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NDSU
8: Frequency Domain Techniquess
iii) K is equal to the scaling used to make
1
GK
ECE 461
tangent to the M-circle. If
1
G
is 2x larger, K1 =2.
Method #2: (useful on tests)
The above procedure is difficult to use since you need to redraw
1
GK
every time you guess K. A trick you can use
to save time follows:
First, note that if K=2,
1
GK
becomes half as large. Equivalently, relative to G1 , the axis looks 2x as large.
G ⎞
Second, note that you only care about the closed-loop gain ⎛⎝ 1+G
⎠ , or in this case, the inverse of this:
⎛ G ⎞
⎝ 1+G ⎠
−1
⎞ = ⎛1 + 1 ⎞
= ⎛⎝ 1+G
G ⎠
G⎠
⎝
This is simply the vector from
1
G
to -1.
Hence, the inverse of the closed-loop gain is simply the distance from
1
G
to -1. The resonance will be when this
distance is the smallest (and its inverse is its largest).
Procedure:
i) Plot G(jω) as real ⎛⎝ G1 ⎞⎠ vs imag ⎛⎝ G1 ⎞⎠ .
ii) Find a point on the - real axis, X, where
The system is stable ( G1 passes to the left of X, meaning the gain will be less than 1 when the phase is 180
degrees).
The distance from 1/G to -1
The minimum value of ⎛⎝ The distance from X to the origin ⎞⎠ is M1m
iii) K will then be X.
JSG
5
rev 02/27/02