Pulse Shaping to Improve Spectral Efficiency

Publish Date: May 07, 2014 | 0 Ratings | 0.00 out of 5 |  PDF

Overview

This tutorial is part of the National Instruments Signal Generator Tutorial series. Each tutorial in this series, will teach you a specific topic of common measurement applications, by explaining the theory and giving practical examples. This tutorial covers how to improve spectral efficiency with pulse shaping. For additional signal generator concepts, refer to the Signal Generator Fundamentals main page.

Table of Contents

  1. Pulse Shaping to Improve Spectral Efficiency
  2. The Sinc-Shaped Pulse
  3.  The Raised Cosine Pulse (Damped Sinc-Shaped Pulse)
  4. Relevant NI products
  5. Buy the Book

1. Pulse Shaping to Improve Spectral Efficiency

Suppose we want to increase transmission speed without reducing accuracy or increasing bandwidth. (As an example, consider trying to develop a faster telephone modem for the Internet. The bandwidth of the telephone line is fixed, so we can't increase the system's bandwidth, and there are minimum requirements for accuracy.) How can we increase transmission speed without reducing accuracy or increasing bandwidth? Let's consider transmitting pulses that require less bandwidth than the rectangular pulse. What pulse shape should we choose? Consider the following requirements for a "good" pulse shape:

  1. We want a "smooth" shape because we know that such a shape has an energy spectrum consisting of lower frequency components (thus minimizing bandwidth).
  2. We need to be sure that the pulse transmitted to represent a particular bit (say, the n th bit) does not interfere at the receiver with the pulse transmitted previously (the pulse for the n – 1st bit) or the pulse to be transmitted next (the pulse for the n + 1st bit). The phenomenon of pulses interfering with each other is known as intersymbol interference (ISI).

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    2. The Sinc-Shaped Pulse

Keeping in mind our two requirements (smooth shape for spectral efficiency and no ISI), consider the pulse shape shown in Figure 3-14. This is the sinc function, which we used extensively in Chapter 2. Note, however, that in Chapter 2 the sinc function was used in the frequency domain, while in Figure 3-14 the sinc function is used in the time domain.


Figure 3-14—A pulse that is sinc-shaped in the time domain.


Figure 3-15—A sinc-shaped time domain pulse has a rectangular-shaped magnitude spectrum.

What is the frequency domain representation of the sinc-shaped pulse in Figure 3-14? Remembering the duality property of the Fourier transform (Equation 2.62), a pulse that is sinc-shaped in the time domain has a rectangular shape magnitude spectrum, as illustrated in Figure 3-15.

Squaring the magnitude spectrum produces the normalized energy spectral density shown in Figure 3-16. All of the pulse's energy lies within the frequency band from 0 to rb /2 Hz, meaning that the sinc-shaped pulse is extremely spectrally efficient. For a sinc shaped pulse,




Figure 3-16—Normalized energy spectral density for sinc-shaped time domain pulse.

Figure 3-16 shows that the sinc-shaped pulse is spectrally compact (our first requirement for a "good" pulse shape). What about avoiding ISI (our second requirement)? Consider the transmitted waveform for a typical data sequence (say, 110100) at a speed of rb bits/sec. As we did with the rectangular pulses, let's transmit a positive sinc-shaped pulse to signify each "1" (see Figure 3-17a) and a negative sinc-shaped pulse to signify each "0" (see Figure 3-17b). The sequence of pulses corresponding to the data sequence "110100" transmitted at rb bits/sec is shown in Figure 3-17c.


Figure 3-17a—A positive sinc-shaped pulse used to represent a "1."


Figure 3-17b—A negative sinc-shaped pulse used to represent a "0."



Figure 3-17c—Sequence of sinc-shaped pulses corresponding to the data sequence "110100."

Let's examine Figure 3-17c. Note the following:

 

 

1. Each pulse is centered within its respective bit period (i.e., the pulse representing the first bit is centered at 0.5/rb, the pulse representing the second bit is centered at 1.5/rb, etc.).
2. The actual signal that will be sent across the channel by the transmitter is the sum of the individual pulses. This signal is shown in Figure 3-17c by a thick, black line marked "total signal."
3. Most of the time the total signal is affected by more than one individual pulse (thus producing ISI), but note that at exactly the center of each bit period, all of the pulses are at 0 volts except for the single pulse representing the corresponding data bit. Thus, at the center of the first bit period, the transmitted signal has an amplitude of A, corresponding only to the amplitude of the first pulse; at the center of the second bit period, the transmitted signal also has an amplitude of A, corresponding only to the amplitude of the second pulse; at the center of the third bit period, the transmitted signal has an amplitude of –A, corresponding only to the amplitude of the third pulse, etc.


