Digital communication formulas

Uniform quantiser with peak-to-peak range $2m_p$ and $L=2^n$ levels.

Error uniform on $(-\Delta /2,\Delta /2)$; valid when the signal is large compared with $\Delta$ and does not overload the quantiser.

Here $S=\overline{m^2}$ is the average signal power and $m_p$ the quantiser's peak value.

Sinusoid of amplitude $m_p$ so $S=m_p^2/2$ and SQNR $=1.5\cdot 4^n$. Each extra bit adds about 6 dB. For a full-scale uniformly distributed signal ($\overline{m^2}=m_p^2/3$) the SQNR is $4^n$, i.e. $6.02n$ dB.

Minimum (ideal Nyquist, sinc pulses) baseband bandwidth for binary PCM; with a raised-cosine pulse of roll-off $\alpha$ it becomes $(1+\alpha )R_b/2$. The first-null bandwidth of rectangular NRZ pulses is $R_b$.

$\mu =255$ in the standard North American/Japanese system. In the limit $\mu \to 0$, $y=x$ (no compression).

For $m(t)=A\sin 2\pi f_{m}t$ this is $2\pi f_{m}A\le \Delta f_s$. If violated, the staircase cannot follow the signal (slope overload).

Equivalent to $p(kT)={\delta }_{k0}$ (normalised so $p(0)=1$). Smallest bandwidth satisfying it: $1/(2T)$ Hz, with $p(t)=\mathrm{sinc}(t/T)$, where $\mathrm{sinc}x=\sin (\pi x)/(\pi x)$ (ideal, not realisable).

Baseband (one-sided) bandwidth for symbol rate $R_s=1/T$. $\alpha$ is the roll-off factor. At $\alpha =0$ it is the ideal $R_s/2$; at $\alpha =1$ it is $R_s$.

For a baseband channel of bandwidth $B$. For a bandpass (double-sideband modulated) channel of bandwidth $B_c$ the limit is $R_s\le B_c/(1+\alpha )$.

With a raised-cosine pulse the passband bandwidth is $(1+\alpha )R_s$, so $\eta ={\log }_{2}M/(1+\alpha )$. Not for FSK, whose bandwidth also depends on the tone spacing.

The two tones must be orthogonal over one bit interval for the stated error probability to hold.

Output sampled at $t=T$. In the frequency domain $H(f)=S^{\ast }(f)e^{-j2\pi fT}$. The peak SNR depends only on the pulse energy, not its shape. A correlation receiver (multiply by $s(t)$ and integrate over $T$) gives the same output at the sampling instant $t=T$.

Sample $r=s_i+n$ with $s_1>s_0$ and noise $n\sim \mathcal{N}(0,{\sigma }^2)$ under both hypotheses. For antipodal levels $s_1=+a$, $s_0=-a$ it is $r_{th}=\frac{{\sigma }^2}{2a}\ln \frac{P(H_0)}{P(H_1)}$. For equal priors the threshold is the midpoint (ML). A less likely $H_1$ moves the threshold toward $s_1$.

Here $d^2=4E_b$ for BPSK. QPSK is two independent BPSK streams on the I and Q carriers, so its per-bit error probability is the same.

Orthogonal tones, $d^2=2E_b$. This needs 3 dB more $E_b/N_0$ than BPSK for the same BER.

Equal priors, threshold midway between the two received levels. Here $E_b$ is the average bit energy: the "1" has energy $2E_b$, the "0" none. If the same formula is written with the energy $E_1$ of the "1", use $Q(\sqrt{E_1/(2N_0)})$.

$S$ is the received signal power and $N=N_{0}B$ the noise power in bandwidth $B$. In dB: $(E_b/N_0{)}_{\text{dB}}={\text{SNR}}_{\text{dB}}-10{\log }_{10}(R_b/B)$.