Understanding Bode Plots for Second Order Systems with ς‚= 0

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The discussion centers on understanding the phase plot of the Bode plot for the transfer function g(s) = 1/(s^2 + 4). The user correctly calculated the magnitude plot but is confused about the phase plot, noting a discrepancy between manual calculations (0°) and MATLAB results (-360° to -180°). It is clarified that each pole contributes a 90° phase shift, resulting in a total of -180° for two poles. The phase shift begins one decade before the corner frequency and ends one decade after, explaining the transition observed in MATLAB. Understanding these phase shifts is crucial for accurately drawing the phase plot for second-order systems with ζ = 0.
ksurabhi
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hey guys,

i have a question regarding bode plot

g(s)= 1/(s2+4)

i did get the magnitude plot correct but i am unable to understand the phase plot. by calculating on paper i got 0° but in MATLAB it changes from -360° to -180°

i haven't understood how the initial phase is -360 which changed to -180 at the corner frequency(2)

please explain. i would really appreciate if you could explain me how to draw the phase plot for second order system with ς‚= 0

thank you in advance
 
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-360 degrees is equal to 0!

Typically each pole will add a 90 degrees phase shift that begins 1 decade before the pole and ends one decade after. Since there are two poles the phase shift will be 180 degrees
 
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