This last observation is extremely important: There is no ISI at exactly the center of each bit period. Using this observation and assuming a noiseless, ideal baseband channel with sufficient bandwidth (a bandwidth of at least rb /2 Hz), the receiver can demodulate the transmitted signal by using the following process:

a. Sample the received signal exactly in the center of each bit period. If a particular sample has a value of A volts (or, in fact, any value greater than or equal to 0 volts), then the corresponding data from the source was a "1"; if the sample value is –A volts (or, in fact, any value less than zero volts), then the corresponding data from the source was a "0."
b. Output the demodulated data (i.e., the series of "1"s and "0"s) to the user.


Figures 3-18a and 3-18b illustrate the complete modulation/demodulation process for the data sequence "110100" using sinc-shaped pulses of amplitudes +2 volts and –2 volts, and using a bit rate of 10,000 bits/sec.

Let's closely inspect Figures 3-18a and 3-18b. Note in Figure 3-18a that since the transmission speed is 10,000 bits/sec, the bit period is 100 m sec, or 1/rb (the bit boundaries are shown using dashed vertical lines). The individual pulses are centered within each bit period, and the width of the main lobe of each individual pulse is 200 m sec or 2/rb. Observe also that there is no ISI in the exact center of each bit period: At 50 m sec only the first pulse is nonzero, at 150 m sec only the second pulse is nonzero, etc. Thus, at 50 m sec the transmitted signal is affected by only the value of the first bit from the source, at 150 m sec the transmitted signal is affected by only the value of the second bit from the source, etc. This concept is key to demodulating the received signal, shown in Figure 3-18b. The receiver has advance knowledge of the transmission speed (in this case, 10,000 bits/sec) and thus knows that the bit period is 100 m sec. By sampling the received waveform in the exact center of the first bit period (i.e., at 50 m sec), the receiver obtains a value of +2 volts and therefore knows that the first bit output by the source was a "1." Similarly, sampling at 150 m sec produces +2 volts, thereby indicating that the second bit was also a "1," sampling at 250 m sec produces –2 volts, indicating that the third bit was a "0," etc. It is very important that the receiver be synchronized with the transmitter so that the received signal is sampled exactly in the center of each bit period. We will discuss synchronization in detail later in this chapter and in Chapter 4.



Figure 3-18a—Modulation process for transmitting the data sequence "110100" at 10,000 bits/sec using sinc-shaped pulses. Note that the thick black waveform is the only signal that is transmitted.


Figure 3-18b—Demodulation process for received signal. Receiver samples signal in the center of each bit period (i.e., at 50 m sec, 150 m sec, 250 m sec, etc.). If sampled value is +2 volts, corresponding data is a "1." If sampled value is –2 volts, corresponding data is a "0."

Let's make one more observation: In Figure 3-14, the width of the main lobe of the sinc-shaped pulse was set to 2/rb; this value is used in Figure 3-17 and in the above example. Is this the optimum width for the main lobe of the sinc-shaped pulse? We know that we want a pulse width that produces no ISI but that does produce the lowest bandwidth possible for a given bit rate. Figure 3-19 shows a general sinc-shaped pulse and its corresponding magnitude spectrum. If the main lobe of the pulse has a width of t, then the magnitude spectrum shows a bandwidth of 1/(2t).


Figure 3-19—Relationship between bandwidth and width of main lobe of sinc-shaped pulse.

We thus want to choose a t value as large as possible without producing ISI. As shown in Figure 3-17, choosing t = 2/rb produces no ISI if the received signal is sampled in the center of each bit period. By making a few diagrams, you can see that while certain smaller values of t (such as t = 1/rb) also produce no ISI if the received signal is sampled in the center of the bit period, there are no larger values of t for which this is true. Thus, t = 2/rb represents the optimum width for the main lobe of the sinc-shaped pulse.

