Given the following feedback system, a. Find the System type b. Find the value of the static error constants c. Find the value of K to have an steady-state error of 0.01 R(s) + E(s) C(s) K 1 s(s2+5s+6) S

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### Feedback Control System Analysis

For the given feedback system, we will perform the following tasks:

1. **Determine the System Type**
2. **Calculate the Static Error Constants**
3. **Find the value of \( K \) for a Steady-State Error of 0.01**

---

#### System Overview:

Below is the block diagram of the given feedback control system:

```plaintext
R(s) --- ( + ) ---> E(s) --> [ K ] ------ ( + ) --> [ 1 / ( s(s^2 + 5s + 6) ) ] ---> C(s)
         - ^                                   |
             |                                 S
             <-------------------------------- < 
```

- **\( R(s) \)**: Reference input
- **\( E(s) \)**: Error signal
- **\( K \)**: Proportional gain
- **\( C(s) \)**: Output signal
- **\( S \)**: Feedback path, simply \( s \) in this context

#### Tasks:

**a. Find the System Type**

The system type is determined by the number of poles at the origin in the open-loop transfer function.

Open-loop transfer function:
\[ G(s)H(s) = \frac{K}{s(s^2 + 5s + 6)} \cdot s \]

\[ = \frac{K}{s(s^2 + 5s + 6)} \]

Taking the denominator:
\[ s(s^2 + 5s + 6) = s^3 + 5s^2 + 6s \]

There is **1 pole** at \( s=0 \), so the system type is **Type 1**.

**b. Find the Value of the Static Error Constants**

For a Type 1 system, we consider the velocity error constant \( \text{Kv} \).

Velocity error constant \( \text{Kv} \):
\[ \text{Kv} = \lim_{s \to 0} s G(s) \]

Open-loop transfer function:
\[ G(s) = \frac{K}{s(s^2 + 5s + 6)} \]

\[ \text{Kv} = \lim_{s \to 0} s \left( \frac{K}{s(s^2 + 5s + 6)} \right) \]

\[ \text{Kv} = \lim_{s
Transcribed Image Text:--- ### Feedback Control System Analysis For the given feedback system, we will perform the following tasks: 1. **Determine the System Type** 2. **Calculate the Static Error Constants** 3. **Find the value of \( K \) for a Steady-State Error of 0.01** --- #### System Overview: Below is the block diagram of the given feedback control system: ```plaintext R(s) --- ( + ) ---> E(s) --> [ K ] ------ ( + ) --> [ 1 / ( s(s^2 + 5s + 6) ) ] ---> C(s) - ^ | | S <-------------------------------- < ``` - **\( R(s) \)**: Reference input - **\( E(s) \)**: Error signal - **\( K \)**: Proportional gain - **\( C(s) \)**: Output signal - **\( S \)**: Feedback path, simply \( s \) in this context #### Tasks: **a. Find the System Type** The system type is determined by the number of poles at the origin in the open-loop transfer function. Open-loop transfer function: \[ G(s)H(s) = \frac{K}{s(s^2 + 5s + 6)} \cdot s \] \[ = \frac{K}{s(s^2 + 5s + 6)} \] Taking the denominator: \[ s(s^2 + 5s + 6) = s^3 + 5s^2 + 6s \] There is **1 pole** at \( s=0 \), so the system type is **Type 1**. **b. Find the Value of the Static Error Constants** For a Type 1 system, we consider the velocity error constant \( \text{Kv} \). Velocity error constant \( \text{Kv} \): \[ \text{Kv} = \lim_{s \to 0} s G(s) \] Open-loop transfer function: \[ G(s) = \frac{K}{s(s^2 + 5s + 6)} \] \[ \text{Kv} = \lim_{s \to 0} s \left( \frac{K}{s(s^2 + 5s + 6)} \right) \] \[ \text{Kv} = \lim_{s
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