Hi Robert,
I understand that you are trying to determine the maximum deviation in magnitude (in dB) and the corresponding frequency at which the two curves, bodemag and linear approximation, exhibit the greatest difference.
To do so:
- you can specify the frequency range of the curve, and as per your given plot range can be written as: “frequencies = logspace(-2, 3, 1000);”
- Then substitute these frequencies in the “orig” curve, which can be done by using: “origMagnitude = abs(freqresp(orig, frequencies));”
- This will return the respective magnitude of the curve and to convert the curve in dB, use “20*log10(origMagnitude)”.
- Similarly substitute the frequency value is the “linApprox” curve using: “linApproxMagnitude = subs(linApprox, x, frequencies);”
- Finally find deviation between the two curves using the following equation: “deviation = squeeze(20*log10(origMagnitude))'-squeeze(double(linApproxMagnitude));”
- To find the maximum value of all the deviations calculated, use the “max()” function.
Final code snippet is given below:
orig = tf([2500],[1 20 2500]);
syms x;
linApprox = piecewise(x<50, 0, x>50, -40*log10(x/50));
fplot(linApprox,'r')
hold on
bodemag(orig,'b')
%%
frequencies = logspace(-2, 3, 1000); % Adjust the range and number of frequencies as needed
% Calculate the magnitude response of the transfer function
origMagnitude = abs(freqresp(orig, frequencies));
% Calculate the magnitude response of the piecewise function
linApproxMagnitude = subs(linApprox, x, frequencies);
% Calculate the deviation between the two magnitude responses
deviation = squeeze(20*log10(origMagnitude))' - squeeze(double(linApproxMagnitude));
% Find the maximum deviation and its corresponding frequency
[maxDeviation, maxDeviationIndex] = max(abs(deviation));
maxDeviationFrequency = frequencies(maxDeviationIndex);
% Display the results
disp("Maximum deviation in magnitude frequency response:");
disp("Deviation: " + maxDeviation);
disp("Frequency: " + maxDeviationFrequency);
Hope this helps!
