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📘 Routh-Hurwitz Criterion – Complete Theory & Worked Examples

Routh-Hurwitz Criterion is a powerful method to determine stability without solving the characteristic equation. It tells how many roots lie in the Right Half Plane.

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🔹 1. Characteristic Equation

General nth order polynomial:

a₀sⁿ + a₁sⁿ⁻¹ + a₂sⁿ⁻² + ... + aₙ = 0

All coefficients must be positive for possible stability.

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🔹 2. Routh Array Formation

First two rows are formed from coefficients:

Row 1 → a₀ a₂ a₄ ... Row 2 → a₁ a₃ a₅ ...

Remaining rows computed using determinant formula.

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🔹 3. Stability Condition

• All elements of first column must be positive.
• Number of sign changes in first column = number of RHP roots.

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🔹 4. Worked Example 1

Given characteristic equation:

s³ + 5s² + 6s + 2 = 0

Step 1: Form Routh Table

s³ | 1 6 s² | 5 2 s¹ | (5×6 − 1×2)/5 = 28/5 s⁰ | 2

Step 2: First Column

1 5 28/5 2

No sign change → Stable system.

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🔹 5. Worked Example 2 (Unstable Case)

s³ + 2s² − s − 2 = 0

Routh Table

s³ | 1 -1 s² | 2 -2 s¹ | (2×(-1) − 1×(-2))/2 = 0 s⁰ | -2

First column: 1 2 0 -2 Sign change from + to - → One RHP root → Unstable.

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🔹 6. Special Case – Zero in First Column

If first element becomes zero: Replace zero with small ε and continue. This indicates possible marginal stability.

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🔹 7. Special Case – Entire Row Zero

Indicates:

• Symmetrical roots
• Imaginary axis roots

Form auxiliary polynomial from previous row and differentiate.

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🔹 8. Gain K Stability Range

Given:

s³ + 3s² + 2s + K = 0

Form Routh array:

s³ | 1 2 s² | 3 K s¹ | (3×2 − 1×K)/3 = (6 − K)/3 s⁰ | K

For stability:

• All first column elements > 0
• K > 0
• 6 − K > 0 → K < 6

Final Stability Range:

0 < K < 6

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🎯 GATE Important Points

• Sign change count directly gives RHP roots
• Gain range questions very common
• Zero row case very important
• Always check coefficient positivity first

Routh = Fastest Way to Check Stability

📘 Routh-Hurwitz – Advanced Worked Examples (Deep Stability Practice)

These examples include higher order systems and special cases. Follow the first-column sign changes carefully.

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🔹 Example 1 – 4th Order Polynomial

Given:

s⁴ + 2s³ + 3s² + 4s + 5 = 0

Step 1: Form Routh Table

s⁴ | 1 3 5 s³ | 2 4 0 s² | (2×3 − 1×4)/2 = 1 (2×5 − 1×0)/2 = 5 s¹ | (1×4 − 2×5)/1 = -6 s⁰ | 5

Step 2: First Column

1 2 1 -6 5 Sign changes: + to - (1 change) - to + (1 change) Total 2 sign changes → 2 RHP roots → Unstable system.

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🔹 Example 2 – Zero in First Column

Given:

s³ + 4s² + 5s + 0 = 0

Step 1: Routh Table

s³ | 1 5 s² | 4 0 s¹ | (4×5 − 1×0)/4 = 5 s⁰ | 0

Last element zero indicates root at origin. System is marginally stable.

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🔹 Example 3 – Entire Row Zero Case

Given:

s⁴ + 2s³ + 3s² + 2s + 1 = 0

Step 1: Routh Table

s⁴ | 1 3 1 s³ | 2 2 0 s² | (2×3 − 1×2)/2 = 2 (2×1 − 1×0)/2 = 1 s¹ | 0 0

Entire row zero.

Step 2: Auxiliary Polynomial

From row above (s² row):

A(s) = 2s² + 1

Differentiate:

dA/ds = 4s

Replace zero row with coefficients of derivative. System has imaginary axis roots → Marginal stability.

