https://file.aiquickdraw.com/mathbot/file/1730454119650koa6srsv.jpg

function [M, P] = myFunction(f)
% Define break frequencies based on transfer function analysis
f1 = 10;    % First break frequency due to zero
f2 = 1000;  % Second break frequency due to poles

% Magnitude calculation (dB)

% Phase calculation (degrees)
if f < 0.1 * f1
P = 0;  % Flat at 0 degrees far below f1
elseif f < 10 * f1
P = 45 * log10(f / (0.1 * f1));  % Transition to +90 degrees
elseif f < 0.1 * f2
P = 90;  % Flat at +90 degrees between f1 and f2
elseif f < 10 * f2
P = 90 - 90 * log10(f / (0.1 * f2));  % Transition to -90 degrees
else
P = -90;  % Flat at -90 degrees after f2
end
end

I see you've uploaded an image with text and provided MATLAB code related to creating a Bode plot for a transfer function. Let's walk through the setup based on the problem requirements and refine the code as needed.

### Problem Breakdown
The transfer function \( H(s) = \frac{s - s_1}{(s - s_2)(s - s_2^*)} \) has complex poles and a real zero. This suggests:
- **Magnitude Plot**: Approximate each break (corner) frequency by a piecewise-linear function.
- **Phase Plot**: Transition between phases based on the effect of each pole and zero.

Given values:
- \( s_1 = -20\pi \): Corresponds to a zero at approximately \( f_1 = 10 \) Hz.
- \( s_2 = 2000\pi e^{j2\pi/3} \): Complex-conjugate poles contributing a break at \( f_2 = 1000 \) Hz.

### MATLAB Function for Piecewise Bode Plot
Let’s expand the code to calculate both the **magnitude** \( M \) (in dB) and **phase** \( P \) (in degrees).

```matlab
function [M, P] = myFunction(f)
    % Define break frequencies based on transfer function analysis
    f1 = 10;    % First break frequency due to zero
    f2 = 1000;  % Second break frequency due to complex poles

    % Magnitude calculation (dB)
    if f < f1
        % Below first break, magnitude starts at 0 dB
        M = 0;
    elseif f < f2
        % Between first and second break, magnitude rises at +20 dB/decade
        M = 20 * log10(f / f1);
    else
        % Beyond second break, magnitude falls at -20 dB/decade (net -20 dB/decade)
        M = 20 * log10(f / f1) - 40 * log10(f / f2);
    end

    % Phase calculation (degrees)
    if f < 0.1 * f1
        % Far below first break frequency
        P = 0;  
    elseif f < 10 * f1
        % Transitioning due to zero, up to +90 degrees
        P = 45 * log10(f / (0.1 * f1));  
    elseif f < 0.1 * f2
        % After zero impact, flat at +90 degrees
        P = 90;
    elseif f < 10 * f2
        % Transitioning due to poles, down to -90 degrees
        P = 90 - 90 * log10(f / (0.1 * f2));  
    else
        % After poles' impact, flat at -90 degrees
        P = -90;  
    end
end
```

### Explanation of the Code
- **Magnitude (M):** The function first evaluates which frequency region \( f \) is in, then applies the corresponding slope in dB.
  - Below \( f_1 \): Magnitude is 0 dB.
  - Between \( f_1 \) and \( f_2 \): Gain increases at \( +20 \) dB/decade.
  - Beyond \( f_2 \): Gain decreases at \( -20 \) dB/decade, considering both poles.

- **Phase (P):** We use piecewise transitions:
  - \( 0^\circ \) far below \( f_1 \).
  - Transition from \( 0^\circ \) to \( +90^\circ \) as \( f \) approaches \( f_1 \).
  - Flat \( +90^\circ \) between \( f_1 \) and \( f_2 \).
  - Transition from \( +90^\circ \) to \( -90^\circ \) as \( f \) approaches \( f_2 \).
  - Flat \( -90^\circ \) beyond \( f_2 \).

Would you like additional details on specific parts of the function, or further clarification on how Bode plots are generated?

### Additional Questions
1. Why does the Bode magnitude plot change slope at each break frequency?
2. How does the phase transition depend on the frequency and the location of poles and zeros?
3. What would happen to the Bode plot if \( s_1 \) or \( s_2 \) were at different values?
4. Can you explain why the piecewise-linear approximation is valid for high frequencies?
5. How would you extend this code to handle additional zeros or poles?

**Tip:** When creating Bode plots, using logarithmic scales for frequency helps to visualize system behavior over a broad range.

A certain system has a transfer function in the Laplace domain given by H(s) = (s - s1) / ((s - s2)(s - s2*)), where s1 = -20π and s2 = 2000πe^(j2π/3). Part A) Sketch by hand the Bode plot (straight line approximation) of the magnitude response for this system. Part B) Sketch by hand the Bode plot (straight line approximation) of the phase response for this system. Write a function in MATLAB that takes f (frequency in Hz) as input and outputs M (magnitude in dB) and P (phase in degrees) based on your hand plots.

Learn how to sketch the Bode plot (magnitude and phase) for a transfer function with complex poles and a real zero. This MATLAB problem involves frequency response analysis and piecewise-linear approximation for Bode plots, suitable for control systems courses at the undergraduate level.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation