Question on Transfer Functions

[doublemint]

Hi,

Just a simple question on Transfer Functions.

If I was give a open loop TF, G, and a feedback loop H, then the closed loop TF is G'=$\frac{G}{1+GH}$.

So my first attempt was to simplify the G' by hand and then plot it in excel which produced a crescent shaped moon.
By simplification, I mean this:
Let G=1/s and H=1/(s+1)
G'=$\frac{G}{1+GH}=\frac{1/s}{1+\frac{1}{s}+\frac{1}{s+1}}$=$\frac{s+1}{s(s+1)+1}$
But, then I go to mathematica and enter the G' without simplification (just $\frac{G}{1+GH}$)and it gave me a totally different graph.

Even though the the G' is the same (mathematically) in both cases, why does the Nyquist (and thus Bode) plots are different?

 

[aralbrec]

By simplification, I mean this:
Let G=1/s and H=1/(s+1)
G'=$\frac{G}{1+GH}=\frac{1/s}{1+\frac{1}{s}+\frac{1}{s+1}}$=$\frac{s+1}{s(s+1)+1}$

That looks fine except the second step looks to be a typo with the second '+' in the denominator meant to be multiplication.

Even though the the G' is the same (mathematically) in both cases, why does the Nyquist (and thus Bode) plots are different?

What kind of graphs did you do? Obviously they should be the same whether you use excel or mathematica.

Bode plots have increasing frequency on the horizontal axis and present the gain and phase of a complex quantity as w varies in a typical graph form. A nyquist plot, on the other hand, is essentially a polar plot that traces the value of a complex quantity in the complex plane as w varies.

Usually the nyquist plot is done on the loop gain GH to find out if there are poles in the right half plane of the overall function G/(1+GH). You can get similar information from a bode plot of GH too but, as mentioned, the bode plot is like a standard graph whereas the nyquist plot is a polar plot.

You can think if their similarity like this: a Nyquist plot is looking to see if a polar plot of GH encircles the point -1 to determine stability. In a bode plot of GH, the magnitude of GH should be less than 1 when the phase is 180 degrees (pi). The point -1∠0 in the nyquist plot occurs at phase 180 degrees and there will be no encirclements of -1 if |GH| < 1 when the phase is 180 degrees. So you can see the gain/phase conditions on GH in the bode plot are the same as the encirclement of -1 test in the nyquist plot.

 

[doublemint]

I did Nyquist.
Yeah you are correct. I just tried the simplified equations I posted and did another graph with functions of more complexity and both Nyquist plots are the same..

I guess I typed something in wrong :(

Thanks for the help though!