Example 3.3Bandwidth and transmission speed using sine-shaped pulses


Suppose an application has accuracy needs that require 100% of the transmitted signal's average power to lie within the bandwidth of the channel. Given an ideal baseband channel, what bandwidth is required to transmit the signal in Figure 3-18?

Solution

From Equation (3.7),

Example 3.4Maximum transmission speed for a fixed-bandwidth system


A baseband channel has a bandwidth of 100 kHz. Accuracy requirements dictate that 100% of the signal's average power must be within the bandwidth of the channel. Determine the maximum transmission speed of the system.

Solution

Again using Equation (3.7),


Example 3.5Comparing the performance of sine-shaped and rectangular pulses


Compare the results of Example 3.4 with Example 3.1.

Solution

Examples 3.1 and 3.4 both use a baseband channel with 100 kHz bandwidth. If accuracy requires that a minimum of 95% of the transmitted signal's average power be within the bandwidth of the channel, then data can be transmitted at a maximum rate of 50,000 bits/sec using rectangular pulses. However, if sinc-shaped pulses are used instead of rectangular pulses, the data can be transmitted four times as fast. Also, since 100% of the signal's average power lies within the channel's bandwidth if sine-shaped pulses are used, the sinc-shaped pulses will produce a received signal with less distortion than the rectangular pulses. We can now see the trade-offs involved in using binary PAM with sinc-shaped pulses rather than rectangular pulses: The sinc-shaped pulses allow a greater transmission speed for a given bandwidth (or a smaller bandwidth for a given transmission speed), but the sinc-shaped pulses require more complex equipment, both at the transmitter (to produce the shape) and at the receiver (to ensure synchronization).


Example 3.6Designing a binary PAM system using sinc-shaped pulses


Consider a transmitter using binary PAM with sinc-shaped pulses of amplitudes 1 volt and –1 volt. Given a transmission speed of 50,000 bits/sec.:

a. Draw the transmitted signal corresponding to the data sequence "10010."
b. If the signal is transmitted across an ideal baseband channel, what is the minimum channel bandwidth required to ensure that the received signal is not distorted?
c. Draw the received signal and indicate how the receiver extracts the digital information for the user.

Solution

 

 

a. The transmitted pulse is shown as the thick, black line in Figure 3-20a.

b.




Figure 3-20a—Transmitted binary PAM waveform for the data sequence "10010" using sinc-shaped pulses at a transmission speed of 50,000 bits/sec.


Figure 3-20b—Demodulation of received binary PAM waveform.


c.
The received signal and the demodulation process are shown in Figure 3-20b. The receiver samples the received signal in the center of each bit period (i.e., at 10 m sec, 30 m sec, 50 m sec, etc.). If a received sample has a value of 1 volt, then the corresponding data delivered to the user is a "1." If a received sample has a value of –1 volt, then the corresponding data delivered to the user is a "0."


Example 3.6 is important because it shows a problem that can occur with binary PAM when we use a sinc-shaped pulse. There is no ISI if the receiver samples the received waveform exactly in the center of each bit period (i.e., at 10 m sec, 30 m sec, 50 m sec, etc.). Let's look at what happens, however, if the receiver is not completely synchronized with the transmitter (this lack of complete synchronization is called timing jitter, or jitter ).

Example 3.7Examining the effects of losing synchronization

Consider the waveform transmitted in Example 3.6: binary PAM with sinc-shaped pulses of amplitudes 1 volt and –1 volt, a transmission speed of 50,000 bits/sec, and a data sequence of "10010." Let's examine the sample at the receiver corresponding to the third bit. If the receiver is perfectly synchronized, it will sample the waveform at exactly 50 m sec (the exact middle of the bit period, where there is no ISI) and the sampled value will be –1 volt. Suppose, however, that the receiver is not perfectly synchronized. Determine the value of the receiver's sample for the following three cases:


a. The receiver is slightly out of synchronization, and sampling occurs at 52 m sec.
b. The receiver is a little farther out of synchronization, and sampling occurs at 54 m sec.
c. The receiver is even farther out of synchronization, and sampling occurs at 56 m sec.


Solution

a.

