A rectangular NRZ random data stream d(t;) of length N-bits has an analytical representation given in (2.1). A pseudo-random data sequence dN(t;) can be
Mathematical Preliminaries 31
generated fromd(t;)by repeating the signal every N bits. SincedN(t;)is periodic with a period ofNT, it can be represented by a Fourier series of the form
dN(t;)=a0()
2 +
1
X
m=1am()cos
2mt NT
+ 1
X
m=1bm()sin
2mt NT
: (2.5)
The coefficients of the Fourier series are random variables and can be extracted from the original signal. Since all harmonics of the fundamental frequency are mutually orthogonal when integrated over the periodNT, random spectral coefficients are determined according to
am()=NT2
Z NT
0
dN(t;)cos
2mt NT
dt (2.6)
bm()=NT2
Z NT
0
dN(t;)sin
2mt NT
dt: (2.7)
These coefficients of the Fourier series can be considered as “dot-products,” or equiv- alently, projections of the data signal onto the orthogonal basis functions. Since the cosine and sine are quadrature signals, they are also mutually orthogonal, and both must be included in the Fourier series expansion, with the relative magnitudes of the coefficientsam()andbm()determining the phase. In the analysis that follows both
am()andbm()will be evaluated directly from the above definitions, and the inter- pretation of the result will be clear. Later the complex form of the Fourier series and negative frequencies will be introduced for analytical convenience.
The process of finding the power spectral density of the random data begins by evalu- atingam()directly. Applying the definition,
am()= NT2
Z NT
0
NX,1
n=0rn()pT(t,nT)cos
2mt NT
dt; (2.8)
interchanging the order of integration and summation,
am()= NT2
NX,1 n=0rn()
Z NT
0
pT(t,nT)cos
2mt NT
dt; (2.9)
and recalling that the pulsepT(t)is rectangular, such that
pT(t,nT)=
1 for nT t(n+1)T
0 elsewhere, (2.10)
am()can now be expressed as the sum of integrals
am()= NT2
NX,1 n=0rn()
Z
(n+1)T nT cos
2mt NT
dt: (2.11)
32 Chapter 2
Evaluating the integrals
am()= m1
NX,1 n=0rn()
sin
2m(n+1) N
,sin
2mn N
: (2.12)
To facilitate manipulation of the sinusoids we define
n4=2mn
N and, =4 m
N :
The result in (2.12) can now be simplified. Leaving in all of the intermediate steps,
am()= m1
NX,1
n=0rn()=ejnej2,ejn (2.13a) am()= m1
NX,1
n=0rn()=ejejnej,e,j (2.13b) am()= m1
NX,1
n=0rn()=ejejn[2jsin()] (2.13c) am()= 2sin()
m
NX,1
n=0rn()=jejejn (2.13d) am()= 2sin()
m
NX,1
n=0rn()cos(n+): (2.13e)
Using the definition of the sinc function
sinc(x)4=sin(x) x ;
then the Fourier series coefficientsam()are given by
am()=N2sinc(m=N)NX,1
n=0rn()cos
2m
N (n+1=2)
: (2.14)
The random coefficientsbm()can be found in a similar manner.
bm()=NT2
Z NT
0
NX,1
n=0rn()pT(t,nT)sin
2mt NT
dt (2.15)
Mathematical Preliminaries 33
This too can be expressed as the sum of integrals
bm()= NT2
NX,1 n=0rn()
Z
(n+1)T nT sin
2mt NT
dt: (2.16)
The result of the integration is
bm()=,m1
NX,1 n=0rn()
cos
2m(n+1) N
,cos
2mn N
: (2.17)
From (2.13) it can be seen thatbm()can be expressed similarly,
bm()=,2sin() m
NX,1
n=0rn()<jejejn ; (2.18)
and after simplifying
bm()=N2sinc(m=N)NX,1
n=0rn()sin
2m
N (n+1=2)
: (2.19)
The pseudo-random rectangular NRZ data stream has now been represented by a Fourier series expansion, where the coefficients given in (2.14) and (2.19) are random variables that depend on the actual data stream. It is desirable to find the average behavior of the data in the frequency domain so that the result would correspond to the output of a spectrum analyzer, averaging several sweeps. Each sweep measures the time-averaged power in a given bandwidth, and the final display is an average over several smaller segments of the complete data signal. To perform this operation analytically, we first need to find the time-averaged power of the random data in a given bandwidth. This power will be a random variable which also depends on the actual data sequence. By averaging over the ensemble of all possible data sequences the statistical average of the time-averaged power2can be determined.
