Electrical – How to graph ω² in a Bode Plot

bode plot

Assume this transfer function. How would you graph \$\omega^2\$?

enter image description here

Apparently, the solution just says you would graph 0.1 by itself for a -20 dB curve. Could someone explain? I also have a screenshot of the magnitude Bode Plots which I am not understanding where w^2 is factored in. Thank you

bode plot

Best Answer

You just calculate it and plot it:

In Matlab: 

>> w=logspace( -1 , 10,10000);
>> xfun = .1*w.^2.*(1 - (w./10).^2 +j*w./2)./((1+j.*w).*(1+j.*w/100).^3);
>> subplot(2,1,1)
>> semilogx(w, 20*log10(abs(xfun)))
>> subplot(2,1,2)
>> semilogx(w, angle(xfun)*180/pi)][1]][1]

enter image description here

You could just use "bode", but that's easier w/ the Laplace transform. Because I didn't, the x-axes are in radians, with no labels. I adjusted the line widths and font sizes for export to a jpg.

If you don't have Matlab, this should work directly in Octave, which is free. If the freq scale is off, feel free to adjust w. If you want freq in Hz, remember that \$\omega=2\pi f\$

As a tip, remember that your exponents need to be ".^", not "^", and your multiplies and divides need to be ".*" and "./" -- all for element-by-element operations.

If you're just interested in \$\omega^2\$, \$20log_{10}\omega^2 = 40log_{10}\omega\$. At \$\omega=0.1\$, this is equal to -40dB, and it will have a slope of 40db/decade. Perhaps even more intuitively, it is 0dB at \$\omega=1\$. It is real, so the angle is 0.

>> w=logspace( -1 , 10,10000);
>> w2=w.^2;
>> subplot(2,1,1)
>> semilogx(w, 20*log10(abs(w2)))
>> subplot(2,1,2)
>> semilogx(w, angle(w2))

enter image description here

This, of course gives you 80dB/decade total going up, above 100 rad/sec, and 80 going down, for a flat line.

Related Topic