Math Problem Statement

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

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.

Solution

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)=ss1(ss2)(ss2)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:

  • s1=20πs_1 = -20\pi: Corresponds to a zero at approximately f1=10f_1 = 10 Hz.
  • s2=2000πej2π/3s_2 = 2000\pi e^{j2\pi/3}: Complex-conjugate poles contributing a break at f2=1000f_2 = 1000 Hz.

MATLAB Function for Piecewise Bode Plot

Let’s expand the code to calculate both the magnitude MM (in dB) and phase PP (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 ff is in, then applies the corresponding slope in dB.

    • Below f1f_1: Magnitude is 0 dB.
    • Between f1f_1 and f2f_2: Gain increases at +20+20 dB/decade.
    • Beyond f2f_2: Gain decreases at 20-20 dB/decade, considering both poles.
  • Phase (P): We use piecewise transitions:

    • 00^\circ far below f1f_1.
    • Transition from 00^\circ to +90+90^\circ as ff approaches f1f_1.
    • Flat +90+90^\circ between f1f_1 and f2f_2.
    • Transition from +90+90^\circ to 90-90^\circ as ff approaches f2f_2.
    • Flat 90-90^\circ beyond f2f_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 s1s_1 or s2s_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.

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Math Problem Analysis

Mathematical Concepts

Control Systems
Bode Plot
Frequency Response Analysis

Formulas

Transfer function: H(s) = (s - s1) / ((s - s2)(s - s2*))
Magnitude (dB): 20 * log10(|H(jω)|)
Phase (degrees): arg(H(jω)) in degrees

Theorems

Bode Plot Straight Line Approximation

Suitable Grade Level

Undergraduate