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Introduces a method for representing bits with an analog signal known as binary phase shift keying.

A commonly used example of a signal set consists of pulses that are negatives of each other ( [link] ).

s 0 t A p T t
s 1 t A p T t
Here, we have a baseband signal set suitable for wireline transmission. The entire bit stream b n is represented by a sequence of these signals. Mathematically, the transmitted signal has the form
x t n n 1 b n A p T t n T
and graphically [link] shows what a typical transmitted signal might be.

The upper plot shows how a baseband signal set for transmitting the bit sequence 0110 . The lower one shows an amplitude-modulated variant suitable for wirelesschannels.

This way of representing a bit stream---changing the bit changes the sign of the transmitted signal---is known as binary phase shift keying and abbreviated BPSK. The name comes from concisely expressing this popular way of communicating digital information. The word "binary" is clear enough (one binary-valued quantity is transmitted during a bitinterval). Changing the sign of sinusoid amounts to changing---shifting---the phase by (although we don't have a sinusoid yet). The word "keying" reflects back to the first electrical communication system, which happened to be digital as well: the telegraph.

The datarate R of a digital communication system is how frequently an informationbit is transmitted. In this example it equals the reciprocal of the bit interval: R 1 T . Thus, for a 1 Mbps (megabit per second) transmission, we must have T 1 μs .

The choice of signals to represent bit values is arbitrary to some degree. Clearly, we do not want to choosesignal set members to be the same; we couldn't distinguish bits if we did so. We could also have made the negative-amplitude pulserepresent a 0 and the positive one a 1. This choice is indeed arbitrary and will have no effect on performance assuming the receiver knows which signal represents which bit. As in all communication systems, we designtransmitter and receiver together.

A simple signal set for both wireless and wireline channels amounts to amplitude modulating a baseband signal set (moreappropriate for a wireline channel) by a carrier having a frequency harmonic with the bit interval.

s 0 t A p T t 2 k t T

s 1 t A p T t 2 k t T

What is the value of k in this example?

k 4 .

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This signal set is also known as a BPSK signal set. We'll show later that indeed both signal sets provide identical performancelevels when the signal-to-noise ratios are equal.

Write a formula, in the style of the baseband signal set , for the transmitted signal as shown in the plot of the baseband signal set that emerges when we use this modulated signal.

x t n n 1 b n A p T t n T 2 k t T

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What is the transmission bandwidth of these signal sets? We need only consider the baseband version as the second is anamplitude-modulated version of the first. The bandwidth is determined by the bit sequence. If the bit sequence isconstant—always 0 or always 1—the transmitted signal is a constant, which has zero bandwidth. Theworst-case—bandwidth consuming—bit sequence is the alternating one shown in [link] . In this case, the transmitted signal is a square wave having a period of 2 T .

Here we show the transmitted waveform corresponding to an alternating bit sequence.

From our work in Fourier series, we know that this signal's spectrum contains odd-harmonics of the fundamental, which hereequals 1 2 T . Thus, strictly speaking, the signal's bandwidth is infinite. In practical terms, we use the 90%-power bandwidth toassess the effective range of frequencies consumed by the signal. The first and third harmonics contain that fraction ofthe total power, meaning that the effective bandwidth of our baseband signal is 3 2 T or, expressing this quantity in terms of the datarate, 3 R 2 . Thus, a digital communications signal requires more bandwidth than the datarate: a 1 Mbps baseband systemrequires a bandwidth of at least 1.5 MHz. Listen carefully when someone describes the transmission bandwidth of digitalcommunication systems: Did they say "megabits" or "megahertz"?

Show that indeed the first and third harmonics contain 90% of the transmitted power. If the receiver uses a front-endfilter of bandwidth 3 2 T , what is the total harmonic distortion of the received signal?

The harmonic distortion is 10%.

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What is the 90% transmission bandwidth of the modulated signal set?

Twice the baseband bandwidth because both positive and negative frequencies are shifted to the carrier by themodulation: 3 R .

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Questions & Answers

how did you get 1640
Noor Reply
If auger is pair are the roots of equation x2+5x-3=0
Peter Reply
Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
MATTHEW Reply
420
Sharon
from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28
Salma
Devon is 32 32​​ years older than his son, Milan. The sum of both their ages is 54 54​. Using the variables d d​ and m m​ to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.
Aaron Reply
find product (-6m+6) ( 3m²+4m-3)
SIMRAN Reply
-42m²+60m-18
Salma
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bill
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bill
-24m+3+3mÁ^2
Susan
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Amira
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Aphelele
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Bajemah
-6m(3mA²+4m-3)+6(3mA²+4m-3) =-18m²A²-24m²+18m+18mA²+24m-18 Rearrange like items -18m²A²-24m²+42m+18A²-18
Salma
complete the table of valuesfor each given equatio then graph. 1.x+2y=3
Jovelyn Reply
x=3-2y
Salma
y=x+3/2
Salma
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Enock
given that (7x-5):(2+4x)=8:7find the value of x
Nandala
3x-12y=18
Kelvin
please why isn't that the 0is in ten thousand place
Grace Reply
please why is it that the 0is in the place of ten thousand
Grace
Send the example to me here and let me see
Stephen
A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg
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state in which quadrant or on which axis each of the following angles given measure. in standard position would lie 89°
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Method
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The Jones family took a 15 mile canoe ride down the Indian River in three hours. After lunch, the return trip back up the river took five hours. Find the rate, in mph, of the canoe in still water and the rate of the current.
cameron Reply
Shakir works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925.
mahnoor Reply
I'm guessing, but it's somewhere around $4335.00 I think
Lewis
12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67. Check: Sales = 3542 Commission 12%=425.04 Pay = 500 + 425.04 = 925.04. 925.04 > 925.00
Munster
difference between rational and irrational numbers
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When traveling to Great Britain, Bethany exchanged $602 US dollars into £515 British pounds. How many pounds did she receive for each US dollar?
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Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?
Zack Reply
d=r×t the equation would be 8/r+24/r+4=3 worked out
Sheirtina
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Source:  OpenStax, Fundamentals of electrical engineering i. OpenStax CNX. Aug 06, 2008 Download for free at http://legacy.cnx.org/content/col10040/1.9
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