Once we have the edge detected signal, we now want to separate the pure tone at the bit rate from the random data-dependent variations. One method is to filter out the unwanted signal with a bandpass filter tuned to the clock bit-rate. The BPF is
Clock Recovery 183
a resonant circuit that will ring in response to an input pulse. The signale(t;)is a
random stream of identical pulses at integer multiples of the bit-period. A pulse will be present when there was a transition in the data, and no pulse will be present when the data does not change states. Clearly this signal can be used to keep a resonator ringing at the bit-rate, provided that the pulse repetition rate is within the bandwidth of the BPF. Since there will be missing pulses whenever no data transition occurs, the ringing will tend to die away during long periods of missing pulses due to dissipation in the resonator. This dissipation will cause both amplitude, and phase modulation in the extracted tone. This effect can easily be seen in the time domain. For a simple second-order BPF with a transfer function of the form
H(s)= 2!ns
s2+2!ns+!2n (4.5)
there is a zero at the origin, and two complex poles, as shown in the pole-zero plot of Fig. 4.21(a). The dissipation of the filter is the real-part of the complex poles;!n, where is the damping ratio, and!nis the undamped natural frequency. We saw in (2.218) and (2.228) that random amplitude and phase modulations were related to the equivalent selectivity of the filter by
am=
1
pQeq (4.6)
for pure amplitude modulation, and
= pQ1eq (4.7)
for pure phase modulation. We can relate the selectivity to the dissipation in this simple filter as the inverse of the integral of the normalized frequency response.
1
Qeq =!njH(1j!n)j2
Z
1
0
jH(j!)j2d!
= Z
1
0
42!^2
1+(42;2)^!2+!^4d !^
=
(4.8)
We can also define a selectivityQ3dB that is the ratio of the 3-dB bandwidth to the center frequency. For a second-order bandpass this can be shown to be
Q3dB= 1
2 (4.9)
184 Chapter 4
Normalized real -part (σ/ωn) Normalized imaginary -part (jω/ωn)
X X
0
-1.5 -1 -0.5 0 0.5 1 1.5
x
x o
0 0.2 0.4 0.6 0.8 1
0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5
Normalized frequency (f/fn)
Normalized Magnitude Squared
(a) (b)
Figure 4.21 Second-order bandpass filter, (a) pole-zero diagram, (b) frequency response of real filter and equivalent ideal filter forQeq=4.
Therefore, the relationship between these two selectivities is
Qeq=1 = 2Q3dB (4.10)
The frequency response of this second-order bandpass filter is shown in Fig. 4.21(b) together with an ideal BPF of normalized bandwidth1=Qeq. In this plotQeq = 4, which correspond to a damping ratio of = 1=4. Taking the inverse Laplace transform we know that the impulse response of the filter has a decaying envelope of the form
envc(t)=e;!nt (4.11)
For a filter that is tuned to the bit-rate, then!n=2BT, and if we normalize time by the bit-period such that
nt=4 t
T ; (4.12)
then the decaying envelope of the impulse response of the filter is env(nt)=exp (;2nt)
=exp
;2nt
Qeq
(4.13)
This can be written in terms of a normalized time constantn, wheren is the number of bit-periods before the envelope decreases to a value of1=e=0:37;
env(nt)=exp
;nt
n
; (4.14)
Clock Recovery 185
and the normalized time constant is given by
n =Qeq
2
=
Q3dB
: (4.15)
The decay in the power envelope is proportional to the voltage squared, and is simply Penv(nt)=exp
;nt
n =2
; (4.16)
which is the same as the result given in [15] withQ=Q3dB.
Physical Interpretation of Quality FactorQ (4.15) can be written in a form that adds physical insight [16, ch. 10, p. 297]. The envelope of the stored energy in the system will have the same functional form as the envelope of the dissipated power. For example, we could consider the signal of interest to be the voltage across a capacitor, in which case the energy stored on this capacitor isE=1=2(CV2), and the envelope of the stored energy can be written as
Eenv(nt)=E0exp
;nt
n =2
; (4.17)
whereE0is the initial stored energy at timet0. Differentiating both sides gives
dEenv(nt) dnt =
;E0 n =2exp
;nt n =2
=
;1
n =2Eenv(nt); (4.18)
which is a first-order differential equation relating the rate of energy dissipation to the total energy stored, from which we observe that
n =2
Eenv(nt)
;dEenv(nt)=dnt
: (4.19)
Therefore the normalized time constantn is twice the ratio of the stored-energy to the energy-lost-per-cycle; substituting for values ofQwe obtain the following physical interpretation for the filter’s quality factor.
Qeq =4
stored energy for thenthcycle
energy lost in thenthcycle
Q3dB =2
stored energy for thenthcycle
energy lost in thenthcycle
(4.20)
Since Qis a constant, the fractional energy-lost-per-cycle is constant and equal to
4=Qeq. Hence for a bandpass filter withQeq =10, the resonator will lose 40% of its stored energy per cycle if no input is applied.
186 Chapter 4
Deviations in Clock Signal Envelope in Terms ofQ Due to energy dissipation in the resonator, the voltage envelope is reduced by 86.5%, and the power envelope is reduced by 98% inQeqclock periods. Statistical analysis shows that the rms envelope deviation for a linear phase filter is1=pQeq. For example, a 3variation in the clock envelope of within 50% requiresQeq 36. However, this result is derived from an ensemble average, and there will be time intervals when the deviation in the clock envelope is significantly worse. The above time domain analysis gives us another means to estimate the selectivity of a BPF needed to meet desired specifications. If we have a requirement that the clock envelope can not drop below 50% of the nominal for
NBconsecutive bits without a transition, then we required
e;NB=n 1=2 =) n NB
ln(2)
; (4.21)
therefore,
Qeq 2NB
ln(2)
=2:89NB'3NB: (4.22)
So as a rule of thumb for an arbitrary BPF, the number of consecutive bits without a transition that can be tolerated before the clock amplitude is cut in half is
NB 'Qeq
3
; (4.23)
andNB 'Qeq=6before the clock power is halved. If all bits are independent and equally likely, then the probability that a sequence ofNBbits will not have a transition
is PNB =Pr [no transition]=2;(NB;1) (4.24)
Therefore, for a given probability, the sequence length is given by
NB =1+;log(PNB)
log(2)
; (4.25)
and the required filter selectivity is therefore
Qeq=2:89;9:6log(PNB): (4.26)
Therefore, a probability of less than10;9 that the clock amplitude will fall below 50% of the nominal value requiresQeq 90. For the same probability that the clock power falls below 50%,Qeqmust be greater than 180. For a 10-Gb/s data signal, the probability of an event of duration 100 ps happening once in ten years of operation is
3:1710;19. This corresponds to a transitionless string of bits of lengthNB =62. The resultingQeqvalues needed are 180 for 50% envelope reduction, and 360 for 50%
Clock Recovery 187
power reduction; theQ3dB values are 283 and 566 respectively. It should be stressed that all of this analysis is approximate because it is assumed that the clock signal was at the nominal value when the string of no transitions started. In reality, shorter strings of data with no transitions will cause the same envelope reduction, provided that the shorter strings occur in rapid succession. This analysis, however, does provide useful information aboutQeqand its relationship between the transition density of data and the amplitude modulation; (4.26) supplements the information derived previously that the rms amplitude and phase modulations are approximately equal to1=pQeq. These results together provide the fundamental guidelines for determining the maximum selectivity of a BPF required to meet a given specification.