5.2 LINE CODES AND THEIR POWER SPECTRA
5.2.2 Power Spectra for Line-Coded Data
It is important to know the spectral occupancy of line-coded data in order to predict the bandwidth requirements for the data transmission system (conversely, given a certain system bandwidth specification, the line code used will imply a certain maximum data rate). We now consider the power spectra for line-coded data assuming that the data source produces a random coin-toss sequence of 1s and 0s, with a binary digit being produced each𝑇 seconds (recall that each binary digit is referred to as a bit, which is a contraction for ‘‘binary digit’’).
Since all waveforms are binary in this chapter, we use𝑇 without the subscript𝑏for the bit period.
To compute the power spectra for line-coded data, we use a result to be derived in Chapter 7, Section 7.3.4, for the autocorrelation function of pulse-train-type signals. While it may be pedagogically unsound to use a result yet to be described, the avenue suggested to the student is to simply accept the result of Section 7.3.4 for now and concentrate on the results to be derived and the system implications of these results. In particular, a pulse-train signal
of the form
𝑋 (𝑡) =
∑∞ 𝑘=−∞
𝑎𝑘𝑝(𝑡 − 𝑘𝑇 − Δ) (5.1)
is considered in Section 7.3.4 where...𝑎−1, 𝑎0, 𝑎1, … , 𝑎𝑘...is a sequence of random variables with the averages
𝑅𝑚=⟨
𝑎𝑘𝑎𝑘+𝑚⟩
𝑚 = 0, ± 1, ± 2, ... (5.2)
The function𝑝(𝑡)is a deterministic pulse-type waveform,𝑇 is the separation between pulses, andΔis a random variable that is independent of the value of𝑎𝑘and uniformly distributed in the interval(−𝑇 ∕2, 𝑇 ∕2). It is shown that the autocorrelation function of such a waveform is
𝑅𝑋(𝜏) =
∑∞ 𝑚=−∞
𝑅𝑚𝑟 (𝜏 − 𝑚𝑇 ) (5.3)
in which
𝑟 (𝜏) = 1 𝑇 ∫
∞
−∞𝑝 (𝑡 + 𝜏) 𝑝 (𝑡) 𝑑𝑡 (5.4)
The power spectral density is the Fourier transform of𝑅𝑋(𝜏), which is 𝑆𝑋(𝑓) =ℑ[
𝑅𝑋(𝜏)]
=ℑ [ ∞
∑
𝑚=−∞
𝑅𝑚𝑟 (𝜏 − 𝑚𝑇 ) ]
=
∑∞ 𝑚=−∞
𝑅𝑚ℑ[𝑟 (𝜏 − 𝑚𝑇 )]
=
∑∞ 𝑚=−∞
𝑅𝑚𝑆𝑟(𝑓) 𝑒−𝑗2𝜋𝑚𝑇𝑓
= 𝑆𝑟(𝑓)
∑∞
𝑚=−∞𝑅𝑚𝑒−𝑗2𝜋𝑚𝑇𝑓 (5.5)
where𝑆𝑟(𝑓) =ℑ[𝑟 (𝜏)]. Noting that𝑟 (𝜏) = 𝑇1∫−∞∞ 𝑝 (𝑡 + 𝜏) 𝑝 (𝑡) 𝑑𝑡 =(1
𝑇
)𝑝 (−𝑡) ∗ 𝑝 (𝑡), we obtain
𝑆𝑟(𝑓) = |𝑃 (𝑓)|2
𝑇 (5.6)
where𝑃 (𝑓) =ℑ[𝑝 (𝑡)].
EXAMPLE 5.1
In this example we apply the above result to find the power spectral density of NRZ. For NRZ, the pulse shape function is𝑝 (𝑡) = Π (𝑡∕𝑇 )so that
𝑃 (𝑓) = 𝑇 sinc (𝑇𝑓) (5.7)
and
𝑆𝑟(𝑓) = 1
𝑇 |𝑇 sinc(𝑇𝑓)|2= 𝑇 sinc2(𝑇𝑓) (5.8)
The time average𝑅𝑚=⟨ 𝑎𝑘𝑎𝑘+𝑚⟩
can be deduced by noting that, for a given pulse, the amplitude is+𝐴 half the time and−𝐴half the time, while, for a sequence of two pulses with a given sign on the first pulse, the second pulse is+𝐴half the time and−𝐴half the time. Thus,
𝑅𝑚=
⎧⎪
⎨⎪
⎩ 1 2 𝐴
2+ 1
2(−𝐴)2= 𝐴2, 𝑚 = 0
1
4 𝐴(𝐴) + 1
4 𝐴(−𝐴) +1
4(−𝐴) 𝐴 +1
4(−𝐴) (−𝐴) = 0, 𝑚≠0 (5.9) Thus, using(5.8)and(5.9)in(5.5), the power spectral density for NRZ is
𝑆NRZ(𝑓) = 𝐴2𝑇 sinc2(𝑇𝑓) (5.10)
This is plotted in Figure 5.3(a) where it is seen that the bandwidth to the first null of the power spectral density is𝐵NRZ= 1∕𝑇 hertz. Note that𝐴 = 1gives unit power as seen from squaring and averaging the time-domain waveform.
