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Bit streams through digital communications receivers.

The receiver interested in the transmitted bit stream must perform two tasks when received waveform r t begins.

  • It must determine when bit boundaries occur: The receiver needs to synchronize with the transmitted signal. Because transmitter and receiver are designed inconcert, both use the same value for the bit interval T . Synchronization can occur because the transmitter begins sending with a reference bitsequence, known as the preamble . This reference bit sequence is usually the alternating sequenceas shown in the square wave example and in the FSK example . The receiver knows what the preamble bit sequence is and uses it to determine when bit boundariesoccur. This procedure amounts to what in digital hardware as self-clocking signaling : The receiver of a bit stream must derive the clock — when bit boundariesoccur — from its input signal. Because the receiver usually does not determine which bit was sent untilsynchronization occurs, it does not know when during the preamble it obtained synchronization. The transmittersignals the end of the preamble by switching to a second bit sequence. The second preamble phase informs the receiverthat data bits are about to come and that the preamble is almost over.
  • Once synchronized and data bits are transmitted, the receiver must then determine every T seconds what bit was transmitted during the previous bit interval. We focus on this aspect of thedigital receiver because this strategy is also used in synchronization.
The receiver for digital communication is known as a matched filter .

Optimal receiver structure

The optimal receiver structure for digital communication faced with additive white noise channels is the depicted matchedfilter.
This receiver, shown in [link] , multiplies the received signal by each of the possible membersof the transmitter signal set, integrates the product over the bit interval, and compares the results. Whichever path throughthe receiver yields the largest value corresponds to the receiver's decision as to what bit was sent during the previousbit interval. For the next bit interval, the multiplication and integration begins again, with the next bit decision made at theend of the bit interval. Mathematically, the received value of b n , which we label b ^ n , is given by
b ^ n i t n 1 T n T r t s i t
You may not have seen the i notation before. i · yields the maximum value of its argument with respect to the index i . i equals the value of the index that yields the maximum.Note that the precise numerical value of the integrator's output does not matter; what does matter is its value relative to theother integrator's output.

Let's assume a perfect channel for the moment: The received signal equals the transmitted one. If bit 0 were sent using thebaseband BPSK signal set, the integrator outputs would be

t n 1 T n T r t s 0 t A 2 T
t n 1 T n T r t s 1 t A 2 T If bit 1 were sent,
t n 1 T n T r t s 0 t A 2 T
t n 1 T n T r t s 1 t A 2 T

Can you develop a receiver for BPSK signal sets that requires only one multiplier-integrator combination?

In BPSK, the signals are negatives of each other: s 1 t s 0 t . Consequently, the output of each multiplier-integrator combination is the negative of theother. Choosing the largest therefore amounts to choosing which one is positive. We only need to calculate one ofthese. If it is positive, we are done. If it is negative, we choose the other signal.

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What is the corresponding result when the amplitude-modulated BPSK signal set is used?

The matched filter outputs are ± A 2 T 2 because the sinusoid has less power than a pulse having the same amplitude.

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Clearly, this receiver would always choose the bit correctly. Channel attenuation would not affect this correctness; it wouldonly make the values smaller, but all that matters is which is largest.

Questions & Answers

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?
Kate Reply
what is the change in momentum of a body?
Eunice Reply
what is a capacitor?
Raymond Reply
Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.
Gautam
A motor travelling at 72km/m on sighting a stop sign applying the breaks such that under constant deaccelerate in the meters of 50 metres what is the magnitude of the accelerate
Maria Reply
please solve
Sharon
8m/s²
Aishat
What is Thermodynamics
Muordit
velocity can be 72 km/h in question. 72 km/h=20 m/s, v^2=2.a.x , 20^2=2.a.50, a=4 m/s^2.
Mehmet
A boat travels due east at a speed of 40meter per seconds across a river flowing due south at 30meter per seconds. what is the resultant speed of the boat
Saheed Reply
50 m/s due south east
Someone
which has a higher temperature, 1cup of boiling water or 1teapot of boiling water which can transfer more heat 1cup of boiling water or 1 teapot of boiling water explain your . answer
Ramon Reply
I believe temperature being an intensive property does not change for any amount of boiling water whereas heat being an extensive property changes with amount/size of the system.
Someone
Scratch that
Someone
temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....
Someone
about the amount of heat stored in the system then in that case since the mass of water in the kettle is greater so more energy is required to raise the temperature b/c more molecules of water are present in the kettle
Someone
definitely of physics
Haryormhidey Reply
how many start and codon
Esrael Reply
what is field
Felix Reply
physics, biology and chemistry this is my Field
ALIYU
field is a region of space under the influence of some physical properties
Collete
what is ogarnic chemistry
WISDOM Reply
determine the slope giving that 3y+ 2x-14=0
WISDOM
Another formula for Acceleration
Belty Reply
a=v/t. a=f/m a
IHUMA
innocent
Adah
pratica A on solution of hydro chloric acid,B is a solution containing 0.5000 mole ofsodium chlorid per dm³,put A in the burret and titrate 20.00 or 25.00cm³ portion of B using melting orange as the indicator. record the deside of your burret tabulate the burret reading and calculate the average volume of acid used?
Nassze Reply
how do lnternal energy measures
Esrael
Two bodies attract each other electrically. Do they both have to be charged? Answer the same question if the bodies repel one another.
JALLAH Reply
No. According to Isac Newtons law. this two bodies maybe you and the wall beside you. Attracting depends on the mass och each body and distance between them.
Dlovan
Are you really asking if two bodies have to be charged to be influenced by Coulombs Law?
Robert
like charges repel while unlike charges atttact
Raymond
What is specific heat capacity
Destiny Reply
Specific heat capacity is a measure of the amount of energy required to raise the temperature of a substance by one degree Celsius (or Kelvin). It is measured in Joules per kilogram per degree Celsius (J/kg°C).
AI-Robot
specific heat capacity is the amount of energy needed to raise the temperature of a substance by one degree Celsius or kelvin
ROKEEB
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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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