Provide a correct and error-free solution. Ensure the method is correct, there are no calculation errors, and the solution is complete and well-framed properly and Don't give direct answers; it must b

Question

Provide a correct and error-free solution. Ensure the method is correct, there are no calculation errors, and the solution is complete and well-framed properly and Don't give direct answers; it must be clear and properly framed and aligned. Avoid using personal pronouns like 'we,' 'you,' and 'I' in the answer. Give the answer as a teacher who is giving the answer to a student and making them understand properly. Don't use Python, please. The solutions will be evaluated based on six key parameters:Relevancy, ensuring closely matches with the question; Completeness , including like calculations, diagrams, procedural steps, & explanatory statements in the solution; Accuracy, ensuring correctness of concepts and methods; Clarity , assessing readability and simplicity of language; Structure , focusing on organization and presentation; and Voice: Determining the response feels natural and human-like rather than robotic. Give a two-line best explanation or interpretation where there is a need for it. Give an overall short final answer last. The complete solution is lacking; please complete the whole answer till the end. Please adhere to and strictly follow all these guidelines anyhow. Tha

VIPSolutions · Accepted Answer

# Solution

Given matrices:
$$
A = \begin{bmatrix}  & 1 \ -8 & -6 \end{bmatrix},\quad B = \begin{bmatrix}  \ 2 \end{bmatrix},\quad C = \begin{bmatrix} 1 & 3 \end{bmatrix},\quad D = 
$$

---

## (i) State Transition Matrix (STM) using Cayley–Hamilton Theorem

### **Step 1: Find the Characteristic Equation of $A$**

The characteristic equation is determined as follows:
$$
|A - \lambda I| = 
$$

$$
\begin{vmatrix}
 - \lambda & 1 \
-8 & -6 - \lambda
\end{vmatrix} = 
$$

$$
(-\lambda)(-6-\lambda) - (-8)(1) = 
$$
$$
\lambda(\lambda + 6) + 8 = 
$$
$$
\lambda^2 + 6\lambda + 8 = 
$$
$$
(\lambda + 4)(\lambda + 2) = 
$$
So, eigenvalues are $\lambda_1 = -4$, $\lambda_2 = -2$.

### **Step 2: Cayley–Hamilton Theorem Application**

By Cayley–Hamilton theorem:
$$
A^2 + 6A + 8I = 
$$
Thus,
$$
A^2 = -6A - 8I
$$

### **Step 3: Express the State Transition Matrix $\Phi(t) = e^{At}$**
For a $2 \times 2$ system, using the Cayley–Hamilton theorem:
$$
e^{At} = \alpha_(t)I + \alpha_1(t)A
$$
where $\alpha_(t)$ and $\alpha_1(t)$ are determined using the eigenvalues:

For distinct eigenvalues $\lambda_1$ and $\lambda_2$,
$$
e^{At} = \frac{e^{\lambda_1 t} - e^{\lambda_2 t}}{\lambda_1 - \lambda_2}A + \frac{\lambda_1 e^{\lambda_2 t} - \lambda_2 e^{\lambda_1 t}}{\lambda_1 - \lambda_2}I
$$

Plugging in $\lambda_1 = -4$, $\lambda_2 = -2$:

$$
e^{At} = \frac{e^{-4t} - e^{-2t}}{-4 + 2}A + \frac{-4 e^{-2t} + 2 e^{-4t}}{-4 + 2}I
$$
$$
= \frac{e^{-4t} - e^{-2t}}{-2}A + \frac{-4e^{-2t} + 2e^{-4t}}{-2}I
$$
$$
= -\frac{1}{2}(e^{-4t} - e^{-2t})A + 2e^{-2t} - e^{-4t})I
$$

So, the state transition matrix is:
$$
\boxed{
e^{At} = [2e^{-2t} - e^{-4t}]I - \frac{1}{2}(e^{-4t} - e^{-2t})A
}
$$

