Today lecture is on stability and how to use Routh table to check for stability. We will use BIBO stability. BIBO stable system is one which has bounded output for all time, when the input is also bounded for all time. Analytically, a system can be determined if it is stable from the convolution definition of system response given by\begin{equation} y\left ( t\right ) =\int _{0}^{t}r\left ( t-\tau \right ) g\left ( \tau \right ) d\tau \tag{1} \end{equation} Where \(r\left ( t\right ) \) is the input, and \(g\left ( t\right ) \) is the impulse response of the system. For bounded input, which means \(\left \vert r\left ( t\right ) \right \vert \leq B\) where \(B\) is some constant that do not depend on time, the output \(y\left ( t\right ) \) magnitude can be now found from (1) as follows\begin{align*} \left \vert y\left ( t\right ) \right \vert & =\left \vert \int _{0}^{t}r\left ( t-\tau \right ) g\left ( \tau \right ) d\tau \right \vert \\ & \leq \int _{0}^{t}\left \vert r\left ( t-\tau \right ) g\left ( \tau \right ) \right \vert d\tau \\ & =\int _{0}^{t}\left \vert r\left ( t-\tau \right ) \right \vert \left \vert g\left ( \tau \right ) \right \vert d\tau \\ & \leq B\int _{0}^{t}\left \vert g\left ( \tau \right ) \right \vert d\tau \end{align*}
Therefore, for \(\left \vert y\left ( t\right ) \right \vert \) to be bounded, which means \(\left \vert y\left ( t\right ) \right \vert \leq C\) where \(C\) is some constant, then we need \(\int _{0}^{t}\left \vert g\left ( \tau \right ) \right \vert d\tau \leq \infty \). This means that a system is BIBO is \(\int _{0}^{\infty }\left \vert g\left ( t\right ) \right \vert dt\leq \infty \) where \(g\left ( t\right ) \) is the impulse response of the system. The following are examples how to use the above to determine BIBO stability.
Example: Given \(G\left ( s\right ) =\frac{1}{s-1}\) check if it is BIBO stable. The impulse response is \(g\left ( t\right ) =e^{t}\). Hence \(\int _{0}^{\infty }\left \vert e^{t}\right \vert dt=\int _{0}^{\infty }e^{t}dt=\left . e^{t}\right \vert _{0}^{\infty }=\infty \) so this is not BIBO stable.
We can also see that this is not stable, since it has one pole in the RHS.
Example: Given \(G\left ( s\right ) =\frac{1}{s+1}\) check if it is BIBO stable. The impulse response is \(g\left ( t\right ) =e^{-t}\). Hence \(\int _{0}^{\infty }\left \vert e^{-t}\right \vert dt=\int _{0}^{\infty }e^{=t}dt=\left . -e^{-t}\right \vert _{0}^{\infty }=1\) so this is BIBO stable.
We can also see that this is stable since it has no poles in the RHS.
Therefore, as long as a system has no poles in the RHS, then it is BIBO stable. One way to check if there are poles in the RHS and how many there are, without actually solving for the roots or without doing the above integration, is to use Routh-Hurwitz table. We will look at three cases. When the first column in the table has no zeros (classical case), and when the first column has a zero, and when a whole row in the table has zeros, and see how to handle each case. We will do this using three examples of each case.
Example 1: Given \(G\left ( s\right ) =\frac{N\left ( s\right ) }{D\left ( s\right ) }\) where \(D\left ( s\right ) =3s^{4}+10s^{3}+5s^{2}+5s+2\). We set up Routh table as follows
| \(s^{4}\) | \(3\) | \(5\) | \(2\) |
| \(s^{3}\) | \(10\) | \(5\) | |
| \(s^{2}\) | \(3.5\) | \(2\) | |
| \(s^{1}\) | \(-\frac{5}{7}\) | \(0\) | |
| \(s^{0}\) | \(2\) | \(0\) | |
Looking at the first numerical column (i.e. second column in the table above), we see there are two sign changes. This means there are 2 poles in the RHS. Which also means this system is not BIBO stable.
Reader: Check the roots using Matlab using roots command and verify the above.
Example 2: Given \(G\left ( s\right ) =\frac{N\left ( s\right ) }{D\left ( s\right ) }\) where \(D\left ( s\right ) =4s^{4}+10s^{3}+5s^{2}+12.5s+5\). We set up Routh table as follows
| \(s^{4}\) | \(4\) | \(5\) | \(5\) |
| \(s^{3}\) | \(10\) | \(12.5\) | |
| \(s^{2}\) | \(0\) | \(5\) | |
Since we have zero at the pivot, then we change it with \(\varepsilon \) and continue as follows
| \(s^{4}\) | \(4\) | \(5\) | \(5\) |
| \(s^{3}\) | \(10\) | \(12.5\) | |
| \(s^{2}\) | \(\varepsilon \) | \(5\) | |
| \(s^{1}\) | \(12.5-\frac{50}{\varepsilon }\) | ||
| \(s^{0}\) | \(5\) | ||
We now take the limit as \(\varepsilon \rightarrow 0\) from above, and see that a sign change between the third and fourth row and then another sign change from the fourth to the fifth row. So this is not sable system.