Figure 3-20a shows the transmitted waveform, which is the sum of five individual pulses corresponding to the five data bits. The pulse representing the first data bit (shown as the thin solid line) is centered in the middle of the first bit period and has zero crossings in the centers of all other bit periods. Mathematically, we can express the first pulse as:

Be sure you understand the reason for each portion of the expression for p1(t ). Similarly, the pulse representing the second bit is:

The negative sign is necessary because the second bit is a "0." The expressions for the third, fourth, and fifth pulses are:

The transmitted waveform is the sum of the individual pulses:

At 52 m sec,

b.
At 54 m sec,

c.
At 56 m sec,


The sampled values at 50, 52, 54, and 56 m sec, which illustrate the received signal's susceptibility to jitter, are illustrated in Figure 3-21. We see that if the receiver is out of synchronization and samples the received signal later than it should (for instance, at 52 m sec instead of 50 m sec), the magnitude of the sampled voltage corresponding to the third bit is reduced. In fact, the farther the receiver is out of synchronization, the closer the third bit's sampled value is to the threshold of zero volts. This is problematic because if the sampled value for the third bit happens to be greater than zero volts, an error will occur: The receiver will demodulate the sample as a "1" instead of a "0." As will be discussed in Chapter 4, achieving and keeping synchronization in the receiver can be difficult and can involve significant equipment complexity.



Figure 3-21—Received waveform's susceptibility to jitter.

In Example 3.7, it appears that the receiver would have to be out of synchronization by at least 6.5 m sec (or 32.5% of a bit period) in order for the third bit to be received in error. The problem becomes worse when we consider that the example assumed no noise and an ideal baseband channel. In practical situations, the received signal will be corrupted by noise and the channel will attenuate the transmitted signal (reduce the signal's amplitude) and possibly distort the transmitted signal. Figure 3-22a shows the same transmitted signal as in Example 3.7 ("10010" transmitted using binary PAM with sinc-shaped pulses at 50,000 bits/sec), but Figure 3-22b shows the received signal after passing through a channel that adds noise and produces losses that attenuate the transmitted signal by 50%. The received signal has been halved (from the attenuation) and "hashed" (from the noise). We are now no longer confident about correctly receiving the third bit if the receiver is out of synchronization by more than approximately 3 m sec.

We have seen that the farther the sampled value of a noiseless signal is from the threshhold (0 volts), the more likely it is that noise will not corrupt the signal badly enough to cause an error. Let's define noise margin as the distance between a noiseless received signal and the threshhold at the instant of sampling. For example, consider the signal shown in Figure 3-21. If this signal is transmitted across an ideal baseband channel (no attenuation occurs), then the noise margin corresponding to the third bit will be 1 volt if the receiver is perfectly synchronized, 0.69 volts if the receiver samples the signal 2 m sec late, 0.36 volts if the receiver samples the signal 4 m sec late, etc. On the other hand, if the signal in Figure 3-21 is transmitted across a channel that causes a 50% attenuation, the noise margin corresponding to the third bit will be 0.5 volts if the receiver is perfectly synchronized, 0.34 volts if the receiver samples the signal 2 m sec late, 0.18 volts if the receiver samples the signal 4 m sec late, etc.

How can we increase noise margin in the presence of jitter (and thus reduce susceptibility to noise when the receiver is not perfectly synchronized)? One way is to increase the amplitude of the transmitted signal (which, of course, requires an increase in the power of the transmitted signal).



Figure 3-22a—Transmitted signal for "10010" using sinc-shaped pulses.


Figure 3-22b—Received signal—the channel is attenuating the transmitted signal levels by 50% and corrupting the signal with noise.

Thus, if we can increase the power of the transmitted signal, we can increase the accuracy of the received signal. Such a trade-off, however, may not be desirable or even feasible in many applications. (For example, increasing transmitted power in a cell phone reduces the operating time until the battery needs recharging and also causes increased interference to other users.)

Another approach to reducing susceptibility to jitter is to use a different pulse shape. Figure 3-23a shows a binary PAM signal representing the same data pattern and transmitted at the same speed as the signal shown in Figure 3-22a, except that rectangular-shaped pulses are used instead of sinc-shaped pulses. In Figure 3-23b, the signal has been passed through a channel adding the same amount of noise and producing the same amount of attenuation as in the signal shown in Figure 3-22b. In comparing Figures 3-22a and b with Figures 3-23a and b, note that transmitting with rectangular pulses rather than sinc-shaped pulses produces a signal with a constant noise margin, even if the receiver is out of synchronization by as much as ±49% of one bit period.


