Modern Control Engineering

480 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method

Constant-Phase-Angle Loci (NCircles). We shall obtain the phase angle ain

terms of XandY. Since

the phase angle ais

If we define

then

Since

we obtain

or

The addition of to both sides of this last equation yields

(7–24)

This is an equation of a circle with center at Y=1(2N)and with radius

For example, if a=30°, then N=tana=0.577, and the center and

the radius of the circle corresponding to a=30° are found to be (–0.5, 0.866)and unity,

respectively. Since Equation (7–24) is satisfied when X=Y=0andX=–1,Y=0

regardless of the value of N, each circle passes through the origin and the –1+j0point.

The constant aloci can be drawn easily, once the value of Nis given. A family of constant

Ncircles is shown in Figure 7–82 with aas a parameter.

It should be noted that the constant Nlocus for a given value of ais actually not the

entire circle, but only an arc. In other words, the a=30° and a=–150° arcs are parts

of the same circle. This is so because the tangent of an angle remains the same if ;180°

(or multiples thereof) is added to the angle.

The use of the MandNcircles enables us to find the entire closed-loop frequency

response from the open-loop frequency response G(jv)without calculating the magni-

tude and phase of the closed-loop transfer function at each frequency. The intersections

3

1

4 +^1 (^2 N)

(^2).

X=-^12 ,

aX+

1

2

b

2

+aY-

1

2N

b

2

=

1

4

+ a

1

2N

b

2

A^14 B+ 1 (2N)^2

X^2 +X+Y^2 -

1

N

Y= 0

N=

Y

X

-

Y

1 +X

1 +

Y

X

a

Y

1 +X

b

=

Y

X^2 +X+Y^2

tan (A-B)=

tanA-tanB

1 +tanA tanB

N=tan ctan-^1 a

Y

X

b -tan-^1 a

Y

1 +X

bd

tana=N

a=tan-^1 a

Y

X

b -tan-^1 a

Y

1 +X

b

/eja=n

X+jY

1 +X+jY

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