I can verify that BodePlot gives the correct answer. You can plot the function yourself with LogLogPlot, and LogLinearPlot as follows: f[x_]:=Abs[4(s-20)/(s (s+0.8))/.s->I x] g[x_]:=Arg[4(s-20)/(s (s+0.8))/.s->I x] LogLogPlot[f[x],{x,0.001,10000}] LogLinearPlot[180/[Pi] g[x],{x,0.1,10000}]

If you have an electrical engineering background, I recommend you reorder the terms as zeros, and poles form. You expression becomes -100 1/s (1-s/20)/(1 + s/0.8)

Let's look at the magnitude of the frequency response first. 1/s gives 20db/dec drop. then you hit a pole at 0.8. now you have 40db/dec drop. then you hit a right-half plane zero. so it bends backwards, returning to 20db/dec drop again.

You can see two knee points at 0.8 and 20.

Now about the phase. a normal 1/s will give you -90 constant phase. but the term is -1/s. so it flips. to become a +90 phase lead. then around the pole (+0.8), you drop 90 degree at -45deg/dec. so you reach 0 degree. then you hit a right-plane zero, which drops you another 90 degree to -90 degree.

So the BodePlot is correct.