Are the error probability of orthogonal signal and on/off signal the same?

Question

I have to write the error probability of orthogonal signal and on/off signal,and now i have finished the error probability of on/off signal,but i don't know how to write the code about error probability of orthogonal signal.

Generally,we will introduce the orthogonal signal and on/off signal with the picture below ,first is orthogonal signal,second is on/off signal.

(a) orthogonal signal

(b) on/off signal

And the definition of orthogonal binary signal is that if we do the inner product with the signal 1 and signal 0 ,the result will become 0. I mean $\int s_0 s_1 dt = 0$

And i found that the on/off binary signal is also $\int s_0 s_1 dt = 0$,so can i say the on/off signal is the orthogonal binary signal too,so their error probability should be the same theoretically?

Answer 1

I found that the on/off binary signal is also $\int s_0 s_1 dt = 0$, so can I say the on/off signal is the orthogonal binary signal too, so their error probability should be the same theoretically?

No! That's a wrong inference: Just because the zero vector (the off signal in your case) is orthogonal to everything doesn't mean everything has the same error probability.

When comparing error probabilities, you first have to agree on a way of making the energy spent on each transmitted bit equal.
Otherwise, you might be comparing an On/off-keying transmitter with 1 kW to a BPSK transmitter with 1 mW output power – you'll notice that this comparison will probably be won by the on/off-keying, no matter whether the other one is better.

So, same average bit energy is what we demand. Since one symbol carries one bit in your example, we can set the average symbol energy to 1 for both cases (without loss of generality, as we can later scale), and compare then.

So, sadly, the book you're learning from doesn't do that energy normalization to begin with.

Let's say for the on-off case, $A_{OOK} = 1$, and $T_b=2$, so that the energy the on-period is $E_{On}=P_{On} T_b = A_{OOK}^2 T_b = 1^2 2 = 2$. Obviously, during the off-period, the power is $P_{Off}=0=E_{Off}$, so that the average energy is $E_{OOK}=1$ (you always assume you send as many 0s as 1s, otherwise your source coding is broken).

For what you call the "orthogonal case", the power is always the same – the sign of the signal doesn't matter to the power, since you square. So, we set $A_{ortho}=\frac1{\sqrt2}$ and get $P_{ortho}=A_{ortho}^2=\frac12$ and thus also $E_{ortho}=1$. So, now we're fair!

That means that the pictures you posted aren't fair: The $A$ from the "orthogonal signalling" picture should be $\sqrt2$ smaller than the one from OOK.

We also see that we need a "stronger" amplifier for the On/off case just to get the same signal power. That alone might be an argument why OOK isn't that good.

Now, the rest of the derivation of what an error probability is depends on whether you assume coherent or incoherent reception – you don't define that, but your textbook will certainly do.

You'll, in any case, notice that the two things have different symbol error rates at the same noise power when you normalize their power.