Figure 3-23a—Transmitted signal for "10010" using rectangular pulses.


Figure 3-23b—Received signal with channel attenuation and noise.

We've now seen that transmitting binary PAM with rectangular-shaped pulses requires much more bandwidth than does transmitting with sinc-shaped pulses, but transmitting with rectangular pulses produces a signal that is much less susceptible to timing jitter at the receiver. The performance-versus-cost parameters of our observations are listed in Table 3-2.



Is there a compromise between the two pulse shapes? Can we find a pulse shape that produces low bandwidth (like the sinc pulse) while providing good protection against timing jitter (like the rectangular pulse)? To find out, let's begin by asking, "What are the physical characteristics of the sinc pulse that cause it to be so susceptible to jitter?" Let's revisit Figure 3-20a (in Example 3.6) and examine what is happening to the transmitted waveform between t = 50 m sec and t = 60 m sec.

As we discussed earlier in this section, at t = 50 m sec all pulses are zero except for the third pulse, which has a value of –1 volt. Thus, at
t = 50 m sec, the transmitted waveform has a value of –1 volt and there is no ISI. At t = 52 m sec, the third pulse has a value of –0.98 volts (still a large negative number), but the values of the first, second, fourth, and fifth pulses are no longer zero. In fact, as we established in Example 3.7, at t = 52 m sec the value of the first pulse is 0.05 volts, the value of the second pulse is 0.09 volts, the value of the fourth pulse is 0.11 volts, and the value of the fifth pulse is 0.05 volts. Thus, the value of the transmitted signal at t = 52 m sec is 0.05 + 0.09 – 0.98 + 0.11 + 0.05 = –0.68 volts.

The ISI from the first, second, fourth, and fifth pulses erodes the noise margin by 0.3 volts, and the situation worsens as t approaches 60 m sec. How can we reduce the amount of ISI? In looking back at Figure 3-20a, we see that we can reduce the ISI from the fourth pulse by making its main lobe thinner, and we can reduce the ISI from the first, second, and fifth pulses by making their tails flatter. In other words, to produce low bandwidth but reduce susceptibility to timing jitter, we should use a pulse that is smooth like a sinc-shaped pulse but that has a narrower main lobe and flatter tails than a sinc pulse.

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3.  The Raised Cosine Pulse (Damped Sinc-Shaped Pulse)


How can we produce a pulse that is smooth like a sinc pulse but has a narrower main lobe and flatter tails? Consider the waveform in Figure 3-24, which is a sinc-shaped pulse multiplied by a damping factor.


Figure 3-24—A damped sinc-shaped pulse.


Figure 3-25 shows the new pulse for various damping factors, from b = 0 (no damping at all) to b = rb /2. Note that the larger the value of b, the narrower the main lobe and the flatter the tails of the pulse. Thus, the larger the value of b, the less the effects of ISI and thus the less susceptible the transmitted signal will be to timing jitter.

What is the trade-off for this reduced susceptibility to jitter? Figure 3-26 shows the magnitude spectrum of the damped sinc-shaped pulse. Note that the bandwidth of the pulse is

Thus, the greater the value of b, the greater the bandwidth of the transmitted signal but the less susceptible the transmitted signal is to timing jitter.

The Fourier transform of the damped sinc-shaped pulse is


Figure 3-25a—Damped sinc-shaped pulse with damping factor b = 0.


Figure 3-25b—Damped sinc-shaped pulse with damping factor .


Figure 3-25c—Damped sinc-shaped pulse with damping factor .


Figure 3-25d—Damped sine-shaped pulse with damping factor .


Figure 3-25e—Damped sinc-shaped pulse with damping factor .

Figure 3-26—A damped sinc-shaped time domain pulse has a raised-cosine-shaped magnitude spectrum.