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EXAMPLE 5.2
The computation of the power spectral density for split phase differs from that for NRZ only in the spectrum of the pulse-shape function because the coefficients𝑅𝑚are the same as for NRZ. The pulse- shape function for split phase is given by
𝑝 (𝑡) = Π
(𝑡 + 𝑇 ∕4 𝑇 ∕2
)
− Π
(𝑡 − 𝑇 ∕4 𝑇 ∕2
)
(5.11) By applying the time delay and superposition theorems of Fourier transforms, we have
𝑃 (𝑓) = 𝑇2 sinc (𝑇
2 𝑓
)𝑒𝑗2𝜋(𝑇 ∕4)𝑓 − 𝑇 2 sinc
(𝑇 2 𝑓
)𝑒−𝑗2𝜋(𝑇 ∕4)𝑓
= 𝑇2 sinc (𝑇
2 𝑓
) (𝑒𝑗𝜋𝑇𝑓∕2− 𝑒−𝑗𝜋𝑇𝑓∕2)
= 𝑗𝑇 sinc (𝑇
2 𝑓
)sin(𝜋𝑇 2 𝑓
)
(5.12) Thus,
𝑆𝑟(𝑓) = 1 𝑇 ||
||𝑗𝑇 sinc (𝑇
2 𝑓
)sin(𝜋𝑇 2 𝑓
)||||
2
= 𝑇 sinc2 (𝑇
2 𝑓
)sin2(𝜋𝑇 2 𝑓
)
(5.13) Hence, for split phase the power spectral density is
𝑆SP(𝑓) = 𝐴2𝑇 sinc2 (𝑇
2 𝑓
)sin2(𝜋𝑇 2 𝑓
)
(5.14) This is plotted in Figure 5.3(b) where it is seen that the bandwidth to the first null of the power spectral density is𝐵SP= 2∕𝑇 hertz. However, unlike NRZ, split phase has a null at𝑓 = 0, which might have favorable implications if the transmission channel does not pass DC. Note that by squaring the time waveform and averaging the result, it is evident that𝐴 = 1gives unit power.
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EXAMPLE 5.3
In this example, we compute the power spectrum of unipolar RZ, which provides the additional challenge of discrete spectral lines. For unipolar RZ, the data correlation coefficients are
𝑅𝑚=
⎧⎪
⎨⎪
⎩
1 2 𝐴2+ 1
2(0)2= 1
2 𝐴2, 𝑚 = 0 1
4(𝐴) (𝐴) +1
4(𝐴) (0) +1
4(0) (𝐴) + 1
4(0) (0) =1
4 𝐴2, 𝑚≠0
(5.15)
The pulse-shape function is given by
𝑝 (𝑡) = Π (2𝑡∕𝑇 ) (5.16)
Therefore, we have
𝑃 (𝑓) = 𝑇 2 sinc
(𝑇 2 𝑓
)
(5.17) and
𝑆𝑟(𝑓) = 1 𝑇 ||
||𝑇 2 sinc
(𝑇 2 𝑓
)||||
2
= 𝑇4 sinc2 (𝑇
2 𝑓 )
(5.18) For unipolar RZ, we therefore have
𝑆URZ(𝑓) = 𝑇 4 sinc2
(𝑇 2 𝑓
) [1 2 𝐴
2+ 1 4 𝐴
2
∑∞ 𝑚=−∞, 𝑚≠0
𝑒−𝑗2𝜋𝑚𝑇𝑓 ]
= 𝑇4 sinc2 (𝑇
2 𝑓 ) [1
4 𝐴
2+ 1 4 𝐴
2
∑∞ 𝑚=−∞
𝑒−𝑗2𝜋𝑚𝑇𝑓 ]
(5.19) where12𝐴2has been split between the initial term inside the brackets and the summation (which supplies the term for𝑚 = 0in the summation). But from (2.121) we have
∑∞ 𝑚=−∞
𝑒−𝑗2𝜋𝑚𝑇𝑓 =
∑∞ 𝑚=−∞
𝑒𝑗2𝜋𝑚𝑇𝑓 = 1 𝑇
∑∞ 𝑛=−∞
𝛿 (𝑓 − 𝑛∕𝑇 ) (5.20)
Thus,𝑆URZ(𝑓)can be written as 𝑆URZ(𝑓) = 𝑇
4 sinc2 (𝑇
2 𝑓 ) [1
4 𝐴
2+ 1 4𝐴2
𝑇
∑∞ 𝑛=−∞
𝛿 (𝑓 − 𝑛∕𝑇 ) ]
= 𝐴2𝑇 16 sinc2
(𝑇 2 𝑓
)+ 𝐴2
16 𝛿(𝑓) + 𝐴162 sinc2 (1
2 ) [𝛿(
𝑓 − 1 𝑇
)+ 𝛿( 𝑓 + 1
𝑇 )]
+ 𝐴2 16 sinc2
(3 2
) [𝛿( 𝑓 − 3
𝑇 )+ 𝛿(
𝑓 + 3 𝑇
)]+⋯ (5.21)
where the fact that𝑌 (𝑓) 𝛿( 𝑓 − 𝑓𝑛)
= 𝑌( 𝑓𝑛)
𝛿( 𝑓 − 𝑓𝑛)
for𝑌 (𝑓)continuous at𝑓 = 𝑓𝑛has been used to simplify the sinc2
(𝑇 2𝑓)
𝛿 (𝑓 − 𝑛∕𝑇 )terms. [Note that sinc2 (𝑛
2
)= 0for𝑛even.]
The power spectrum of unipolar RZ is plotted in Figure 5.3(c) where it is seen that the bandwidth to the first null of the power spectral density is𝐵URZ= 2∕𝑇 hertz. The reason for the impulses in the spectrum is because the unipolar nature of this waveform is reflected in finite power at DC and harmonics of1∕𝑇hertz. This can be a useful feature for synchronization purposes.
Note that for unit power in unipolar RZ,𝐴 = 2because the average of the time-domain waveform squared is 1
𝑇
[1 2
(𝐴2 𝑇2 + 02 𝑇2)
+1202𝑇]
= 𝐴42.
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EXAMPLE 5.4
The power spectral density of polar RZ is straightforward to compute based on the results for NRZ. The data correlation coeffients are the same as for NRZ. The pulse-shape function is𝑝 (𝑡) = Π (2𝑡∕𝑇 ), the same as for unipolar RZ, so𝑆𝑟(𝑓) = 𝑇4 sinc2
(𝑇 2𝑓)
. Thus, 𝑆PRZ(𝑓) = 𝐴2𝑇
4 sinc2 (𝑇
2 𝑓 )
(5.22) The power spectrum of polar RZ is plotted in Figure 5.3(d) where it is seen that the bandwidth to the first null of the power spectral density is𝐵PRZ= 2∕𝑇 hertz. Unlike polar RZ, there are no discrete spectral lines. Note that by squaring and averaging the time-domain waveform, we get 1
𝑇
(𝐴2 𝑇2 + 02 𝑇2)
= 𝐴22, so 𝐴 =√
2for unit average power.
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EXAMPLE 5.5
The final line code for which we will compute the power spectrum is bipolar RZ. For𝑚 = 0, the possible 𝑎𝑘𝑎𝑘products are𝐴𝐴 = (−𝐴) (−𝐴) = 𝐴2---each of which occurs14the time and(0) (0) = 0which occurs
1
2 the time. For𝑚 = ±1, the possible data sequences are (1, 1), (1, 0), (0, 1), and (0, 0) for which the possible𝑎𝑘𝑎𝑘+1products are−𝐴2,0,0, and0, respectively, each of which occurs with probability14. For 𝑚 > 1the possible products are𝐴2and−𝐴2, each of which occurs with probability18, and±𝐴 (0), and (0) (0), each of which occur with probability14.Thus, the data correlation coefficients become
𝑅𝑚=
⎧⎪
⎪⎨
⎪⎪
⎩
1 4 𝐴2+ 1
4(−𝐴)2+ 1 2(0)2= 1
2 𝐴2, 𝑚 = 0 1
4(−𝐴)2+ 1
4(𝐴) (0) +1
4(0) (𝐴) +1
4(0) (0) = − 𝐴2
4 , 𝑚= ±1 1
8 𝐴
2+ 1 8
(−𝐴2)
+ 14(𝐴) (0) +1
4(−𝐴) (0) +1
4(0) (0) = 0, |𝑚|> 1
(5.23)
The pulse-shape function is
𝑝 (𝑡) = Π (2𝑡∕𝑇 ) (5.24)
Therefore, we have
𝑃 (𝑓) = 𝑇2 sinc (𝑇
2 𝑓 )
(5.25) and
𝑆𝑟(𝑓) = 1 𝑇 ||
||𝑇 2 sinc
(𝑇 2 𝑓
)||||
2
= 𝑇4 sinc2 (𝑇
2 𝑓 )
(5.26)
Therefore, for bipolar RZ we have 𝑆BPRZ(𝑓) = 𝑆𝑟(𝑓)
∑∞ 𝑚=−∞
𝑅𝑚𝑒−𝑗2𝜋𝑚𝑇𝑓
= 𝐴2𝑇 8 sinc2
(𝑇 2 𝑓
) (1 − 1
2 𝑒𝑗2𝜋𝑇𝑓− 1 2 𝑒−𝑗2𝜋𝑇𝑓
)
= 𝐴2𝑇 8 sinc2
(𝑇 2 𝑓
)[1 − cos (2𝜋𝑇𝑓)]
= 𝐴2𝑇 4 sinc2
(𝑇 2 𝑓
)sin2(𝜋𝑇𝑓) (5.27)
which is shown in Figure 5.3(e).
