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### Question: For an open-loop system shown in Figure 2, determine if the open-loop system is stable or not. If the system is unstable, identify how many poles are in the right half-plane, the left half-plane, and on the imaginary axis (i.e., ( j\omega )-axis). *(30 points)* [ G(s) = \frac{-8}{s^4 + s^3 + 2s^2 + 2s + 3} ]

Question:

### Question: For an open-loop system shown in Figure 2, determine if the open-loop system is stable or not. If the system is unstable, identify how many poles are in the right half-plane, the left half-plane, and on the imaginary axis (i.e., ( j\omega )-axis). *(30 points)* [ G(s) = \frac{-8}{s^4 + s^3 + 2s^2 + 2s + 3} ]

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Asked by: Jack
Created at: 2025-10-22 08:29:34

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# Stability Analysis of the Open-Loop System Given transfer function: \[ G(s) = \frac{-8}{s^4 + s^3 + 2s^2 + 2s + 3} \] --- ## Step 1: Determine the Characteristic Equation The denominator of \( G(s) \) is the characteristic polynomial: \[ D(s) = s^4 + s^3 + 2s^2 + 2s + 3 \] --- ## Step 2: Stability Criterion - The system is **stable** if **all poles** (roots of \( D(s) \)) are in the **left half-plane** (LHP). - The system is **unstable** if **any pole** is in the **right half-plane** (RHP). - Poles on the imaginary axis (jω-axis) indicate marginal stability. --- ## Step 3: Analyze the Polynomial using Routh-Hurwitz Criterion ### Routh Array: | Row | Coefficients | |-------|-----------------------------------| | \( s^4 \) | 1 | 2 | 3 | | \( s^3 \) | 1 | 2 | | | \( s^2 \) | \( \frac{(1)(2) - (1)(3)}{1} = -1 \) | \( \frac{(1)() - (1)(2)}{-1} = 2 \) | | | \( s^1 \) | \( \frac{(-1)(2) - (1)()}{-1} = 2 \) | \( \frac{(-1)() - (-1)(2)}{2} = 1 \) | | | \( s^ \) | 3 | | | --- ## Step 4: Sign Analysis - First column of Routh array: 1, 1, -1, 2, 3 - Sign changes: - From \( s^4 \) to \( s^3 \): 1 to 1 → no change - From \( s^3 \) to \( s^2 \): 1 to -1 → **change** (indicates RHP pole) - From \( s^2 \) to \( s^1 \): -1 to 2 → change - From \( s^1 \) to \( s^ \): 2 to 3 → no change Total sign changes: **2** --- ## Step 5: Interpretation - The number of poles in RHP = **number of sign changes** = **2** - Remaining poles are in the LHP (since total degree = 4) **Number of poles:** | Region | Poles Count | |---------|--------------| | RHP | 2 | | LHP | 2 | | On \( j\omega \)-axis | | --- ## **Conclusion:** - **The system is unstable** because there are poles in the RHP. - **Number of RHP poles:** 2 - **Number of LHP poles:** 2 - **Poles on the imaginary axis:** --- ## Final note: The presence of two RHP poles confirms the open-loop system is unstable.

Question:

please give me the solution in a neat manner and please give clear routh array

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Asked by: Jack
Created at: 2025-10-22 08:30:59