We call this particular shape a raised cosine. In examining Figure 3-26 and Equation (3.9), note that the raised cosine has the following properties:


1. The raised cosine has a constant value of A/rb for |f | £ (rb /2) – b and a constant value of zero for |f | ³ (rb /2) +b.
2. In the region between these two constants (i.e., for (rb /2) – b < |f |< (rb /2) +b ), the waveform is odd-symmetric about the points |f | = rb /2.
3. b = 0 corresponds to a rectangular shape in the frequency domain and a sinc-shaped pulse in the time domain.
4. Bandwidth = (rb /2) +b, so larger values of b require larger bandwidth. Note from Figure 3-25 that larger values of b provide more damping (i.e., flatter tails and a narrower main lobe) for the corresponding time domain waveform, thus reducing ISI. Choosing a value for b therefore involves trading off increased bandwidth for reduced susceptibility to timing jitter (we'll illustrate this point in Examples 3.8 and 3.9).



The fourth observation is especially important because it allows us to optimize system performance for particular applications. Often, we are interested in expressing the additional bandwidth involved in the trade-off not in terms of Hz, but rather as a percentage of the minimum possible bandwidth, rb /2. With this in mind, let's define a quantity called rolloff factor (a):


Since b varies between 0 and rb /2, a varies between 0 and 1. We can now express the bandwidth as

Using rolloff factor (a) instead of b allows us to express the trade-off of additional bandwidth for less susceptibility to jitter in a manner that is independent of transmission speed.

So far, we've investigated binary PAM using three different pulse shapes: rectangular pulses, sinc-shaped pulses, and damped sinc-shaped pulses. General practice in industry and the profession is to describe the damped sinc-shaped pulse by the name of its frequency domain shape rather than its time domain shape. The shape is thus called a raised cosine pulse, and the term raised cosine pulse shaping refers to the pulse that looks like a raised cosine in the frequency domain but a damped sinc shape in the time domain. Note that this is not consistent with the terminology for the other pulse shapes (which we call rectangular and sinc shaped, based on their time domain shapes), but it has become standard practice.

We can use the following notation to mathematically describe each of the pulse shapes:


p (t ) = time domain representation of a pulse
P(f ) = frequency domain representation of a pulse
t = width of the pulse in the time domain. For the rectangular pulse, tois the span of all nonzero values. For the sinc-shaped and raised cosine pulse shapes,
tois the width of the main lobe of the pulse (the portion of the pulse between the first zero crossing to the left of the y axis and the first zero crossing to the right of the y axis).


We know that the optimum pulse width for the rectangular pulse is t = 1/rb , and the optimum pulse width for sinc-shaped and raised cosine pulses is t = 2/rb. We're now ready to work some examples illustrating how different rolloff factors affect the characteristics of the raised cosine pulse.

Example 3.8Plotting the raised cosine pulse for various values of a


Using one set of axes, plot the time domain representation of the raised cosine pulse for each of the following rolloff factors. Pulse width should be set to the optimum value for a transmission speed of 100,000 bits/sec, and each plot should extend from at least
t = –3/rb, to t = 3/rb,


a. a = 0
b. a = 0.25
c. a = 0.5
d. a = 0.75
e. a = 1


Solution


Plots for the five rolloff factors are shown in Figure 3-27.

Figure 3-27—A comparison of the time domain representation of the raised cosine pulse for five different rolloff factors.


Example 3.9Plotting PAM waveforms with raised cosine pulses


A source outputs data at the rate of 50,000 bits/sec. The transmitter uses binary PAM with raised cosine pulse shaping of optimum pulse width. Each pulse has a maximum amplitude of 1 volt. For each of the five rolloff factors in Example 3.8:


1. Draw the transmitted waveform for the data pattern "10010."
2. Determine the bandwidth of the transmitted waveform.


Solution

1.

2.

a. Bandwidth = 25,000(1 + 0) = 25 kHz
b. Bandwidth = 25,000(1 + 0.25) = 31.25 kHz
c. Bandwidth = 25,000(1 + 0.5) = 37.5 kHz
d. Bandwidth = 25,000(1 + 0.75) = 43.75 kHz
e. Bandwidth = 25,000(1 + 1) = 50 kHz


Figure 3-28—Transmitted binary PAM waveform for the data sequence "10010" using sinc-shaped pulses (raised cosine pulse shaping, a = 0) at a transmission speed of 50,000 bits/sec. Note that this is the same plot as in Figure 3-20a.


Figure 3-29—"10010" with raised cosine pulse shaping, a = 0.25.


Figure 3-30—"10010" with raised cosine pulse shaping, a = 0.5.



Figure 3-31—"10010" with raised cosine pulse shaping, a = 0.75.


Figure 3-32—"10010" with raised cosine pulse shaping, a = 1.


Example 3.10Susceptibility to jitter using raised cosine pulses

 

1. Consider the data pattern "10010" transmitted at 50,000 bits/sec using PAM with raised cosine pulse shaping of optimum width and a rolloff factor of a = 0. Pulse amplitudes are 1 volt and –1 volt. If the signal is transmitted over a channel that produces a 50% attenuation in voltage, determine the noise margin of the received signal corresponding to the third bit for each of the following cases:

  1. Receiver is perfectly synchronized.
  2. Receiver is out of synchronization by +2 msec (i.e., receiver is sampling the signal 2 msec later than the optimum time).
  3. Receiver is out of synchronization by +4 msec. (i.e., receiver is sampling the signal 4 msec later than the optimum time).
  4. Receiver is out of svnchronization by +6 msec (i.e., receiver is sampling the signal 6 msec later than the optimum time).
  5. Receiver is out of synchronization by more than +6 msec; what happens to the demodulated bit?


2. Repeat Part 1, except use a rolloff factor of a = 0.5.

3. Repeat Part 1, except use a rolloff factor of a = 1.

Solution

Note that the data pattern and transmission speed in this example are the same as in Example 3.9. Therefore, the transmitted waveforms corresponding to a = 0, a = 0.5, and a = 1 are the same as those shown in Figures 3-28, 3-30, and 3-32, respectively. Fifty percent attenuation means that the signal at the receiver will have half the amplitude of the transmitted signal, so the received signals for a = 0, a = 0.5, and a = 1 are shown in Figures 3-33a, b, and c, respectively. Table 3-3 shows the noise margins at the receiver for the third bit if the receiver is perfectly synchronized, if the receiver is out of synchronization by +2 msec, if the receiver is out of synchronization by +4 msec, and if the receiver is out of synchronization by +6 msec. Table 3-3 shows that increasing the rolloff factor significantly reduces the effects of the receiver being out of synchronization (but at the cost of additional bandwidth). This concept is also illustrated in Figure 3-33d.


Table 3-3—Noise Margin for Third Bit of Received Signal Using Raised Cosine Pulse Shaping with Various Rolloff Factors

 

Receiver Synchronization Rlloff Factor (in Volts)
a = –0——a =–0.5—–a =–1
Perfect 0.50———0.50———0.50
+2 msec out of sync 0.34———0.40———0.46
+4 msec out of sync 0.18———0.28———0.38
+6 msec out of sync 0.014——–0.14———0.28


If the receiver is too far out of synchronization, the noise margin is completely eliminated and the probability that the bit will be demodulated in error is very high. In this example, if a = 0 and the receiver is out of synchronization by more than 6 msec, the noiseless sampled value for the third bit will be positive, and hence in error, even in the absence of noise. For a = 0.5, the receiver can be out of synchronization by 8 msec before a similar point is reached, and for a = 1, the receiver can be out of synchronization by almost 10 msec before a similar point is reached.

We now have the capability to quantify and analyze many of the performance-versus-cost trade-offs involved in designing a communication system for a baseband channel. Table 3-4 summarizes some comparisons concerning the three different baseband binary PAM techniques. Review Table 3-4, asking yourself how each of the performance-versus-cost parameters (equipment complexity, bandwidth, transmitted signal power, transmission speed, accuracy, and reliability) are quantitatively involved in determining the optimum pulse shape for a particular application.


Figure 3-33a—Received signal, a = 0, bandwidth = 25 kHz.


Figure 3-33b—Received signal, a = 0.5, bandwidth = 37.5 kHz.


Figure 3-33c—Received signal, a = 1, bandwidth = 50 kHz.



Figure 3-33d—Illustrating the influence of rolloff factor on noise margin.


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4. Relevant NI products


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For the complete list of tutorials, return to the NI Signal Generator Fundamentals Main page.

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5. Buy the Book

Publication Information
Author: Harold P.E. Stern and Samy A. Mahmoud Book: Communication Systems: Analysis and Design
Copyright: 2004 ISBN: 0-13-040268-0

 

Purchase Communication Systems: Analysis and Design from Prentice Hall

 

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Excerpt from the book published by Prentice Hall Professional (http://www.phptr.com). Copyright Prentice Hall Inc. 2006. All rights reserved.

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