Bit streams through digital communications receivers.
The receiver interested in the transmitted bit stream must
perform two tasks when received waveform
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
. 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
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
, which we label
, is given by
You may not have seen the
notation before.
yields the maximum value of its argument with respect to the index
.
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
If bit 1 were sent,
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:
. 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.
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
calculate molarity of NaOH solution when 25.0ml of NaOH titrated with 27.2ml of 0.2m H2SO4
