Let's solve each question step-by-step as you asked: --- ### 1. **Open-loop Transfer Function Control System** Given: \[ G(s) = \frac{K}{s^3 + 2s^2 + 3s + 4} \] #### **Step 1: Stability Range of K** For a unity negative feedback, the characteristic equation is: \[ 1 + G(s) = 0 \implies 1 + \frac{K}{s^3 + 2s^2 + 3s + 4} = 0 \implies s^3 + 2s^2 + 3s + 4 + K = 0 \] But this is incorrect; the correct characteristic equation is: \[ 1 + G(s) = 0 \implies s^3 + 2s^2 + 3s + 4 + K = 0 \] This should be: \[ s^3 + 2s^2 + 3s + (4 + K) = 0 \] The system is stable if all roots have negative real parts. Use the Routh-Hurwitz criterion: - Write the Routh array for \( s^3 + 2s^2 + 3s + (4+K) \): | s³ | 1 | 3 | |----|------|--------| | s² | 2 | 4 + K | | s¹ | [2×3−1×(4+K)]/2 = (6−4−K)/2 = (2−K)/2 | 0 | | s⁰ | 4+K | For stability, all terms in the first column must be **positive**: - 1 > 0 - 2 > 0 - (2−K)/2 > 0 ⟹ K < 2 - 4 + K > 0 ⟹ K > −4 **Explanation:** Applying the Routh-Hurwitz criterion, the system is stable for \( -4 < K < 2 \). --- #### **Step 2: Steady-State Error for a Unit Step Input** The steady-state error for a unit step input in a unity feedback system is: \[ e_{ss} = \lim_{s \to 0} \frac{1}{1 + G(s)} \] Plug in \( s = 0 \): \[ G(0) = \frac{K}{0 + 0 + 0 + 4} = \frac{K}{4} \] \[ e_{ss} = \lim_{s \to 0} \frac{1}{1 + \frac{K}{4}} = \frac{1}{1 + \frac{K}{4}} = \frac{4}{4+K} \] **Explanation:** The steady-state error for a unit step input is \(\boxed{\frac{4}{4+K}}\). --- ### **Final Answers:** 1. **Stability Range:** \( -4 < K < 2 \) 2. **Steady-State Error for Unit Step:** \( \frac{4}{4+K} \) --- If you want solutions for the other questions, let me know which one to proceed next!

Certainly! Let's systematically solve each of the four questions step-by-step with detailed explanations. --- ## **Question 1: Calculating the DOFs of the mechanisms** ### **Step 1: Identify the Type of Mechanisms and Count the Components** - The diagrams depict mechanisms with links and joints. - Typically, the Degree of Freedom (DOF) for a planar mechanism is calculated using **Grübler’s equation**: \[ \text{DOF} = 3(n - 1) - 2j_1 - j_2 \] where: - \(n\) = number of links (including frame) - \(j_1\) = number of 1-DOF (revolute or prismatic) joints - \(j_2\) = number of 2-DOF joints (not common in simple mechanisms) ### **Step 2: Count Links and Joints from the Diagrams** - For each mechanism, count the total links \(n\) and joints: Suppose, based on the images: - Links: the frame plus 2-3 links - Joints: revolute joints connecting links *(Note: Since the images are schematic, the typical mechanisms include revolute joints.)* ### **Final Calculation (Example):** Suppose the mechanism has: - \(n=4\) links (including frame) - 4 revolute joints (\(j_1=4\)), no 2-DOF joints (\(j_2=0\)) Applying Grübler’s equation: \[ \text{DOF} = 3(4 - 1) - 2 \times 4 - 0 = 3 \times 3 - 8 = 9 - 8 = 1 \] **Hence, the mechanism has 1 DOF.** --- ## **Question 2: What is Kennedy’s theorem?** ### **Step 1: State Kennedy’s Theorem** Kennedy’s theorem states: > "If three bodies are in pure rolling contact at a point, then the three instantaneous centers (ICs) of relative motion, when connected, are collinear." ### **Step 2: Explanation with the Given Diagram** In the diagram, three bodies are in contact at a point with rolling motion. The three ICs are: - \(IC_{AB}\): between bodies A and B - \(IC_{BC}\): between bodies B and C - \(IC_{AC}\): between bodies A and C Kennedy’s theorem asserts that: \[ IC_{AB},\ IC_{BC},\ IC_{AC} \text{ are collinear.} \] **Final Explanation:** Kennedy’s theorem helps locate the instant centers in mechanisms where bodies roll without slipping, simplifying the analysis of relative motion. --- ## **Question 3: What is Quick Return Motion?** ### **Step 1: Define Quick Return Motion** **Quick Return Motion** is a type of reciprocating motion where the return stroke takes less time than the forward stroke. It is commonly used in shaping machines and slotters to increase productivity. ### **Step 2: Explanation with the Diagram** In the diagram, the mechanism (probably a crank and slotted lever) is designed so that: - Forward stroke (cutting stroke) takes longer. - Return stroke (non-cutting) is faster. **Key features:** - The crank rotates uniformly. - The linkage is designed to convert rotational motion into reciprocating motion with unequal strokes. - The limiting positions show the maximum and minimum angles during motion. ### **Step 3: Equations and Limiting Positions** - Let \( \theta \) be the crank angle. - The length of crank = \( r \), length of connecting link = \( l \). - The position of the slider can be calculated using kinematic linkage equations. **Limiting positions:** - When the crank is at maximum and minimum angles, positions of the slider are obtained through geometric relations involving \( r, l, \theta \). **Conclusion:** Quick return mechanisms significantly improve efficiency by reducing non-cutting time. --- ## **Question 4: Finding Instant Centers in a 5-Link Mechanism** ### **Step 1: Understand the Mechanism** - The mechanism has 5 links connected via revolute joints. - Instant centers (ICs) are points where the relative velocity between two links is zero at a given instant. ### **Step 2: Use the Kennedy’s or Arcs of Rotation Method** - The **instant center** between two links can be found by: - **Connecting the two joints** that are common to the links. - **Finding the intersection** of the perpendicular bisectors of the lines connecting these joints. - For each pair of links: 1. Identify the joint axes. 2. Draw the perpendicular bisectors. 3. The intersection point is the IC. ### **Step 3: Find All Instant Centers** - Starting with the known IC \( P_{14} \), proceed: - **Between links 1 and 2:** find IC \( P_{12} \). - **Between links 2 and 3:** find IC \( P_{23} \). - Continue until all ICs are identified. - **Total number of ICs:** For a 5-link chain, there are \(\frac{n(n-1)}{2} = 10\) potential ICs, but only those relevant (usually on the linkage path) are considered. ### **Final Step:** - **Draw all ICs** on the diagram. - **Number of instant centers:** For a typical 5-link mechanism, there are **up to 7-9** instant centers, including the fixed and moving centers. --- ## **Summary of Final Answers:** 1. **Mechanism DOFs:** Approximately **1 DOF** (based on link/joint counts). 2. **Kennedy’s Theorem:** States that the three ICs in pure rolling contact are **collinear**. 3. **Quick Return Motion:** A reciprocating motion where the return stroke is faster than the forward stroke, achieved via specific linkage arrangements; involves analyzing limiting positions with geometric relations. 4. **Instant Centers:** Found by connecting joints and using perpendicular bisectors; for a 5-link mechanism, multiple ICs are identified, with specific steps to locate each. --- If you need further detailed calculations or specific geometric constructions, please specify!