For a deterministic signal of the form,
f(t)=amcos(2fmt)+bmsin(2fmt); (2.20)
2Thus far we have not defined the units ofamandbm. However, if we want to talk about power, then they clearly must have units proportional to
p
Watts. Normally we will consider the signalf(t)to be a either a current or a voltage. Therefore, a resistance must be associated with the coefficients to obtain a power. If we associate a1resistor with each coefficient, then for a voltage signal the units are Volts=
p
1= p
Watts, and for a current the units are Amps
p
1= p
Watts. However, usually we will ignore the1normalization and still talk about the power when the units are actually Amps2or Volts2and not Watts. Although this is a misnomer, we will use the word “power,” when it should be kept in mind that we actually mean the power dissipated in a1resistor.
34 Chapter 2
the time-averaged powerPmis equal to
Pm= a2m+b2m
2
: (2.21)
A periodic deterministic signal can be represented by a Fourier series, such that
g(t)=a0
2 +
1
X
m=1amcos (2fmt)+X1
m=1bmsin(2fmt): (2.22)
Since the basis functions are mutually orthogonal, then the power in themthharmonic
is also given by (2.21), except at dc where the average power is
P0= a20
4
: (2.23)
To facilitate power calculations it is convenient to express the Fourier series coefficients as the real and imaginary parts of a complex number. From (2.14) and (2.19) it can be seen thatam()andbm()can be expressed in the following form
am()=N2sinc(m=N)<
(NX,1
n=0rn()e,je,jn
)
(2.24a)
bm()=,N2sinc(m=N)=
(NX,1
n=0rn()e,je,jn
)
: (2.24b)
Defining a complex Fourier coefficient as
cm()=4 am(),jbm()
2
; (2.25)
then
cm()= N1sinc(m=N)
(NX,1
n=0rn()e,je,jn
)
: (2.26)
The squared magnitude ofcm()is proportional to the average power,
jcm()j2=cm()cm()= am()2+bm()2
4
(2.27) so that
Pm()=
jcm()j2 form=0
2jcm()j2 form6=0. (2.28)
Mathematical Preliminaries 35
Now the magnitude ofcm()can be determined from (2.26).
jcm()j2=
1
Nsinc(m=N)
2NX,1
n=0rn()e,je,jnNX,1
n=0rn()ejejn; (2.29)
and writing this as a double summation we have
jcm()j2=
1
Nsinc(m=N)
2NX,1 n=0
NX,1
k=0rn()rk()ej(n,k): (2.30)
Ensemble Expectations of the Average Power Continuing with the analysis we want to find the ensemble average ofPm(). To do this requires a knowledge of the statistics of the random variablern(). Sincern()represents the polarity of the pulse, it can only take on values of +1 or -1. It will be assumed, unless otherwise specified, that the data is equally likely to be positive as it is negative. Therefore,
rn()=
+1 with Probability 1/2
,1 with Probability 1/2. (2.31)
As a result, the mean of the data is zero,
E[rn()]=0: (2.32)
It is further assumed that all data bits are uncorrelated, so that a knowledge of one, or more bits, gives no information about the value of any other bit. Therefore,
E[rn()rk()]=
1 forn=k
0 forn6=k (2.33)
Now the expected value ofjcm()j2can be determined.
E[jcm()j2]=
N1sinc(m=N)
2NX,1 n=0
NX,1
k=0E[rn()rk()]ej(n,k) (2.34)
The inner sum vanishes for all values ofk6=n, so that the double sum can be replaced by a single sum.
E[jcm()j2]=
1
Nsinc(m=N)
2NX,1 n=0
(1)ej0 (2.35)
= 1
Nsinc2(m=N) (2.36)
36 Chapter 2
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0 0.5 1 1.5 2 2.5
Normalized Power in Bandwidth B /16
Normalized Frequency (f / B ) T
T
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0 0.5 1 1.5 2 2.5
Normalized Cumulative Power
Normalized Frequency (f / B ) T
(a) (b)
Figure 2.5 (a) Power of Fourier series constituent tones forN =16, (b) Cumulative power.
Letting the mean of the of the time-averaged power be defined by
Pm=4E[Pm()];
then
Pm=
8
>
<
>
: 1
N form=0
2
Nsinc2(m=N) form6=0. (2.37) Pm is plotted in Fig. 2.5a for the case ofN =16bits. We recall that this is a plot of the expected value, over all possible periodic pseudo-random sequences with a period of 16 bits, of the time averaged power in each of the harmonics of the Fourier series representation of NRZ data. We can use Parseval’s Theorem that equates the average power in the time and frequency domains to check this result.
Ptime4=E
"
Z NT
0
d2N(t;)dt
#
=1 (2.38)
Pfreq4=N1 +N2
1
X
m=1sinc
2
(m=N)=1; (2.39)
and we see that the expected value of the time-averaged power is the same in both the time and frequency domains as anticipated. The cumulative power of the coefficients is plotted in Fig. 2.5b, where it can be seen that 80% of the expected signal power lies within a bandwidth ofBT=2. Since the frequency increment in the Fourier Series expansion is
f =BT=N=1=NT; (2.40)
Mathematical Preliminaries 37
a spectral-density coefficient can be defined that gives the power in a given harmonic, divided by the frequency spacing;
Pm
f =
T form=0
2Tsinc2(m=N) form6=0. (2.41)
Interpretation of Results
It is interesting to note that the power spectrum of the random NRZ data contains a component at dc, even though the data is balanced around zero, and is equally likely to be positive, as it is to be negative. The emergence of a dc component arises because
Pmwas derived using an ensemble expectation operator which averages the square of one particular sample function of the random process. Therefore, we need a moment to clarify what is actually meant by the presence of a “dc” term. The interpretation of the results is, for every possible combination of sequences N-bits long, there will not always be an equal number of positive and negative pulses (“ones” and “zeros”) in the bit sequence. The average dc value of this sequence will be the difference between the number of “ones” and “zeros”1=0divided by the total bit lengthN.
Naturally, any deterministic, periodic sequence will have a well defined dc value. If the number of positive pulses is equal to the number of negative pulses, then1=0=0, and the dc value is also zero. However, if the sequence is very long, and is broken into several subintervals of lengthNs, then each of these subintervals will have an average value that may not be zero. Therefore it can be considered to have a “dc” component, if the idea of “dc” is interpreted to mean a frequency that varies slower than can be observed in the given measurement interval. For example: if we have a 65.5 KHz clock and a pseudo-random sequence of length,N =216,1, then the sequence will repeat once every second. If the “dc” value is measured once every 0.1 s, squared, and averaged over a one-second interval, then a non-zero result can occur.
Finding the dc Power in the Time-Domain Instead of findingP0as in (2.37) we could do it in the time domain. The average dc power for the sample function corresponding to the random event1is then
P0(1)=
1=0(1) N
2; (2.42)
and the expected value of the dc power for all possible sample functions is
P0=E
"
1=0()
N
2
#
: (2.43)
38 Chapter 2
Data rate = BT
Clock Frequency = BT /2
+ + + + - - - + - - +
Figure 2.6 Random NRZ data and a tone with a frequency of one-half of the bit-rate.
This operation implies that we need to find the difference between the positive and negative pulses for all possible sequences and average the square. This seems an ominous task and it illustrates the power of the using the expectation operator and the statistical correlation between bits.P0can be obtained as follows;
P0=E
"
1
N
NX,1 n=0rn()
#
2
(2.44)
= 1
N2
NX,1 n=0
NX,1
k=0E[rn()rk()] (2.45)
=1=N; (2.46)
which is the same result obtained in the frequency domain. The result also shows that the dc value does approach zero, for equally likely1signals, as the length of the periodNT is increased.
The reason for devoting a lengthy discussion to the power in the dc component is because the expected value of all harmonic power componentsPm have a similar interpretation. The mean of the random Fourier coefficientsam()andbm()are in fact all equal to zero as can be seen from (2.14) and (2.19). Fig. 2.6 illustrates how a tone at half of the bit-rate lines up in phase with the data half of the time and is out of phase with the data the other half. Hence, there is no frequency component at any of the harmonics on average. However, any individual sample function will have bits that line-up in the appropriate order such that there is a residual component at, for instance, the3rd, or the7th, harmonic. When these values are squared and averaged over all possible sample functions, a non-zero average powerPmis obtained.
This has been a rather long journey to confirm the result that was anticipated in Fig. 2.4, that the PSD is a sinc2 function in the ensemble average. In the sections to follow, these results will be considered within a larger framework so that similar results can be obtained much more readily.
Mathematical Preliminaries 39
Data
Clock
+ + + + + + + +
Edge
Figure 2.7 NRZ data, edge-detected data, and a tone at the bit-rate that is in phase- alignment with the edge-detected signal.