I want to show the bit error performance of a communication link using QPSK in a system containing a channel and antennas. The resulting received SNR is calculated by:

$$ SNR(f) = \frac{P_\text{t} \cdot H(f)}{\text{k}_\text{B} \cdot T \cdot B \cdot F} $$

$F$ is the total noise figure of the system and $H(f)$ is the total transfer function (free-space loss of channel and antenna).

Now the error probability of QPSK is defined as: $$ P_\text{b} = Q\biggr(\sqrt{\frac{2E_\text{b}}{N_\text{0}}}\biggl) $$

Question is now how I can obtain $\frac{E_\text{b}}{N_\text{0}}$ from the receiver SNR. I want to simulate for a bandwidth of $B=10 GHz$ and want to make a comparison between different channels. Is that true that the relationship is (got this from Wikipedia): $$ SNR(f) \cdot \frac{B}{f_\text{b}} = \frac{E_\text{b}}{N_\text{0}} $$ where $R_\text{b}$ is the data rate (e.g. 10Gbit/s). Is my assumption correct? Many thanks in advance for any comment!

  • $\begingroup$ Do you need to use that formula for SNR? Is $H(f)$ constant over the band you're interested in? $\endgroup$
    – MBaz
    Jul 8, 2023 at 18:49
  • $\begingroup$ No, H(f) isn't constant but I'm only interested to calculate SNR at specific frequency. So we can assume that $$SNR_ = \frac{P_\text{t} \cdot H}{\text{k}_\text{B} \cdot T \cdot B \cdot F_\text{G}}$$ and neglect the frequency dependencies. $\endgroup$
    – finalr
    Jul 8, 2023 at 20:24
  • 1
    $\begingroup$ If the bandwidth is 10 GHz and the channel response is not constant, then the channel will distort the signal and you'll need an equalizer. The error probability is no longer given by the formula you posted. $\endgroup$
    – MBaz
    Jul 8, 2023 at 21:03

1 Answer 1


The equation given by the OP is correct when the noise is Additive White Gaussian Noise (AWGN) within the bandwidth of interest with $H(f)$ constant and linear phase over that bandwidth (referred to as an AWGN channel). However the bandwidth $B$ would be 5 GHz, not 10 GHz if the waveform is QPSK with a bit rate of $R_b= 10$ Gbit/s.

The bandwidth $B$ is the receiver noise bandwidth. If the Bit Rate is 10 Gbit/s and QPSK, then the Symbol Rate is 5 G Symbols/s as we send 2 bits for every symbol. If the waveform is pulse shaped (as it should be if we care about spectral efficiency) then the spectral occupancy will be close to the Symbol Rate (5 GHz). This will be the noise bandwidth in the optimum receiver under conditions of AWGN noise, as the bandwidth of the matched filter.

The power spectral densities shown below for a simulated QPSK waveform at 10 Gbit/s (with Raised Cosine roll-off factor of 0.3) demonstrate the relationships between the symbol rate $R_s$ and occupied bandwidth, as well as the noise floor of the power spectral density assuming AWGN with relation to $N_o$ for both one-sided and two-sided spectrums. Not understanding one-sided vs two-sided spectrums is a source of confusion when $N_o$ is used. For more details on one-sided vs two-sided spectrums, please see this post.

Spectral Density Plots

Note: Even if the waveform was not pulse shaped we will get the same result for $B$ and the same equation under condition of AWGN, but the pulse shaped waveform provides easier intuition as the resulting spectrum is closer to a brickwall shape. Without pulse-shaping, the waveform will then have a Sinc shaped spectrum (Sinc squared shaped power spectral density). For a 5 GSymbol/Sec symbol rate, the null to null bandwidth would be 10 GHz, and the equivalent noise bandwidth will be 5 GHz. This may be understood from the example of a bandpass Sinc filter: a bandpass filter with a Sinc frequency response that has a main-lobe with a null to null bandwidth of $B$ Hz will have the same power when filtering white noise as a a brick-wall bandpass filter with a bandwidth of $B/2$ Hz. This is what "equivalent noise bandwidth" means.


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