# Why is the relationship between Es/N0 and SNR different for complex and real signals?

I try to use the AWGN Channel model of MATLAB to build my simulation model, but the explanation of the relationship between Es/N0 and SNR in the official manual of MATLAB makes me confused.

It says that :

\begin{align} E_s/N_0 \ \text{(dB)}&= 10\log_{10}(T_{sym}/T_{samp})+SNR\ \text{(dB)}\quad\text{for complex input signals}\\ E_s/N_0 \ \text{(dB)}&= 10\log_{10}(0.5T_{sym}/T_{samp})+SNR\ \text{(dB)}\quad\text{for real input signals} \end{align}

I wonder why there is a $$3\ \rm dB$$ difference whether the signal is complex or real.

• You sure those aren't backwards? I would think you'd see the 0.5 factor in the complex case since you don't want to double the noise b/c of the real and imaginary components. Aug 14 '20 at 10:38
• I not sure if I get your point. Do you mean that the official manual of MATLAB is wrong? To be honest, I think the 0.5 factor should appear in the complex case, too. Because for a specific noise power (variance), the bandwidth of a baseband complex signal is B, and the bandwidth of a passband real signal is 2B, just as showed in the MATLAB manual, thus the noise spectral density in complex case is twice in the real case. Aug 14 '20 at 13:05

In my opinion, no matter it's complex signal or real signal, the SNR equals to the input signal power, S, divided by the noise power, N. The difference is, for complex signal,

S = Si + Sq, (Si = Sq, no consideration of IQ imbalance)


and for real signal,

S = Si or S = Sq


Then, no matter it's complex or real,

Es = S*Tsym


and,

N0 = N*Tsamp


So,

Es/N0 = 10*log10(Tsym/Tsamp) + SNR


Then, the question is why there is a 0.5 factor exists in the Es/N0 formula of real signal as the documentation of MATLAB AWGN CHANNEL model illustrated. I suppose it's because the AWGN CHANNEL just simply think the signal power that is specified through SNR or Eb/N0 or Es/N0 in the AWGN model configuration panel is the power of a complex signal. So, if one want to specify a SNR of 10dB in case of a real signal, he/she has to set the SNR to 13dB in the configuration panel.

When you double the sampling rate of the signal you are receiving and sampling, $$E_s$$ doubles, because the frequency domain amplitude doubles as the number of samples, $$N$$, doubles $$(AN)^2 / N = E_{\omega}$$ vs. $$(2AN)^2 / 2N = 2E_{\omega}$$, but then so does $$N_0$$, and therefore you end up with the same $$E_s/N_0$$

When you double the sampling rate, SNR halves, because the sampling bandwidth (bandwidth of the noise) doubles, so you are picking up double the noise (because the bandwidth thats being sampled is double), $$N_0$$ also doubles because double the number of samples results in double the frequency domain amplitude, so noise power is quadruple, but signal power is only double, as $$E_s$$ doubles but the signal bandwidth does not, so SNR is $$2/4 = 1/2$$ of the $$E_s/N_0$$. $$E_s/N_0$$ remains constant regardless of sampling rate.

The energy of a real cos is half the energy of a complex exponential with the same amplitude, $$E_r = E_c/2$$. If you have a real bandpass signal then the noise is just $$N_0/2$$ but if it is complex then the PSD of the noise is $$N_0$$, because it affects the real and the imaginary part of the complex exponentials in the time domain signal, which result in a sum of their frequency domain components. For a real signal of sample rate 20MHz you end up with $$N_r = N_0/2$$ noise and $$E_r/2$$ energy for 10MHz. For a complex signal of sample rate 20MHz you end up with $$N_c = N_0$$ noise and $$E_c$$ energy for 20MHz. For the real signal, $$SNR = E_r/N_r$$. For the complex signal $$SNR = E_c/N_c$$. Therefore $$E_s/N_0$$ is the same as SNR, if you denote $$E_s = E_c$$ and $$E_s = E_r$$ in the separate cases. SNR of the complex signal is double the real SNR, as is $$E_s/N_0$$ double the real. In order for the property in question to hold, you'd have fix SNR as the complex SNR, only then will the real $$E_s/N_0$$ be half of it.

My other more likely theory is that SNR the power of the signal within the sampling bandwidth over the power of the noise within the sampling bandwidth, whereas $$E_s/N_0$$ is the energy within the useful symbol bandwidth over the energy of the noise within the symbol bandwidth. Then you get $$E_c/N_c = SNR_c$$ and $$E_r/N_r = SNR_r/2$$ and it holds. Essentially $$E_r$$ is now actually $$E_r/2$$ because you don't sum the negative signal component aliased in the second half of the sampling bandwidth, but you do sum the noise part of that component to the noise. It implies that $$E_s$$ implies some knowledge of the symbol and discludes the signal alias but not the noise, whereas SNR measures the raw frequency spectrum power received. But $$E_s$$ should really include the signal alias and indeed Parseval's discrete frequency domain equation does; $$E_s$$ is being used to represent half the energy of the signal instead of the energy of the signal, which doesn't really make sense. But maybe that's the distinction between $$E_s$$ as in energy of a signal and $$E_s$$ energy of a symbol.

• Whoa, thanks for you answer. I think your answer is almost correct. But there are some points I cannot agree with. <br/> 1. Sep 11 at 15:47
• 1. You say that the useful energy is half is because of the useful bandwidth is half. I guess you want to express that the useful energy in a real time channel is half that of a complex channel. But I think no matter for a real signal or for a complex signal, the whole signal power is useful power, except the noise power in it. That's to say, for a real signal, the signal power equals to the real part of the signal which is useful energy as you called it, because there is no imagine part. Sep 11 at 16:05
• For a complex signal, it's true that the power of the real part is half of the whole signal power. But the power of the imagine part is part of the useful energy, too. Sep 11 at 16:06
• 2. I don't think the SNR would halve if the sampling rate is doubled. For a specific signal, the signal power, noise power and signal bandwidth are determined, they can't be changed by increasing the sampling rate. I think the sampling bandwidth is not equivalent to the signal bandwidth. When the sampling bandwidth is doubled, for this digital system, N0 is halved, which makes the noise power is unchanged. Thus, the SNR is doubled in my opinion when the sampling rate is doubled. Sep 11 at 16:17
• You can find a evidence by doing a sampling noise experiment. Suppose you generate a noise sequence of zero mean, and sigma standard variance. Then you can get two different sub-sequences by selecting one data in different interval. Then, calculate the variance of the two sub-sequences, you will find they both are the square of sigma, which is the power of the noise. Sep 11 at 16:22