It was determined in the previous section that the PSD of NRZ data has a null at the bit-rate, but clearly, if bits are being transmitted at a rate ofBT, then there must be information about the bit-rate contained within the data signal itself. Since any linear operation on the data will always contain a null at the bit-rate, it is appropriate to consider what types of nonlinear operation can be performed on the data that will generate a tone at eitherBT or a multiple ofBT. One such nonlinear operation is edge-detection. Since the data does not return to zero in the bit-interval, there is no discernible timing information contained in the data, unless adjacent bits are different.
Therefore the timing information is contained in the transitions between adjacent, non-identical, bits. However, since this transition is equally likely to be positive as negative, at any timet=nT for[n=1;2;3:::], this random phase reversal, as was illustrated in Fig. 2.6, prevents the accumulation of a steady-state resonance if the data signal were applied directly to the input of a bandpass filter. It was found that a residual signal at half the bit-rate exists simply due the the random placement of bits in the proper phase, but this frequency component is small and indistinguishable from all of the other frequency components in the data signal. Noticing that the random phase reversals are what prevents a resonant circuit atBT=2from sustaining oscillation, we can devise an operation that removes the random phase reversals, and a signal with a strong clock component can be derived. Consider generating a positive pulse for a fraction of the bit period every time that a transition of the data is encountered.
This operation is illustrated in Fig. 2.7. It can be seen that the edge-detected pulses
eN(t;)are always in-phase with a resonant signal at the bit-rate;eN(t;)could be
used as the input to a resonant circuit, and a sustained oscillation at a frequency ofBT
would result. Therefore, one would expect to find a strong component in the frequency domain description ofeN(t;)atBT. The task at hand is to determine the PSD of
eN(t;)and predict the power in the derived clock signal.
Functional Form of the Edge-Detected Signal The signaleN(t;)can be expressed in the same manner as the NRZ data. The fundamental pulse shapeeT(t)is shown in
40 Chapter 2
eT(t)
τp
1
t T
Figure 2.8 Pulse shape for edge-detected data
Fig. 2.8, where
eT(t)=
1 for0tp
0 elsewhere. (2.47)
A time limited signal derived from an N-bit data sequence is then given by
e(t;)=NX,1
n=0sn()eT(t,nT); (2.48)
wheresn()is a random variable derived from two adjacent data bits. If the data bits are identical thensn()=0, andsn()=1when the adjacent bits are different. It is clear that adjacent data bits are equally likely to be identical as they are different so
that sn()=
1 with Probability 1/2
0 with Probability 1/2 (2.49)
The random variablesn()can be written in terms of a random variableqn().
sn()=1=2(1+qn()); (2.50)
whereqn() has statistics that are identical with the original data polarity random variablern().
qn()=
+1 with Probability 1/2
,1 with Probability 1/2, (2.51)
and
E[qn()]=0 (2.52)
E[qn()qk()]=
1 forn=k
0 forn6=k. (2.53)
Therefore, e(t;)can be written as the sum of a deterministic signal and a random signal. This decomposition into two parts is illustrated in Fig. 2.9, and the signal can be expressed analytically as,
e(t;)=1
2
NX,1
n=0eT(t,nT)+1
2
NX,1
n=0qn()eT(t,nT): (2.54)
Mathematical Preliminaries 41
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
0 1
0 1/2
0 1/2 -1/2 e(t, )
eD(t) eR(t, ) .
.
Figure 2.9 Decomposition of a edge-detected NRZ random signal into a deterministic and random parts.
The signaleN(t;)is now derived by repeatinge(t;)indefinitely.
Power Spectrum of the Edge-Detected Signal SinceeN(t;)is periodic with a period ofNT, it can be represented by a Fourier series with a fundamental frequency of
BT=N. Obtaining the Fourier series coefficients is a linear operation, so we can utilize superposition to determine the coefficients due to the deterministic part and random parts of the signal separately. We can write
eN(t;)=eND(t)+eNR(t;); (2.55)
and the deterministic portion can be written as
eND(t)= aD0
2 +
1
X
M=1aDMcos
2Mt T
+ 1
X
M=1bDMsin
2Mt T
: (2.56)
EvaluatedaDMdirectly from (2.6)
aDM =T2
Z T
0 1
2
eT(t)cos
2Mt T
dt (2.57a)
= 1
T
Z p
0 cos
2Mt T
dt (2.57b)
= 1
2M sin
2Mp
T
(2.57c)
= 1
M sin
Mp
T
cos
Mp
T
; (2.57d)
and defining a pulse-width parameter
p=4p=T; (2.58)
42 Chapter 2
Normalized Power
Normalized Frequency (f / B ) T 0
0.02 0.04 0.06 0.08 0.1 0.12 0.14
0 1 2 3 4 5
Pulse-Width = 0.5T
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14
0 1 2 3 4 5
Normalized Power
Normalized Frequency (f / B ) T
Pulse-Width = 0.4T
(a) (b)
Figure 2.10 The power in harmonics of the deterministic portion of an edge-detected NRZ data signal: (a)p=0:5, (b)p=0:4. The dotted line is the sinc function envelope that is controlled by the shape of the edge-detected pulses.
