give answer in 2 step with explanation at the end of each step and final answer at the end:Consider a robotic system whose position vector is denoted by x = [xy, x,]7. The robot is
instrumented with

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:Consider a robotic system whose position vector is denoted by x = [xy, x,]7. The robot is
instrumented with a sensing system whose feedback features y satisfy the following relation:
_ _ [sin(xy) + xy x; + X2%1 + 2
y=f= vais — sin(x; + x,) — 34
(a) Derive the interaction matrix L = df /x for this system (12 marks)
(b) Explain (with equations and schematic control loop) how you would design a sensorimotor
controller X to drive y into a desired constant reference yg. (8 marks)

VIPSolutions · Accepted Answer

Let's break down the solution step by step:

---

## Step 1: Derive the Interaction Matrix $ \mathbf{L} = \frac{\partial \mathbf{f}}{\partial \mathbf{x}} $

Given:
$$
\mathbf{x} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix}, \quad
\mathbf{y} = \mathbf{f}(\mathbf{x}) = 
\begin{bmatrix}
\sin(x_1) + x_1 x_1 + x_2 x_1 + 21 \
\cos(x_2) - \sin(x_1 + x_2) - 34
\end{bmatrix}
$$

### Compute the Jacobian (Interaction Matrix)

The interaction matrix $ \mathbf{L} $ is the Jacobian of $ \mathbf{f}(\mathbf{x}) $:

$$
\mathbf{L} = \frac{\partial \mathbf{f}}{\partial \mathbf{x}} =
\begin{bmatrix}
\frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \
\frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2}
\end{bmatrix}
$$

#### For $ f_1(x_1, x_2) = \sin(x_1) + x_1^2 + x_2 x_1 + 21 $:

- $ \frac{\partial f_1}{\partial x_1} = \cos(x_1) + 2x_1 + x_2 $
- $ \frac{\partial f_1}{\partial x_2} = x_1 $

#### For $ f_2(x_1, x_2) = \cos(x_2) - \sin(x_1 + x_2) - 34 $:

- $ \frac{\partial f_2}{\partial x_1} = -\cos(x_1 + x_2) $
- $ \frac{\partial f_2}{\partial x_2} = -\sin(x_2) - \cos(x_1 + x_2) $

**So, the interaction matrix is:**

$$
\boxed{
\mathbf{L} =
\begin{bmatrix}
\cos(x_1) + 2x_1 + x_2 & x_1 \
-\cos(x_1 + x_2) & -\sin(x_2) - \cos(x_1 + x_2)
\end{bmatrix}
}
$$

**Explanation:**  
This matrix relates small changes in the robot's position vector $ \mathbf{x} $ to small changes in sensory feedback $ \mathbf{y} $.

---

## Step 2: Design a Sensorimotor Controller to Drive $ \mathbf{y} $ to $ \mathbf{y}_d $

We want to design a control law for $ \dot{\mathbf{x}} $ so that $ \mathbf{y} \to \mathbf{y}_d $ (desired feature vector).

### Error Definition

$$
\mathbf{e} = \mathbf{y} - \mathbf{y}_d
$$

### Control Law (Visual Servoing)

The time derivative of $ \mathbf{y} $:

$$
\dot{\mathbf{y}} = \mathbf{L} \dot{\mathbf{x}}
$$

To drive the error to zero, set a desired dynamics (e.g., exponential decay):

$$
\dot{\mathbf{e}} = -\lambda \mathbf{e}
$$

Since $ \dot{\mathbf{e}} = \dot{\mathbf{y}} = \mathbf{L} \dot{\mathbf{x}} $, we have:

$$
\mathbf{L} \dot{\mathbf{x}} = -\lambda (\mathbf{y} - \mathbf{y}_d)
$$

Assuming $ \mathbf{L} $ is invertible (or using pseudo-inverse if not):

$$
\dot{\mathbf{x}} = -\lambda \mathbf{L}^+ (\mathbf{y} - \mathbf{y}_d)
$$

Where $ \lambda >  $ is a gain, and $ \mathbf{L}^+ $ is the (pseudo-)inverse of $ \mathbf{L} $.

**Schematic Control Loop:**

1. Measure $ \mathbf{x} $, compute $ \mathbf{y} = \mathbf{f}(\mathbf{x}) $.
2. Compute error: $ \mathbf{e} = \mathbf{y} - \mathbf{y}_d $.
3. Compute interaction matrix $ \mathbf{L} $.
4. Compute control law: $ \dot{\mathbf{x}} = -\lambda \mathbf{L}^+ \mathbf{e} $.
5. Apply $ \dot{\mathbf{x}} $ to the robot.
6. Repeat.

**Explanation:**  
This control strategy ensures that the sensory feedback $ \mathbf{y} $ converges exponentially to the desired value $ \mathbf{y}_d $ by using the interaction matrix to map sensory errors to corrective motions.

---

## **Final Answers**

### **(a) Interaction Matrix**
$$
\boxed{
\mathbf{L} =
\begin{bmatrix}
\cos(x_1) + 2x_1 + x_2 & x_1 \
-\cos(x_1 + x_2) & -\sin(x_2) - \cos(x_1 + x_2)
\end{bmatrix}
}
$$

### **(b) Sensorimotor Controller**
$$
\boxed{
\dot{\mathbf{x}} = -\lambda \mathbf{L}^+ (\mathbf{y} - \mathbf{y}_d)
}
$$
Where $ \lambda >  $ and $ \mathbf{L}^+ $ is the pseudo-inverse of $ \mathbf{L} $. This drives $ \mathbf{y} $ to $ \mathbf{y}_d $.

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