Let's **analyze the given system** and each option: ### **Given Transfer Function** \[ G_{ol}(s) = \frac{5K}{s} \] This is a **type-1 integrator** (a pole at the origin). --- ## **1. Stability of Closed-Loop System** For a **unity feedback** system: \[ G_{cl}(s) = \frac{G_{ol}(s)}{1 + G_{ol}(s)} = \frac{\frac{5K}{s}}{1 + \frac{5K}{s}} = \frac{5K}{s + 5K} \] - **Pole**: \( s = -5K \) - For \( K > \), the pole is in the left half-plane (**stable**). - For \( K = \), the system is at the origin (**marginally stable**). - For \( K < \), the pole is in the right half-plane (**unstable**). **Therefore:** - "The closed-loop system is always stable for any \( K \)" is **incorrect**. - "The closed-loop system is boundary stable" is **incorrect** for all \( K \), but **marginally stable** for \( K = \). --- ## **2. Phase Margin** For the open-loop transfer function \( \frac{5K}{s} \): - **At unity gain crossover (\(|G_{ol}(j\omega)| = 1\))**: \[ \left| \frac{5K}{j\omega} \right| = 1 \implies \omega_{gc} = 5K \] - **Phase at this frequency**: Always \(-90^\circ\) (since \( \frac{1}{j\omega} \) gives phase \(-90^\circ\)). - **Phase margin**: \(180^\circ - 90^\circ = 90^\circ\) **So, "The phase margin is always 90° for any \( K \)" is TRUE.** --- ## **3. Gain Margin** - **Phase crossover**: Where phase = \(-180^\circ\). But for an integrator, the phase is always \(-90^\circ\) (never reaches \(-180^\circ\)). - **Gain margin**: Infinite (since the phase never reaches \(-180^\circ\)), so the statement about gain margin being 3 dB for \( K = 1 \) is **incorrect**. --- ## **Summary Table** | Statement | True/False | Reason | |-----------|:----------:|--------| | The closed-loop system is always stable for any \( K \). | False | Only for \( K > \) | | The closed-loop system is boundary stable. | False | Only for \( K = \) | | **The phase margin is always 90° for any \( K \).** | **True** | Integrator always gives 90° PM | | The gain margin is 3 dB for \( K = 1 \). | False | Gain margin is infinite | --- ## **Final Answer** > **The correct statement is:** > > **The phase margin is always 90° for any \( K \).** --- ### **Let me know if you want to see the derivation or details for any specific option!**

# Probability of Device Working To determine the probability of a device working when components A, B, or C can fail independently, we first need to find the probability of each component failing and then calculate the probability of the device functioning. ### **1. Given Probabilities** - **Probability of A failing**: \( P(A) = 0.01 \) - **Probability of B failing**: \( P(B) = 0.03 \) - **Probability of C failing**: \( P(C) = 0.04 \) ### **2. Probability of Each Component Working** The probability of each component working is the complement of the failure probability: - \( P(A \text{ works}) = 1 - P(A) = 1 - 0.01 = 0.99 \) - \( P(B \text{ works}) = 1 - P(B) = 1 - 0.03 = 0.97 \) - \( P(C \text{ works}) = 1 - P(C) = 1 - 0.04 = 0.96 \) ### **3. Total Probability of Device Working** Since the components fail independently, the probability of the device working (i.e., none of the components A, B, or C fail) is calculated as: \[ P(\text{Device works}) = P(A \text{ works}) \times P(B \text{ works}) \times P(C \text{ works}) \] Substituting in the probabilities: \[ P(\text{Device works}) = 0.99 \times 0.97 \times 0.96 \] ### **4. Calculation** Calculating the above expression: \[ P(\text{Device works}) = 0.99 \times 0.97 \times 0.96 \approx 0.941088 \] ### **5. Conclusion** The probability that the device works is approximately: \[ P(\text{Device works}) \approx 0.9411 \text{ or } 94.11\% \] ### **Summary** - **Probability of Device Working**: \( \approx 94.11\% \) - **Probability of Device Failing**: \( 1 - P(\text{Device works}) \approx 5.89\% \) Let me know if you need any further clarification or assistance!