Eb (energy per bit)

On Mon, 05 Aug 2013 22:13:08 -0400, Randy Yates
<yates@digitalsignallabs.com> wrote:

>Randy Yates <yates@digitalsignallabs.com> writes:
>
>> Randy Yates <yates@digitalsignallabs.com> writes:
>>
>>> eric.jacobsen@ieee.org (Eric Jacobsen) writes:
>>>
>>>> On Mon, 05 Aug 2013 19:14:00 -0500, Tim Wescott
>>>> <tim@seemywebsite.really> wrote:
>>>>
>>>>>On Mon, 05 Aug 2013 18:56:37 +0000, Eric Jacobsen wrote:
>>>>>
>>>>>> On Mon, 05 Aug 2013 14:39:44 -0400, Randy Yates
>>>>>> <yates@digitalsignallabs.com> wrote:
>>>>>> 
>>>>>>>Consider simple BPSK. Should the computation of Eb given the amplitude
>>>>>>>of an ideal bit include the effect of pulse-shaping?
>>>>>> 
>>>>>> It generally doesn't.  Eb is just the total power divided by the bit
>>>>>> rate.   The pulse shape doesn't matter, since since things like Eb/No
>>>>>> are power efficiency metrics.   If you make a crappy Rx filter that
>>>>>> causes a lot of ISI, you'll lose power efficiency.
>>>>>
>>>>>Actually, if Eb is the total power divided by the bit rate, then as long 
>>>>>as each bit's pulse is orthogonal to all the other bit's pulses (or is so 
>>>>>on average), then Eb _is_ taking the pulse shape into account.
>>>>
>>>> In a sense, yes.
>>>>
>>>> But measuring Eb, which is what Randy asked about, ...
>>>
>>> Ah, but that is not what I asked about. I asked about *computing* Eb. 

Yes, a mis-statement on my part.

>>> Let me ask it another way. Let's say I want to generate an ideal,
>>> pulse-shaped BPSK signal with a specified Eb. How do I know how to
>>> relate the signal's peak value (we can call it A) to Eb? Does that
>>> relationship depend on the pulse shape?

Not if the receive filter matches the transmit filter, in a
matched-filter sense.   The bits don't really exist until the slicer,
and the slicer only processes (if the system is working properly) the
sampling instant where there is no ISI, and it cannot, therefore, tell
any difference between pulse shapes since ALL matched filter pulse
shapes will provide the same output at that instant.

>> Or, given a pulse-shaped ideal BPSK signal with peak A, what is its Eb?

Compute power using A, then divide by the symbol period, assuming Eb
is for the single bit coming from the BPSK signal (as opposed to after
FEC or other overhead).

A little on why this works below.

>In these last two posts, instead of saying the _signal_, I should have
>stated the _pulse_. That is, A is the peak value of the pulse.

This can be confusing, because the signal peaks in matched filter
systems that use pulse shaping in the modulator happen between the
symbol sampling instants, so the "peaks" will be higher than the
desired (or noise-free) sliced symbol amplitude.   This can be seen in
the figures of one of the articles I linked.

However, if one transmits unfiltered, NRZ, pulses and uses an
integrate-and-dump filter in the receiver, it is a matched-filter
system but the transmit signal has a constant amplitude during the
entire symbol period.   The slicer output will be the integral of that
amplitude over the symbol period with the sign dependent on the
transmitted bit.   Remember, every matched filter system has the same
output value at the sampling instant at the match point for a given
input power, and the signal power of the unfiltered, NRZ BPSK signal
is easy to compute and conceptualize.   This means that whatever you
figure out for the easy, NRZ, case, will hold for the pulse-shaped
case as well. 

Assume that the transmitted signal is of unity amplitude, i.e., A = 1.
The symbol at the matched filter output can also be interpreted at A =
1 (since the scaling, path loss, receiver gain, etc., is arbitrary),
and can be used to compute the input power.   This would also be true
for the ideal, lossless signal transmission case, which is not
realizable.

The upshot is that "Eb" at the slicer output for BPSK can be
arbitrarily chosen as "1", since there is one bit per symbol, as long
as the same scaling is used for computing No.  In other words, the
desired sliced symbol magnitude is Eb = 1, and the sliced error
vectors from the received symbols can be used with the same amplitude
scaling to compute No (in the normal way that power is computed).

