# Is a negative Es/No for an SCPC RF signal possible, and what would such a value signify?

I apologize if this isn't the correct Stack on which to ask this question, but it seemed like the best fit. I've tried to Google an answer, but I haven't been able to find anything that applies directly to Es/No, or which is written in a way I can understand (ie, the mathematical formulas on Wikipedia are beyond me at the moment).

I have worked with legacy satellite communications equipment for years, and I am typically used to seeing receive signal performance measured as a positive Eb/No or Es/No; any carrier visible on a spectrum analyzer would have--in my experience--a positive Es/No value.

However, I was recently shown an SCPC system where the carrier's Es/No was reported as a negative value, but I can't get an explanation as to why. Is a negative Es/No value even possible? Specifically, is such a value possible for a carrier which appears to be above the noise floor (as seen on a spectrum analyzer)?

If it's possible, then what would such negative values signify?

I appreciate any insight anyone can offer, even if it's just to point me to another Stack.

Assuming the number is reported in dB, a negative value of $E_s/N_0$ means is that the energy per symbol $E_s$ is less than the noise spectral density $N_0$.
Even then, a negative $E_s/N_0$ seems hard to obtain if the signal's power is larger than the noise power (as measured on the spectrum analyzer). Assuming that the maximum symbol rate is used ($E_s=2B$, where $B$ is the bandwidth), and that the noise power $\sigma^2=N_0B$, then $$\frac{E_s}{N_0} = \frac{P}{2\sigma^2},$$ where $P$ is the signal power.
So, under these assumptions, a negative $E_s/N_0$ (in dB) means that the signal's power is less than twice the noise power.
• If you want me to add detail to the answer, just let me know where. I'd be surprised to see this ratio reported in anything but dB. A few pointers about the math: $E_s$ is the energy (in joules) spent per transmitted symbol (pulse), so the signal power is $E_s$ times the number of pulses per second ($R$), which has to be less than $2B$. $N_0$ is the noise power density (power per hertz), so the total power is $\sigma^2=N_0B$. So, you have $$\frac{E_s}{N_0}=\frac{\frac{P}{R}}{\frac{\sigma^2}{B}}=\frac{P}{2\sigma^2}.$$ – MBaz Jan 12 '15 at 23:17