Routh Array Special Case 2: Row of Zeros
📌 Special Case 2 in the Routh-Hurwitz (RH) criteria occurs when all elements of a row in the Routh array are zero (a row of zeros).
❌ The presence of a row of zeros means the subsequent row's terms cannot be determined (they become infinite), causing the standard Routh test to fail.
🔄 To resolve this, an auxiliary polynomial, $A(s)$, is formed using the coefficients of the row immediately above the row of zeros.

Solving the Row of Zeros Using the Auxiliary Polynomial
1. 🛠️ The auxiliary polynomial $A(s)$ is constructed using the coefficients from the row above the zeros, ensuring powers of $s$ are always even (e.g., $s^4, s^2, s^0$).
2. 📈 The next step involves taking the derivative of the auxiliary polynomial with respect to $s$, denoted as $\frac{d A(s)}{d s}$.
3. 🔄 The row of zeros in the Routh array is then replaced by the coefficients derived from $\frac{d A(s)}{d s}$, allowing the array computation to continue.

Auxiliary Equation and Stability Implications
💡 Setting the auxiliary polynomial to zero, $A(s) = 0$, yields the auxiliary equation, whose roots determine stability when a row of zeros is present.
🔄 The roots of the auxiliary equation are always symmetric with respect to the origin on the $s$-plane.
📉 If a row of zeros exists, the system will not be stable; it can only be marginally stable or unstable, depending on the auxiliary equation's roots.

Example Application and Stability Determination
🔍 For the characteristic equation $s^6 + 2s^5 + 8s^4 + 12s^3 + 20s^2 + 16s + 16 = 0$, a row of zeros appeared at the $s^3$ row.
➕ The auxiliary polynomial derived was $A(s) = 2s^4 + 12s^2 + 16$.
📈 The roots of $A(s)=0$ were found to be $\pm j\sqrt{2}$ and $\pm j2$, meaning four poles lie on the imaginary axis.
✅ Since there were zero sign changes in the first column (below the modified array) and the auxiliary roots were purely imaginary, the system is marginally stable (two poles in LHP, four on the $j\omega$ axis).

Key Points & Insights
➡️ Whenever a row of zeros occurs, stability hinges on the roots of the auxiliary equation in addition to the sign changes in the first column.
➡️ Roots of the auxiliary equation always appear symmetric to the origin; possibilities include pairs of poles in LHP/RHP (unstable) or pairs of complex conjugates on the $j\omega$ axis (marginally stable).
➡️ If the number of sign changes in the first column is zero, the number of Right Half Plane (RHP) poles equals zero.
➡️ A system exhibiting a row of zeros in the Routh array cannot be asymptotically stable.

📸 Video summarized with SummaryTube.com on Nov 12, 2025, 13:50 UTC