why draw the plot this way?

consider a system in its Fourier transform form:
1+j*w*T/ (1+j*w*T1) and its nonminimum phase counterpart: 1-j*w*T/
(1+j*w*T1) where 0<T<T1
in control textbooks, the bode plot is draw like this: the amplitude is the same in the bode plot, while the first system's phase curve is within 0 to -90. Due to nonmininmum phase characteristics of the nonminimum phase sytem, the phase curve is 0 to -180. However the Matlab always draw the nonminimum phase sytem's phase curve within 180 to 0. Why? we all know that nonmininmum phase system contains a large scale of the phase lag, why does Matlab use this kind of fashion?
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On Sat, 20 Jun 2009 19:24:48 -0700, workaholic wrote:

There's a bug in Matlab's Bode plotter?
Have you tried doing it manually? I.e. get the frequency response into a vector, then plot the vector's amplitude and phase vs. frequency?
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Although I haven't done it manually, I have a chart on the Modern Control Engineering, 3rd Edition, Ogata. The illustration on it is just what I stated in my last post. the nonminimum phase counterpart's phase may also begin in approximately 0 deg and end at -180 deg. I do not know why Matlab begin with 180.
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1/(1+jwT1)=> 0<phase2<-90 T1>T=> (1-jwT)/(1+jwT1)=> 270-90<phase1+phase2<360 => 180<phase1+phase2<0
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Your analysis is good but what does it mean? 0<phase2<-90 180<phase1+phase2<0
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