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
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe36fb7f7-6381-4add-b52b-9085dfd27844%2F1bff32f7-778a-4e7a-9e9c-933ec8370b59%2F1x7v358_processed.jpeg&w=3840&q=75)
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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