May I please get help with this matlab exercise. Took snippets of example in the images. Thank you.1: Obtain the root locus plot of G(S)H(S) = k(S+3)(S+4)/(S+5)(S+6); Determine the poles for gain k = 1, 3,5, 7,10 Determine the range of k for stability. Determine the value of k where the overshoot is 5%. And plot the step response. Include screen shots of code and output in your report ### Root Locus Analysis#### First Graph: Root Locus of \( G(S)H(S)=K(s-2)/(s^3+2s^2+4s+8) \)- **Axes:** - **Real Axis (seconds\(^{-1}\))** along the horizontal - **Imaginary Axis (seconds\(^{-1}\))** along the vertical- **System Information (Top of Graph):** - **Gain:** 0 - **Pole:** \(-2\) - **Damping:** 1 - **Overshoot (%):** 0 - **Frequency (rad/s):** 2- **Additional System Information (Middle of Graph):** - **Gain:** 0 - **Pole:** \(-1.55e-15 + 2i\) and \(-1.55e-15 - 2i\) - **Damping:** 7.77e-16 - **Overshoot (%):** 100 - **Frequency (rad/s):** 2- **Graph Features:** - The root locus shows the path of the poles in the complex plane as the gain \( K \) varies. - The locus starts at poles of the open-loop transfer function and ends at the zeros or infinity.#### Value of Poles for \( k=1 \)#### Second Graph: \( G(S)H(S)=K(s-2)/(s^3+2s^2+4s+8) \)- **Axes:** - **Real Axis (seconds\(^{-1}\))** along the horizontal - **Imaginary Axis (seconds\(^{-1}\))** along the vertical- **System Information (Middle of Graph, Top Box):** - **Gain:** 1.01 - **Pole:** \(-1.43\) - **Damping:** 1 - **Overshoot (%):** 0 - **Frequency (rad/s):** 1.43- **Additional System Information (Middle of Graph, Bottom Box):** - **Gain:** 1.09 - **Pole:** \(-0.312 + 2.031i\) - **Damping:** 0.152 - **Overshoot (%):** 61. The image provides a mathematical representation and analysis of a transfer function in control systems, along with MATLAB code and a root locus plot.**Transfer Function:**\[ G(s)H(s) = \frac{K(s-2)}{(s+2)(s+2j)(s-2j)} \]**Instruction:**First simplify the transfer function to get the numerator and denominator coefficients.**MATLAB Code:**```matlabNum = [1 -2]denum = [1 2 4 8]sys = tf(num, denum)rlocus(sys)```**Root Locus Plot:**- The root locus plot visualizes the roots of the characteristic equation as the gain \( K \) varies.- The horizontal axis represents the Real Axis (seconds\(^{-1}\)).- The vertical axis represents the Imaginary Axis (seconds\(^{-1}\)).- The plot shows paths traced by the system poles: - The green and red curves indicate the locus of poles on the complex plane as gain \( K \) changes. - The blue line crossing through origin suggests the initial positions of poles when \( K = 0 \).**Value of Poles for \( k = 0 \):**(The specific values would typically be calculated using MATLAB or other computational tools to analyze the poles at \( K=0 \).)

Answered: Obtain the root locus plot of…

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May I please get help with this matlab exercise. Took snippets of example in the images. Thank you.

1: Obtain the root locus plot of

G(S)H(S) = k(S+3)(S+4)/(S+5)(S+6);

Determine the poles for gain k = 1, 3,5, 7,10

Determine the range of k for stability.

