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Consider the problem of transmitting just two SRRC pulses through an noiseless Channel. The two transmit bits are {+/-1}. In plots below, the individual transmit pulses are shown in black, while the actual transmit sequence is shown in blue. Now consider transmitting just two pulses.

Case 1 : {+1 , +1 } enter image description here

The total energy of the transmit sequence is : 32.341872 Joules

The energy per transmit pulse is : (transmit energy)/2 = 16.170936 Joules

Case 2 : { +1, -1} enter image description here

The transmit energy is : 23.758018 Joules

The energy per transmit pulse is : (transmit energy)/2 = 11.879009 Joules

Since the transmitted SRRC pulses overlap, the transmitted sequence energy (as well as the energy per pulse) varies depending on the transmit bits, as the numbers suggest. For longer binary transmit sequences, the transmit sequence energy would have multiple levels of variation. How do we reconcile with the fact that the transmit bit energy (energy per pulse) is constant.

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3 Answers 3

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How do we reconcile with the fact that the transmit bit energy (energy per pulse) is constant.

We don't – it isn't constant, as you've shown! (Note that you're also making a bit of a mistake there: you look at two pulses in isolation. Do it for sequences of a length approaching infinity, and you can actually calculate a bit energy as limit of the total energy divided by the number of bits. The point really is that you can see the time-continuous energy depends on the actual bits you transmit, so picking two possible bits and claiming that to be "bit energy" is more than slightly misleading. Still, you're right, the power of a transmitter depends on the actual bits transmitted. That's kind of obvious if you think about it: replace your BPSK with a constellation having points {-1.0, -0.5, +0.5, +1}, and it is logical that if your information source only sends ±0.5, you're going to get less power than if your information source is uniformly distributed.)

What is constant is the energy of the pulse after sampling the output of a matched filter, iff that filter pair fulfills the Nyquist zero-ISI criterion.

When that is the case, we can convolve the reception with the matched filter, and at the symbol instants, we get the pulse's energy (a constant) multiplied with the pulse shape's energy (also a constant).

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You must have made a mistake in your calculations. Both waveforms have the same energy. (I disagree with @Marcus' answer).

Let's say the pulse is $p(t)$ and it has energy $E$. Then, the energy of $p(t)+p(t-T)$, where $T$ is the pulse interval, is given by $$ \begin{align} \int_{-\infty}^\infty \left(p(t) + p(t-T) \right)^2 \text{d}t &= \int_{-\infty}^\infty (p(t))^2 \text{d}t \\ &+ \int_{-\infty}^\infty (p(t-T))^2 \text{d}t \\ &+ \int_{-\infty}^\infty 2p(t)p(t-T) \text{d}t \\ &= E + E = 2E. \end{align} $$ The integral $\int_{-\infty}^\infty 2p(t)p(t-T) \text{d}t = 0$ because the pulses are orthogonal when time-shifted by integer multiples of $T$.

It's easy to see that the result will be the same when the pulse amplitudes are $1, -1$ instead of $1, 1$.

Note: Normally in digital communications, $T$ must be chosen so that $p(t-kT)$ and $p(t-lT)$ for integers $k \neq l$ are orthogonal. If they are not, then (as you noticed) the energy of the waveform is no longer equal to the sum of the energy of the individual pulses. However, the normal communications receiver will no longer work properly.

There is a field within digital communications called "Faster then Nyquist" or "non-orthogonal signaling", where pulses are no longer orthogonal, but this is a more advanced topic.

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  • $\begingroup$ exactly! These are non-orthogonal pulses; which can' be super OK if they become orthogonal after the matched filter! Remember RC vs RRC! $\endgroup$ Mar 3 at 22:01
  • $\begingroup$ I would put it this way: RC pulses are zero ISI but are non-orthogonal, whereas RRC pulses are orthogonal but not zero ISI. (I'm referring to the pulses by themselves, before filtering.) $\endgroup$
    – MBaz
    Mar 3 at 22:17
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Thanks for the answers Marcus and Mbaz. Suppose we use the energy of the transmitted (BPSK) sequence to detect its presence, the transmitted energy was found to vary with the bit pattern. Each bit was represented by an RC pulse with $\alpha = 0.7$. A sample of the transmit signal (consisting of 1000 pulses) and its energy distribution is shown below. enter image description here

enter image description here

Mbaz has used the fact that $\int_{-\infty}^{\infty} p(t)p(t-T) = 0$. To verify that I considered two truncated, shifted since pulses (shifted by T). As a function of their time span of the truncated sinc pulses, the variation in the Auto-correlation function between two shifted sinc is shown below. It has finite non-zero auto-correlation for for truncated sinc pulses, which decreases as the time span of the pulses are increased, suggesting lack of orthogonality. Therefore, the energy of the transmitted pulse sequence is the sum of the individual energies only when the pulses are perfectly orthogonal. Hence, when orthogonality is lost due to truncation, random combinations of these pulses and results in the energy also being random ! enter image description here

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  • $\begingroup$ Please edit this into the question. It does not appear to be an answer to your question directly. $\endgroup$
    – Peter K.
    Mar 5 at 19:12
  • $\begingroup$ "Therefore, the energy of the transmitted pulse sequence is the sum of the individual energies only when the pulses are perfectly orthogonal." -- This is correct, and I pointed this out in my answer. "Hence, when orthogonality is lost due to truncation, random combinations of these pulses and results in the energy also being random!" -- Yes, nothing is actually ideal. In practice, though, if truncation leads to more than a fraction of a percent difference, then you're doing something wrong. $\endgroup$
    – MBaz
    Mar 7 at 20:40

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