---

## (ii) Controllability and Observability Matrices and Ranks

### **Controllability Matrix**

$$
\mathcal{C} = [B \quad AB]
$$

Calculate $AB$:
$$
AB = A \cdot B = \begin{bmatrix}  & 1 \ -8 & -6 \end{bmatrix} \begin{bmatrix}  \ 2 \end{bmatrix} = \begin{bmatrix} 2 \ -12 \end{bmatrix}
$$

So,
$$
\mathcal{C} = \begin{bmatrix}  & 2 \ 2 & -12 \end{bmatrix}
$$

Rank of $\mathcal{C}$:  
The determinant:
$$
\det(\mathcal{C}) = ()(-12) - (2)(2) = -4 \neq 
$$
Thus, rank is **2** (full rank for a 2-state system).

---

### **Observability Matrix**

$$
\mathcal{O} = \begin{bmatrix} C \ CA \end{bmatrix}
$$
Calculate $CA$:
$$
CA = C \cdot A = \begin{bmatrix} 1 & 3 \end{bmatrix} \begin{bmatrix}  & 1 \ -8 & -6 \end{bmatrix} = \begin{bmatrix} 1 \cdot  + 3 \cdot -8 & 1 \cdot 1 + 3 \cdot -6 \end{bmatrix} = \begin{bmatrix} -24 & -17 \end{bmatrix}
$$

So,
$$
\mathcal{O} = \begin{bmatrix} 1 & 3 \ -24 & -17 \end{bmatrix}
$$

Rank of $\mathcal{O}$:
$$
\det(\mathcal{O}) = (1)(-17) - (3)(-24) = -17 + 72 = 55 \neq 
$$
Thus, rank is **2** (full rank).

---

### **Interpretation**  
Both controllability and observability matrices have full rank, so the system is completely controllable and observable.

---

## (iii) Transfer Function $G(s)$

The transfer function for a state-space system is:
$$
G(s) = C(sI - A)^{-1}B + D
$$

### **Step 1: Calculate $(sI - A)$**
$$
sI - A = \begin{bmatrix} s & -1 \ 8 & s+6 \end{bmatrix}
$$

### **Step 2: Find the Inverse**
$$
(sI - A)^{-1} = \frac{1}{(s)(s+6) + 8} \begin{bmatrix} s+6 & 1 \ -8 & s \end{bmatrix}
$$
$$
= \frac{1}{s^2 + 6s + 8} \begin{bmatrix} s+6 & 1 \ 8 & s \end{bmatrix}
$$

### **Step 3: Compute $G(s)$**
$$
G(s) = C(sI - A)^{-1}B
$$

$$
(sI - A)^{-1} B = \frac{1}{s^2 + 6s + 8} \begin{bmatrix} s+6 & 1 \ 8 & s \end{bmatrix} \begin{bmatrix}  \ 2 \end{bmatrix}
= \frac{1}{s^2 + 6s + 8} \begin{bmatrix} 1 \cdot 2 \ s \cdot 2 \end{bmatrix}
= \frac{1}{s^2 + 6s + 8} \begin{bmatrix} 2 \ 2s \end{bmatrix}
$$

Now,
$$
C \cdot \text{(result)} = [1~3] \begin{bmatrix} 2 \ 2s \end{bmatrix} = 2 + 6s
$$

So,
$$
G(s) = \frac{2 + 6s}{s^2 + 6s + 8}
$$

---

## **Summary of Results**

1. **STM**:  
   $$
   e^{At} = [2e^{-2t} - e^{-4t}]I - \frac{1}{2}(e^{-4t} - e^{-2t})A
   $$

2. **Controllability/Observability**:  
   Both matrices have full rank (**2**), indicating the system is controllable and observable.

3. **Transfer Function**:  
   $$
   G(s) = \frac{2 + 6s}{s^2 + 6s + 8}
   $$

---

**Interpretation**:  
This system is fully controllable and observable, and its transfer function indicates a second-order system with zeros and poles determined by the numerator and denominator, respectively.

✨ VIPSolutions

Question:

Final Answer:

📢 Instructions:

Study Help, Fast Answers

✨ VIPSolutions

Question:

Final Answer:

📢 Instructions:

Related Answers

Study Help, Fast Answers