Let's analyze the transfer function and each statement: ## **Given Transfer Function** \[ G(s) = \frac{1+s}{(s+2)(s-5)} \] --- ### **1. The system is of PDT2 characteristic.** - **PDT2** refers to a second-order system with two real poles (the "2") and potentially a zero. - Here, the denominator is \((s+2)(s-5)\), so there are two real poles: \(s = -2\) and \(s = 5\). - The numerator \(1 + s\) introduces a zero at \(s = -1\). - **So, yes, this fits a second-order system with a real zero and two real poles.** ✔️ **This statement is correct.** --- ### **2. The system is stable.** - **Stability** depends on the location of the poles: - Poles: \(s = -2\) (left-half), \(s = 5\) (right-half). - Because of the pole at \(s = 5\) (right-half plane), the system is **unstable**. - **This statement is incorrect.** --- ### **3. The system is not causal.** - **Causality**: For physical systems, the order of the numerator must be less than or equal to the denominator. - Degree of numerator: 1 (\(s\)) - Degree of denominator: 2 (\(s^2\)) - Since \(1 < 2\), the system is **causal**. - **This statement is incorrect.** --- ### **4. The step response of the system converges to \(-\frac{1}{10}\) when \(t \to \infty\).** - The final value of the step response can be found using the **Final Value Theorem**: \[ \lim_{t \to \infty} y(t) = \lim_{s \to } s G(s) \cdot \frac{1}{s} \] \[ = \lim_{s \to } G(s) = \frac{1+}{(+2)(-5)} = \frac{1}{2 \cdot (-5)} = -\frac{1}{10} \] - **This statement is correct** (mathematically), but since the system is unstable, the actual response diverges due to the unstable pole. Final Value Theorem does not apply to unstable systems. --- ## **Summary Table** | Statement | True/False | Explanation | |-----------|------------|-------------| | PDT2 characteristic | ✔️ | 2 real poles, 1 zero | | Stable | ❌ | Pole at \(s=5\) (RHP) | | Not causal | ❌ | Numerator degree < denominator degree | | Step response converges to \(-\frac{1}{10}\) | ❌ | Theoretically correct, but system is unstable | --- ## **Key Points** - The system is **PDT2**. - The system is **unstable** and **causal**. - The step response does **not** actually converge due to instability, despite what the calculation suggests.

To determine how many choices must be eliminated for guessing to be advantageous in a test with the given scoring system, we can use the concept of expected value. ## **Scoring System** - Correct answer: \(7\) points - Incorrect answer: \(0\) points - Blank answer: \(2\) points ## **Expected Value Calculation** Let \(n\) be the total number of choices, and \(k\) be the number of choices eliminated. ### **Remaining Choices** The remaining choices after eliminating \(k\) choices will be \(n - k\). ### **Probability of Correct Answer** The probability of guessing correctly is: \[ P(\text{Correct}) = \frac{1}{n - k} \] ### **Probability of Incorrect Answer** The probability of guessing incorrectly is: \[ P(\text{Incorrect}) = \frac{n - 1 - k}{n - k} \] ### **Expected Value Formula** The expected value \(E\) for guessing is given by: \[ E = P(\text{Correct}) \cdot (\text{Score for Correct}) + P(\text{Incorrect}) \cdot (\text{Score for Incorrect}) + P(\text{Blank}) \cdot (\text{Score for Blank}) \] Considering one option is blank: - Probability of blanking (not choosing any option) is \(1 - P(\text{Correct}) - P(\text{Incorrect})\). Substituting the values: \[ E = \frac{1}{n - k} \cdot 7 + \frac{(n - 1 - k)}{n - k} \cdot 0 + \left(1 - \frac{1}{n - k} - \frac{(n - 1 - k)}{n - k}\right) \cdot 2 \] This simplifies to: \[ E = \frac{7}{n - k} + 2 \cdot \left(1 - \frac{1}{n - k} - \frac{(n - 1 - k)}{n - k}\right) \] \[ E = \frac{7}{n - k} + 2 \cdot \left(1 - \frac{(n - 1)}{n - k}\right) \] \[ E = \frac{7}{n - k} + \frac{2(k + 1)}{n - k} \] \[ E = \frac{7 + 2(k + 1)}{n - k} \] ### **Condition for Advantageous Guessing** We want the expected value to be greater than \(2\): \[ \frac{7 + 2(k + 1)}{n - k} > 2 \] ### **Solving the Inequality** Now we substitute \(n = 5\) (total choices): \[ \frac{7 + 2(k + 1)}{5 - k} > 2 \] Cross-multiplying gives: \[ 7 + 2(k + 1) > 2(5 - k) \] \[ 7 + 2k + 2 > 10 - 2k \] \[ 4k > 1 \quad \Rightarrow \quad k > \frac{1}{4} \] Since \(k\) must be a whole number, the smallest integer satisfying this condition is: \[ k \geq 1 \] ### **Conclusion** At least **1 choice** must be eliminated for guessing to be advantageous (expected value > 2).