U. Glaeser

induced jitter for a set of loop parameters. The peak values and the point at which they occur are
summarized in Table 10.3.
The closed-loop frequency response of the PLL for different values of ζ and for frequencies normalized
to ω N is shown in Fig. 10.6. This plot shows that the PLL is a low-pass filter to phase noise at frequencies
below ω N. Phase noise at frequencies below ω N passes through the PLL unattenuated. Phase noise at
frequencies above ω N is filtered with slope of −20 dB per decade. For small values of ζ, the filter cutoff
at ω N is sharper with initial slopes as high as −40 dB per decade. However, for these values of ζ, the phase
noise is amplified at frequencies near ω N. This phase noise amplification or peaking increases, along with
the initial cutoff slope, for decreasing values of ζ. This phase noise amplification can have adverse affects
on the output jitter of the PLL. It is important to notice that because of the zero in the closed-loop
© 2002 by CRC Press LLC
TABLE 10.3 Peak Values of Second-Order PLL Magnitude, Impulse, Step,
and Ramp Responses
Magnitude Frequency Response (for all ζ):
ω1 = (ωN/(2 ⋅ ζ)) ⋅ ((1 + 8 ⋅ ζ 2 ) 0.5 − 1) 0.5
|H(jω1)| = (N ⋅ (1 + 8 ⋅ ζ 2 ) 0.25 )�
(1 + (1 − 1/(2 ⋅ ζ 2 ) − 1/(8 ⋅ ζ 4 )) ⋅
((1 + 8 ⋅ ζ 2 ) 0.5 − 1) + 1/(2 ⋅ ζ 2 )) 0.5
Step Response:
ζ > 1:
t1 = (1/(ω N ⋅ (ζ 2 − 1) 0.5 )) ⋅
log (2 ⋅ ζ ⋅ (ζ + (1 − ζ 2 ) 0.5 ) − 1)
s(t1) = s(t = t1) ζ = 1:
t1 = 2/ω N
s(t1) = N ⋅ (1 + 1�e 2 )
0 < ζ < 1:
t1 = (π − 2 ⋅ sin −1 (ζ))/(ω N ⋅ (1 − ζ 2 ) 0.5 )
s(t1) = N ⋅ (1 + e ((2 ⋅ sin −1 (ζ) − π) ⋅ (ζ�(1 − ζ 2 ) 0.5 ))
)
Ramp Response:
ζ > 1:
t1 = (1/(2 ⋅ ω N ⋅ (ζ 2 − 1) 0.5 )) ⋅
log (2 ⋅ ζ ⋅ (ζ + (1 − ζ 2 ) 0.5 ) − 1)
r′(t1) = r′(t = t1) ζ = 1:
t1 = 1/ω N
r′(t1) = − N�(e ⋅ ω N)
0 < ζ < 1:
t1 = cos −1 (ζ)�(ω N ⋅ (1 − ζ 2 ) 0.5 )
r′(t1) = − N�ω N ⋅ e (cos −1 (ζ) ⋅ (ζ�(1 − ζ 2 ) 0.5 ))
Slow Step Response:
0 < ζ < 1:
t1 = (1/x) ⋅
tan −1 ((−x + z ⋅ y ⋅ sin (x ⋅ dt) + z ⋅ x ⋅ cos (x ⋅ dt)) �
(−y + z ⋅ y ⋅ cos (x ⋅ dt) + z ⋅ x ⋅ sin (x ⋅ dt)))
for t1 > dt, otherwise given by t1 for r′(t)
where:
x = ω N ⋅ (1 − ζ 2 )
y = ζ ⋅ ω N
(ζ ⋅ ωN ⋅ dt)
z = e
d′(t1) = d′(t = t1) Note that ω 1 or t 1 is the frequency or time where the response from Table 10.2 is
maximized.