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🔹 Example 4 – Gain Range (4th Order)

Given:

s⁴ + 5s³ + 8s² + 4s + K = 0

Step 1: Routh Table

s⁴ | 1 8 K s³ | 5 4 0 s² | (5×8 − 1×4)/5 = 36/5 (5K − 1×0)/5 = K s¹ | [(36/5×4 − 5K)/ (36/5)] s⁰ | K

For stability: All first column > 0 Conditions:

• K > 0
• 36/5 > 0 (already positive)
• 36/5×4 − 5K > 0

Solve:

K < 144/25

Final range:

0 < K < 5.76

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🔹 Example 5 – Number of RHP Roots

Given:

s⁴ + 3s³ + 2s² − s − 1 = 0

After Routh calculation: First column: 1 3 5/3 -2 -1 Sign change: + to - (1 change) Thus: 1 RHP root → Unstable system.

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🎯 Final Advanced Stability Tips

• Count sign changes carefully
• Zero in first column → use ε
• Entire row zero → auxiliary polynomial
• Gain range questions very common in GATE

Routh-Hurwitz = Fast Stability Decision Tool

📘 Routh-Hurwitz – Advanced Worked Examples (Deep Stability Practice)

These examples include higher order systems and special cases. Follow the first-column sign changes carefully.

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🔹 Example 1 – 4th Order Polynomial

Given:

s⁴ + 2s³ + 3s² + 4s + 5 = 0

Step 1: Form Routh Table

s⁴ | 1 3 5 s³ | 2 4 0 s² | (2×3 − 1×4)/2 = 1 (2×5 − 1×0)/2 = 5 s¹ | (1×4 − 2×5)/1 = -6 s⁰ | 5

Step 2: First Column

1 2 1 -6 5 Sign changes: + to - (1 change) - to + (1 change) Total 2 sign changes → 2 RHP roots → Unstable system.

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🔹 Example 2 – Zero in First Column

Given:

s³ + 4s² + 5s + 0 = 0

Step 1: Routh Table

s³ | 1 5 s² | 4 0 s¹ | (4×5 − 1×0)/4 = 5 s⁰ | 0

Last element zero indicates root at origin. System is marginally stable.

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🔹 Example 3 – Entire Row Zero Case

Given:

s⁴ + 2s³ + 3s² + 2s + 1 = 0

Step 1: Routh Table

s⁴ | 1 3 1 s³ | 2 2 0 s² | (2×3 − 1×2)/2 = 2 (2×1 − 1×0)/2 = 1 s¹ | 0 0

Entire row zero.

Step 2: Auxiliary Polynomial

From row above (s² row):

A(s) = 2s² + 1

Differentiate:

dA/ds = 4s

Replace zero row with coefficients of derivative. System has imaginary axis roots → Marginal stability.

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🔹 Example 4 – Gain Range (4th Order)

Given:

s⁴ + 5s³ + 8s² + 4s + K = 0

Step 1: Routh Table

s⁴ | 1 8 K s³ | 5 4 0 s² | (5×8 − 1×4)/5 = 36/5 (5K − 1×0)/5 = K s¹ | [(36/5×4 − 5K)/ (36/5)] s⁰ | K

For stability: All first column > 0 Conditions:

• K > 0
• 36/5 > 0 (already positive)
• 36/5×4 − 5K > 0

Solve:

K < 144/25

Final range:

0 < K < 5.76

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🔹 Example 5 – Number of RHP Roots

Given:

s⁴ + 3s³ + 2s² − s − 1 = 0

After Routh calculation: First column: 1 3 5/3 -2 -1 Sign change: + to - (1 change) Thus: 1 RHP root → Unstable system.

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🎯 Final Advanced Stability Tips

• Count sign changes carefully
• Zero in first column → use ε
• Entire row zero → auxiliary polynomial
• Gain range questions very common in GATE

Routh-Hurwitz = Fast Stability Decision Tool