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# Stability Analysis of the Open-Loop System Given transfer function: \[ G(s) = \frac{-8}{s^4 + s^3 + 2s^2 + 2s + 3} \] --- ## Step 1: Characteristic Equation The characteristic polynomial is derived from the denominator: \[ D(s) = s^4 + s^3 + 2s^2 + 2s + 3 \] --- ## Step 2: Stability Criterion - **Stable**: All poles in the **left half-plane** (LHP). - **Unstable**: Any pole in the **right half-plane** (RHP). - **Marginal Stability**: Poles on the imaginary axis (jω-axis). --- ## Step 3: Routh-Hurwitz Criterion ### Routh Array Construction | Row | Coefficients | |-------|-----------------------------------------| | \( s^4 \) | \( 1 \) | \( 2 \) | \( 3 \) | | \( s^3 \) | \( 1 \) | \( 2 \) | | | \( s^2 \) | \( \frac{(1)(2) - (1)(3)}{1} = -1 \) | \( \) | | \( s^1 \) | \( \frac{(-1)(2) - (1)(0)}{-1} = 2 \) | \( \) | | \( s^0 \) | \( 3 \) | | ### Routh Array | Row | First Column | Second Column | |-------|--------------|----------------| | \( s^4 \) | 1 | 2 | | \( s^3 \) | 1 | 0 | | \( s^2 \) | -1 | 2 | | \( s^1 \) | 2 | 0 | | \( s^0 \) | 3 | | --- ## Step 4: Sign Analysis - First column values: 1, 1, -1, 2, 3 - **Changes in sign**: - \( s^4 \) to \( s^3 \): 1 to 1 → no change - \( s^3 \) to \( s^2 \): 1 to -1 → **change** (indicates RHP pole) - \( s^2 \) to \( s^1 \): -1 to 2 → **change** - \( s^1 \) to \( s^0 \): 2 to 3 → no change **Total sign changes:** **2** --- ## Step 5: Interpretation - **Poles in RHP:** Number of sign changes = **2** - **Remaining poles in LHP:** \( 4 - 2 = 2 \) **Summary of Poles:** | Region | Poles Count | |--------------------------|--------------| | Right Half-Plane (RHP) | 2 | | Left Half-Plane (LHP) | 2 | | On \( j\omega \)-axis | 0 | --- ## Conclusion - **The system is unstable** due to the presence of poles in the RHP. - **Number of poles in RHP:** 2 - **Number of poles in LHP:** 2 - **Poles on the imaginary axis:** 0 This analysis confirms that the open-loop system is unstable.

Question:

please give me the entire routh table

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Asked by: Jack
Created at: 2025-10-22 08:32:38

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# Complete Routh-Hurwitz Table To analyze the stability of the open-loop system, we will construct the complete Routh array for the characteristic polynomial: \[ D(s) = s^4 + s^3 + 2s^2 + 2s + 3 \] ## Routh Array Construction ### Step 1: Organizing Coefficients The coefficients of the polynomial are arranged as follows: - \( s^4: 1 \) - \( s^3: 1 \) - \( s^2: 2 \) - \( s^1: 2 \) - \( s^0: 3 \) ### Step 2: Routh Array | Row | First Column | Second Column | |-------|--------------|----------------| | \( s^4 \) | \( 1 \) | \( 2 \) | | \( s^3 \) | \( 1 \) | \( 0 \) | | \( s^2 \) | \( \frac{(1)(2) - (1)(3)}{1} = -1 \) | \( \frac{(1)(0) - (1)(2)}{-1} = 2 \) | | \( s^1 \) | \( \frac{(-1)(2) - (1)(0)}{-1} = 2 \) | \( 0 \) | | \( s^0 \) | \( 3 \) | | ### Detailed Calculation for Each Row 1. **For \( s^2 \)**: - First column: \[ \frac{(1)(2) - (1)(3)}{1} = 2 - 3 = -1 \] - Second column: \[ \frac{(1)(0) - (1)(2)}{-1} = \frac{0 - 2}{-1} = 2 \] 2. **For \( s^1 \)**: - First column: \[ \frac{(-1)(2) - (1)(0)}{-1} = \frac{-2}{-1} = 2 \] - Second column: - No calculation needed as there are no entries. 3. **For \( s^0 \)**: - The last row simply carries down the constant term: - \( 3 \) ### Final Routh Array | Row | First Column | Second Column | |-------|--------------|----------------| | \( s^4 \) | \( 1 \) | \( 2 \) | | \( s^3 \) | \( 1 \) | \( 0 \) | | \( s^2 \) | \( -1 \) | \( 2 \) | | \( s^1 \) | \( 2 \) | \( 0 \) | | \( s^0 \) | \( 3 \) | | --- ## Sign Analysis - First column values: \( 1, 1, -1, 2, 3 \) - **Sign changes**: - From \( s^4 \) to \( s^3 \): no change (1 to 1) - From \( s^3 \) to \( s^2 \): change (1 to -1) - From \( s^2 \) to \( s^1 \): change (-1 to 2) - From \( s^1 \) to \( s^0 \): no change (2 to 3) **Total sign changes:** **2** --- ## Conclusion - **Poles in RHP:** 2 (indicates instability) - **Poles in LHP:** 2 - **Poles on the imaginary axis:** 0 The open-loop system is unstable due to the presence of poles in the right half-plane.

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