Example 3: Given \(G\left ( s\right ) =\frac{N\left ( s\right ) }{D\left ( s\right ) }\) where \(D\left ( s\right ) =s^{6}+2s^{5}+8s^{4}+12s^{3}+20s^{2}+20s^{2}+16s+16\). We set up Routh table as follows
| \(s^{6}\) | \(1\) | \(8\) | \(20\) | \(16\) |
| \(s^{5}\) | \(2\) | \(12\) | \(16\) | |
| \(s^{4}\) | \(2\) | \(12\) | \(16\) | |
| \(s^{3}\) | \(0\) | \(0\) | ||
Since we have row of zeros. To handle this, we take the polynomial from the row above, which is \(A\left ( s\right ) =2s^{4}+12s^{2}+16\) and take its derivative, giving \(A^{\prime }\left ( s\right ) =8s^{3}+24s\), and use this to replace the row of zeros, so we end up with
| \(s^{6}\) | \(1\) | \(8\) | \(20\) | \(16\) |
| \(s^{5}\) | \(2\) | \(12\) | \(16\) | |
| \(s^{4}\) | \(2\) | \(12\) | \(16\) | |
| \(s^{3}\) | \(8\) | \(24\) | ||
And now we continue as before
| \(s^{6}\) | \(1\) | \(8\) | \(20\) | \(16\) |
| \(s^{5}\) | \(2\) | \(12\) | \(16\) | |
| \(s^{4}\) | \(2\) | \(12\) | \(16\) | |
| \(s^{3}\) | \(8\) | \(24\) | ||
| \(s^{2}\) | \(6\) | \(16\) | ||
| \(s\) | \(\frac{8}{3}\) | |||
| \(s^{0}\) | \(16\) | |||
So there is no sign change, so this is stable.
Another example. Given this system
The above is called master/slave controller design. Where \(G=\frac{K}{\left ( s+1\right ) \left ( s+2\right ) }\) and we want to find if the transfer function from any input to any output is BIBO stable or not. We use Mason rule to obtain the denominator, which is Mason \(\Delta \). The signal graph is
Mason delta is \[ \Delta =1-\left ( -G+G+G^{2}\right ) +\left ( -GG\right ) =1-2G^{2}\]
Hence for \(G=\frac{K}{\left ( s+1\right ) \left ( s+2\right ) }\) then \(\Delta =s^{4}+6s^{3}+13s^{2}+12s+4-2k^{2}\) and now we setup Routh table
| \(s^{4}\) | \(1\) | \(13\) | \(4-2k^{2}\) |
| \(s^{3}\) | \(6\) | \(12\) | |
| \(s^{2}\) | \(11\) | \(4-2k^{2}\) | |
| \(s^{1}\) | \(\frac{108+12k^{2}}{11}\) | ||
| \(s^{0}\) | \(4-2k^{2}\) | ||
Therefore for no sign change we need \(\frac{108+12k^{2}}{11}>0\) which is always true, and we want also \(4-2k^{2}>0\) or \(\left \vert k\right \vert <\sqrt{2}\) as the condition for stability.
Final example. For \(G\left ( s\right ) =\frac{1}{s^{4}+3s^{3}+k_{2}s^{2}+4s+k_{1}}\) find conditions for stability.
| \(s^{4}\) | \(1\) | \(k_{2}\) | \(k_{1}\) |
| \(s^{3}\) | \(3\) | \(4\) | |
| \(s^{2}\) | \(\frac{3k_{2}-4}{3}\) | \(k_{1}\) | |
| \(s^{1}\) | \(\frac{4\left ( \frac{3k_{2}-4}{3}\right ) -3k_{1}}{\frac{3k_{2}-4}{3}}\) | ||
| \(s^{0}\) | \(k_{1}\) | ||
The condition is \(\frac{3k_{2}-3}{3}>0\) or \(k_{2}>\frac{4}{3}\) and \(k_{1}>0\). We see the region of stability to be the following
Reader For \(G\left ( s\right ) =\frac{1}{s^{2}\left ( s+1\right ) \left ( s+2\right ) }\) and \(H\left ( s\right ) =2s+k\) find \(k\) for stable system in the following