the Fourier coefficientsaDMfor the deterministic part are
aDM =psinc(Mp)cos (Mp): (2.59a)
The quadrature coefficientsbDMare found similarly to be
bDM=psinc(Mp)sin(Mp): (2.59b)
Clearly the power in each harmonic for the deterministic portion is
PDM=
(p2
4
forM =0
p2
2
sinc2(Mp) forM 6=0 (2.60)
where the frequency increment between harmonics is nowBT instead ofBT=Nas in
the previous section. Comparing the average power in the time and frequency domains,
1
X
M=0PDM= p
4
"
1
X
M=1
2psinc2(Mp)+p
#
=
p
4
: (2.61)
This is equivalent to the time-domain result obtained by averaging the square of the deterministic pulse over one period T, which has a magnitude of1=2over the time interval[0;pT]. The average power in the harmonics is plotted in Fig. 2.10 forp=0:5
andp=0:4. Notice that particular harmonics of the bit-rate can be nulled by choosing the value of p appropriately. Forp=1=2:5=0:4, there is a spectral null at multiples of2:5BTso that every5thharmonic is zero. For the case ofp=1=2, all even harmonic are nulled.
Mathematical Preliminaries 43
φ0 φ0
modN(m) = 0 modN(m) = 0 5 vectors FOR N = 5
φ0
φ0 φ0
4 vectors 2 vectors
2 vectors
modN(m) = 0 modN(m) = 0 modN(m) = 0 m even
FOR N = 4
m odd
Figure 2.11 Graphical illustration showing the sum of cosines and sines as projects onto
“x” and “y” axes of unit magnitude vectors.
We now want to add these deterministic Fourier coefficients with those obtained from the random portion of the signal. A problem arises in that the the random portion has a period that is N-times longer than the deterministic part. This can be fixed by stuffing zeros in the coefficients for the deterministic part at frequencies that are not a multiple ofBT. Alternatively we could have obtained this result directly by using a period of
NT instead ofT. In this case the Fourier series coefficients for the deterministic part are
aDm= p Nsinc
pm N
NX,1 n=0
cos
2m
N (n+p=2)
(2.62a)
bDm= p Nsinc
pm N
NX,1 n=0
sin
2m
N (n+p=2)
(2.62b) The two summation terms in (2.62) are the sums of the projections onto the “x” and
“y” axes respectively ofNequally spaced vectors around the unit circle with an initial phase offset of0(m)=pm=N radians. This sum can be illustrated graphically, as in Fig. 2.11. Since the total phase between successive vectors is2m=N, each new
vector is traced by traversing the unit circle Integer(m=N)times, and then adding a phase increment of2times the remainder ofm=N. For the case of N-odd, there are always N equally spaced vectors, and for N-even, the vectors continue to double-up, depending on whether N is divisible by a power of 2. In both cases the distribution of the vectors is symmetric, so that if these vectors were the spokes of a bicycle wheel that were balanced horizontally on its hub, then hanging an equal weight from from each of the spokes will keep the wheel in a level position. Due to this symmetry, the vector sum is equal to zero in these cases. If the spokes were distributed in an unsymmetrical way, then the wheel would tip over in the direction of the vector sum of the spokes. For the above summations, only in the case wheremis a multiple ofN (modN(m)=0)
do all the vectors line up on the offset angle and their sum accumulates to a non-zero
44 Chapter 2
value. Therefore the Fourier coefficients are given by
aDm=
Npsinc
,pmN
Ncos,pmN for modN(m)=0
0 for modN(m)6=0 (2.63a)
bDm=
Npsinc
,pmN
Nsin,pmN for modN(m)=0
0 for modN(m)6=0. (2.63b)
Clearly the average powerPDmis the same asPDMgiven in (2.60), whereM =m=N.
Now we need to determine the Fourier coefficients for the random portion ofeN(t;).