Determine the value of k where the overshoot is 5%. And plot the step response. Include screen shots of code and output in your report

### Root Locus Analysis

#### First Graph: Root Locus of \( G(S)H(S)=K(s-2)/(s^3+2s^2+4s+8) \)

- **Axes:**
- **Real Axis (seconds\(^{-1}\))** along the horizontal
- **Imaginary Axis (seconds\(^{-1}\))** along the vertical

- **System Information (Top of Graph):**
- **Gain:** 0
- **Pole:** \(-2\)
- **Damping:** 1
- **Overshoot (%):** 0
- **Frequency (rad/s):** 2

- **Additional System Information (Middle of Graph):**
- **Gain:** 0
- **Pole:** \(-1.55e-15 + 2i\) and \(-1.55e-15 - 2i\)
- **Damping:** 7.77e-16
- **Overshoot (%):** 100
- **Frequency (rad/s):** 2

- **Graph Features:**
- The root locus shows the path of the poles in the complex plane as the gain \( K \) varies.
- The locus starts at poles of the open-loop transfer function and ends at the zeros or infinity.

#### Value of Poles for \( k=1 \)

#### Second Graph: \( G(S)H(S)=K(s-2)/(s^3+2s^2+4s+8) \)

- **Axes:**
- **Real Axis (seconds\(^{-1}\))** along the horizontal
- **Imaginary Axis (seconds\(^{-1}\))** along the vertical

- **System Information (Middle of Graph, Top Box):**
- **Gain:** 1.01
- **Pole:** \(-1.43\)
- **Damping:** 1
- **Overshoot (%):** 0
- **Frequency (rad/s):** 1.43

- **Additional System Information (Middle of Graph, Bottom Box):**
- **Gain:** 1.09
- **Pole:** \(-0.312 + 2.031i\)
- **Damping:** 0.152
- **Overshoot (%):** 61.

The image provides a mathematical representation and analysis of a transfer function in control systems, along with MATLAB code and a root locus plot.

**Transfer Function:**
\[ G(s)H(s) = \frac{K(s-2)}{(s+2)(s+2j)(s-2j)} \]

**Instruction:**
First simplify the transfer function to get the numerator and denominator coefficients.

**MATLAB Code:**
```matlab
Num = [1 -2]
denum = [1 2 4 8]
sys = tf(num, denum)
rlocus(sys)
```

**Root Locus Plot:**
- The root locus plot visualizes the roots of the characteristic equation as the gain \( K \) varies.
- The horizontal axis represents the Real Axis (seconds\(^{-1}\)).
- The vertical axis represents the Imaginary Axis (seconds\(^{-1}\)).
- The plot shows paths traced by the system poles:
- The green and red curves indicate the locus of poles on the complex plane as gain \( K \) changes.
- The blue line crossing through origin suggests the initial positions of poles when \( K = 0 \).

**Value of Poles for \( k = 0 \):**
(The specific values would typically be calculated using MATLAB or other computational tools to analyze the poles at \( K=0 \).)

Expert Solution

Step 1

MATLAB CODE:

clear all
s = tf('s');
GH = ((s+3)*(s+4))/((s+5)*(s+6))
rlocus(GH)
fprintf('Poles of the open loop transfer function:\n')
s =pole(GH)
k = [1 3 5 7 10];
for i=1:length(k)
    CLTF = (1+k(i)*GH);
    fprintf('Poles of closed loop transfer function for k = %d are:\n',k(i));
    s = zero(CLTF)
    
end

COMMAND WINDOW OUTPUT:

GH =
 
  s^2 + 7 s + 12
  ---------------
  s^2 + 11 s + 30
 
Continuous-time transfer function.

Poles of the open loop transfer function:

s =

   -6.0000
   -5.0000

Poles of closed loop transfer function for k = 1 are:

s =

  -4.5000 + 0.8660i
  -4.5000 - 0.8660i

Poles of closed loop transfer function for k = 3 are:

s =

  -4.0000 + 0.7071i
  -4.0000 - 0.7071i

Poles of closed loop transfer function for k = 5 are:

s =

  -3.8333 + 0.5528i
  -3.8333 - 0.5528i

Poles of closed loop transfer function for k = 7 are:

s =

  -3.7500 + 0.4330i
  -3.7500 - 0.4330i

Poles of closed loop transfer function for k = 10 are:

s =

  -3.6818 + 0.2839i
  -3.6818 - 0.2839i

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