Note that by squaring the time-domain waveform and accounting for it being 0 for the time when logic 0s are sent and it being 0 half the time when logic 1s are sent, we get for the power
1 𝑇
[1 2
(1 2 𝐴2𝑇
2+ 1 2(−𝐴)2𝑇
2 + 02𝑇 2
)+ 1 202𝑇]
= 𝐴2
4 (5.28)
so𝐴 = 2for unit average power.
■ Typical power spectra are shown in Figure 5.3 for all of the data modulation formats shown in Figure 5.2, assuming a random (coin toss) bit sequence. For data formats lacking power spectra with significant frequency content at multiples of the bit rate,1∕𝑇 ,nonlinear operations are required to generate power at a frequency of1∕𝑇 Hz or multiples thereof for symbol synchronization purposes. Note that split phase guarantees at least one zero crossing per bit interval, but requires twice the transmission bandwidth of NRZ. Around 0 Hz, NRZ possesses significant power. Generally, no data format possesses all the desired features listed in Section 5.2.1, and the choice of a particular data format will involve trade-offs.
COMPUTER EXAMPLE 5.1
A MATLAB script file for plotting the power spectra of Figure 5.3 is given below.
% File: c5ce1.m
% clf ANRZ = 1;
T = 1;
f = -40:.005:40;
SNRZ = ANRZˆ2*T*(sinc(T*f)).ˆ2;
areaNRZ = trapz(f, SNRZ) % Area of NRZ spectrum as check ASP = 1;
SSP = ASPˆ2*T*(sinc(T*f/2)).ˆ2.*(sin(pi*T*f/2)).ˆ2;
areaSP = trapz(f, SSP) % Area of split-phase spectrum as check AURZ = 2;
SURZc = AURZˆ2*T/16*(sinc(T*f/2)).ˆ2;
areaRZc = trapz(f, SURZc) fdisc = -40:1:40;
SURZd = zeros(size(fdisc));
SURZd = AURZˆ2/16*(sinc(fdisc/2)).ˆ2;
areaRZ = sum(SURZd)+areaRZc % Area of unipolar return-to-zero spect as check
1 0.5
0–5 – 4 –3 –2 –1 0 1 2 3 4 5
1 0.5
0–5 – 4 SNRZ(f)SSP(f)SURZ(f)SPRZ(f)SBPRZ(f)
–3 –2 –1 0 1 2 3 4 5
1 0.5 0
–5 – 4 –3 –2 –1 0 1 2 3 4 5
1 0.5
0–5 – 4 –3 –2 –1 0 1 2 3 4 5
1 0.5
0–5 – 4 –3 –2 –1 0
T f
2 3 4 5
1
(a)
(b)
(c)
(d)
(e)
Figure 5.3
Power spectra for line-coded binary data formats.
APRZ = sqrt(2);
SPRZ = APRZˆ2*T/4*(sinc(T*f/2)).ˆ2;
areaSPRZ = trapz(f, SPRZ) % Area of polar return-to-zero spectrum as check
ABPRZ = 2;
SBPRZ = ABPRZˆ2*T/4*((sinc(T*f/2)).ˆ2).*(sin(pi*T*f)).ˆ2;
areaBPRZ = trapz(f, SBPRZ) % Area of bipolar return-to-zero spectrum as check
subplot(5,1,1), plot(f, SNRZ), axis([-5, 5, 0, 1]), ylabel(’S N R Z(f)’) subplot(5,1,2), plot(f, SSP), axis([-5, 5, 0, 1]), ylabel(’S S P(f)’) subplot(5,1,3), plot(f, SURZc), axis([-5, 5, 0, 1]), yla- bel(’S U R Z(f)’)
hold on
subplot(5,1,3), stem(fdisc, SURZd, ’ˆ’), axis([-5, 5, 0, 1])
subplot(5,1,4), plot(f, SPRZ), axis([-5, 5, 0, 1]), ylabel(’S P R Z(f)’)
subplot(5,1,5), plot(f, SBPRZ), axis([-5, 5, 0, 1]), xlabel(’Tf’), ylabel(’S B P R Z(f)’)
% End of script file
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