Chapter 3 Stability
The following result is called the Routh-Hurwitz stability criterion (note that we assume that the coeficients of the polynomial are real, which will be true for the characteristic polynomial of \(A\) if \(A\) is real).
-
Theorem 3.5. Let \(p\) be given by
\[ p(s)=\sum _{j=0}^na_js^j=a_0+a_1s+\ldots +a_{n-1}s^{n-1}+a_ns^n, \]
where \(a_0,\ldots ,a_n\in \mathbb {R}\) with \(a_n>0\). Let \(H\) be the \(n\times n\) matrix given by
\[ H_{ij}=\begin {cases} a_{n-(2i-j)}&0\leq 2i-j\leq n,\\ 0&\text {otherwise}, \end {cases} \]
i.e.,
\[ H=\bbm { a_{n-1}&a_n&0&0&\cdots &0&0&0 \\a_{n-3}&a_{n-2}&a_{n-1}&a_n&\cdots &0&0&0 \\a_{n-5}&a_{n-4}&a_{n-3}&a_{n-2}&\cdots &0&0&0 \\\vdots &\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots \\0&0&0&0&\cdots &a_2&a_3&a_4 \\0&0&0&0&\cdots &a_0&a_1&a_2 \\0&0&0&0&\cdots &0&0&a_0 }. \]
Define \(H_k\) for \(k=1,\ldots ,n\) as the principal subdeterminants of \(H\):
\(\seteqnumber{0}{3.}{0}\)\begin{gather*} H_1:=a_{n-1},\qquad H_2:=\det \bbm {a_{n-1}&a_n\\a_{n-3}&a_{n-2}},\\ H_3:=\det \bbm {a_{n-1}&a_n&0\\a_{n-3}&a_{n-2}&a_{n-1}\\a_{n-5}&a_{n-4}&a_{n-3}},\, \ldots ,\, H_n:=\det (H)=a_0H_{n-1}. \end{gather*} The polynomial \(p\) is stable if and only if \(H_k>0\) for all \(k\in \{1,\ldots ,n\}\).
-
Example 3.6. Consider the general second degree polynomial
\[ p(s)=a_2s^2+a_1s+a_0, \]
with \(a_2>0\). The corresponding matrix is
\[ H=\bbm {a_1&a_2\\0&a_0}, \]
and the principal subdeterminants are
\[ H_1=a_1,\qquad H_2=a_0H_1. \]
The necessary and sufficient condition for stability therefore is:
\[ a_1>0\qquad a_0>0. \]
-
Example 3.7. Consider the general third degree polynomial
\[ p(s)=a_3s^3+a_2s^2+a_1s+a_0, \]
with \(a_3>0\). The corresponding matrix is
\[ H=\bbm {a_2&a_3&0\\a_0&a_1&a_2\\0&0&a_0}, \]
and the principal subdeterminants are
\[ H_1=a_2,\qquad H_2=\det \bbm {a_2&a_3\\a_0&a_1}=a_2a_1-a_0a_3,\qquad H_3=a_0H_2. \]
The necessary and sufficient condition for stability therefore is:
\[ a_2>0\qquad a_0>0,\qquad a_2a_1-a_0a_3>0. \]
-
Example 3.8. Consider the general fourth degree polynomial
\[ p(s)=a_4s^4+a_3s^3+a_2s^2+a_1s+a_0, \]
with \(a_4>0\). The corresponding matrix is
\[ H=\bbm {a_3&a_4&0&0\\a_1&a_2&a_3&a_4\\0&a_0&a_1&a_2\\0&0&0&a_0}, \]
and the principal subdeterminants are
\(\seteqnumber{0}{3.}{0}\)\begin{gather*} H_1=a_3,\qquad H_2=\det \bbm {a_3&a_4\\a_1&a_2}=a_3a_2-a_1a_4,\\ H_3=\det \bbm {a_3&a_4&0\\a_1&a_2&a_3\\0&a_0&a_1}=a_1H_2-a_3^2a_0,\qquad H_4=a_0H_3. \end{gather*} The condition that \(H_1,H_2,H_3,H_4>0\) can be rewritten in a more convenient way using the necessary condition that all coefficients of the polynomial must be positive. The condition that \(H_3>0\) combined with the condition that \(a_0,a_1>0\) gives that \(H_2>0\). Therefore the necessary and sufficient condition for stability is:
\[ a_3,a_1,a_0>0,\qquad a_3a_2a_1-a_4a_1^2-a_3^2a_0>0. \]
3.1 Examples
-
Example 3.9. The first order scalar differential equation
\[ \dot {x}+x=0, \]
is asymptotically stable. This can be seen directly from the definition: the solutions are \(x(t)=x^0\e ^{-t}\) and all of these converge to zero as \(t\to \infty \). It can also be seen from the characteristic polynomial which is \(s+1\), which has as its only root \(s=-1\) which has negative real part, so that the characteristic polynomial is stable.
