AN535 Datasheet, PDF (4/12 Pages) Motorola, Inc – PHASE LOCKED LOOP DESIGN FUNDAMENTALS

AN535
Freescale Semiconductor, Inc.
The position of the intersection according to the Rule 4 is:
s
+
SP
#P
*
*
SZ
#Z
+
(*
4
* 0) *
2*0
(0)
s + –2
( 28 )
The breakaway point, as defined by Rule 6, can be found by
first writing the characteristic equation.
The response of this type 1, second order system to a step
input, is shown in Figure 4. These curves represent the phase
response to a step position (phase) input for various damping
ratios. The output frequency response as a function of time to
a step velocity (frequency) input is also characterized by the
same set of figures.
C.E. + 1 ) G(s) H(s) + 0
+
1
)
s(s
K
)
4)
+
s2
)
4s
)
K
+
0
Now solving for K yields
( 29 )
1.9
1.8
ζ = 0.1
1.7
1.6
1.5
0.2
1.4
0.3
K = –s² –4s
( 30 )
1.3
1.2
Taking the derivative with respect to s and setting it equal to
zero, then determines the breakaway point.
1.1
1.0
dK
ds
+
d
ds
(*
s2
*
4s)
( 31 )
0.9
0.8
dK
ds
+
*
2s
*
4
+
0
or
0.7
( 32 )
0.6
0.5
0.4
0.6
0.7
0.5
0.8
1.0
1.5
2.0
s = –2
0.4
( 33 )
0.3
is the point of departure. Using this information, the root locus
can be plotted as in Figure 3.
The second order characteristic equation, given by
Equation 29, has be normalized to a standard form²
s² + 2ζω_ns + ω²ⁿ
( 34 )
where the damping ratio ξ = COS φ (0° ≤ φ ≤ 90°) and ω_n is the
natural frequency as shown in Figure 3.
K1
jω
ASYMPTOTE = π/2
0.2
0.1
0
0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10 11 12 13
ω_nt
Figure 4. Type 1 Second Order Step Response
The overshoot and stability as a function of the damping
ratio ξ is illustrated by the various plots. Each response is
plotted as a function of the normalized time ω_nt. For a given ξ
and a lock-up time t, the ω_n required to achieve the desired
results can be determined. Example:
CENTER OF GRAVITY
K=0
ω_n
φ
σ
Assume
ξ = 0.5
error < 10%
for t > 1ms
-4
-2
BREAKAWAY POINT
K=0
From ξ = 0.5 curve error is less than 10% of final value for all
time greater than ω_nt = 4.5. The required ω_n can then be found
by:
ASYMPTOTE = 3π/2
K1
Figure 3. Type 1 Second Order Root Locus Contour
ω_nt = 4.5
or
w_n
+
4.5
t
+
4.5
0.001
+
4.5kradńs
( 35 )
( 36 )
4
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