(s^2 + 1 - s) = 0 \]

Therefore, the poles are

  • \(s = -1\)

  • \(s = \frac{-1 \pm \sqrt{5}}{2}\)

Unfortunately, one of the poles \(\left( s = \frac{-1 + \sqrt{5}}{2} \right)\) has positive real part \(\to\) unstable! In this case, we cannot apply the Frequency Response Theorem.

Warning

When applying the Frequency Response Theorem, don’t forget to first check the stability of the system!