-
Example 3.10. For the second order scalar differential equation \(\ddot {q}+2\zeta \dot {q}+q=0\) (here the damping ratio \(\zeta \geq 0\)) we have the matrix
\[ A=\bbm {0&1\\-1&-2\zeta }, \]
which has characteristic polynomial (note that we can obtain this without computation from the second order form)
\[ s^2+2\zeta s+1. \]
The Routh–Hurwitz condition for stability is that all coefficients should be positive. So if the damping ratio \(\zeta \) is positive, then the second order scalar differential equation is asymptotically stable, but if it is zero (i.e. the system is undamped), then the second order scalar differential equation is not asymptotically stable.
3.2 Case study: a suspension system*
We will show that the matrix
\[ A_\cl =\bbm {0&k_{us}&0&0\\ \frac {-1}{m_{us}}&\frac {-d}{m_{us}}&\frac {-k_s}{m_{us}}&\frac {d}{m_{us}}\\ 0&1&0&-1\\ 0&\frac {d}{m_s}&\frac {k_s}{m_s}&\frac {-d}{m_s} }, \]
obtained from the fixed structure suspension system is asymptotically stable (provided of course that all physical parameters are positive).
The characteristic polynomial of \(A_\cl \) multiplied by \(m_sm_{us}\) is:
\[ m_sm_{us}s^4+d[m_s+m_{us}]s^3+[k_sm_s+k_sm_{us}+k_{us}m_s]s^2+dk_{us}s+k_sk_{us}. \]
We apply the Routh–Hurwitz criterion for the degree four case derived above to this polynomial. Since we assume that all physical parameters are positive, we clearly have that all coefficients are positive. Therefore the only condition to check is \(a_3a_2a_1-a_4a_1^2-a_3^2a_0>0\). We have
\[ \begin {aligned} a_3a_2a_1&=d^2[m_s+m_{us}][k_sm_s+k_sm_{us}+k_{us}m_s]k_{us}\\ &=d^2k_{us}\left ([m_s+m_{us}]^2k_s+m_s^2k_{us}+m_sm_{us}k_{us}\right ), \end {aligned} \]
and
\[ \begin {aligned} a_4a_1^2&=m_sm_{us}d^2k_{us}^2\\ a_3^2a_0&=d^2k_{us}[m_s+m_{us}]^2k_s, \end {aligned} \]
so that cancelling like terms we obtain the condition
\[ d^2k_{us}^2m_s^2>0, \]
which of course is true. Therefore, the matrix \(A_\cl \) is asymptotically stable.
The matrix
\[ A=\bbm {0&k_{us}&0&0\\ \frac {-1}{m_{us}}&0&0&0\\ 0&1&0&-1\\ 0&0&0&0 }, \]
for the general design of a suspension system is not asymptotically stable. Since it has a zero column, \(0\) is an eigenvalue of the matrix and therefore the matrix is not asymptotically stable.
3.3 Case study: control of a tape drive*
The matrix
\[ A=\bbm { -\frac {d_1}{M_1}&0&\frac {1}{M_1}\\ 0&-\frac {d_2}{M_2}&-\frac {1}{M_2}\\ -k&k&0 }, \]
is asymptotically stable. Its characteristic polynomial is
\(\seteqnumber{0}{3.}{0}\)\begin{align*} \det \bbm { s+\frac {d_1}{M_1}&0&-\frac {1}{M_1}\\ 0&s+\frac {d_2}{M_2}&\frac {1}{M_2}\\ k&-k&s } &=\left (s+\frac {d_1}{M_1}\right )\det \bbm {s+\frac {d_2}{M_2}&\frac {1}{M_2}\\-k&s} -\frac {1}{M_1}\det \bbm {0&s+\frac {d_2}{M_2}\\k&-k} \\ &=\left (s+\frac {d_1}{M_1}\right )\left (s^2+\frac {d_2}{M_2}s+\frac {k}{M_2}\right ) +\frac {k}{M_1}\left (s+\frac {d_2}{M_2}\right ) \\ &=s^3+\left (\frac {d_1}{M_1}+\frac {d_2}{M_2}\right )s^2+\left (\frac {k}{M_1}+\frac {k}{M_2}+\frac {d_1d_2}{M_1M_2}\right )s+\frac {(d_1+d_2)k}{M_1M_2}. \end{align*} Multiplying though by \(M_1M_2\) gives
\[ M_1M_2s^3+(d_1M_2+d_2M_1)s^2+(kM_2+kM_1+d_1d_2)s+(d_1+d_2)k. \]
We apply the Routh–Hurwitz criterion for the degree three case derived above to this polynomial. Since we assume that all physical parameters are positive, we clearly have that all coefficients are positive. Therefore the only condition to check is that \(a_2a_1-a_0a_3>0\). This is
\[ (d_1M_2+d_2M_1)(kM_2+kM_1+d_1d_2)-(d_1+d_2)kM_1M_2>0. \]
Multiplying out the left-hand side, this is
\[ d_1kM_2^2+d_1kM_1M_2+d_1^2d_2M_2+d_2kM_1M_2+d_2kM_1^2+d_1d_2^2M_1-d_1kM_1M_2-d_2kM_1M_2>0, \]
which upon cancelling like terms is
\[ d_1kM_2^2+d_1^2d_2M_2+d_2kM_1^2+d_1d_2^2M_1>0, \]
which is true.