This will be simple, however, since the form of the signal is very similar to the form of the random NRZ data itself. Recall thateNR(t;)is given by,
eNR(t;)=1
2
NX,1
n=0qn()eT(t,nT): (2.64)
whereas the data itself is
dN(t;)=NX,1
n=0rn()pT(t,nT): (2.65)
It was shown in (2.11) that the Fourier coefficientam()for NRZ data is
am()= NT2
NX,1 n=0rn()
Z
(n+1)T nT cos
2mt NT
dt; (2.66)
whereasaRm()for the random part of the edge detected signal is
aRm()=NT1
NX,1 n=0qn()
Z
(n+p)T nT cos
2mt NT
dt: (2.67)
It was also shown in (2.13e) thatam()could be simplified to
am()= 2sin() m
NX,1
n=0rn()cos (n+); (2.68)
where = m=N. A similar result can be obtained foraRm(). Defining p =4
pm=N, then
aRm()= sin(p)
m
NX,1
n=0qn()cos (n+p): (2.69)
Mathematical Preliminaries 45
Therefore the expressions for the Fourier coefficient ofeNR(t;)are aRm()= p
Nsinc(pm=N)NX,1
n=0qn()cos2m
N (n+p=2): (2.70a) bRm()= p
Nsinc(pm=N)NX,1
n=0qn()sin
2m
N (n+p=2)
; (2.70b)
and sinceqn()has identical statistics ofrn(), then the expected value of the time averaged power is
PRm =
(p2
4N form=0
p2
2Nsinc2(pm=N) form6=0. (2.71)
Now we have derived the power in the deterministic and random parts. All that remains is to find the average power in the total signal. From superposition we know that
am()=aDm+aRm()
bm()=bDm+bRm(): (2.72)
The time averaged power is then
Pm()= 1
2
(a2Dm+2aDmaRm()+a2Rm())
+ 1
2
(b2Dm+2bDmbRm()+b2Rm()): (2.73)
Taking the expected value gives
Pm=PDm+PRm+aDmE[aRm()]+bDmE[bRm()];
and since the expected values ofaRm()andbRm()are zero, then the total power is just the sum of the two individual powers,
Pm=PDm+PRm: (2.74)
Therefore, the expected value of the time averaged power per harmonic of the edge detected signal is
Pm=
8
>
>
<
>
>
:
p2
4
[1+1=N] form=0
p2
2
[1=N]sinc2(pm=N) form6=0, modN(m)6=0
p2
2
[1+1=N]sinc2(pm=N) form6=0 modN(m)=0.
(2.75)
46 Chapter 2
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14
0 1 2 3 4 5
Normalized Power
Normalized Frequency (f / B ) T
Pulse-Width = 0.5T
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14
0 1 2 3 4 5
Normalized Power
Normalized Frequency (f / B ) T
Pulse-Width = 0.4T
(a) (b)
Figure 2.12 The power in harmonics of an edge-detected NRZ data signal for: (a)p=0:5, (b)p=0:4.
Power in Bandwidth B /32 (dB)
Normalized Frequency (f / B ) T
T
-100 -80 -60 -40 -20 0
0 1 2 3 4 5 6 7 8 9 10
Power in Bandwidth B /32 (dB)
Normalized Frequency (f / B ) T
T
-100 -80 -60 -40 -20 0
0 1 2 3 4 5 6 7 8 9 10
(a) (b)
Figure 2.13 Simulated and calculated power in harmonics of an edge-detected NRZ data signal for: (a)p=0:5, (b)p=0:3438.
This power spectrum is plotted in Fig. 2.12 forN =16. It can be seen that the large spikes at multiples of the bit-rate are due to the deterministic part, and the power in the random part is spread more uniformly over all frequencies. This analytical expression can be verified in simulation. A discrete-time rectangular NRZ data sequence was generated using a a sampling interval of 32 samples-per-bit. A pulse of widthpT was
generated whenever a transition in the data occurred. A Discrete Fourier Transform DFT was taken from the edge-detected data. Since the frequency interval for the DFT isf =BT=32, then the PSD from the simulated data was compared to the calculated value forN =32. The results are plotted in Fig. 2.13, where it can be seen that the simulated value is coincident with the calculated value. This result is consistent with expectations based on arguments about resonant circuits. Once the random phase
Mathematical Preliminaries 47
reversals have been removed from the data, a sustained oscillation can appear at the output of a bandpass filter tuned to the data-rate. The edge-detected signal must, therefore, have a strong spectral component atBT. This was indeed found to be true, and it was also found that by varying the pulse-widthpT, the relative magnitudes of the harmonics of the clock could be altered. Specifically, forp=1=2, all even clock harmonics are nulled.