Computing Eb with some sort of physical units applied depends on
accurately knowing the receiver gain, which usually means having a
carefully calibrated AGC and a means to accurately sense the AGC
level.  It is possible to then also compute the noise power from the
sliced error vectors and get an estimate of SNR or Eb/No or Es/No or
whatever.  

None of it depends on the pulse shape used.   We've done this a lot
where we provided estimated recieved power level and SNR (or similar
metrics) from the slicer statistics, and it is independent of which
programmable pulse shape might be loaded at the time as long as the
transmitter and receiver filters are properly matched.

The only difference for higher-order modulations such as QAM is
accounting for the average value of the constellation points instead
of the magnitude of the constant-modulus signals, and scaling for the
number of bits per symbol to get Eb instead of Es.

I hope that all makes sense. 

>-- 
>Randy Yates
>Digital Signal Labs
>http://www.digitalsignallabs.com

Eric Jacobsen
Anchor Hill Communications
http://www.anchorhill.com

>Consider simple BPSK. Should the computation of Eb given the amplitude
>of an ideal bit include the effect of pulse-shaping?
>-- 
>Randy Yates
>Digital Signal Labs
>http://www.digitalsignallabs.com
>

if p(t) is the pulse shape then

Eb = integral from -inf to +inf of [p(t)]^2 dt ... this is right out of
Proakis.

This assumes of course that p(t) is the shape of the "bit" in question.

-Doug	 

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On Monday, August 5, 2013 8:37:23 PM UTC-5, Randy Yates wrote:

> Let me ask it another way. Let's say I want to generate an ideal,
> 
> pulse-shaped BPSK signal with a specified Eb. How do I know how to
> 
> relate the signal's peak value (we can call it A) to Eb? Does that
> 
> relationship depend on the pulse shape?

and

> If you want to get picky about units, assume
> A is in volts and we are driving a resistance
> of 1 ohm.

Ah, at last a question that I know how to answer.

Assuming the baseband signal p(t) is a rectangular 
pulse of amplitude 1 and duration T seconds, in
fact, zero outside the interval [0,T), the simplest
form of a BPSK signal is

sum_k (-1)^{bk} sqrt{2 Eb/T} p(t-kT)sin(2 pi fc t)

where Eb is the signal energy in joules, the carrier
frequency fc is n/T, that is, the n-th harmonic of 
the channel bit rate, and bk is the k-th bit transmitted
on the channel.

The above result is usable, though not completely exact,
even if the carrier frequency is not a harmonic of the
channel bit rate as long as fc >> 1/T. Note, however,
that at the switching instants mT when bit changes occur,
the signal changes discontinuously from one nonzero value
to another which is impossible to achieve in practice and
so the model is inexact on these grounds as well.

The above result is also usable for nonrectangular pulses
p(t) as long as they nonzero only on [0,T), are relatively
smooth, and enjoy the property that 

integral from 0 to T [p(t)]^2 dt = T

but in this case, we must insist on fc >> 1/T because
the analysis has to come to grips with an integral of

[p(t)]^2 cos(2 pi 2fc t)

over [0,T) and the argument is that [p(t)]^2 is slowly
changing and thus can be regarded as essentially constant
over one period of the sinusoid; and therefore, if we
break up the integral over [0,T) into a sum of integrals
over single periods of the sinusoid, each little integral,
being the integral of (essentially) a constant times a
sinusoid over one period of the sinusoid, is (essentially)
zero.

Dilip Sarwate

>The above result is also usable for nonrectangular pulses
>p(t) as long as they nonzero only on [0,T), are relatively
>smooth, and enjoy the property that 
>
>integral from 0 to T [p(t)]^2 dt = T
>
>Dilip Sarwate
>
>

Yes ... if p(t) is wider than [0,T), then you must integrate 0 to T and
find the average over all bit sequences that are at least as long as p(t)
to get Eb	 

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On Tuesday, August 6, 2013 10:38:32 AM UTC-5, DougB wrote:

> 
> Yes ... if p(t) is wider than [0,T), then you must integrate 0 to T and
> 
> find the average over all bit sequences that are at least as long as p(t)
> 
> to get Eb	 


Except that the signal is no longer a constant
amplitude BPSK signal, and some people might
question whether this is indeed BPSK, or, as
Eric Jacobsen pointed out, will work at all as
intended if the transmitter yanks up the amplitude
to get the maximum power out the door.

Dilip Sarwate