Pulse Code Modulation

1 # OVERVIEW OF THE FOURTH MAIN CHAPTER #
2 Principle and block diagram
3 Sampling and signal reconstruction
4 Natural and discrete sampling
5 Frequency domain view of natural sampling
6 Frequency domain view of discrete sampling
7 Quantization and quantization noise
8 PCM encoding and decoding
9 Signal-to-noise power ratio
10 Influence of transmission errors
11 Estimation of SNR degradation due to transmission errors
12 Nonlinear quantization
13 Compression and expansion
14 Exercises for the chapter

# OVERVIEW OF THE FOURTH MAIN CHAPTER #

The fourth chapter deals with the digital modulation methods »Amplitude Shift Keying« $\rm (ASK)$, »Phase Shift Keying« $\rm (PSK)$ and »Frequency Shift Keying« $\rm (FSK)$ as well as some modifications derived from them. Most of the properties of the analog modulation methods mentioned in the last two chapters still apply. Differences result from the now required »decision component« of the receiver.

We restrict ourselves here essentially to the »system-theoretical and transmission aspects«. The error probability is given only for ideal conditions. The derivations and the consideration of non-ideal boundary conditions can be found in the book "Digital Signal Transmission".

In detail are treated:

the »Pulse Code Modulation« $\rm (PCM)$ and its components "Sampling" – "Quantization" – "Coding",
the »linear modulation« $\rm (ASK)$, $\rm (BPSK)$ and $\rm (DPSK)$ and associated demodulators,
the »quadrature amplitude modulation« $\rm (QAM)$ and more complicated signal space mappings,
the »Frequency Shift Keying« $\rm (FSK$) as an example of non-linear digital modulation,
the FSK with »continuous phase matching« $\rm (CPM)$), especially the $\rm (G)MSK$ method.

Principle and block diagram

Almost all modulation methods used today work digitally. Their advantages have already been mentioned in the first chapter of this book. The first concept for digital signal transmission was already developed in 1938 by Alec Reeves and has also been used in practice since the 1960s under the name "Pulse Code Modulation" $\rm (PCM)$. Even though many of the digital modulation methods conceived in recent years differ from PCM in detail, it is very well suited to explain the principle of all these methods.

The task of the PCM system is

to convert the analog source signal $q(t)$ into the binary signal $q_{\rm C}(t)$ – this process is also called A/D conversion,
transmitting this signal over the channel, where the receiver-side signal $v_{\rm C}(t)$ is also binary because of the decision maker,
to reconstruct from the binary signal $v_{\rm C}(t)$ the analog (continuous-value as well as continuous-time) sink signal $v(t)$ ⇒ D/A conversion.

Principle of Pulse Code Modulation $\rm (PCM)$

$q(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ Q(f)$ ⇒ source signal (from German: "Quellensignal"), analog
$q_{\rm A}(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ Q_{\rm A}(f)$ ⇒ sampled source signal (from German: "abgetastet" ⇒ "A")
$q_{\rm Q}(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ Q_{\rm Q}(f)$ ⇒ quantized source signal (from German: "quantisiert" ⇒ "Q")
$q_{\rm C}(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ Q_{\rm C}(f)$ ⇒ coded source signal (from German: "codiert" ⇒ "C"), binary
$s(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ S(f)$ ⇒ transmitted signal (from German: "Sendesignal"), digital
$n(t)$ ⇒ noise signal, characerized by the power-spectral density ${\it Φ}_n(f)$, analog $r(t)= s(t) \star h_{\rm K}(t) + n(t)$ ⇒ received signal, $h_{\rm K}(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ H_{\rm K}(f)$, analog
Note: spectrum $R(f)$ is due to stochastic component $n(t)$ not specifiable
$v_{\rm C}(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ V_{\rm C}(f)$ ⇒ signal after decision, binary
$v_{\rm Q}(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ V_{\rm Q}(f)$ ⇒ signal after PCM decoding, $M$–level
Note: on the receiver side, there is no counterpart to "Quantization"
$v(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ V(f)$ ⇒ sink signal, analog

Further it should be noted to this PCM block diagram:

The PCM transmitter ("A/D converter") is composed of three function blocks Sampling - Quantization - PCM Coding which will be described in more detail in the next sections.

The gray-background block "Digital Transmission System" shows "transmitter" (modulation), "receiver" (with decision unit), and "analog transmission channel" ⇒ channel frequency response $H_{\rm K}(f)$ and noise power-spectral density ${\it Φ}_n(f)$.

This block is covered in the first three chapters of the book Digital Signal Transmission. In chapter 5 of the same book, you will find Digital Channel Models that phenomenologically describe the transmission behavior using the signals $q_{\rm C}(t)$ and $v_{\rm C}(t)$.

Further, it can be seen from the block diagram that there is no equivalent for "quantization" at the receiver-side. Therefore, even with error-free transmission, i.e., for $v_{\rm C}(t) = q_{\rm C}(t)$, the analog sink signal $v(t)$ will differ from the source signal $q(t)$.

As a measure of the quality of the digital transmission system, we use the Signal-to-Noise Power Ratio ⇒ in short: Sink-SNR as the quotient of the powers of source signal $q(t)$ and fault signal $ε(t) = v(t) - q(t)$:

$$\rho_{v} = \frac{P_q}{P_\varepsilon}\hspace{0.3cm} {\rm with}\hspace{0.3cm}P_q = \overline{[q(t)]^2}, \hspace{0.2cm}P_\varepsilon = \overline{[v(t) - q(t)]^2}\hspace{0.05cm}.$$

Here, an ideal amplitude matching is assumed, so that in the ideal case (that is: sampling according to the sampling theorem, best possible signal reconstruction, infinitely fine quantization) the sink signal $v(t)$ would exactly match the source signal $q(t)$.

We would like to refer you already here to the three-part (German language) learning video "Pulse Code Modulation" which contains all aspects of PCM. Its principle is explained in detail in the first part of the video.

Sampling and signal reconstruction

Sampling - that is, time discretization of the analog signal $q(t)$ - was covered in detail in the chapter Discrete-Time Signal Representation of the book "Signal Representation." Here follows a brief summary of that section.

Time domain representation of sampling

The graph illustrates sampling in the time domain. The (blue) signal $q(t)$ is time continuous, the (green) signal sampled at a distance $T_{\rm A}$ is discrete-time. Here:

The sampling can be calculated by multiplying the analog signal $q(t)$ by the Diracpulse in time domain ⇒ $p_δ(t)$ represent:

$$q_{\rm A}(t) = q(t) \cdot p_{\delta}(t)\hspace{0.3cm} {\rm with}\hspace{0.3cm}p_{\delta}(t)= \sum_{\nu = -\infty}^{\infty}T_{\rm A}\cdot \delta(t - \nu \cdot T_{\rm A}) \hspace{0.05cm}.$$.

The weight of the Dirac function at $t = ν - T_{\rm A}$ is equal to $T_{\rm A} - q(ν - T_{\rm A})$. Since the Dirac function $δ(t)$ has the unit $\rm 1/s$ thus $q_{\rm A}(t)$ has the same unit as $q(t)$, for example "V".

The Fourier transform of the Dirac pulse $p_δ(t)$ is also a Dirac pulse (but now in the frequency domain) ⇒ $P_δ(f)$, where the spacing of the individual Dirac lines $f_{\rm A} = 1/T_{\rm A}$ is. All momentum weights of $P_δ(f)$ are $1$:

$$p_{\delta}(t)= \sum_{\nu = -\infty}^{+\infty}T_{\rm A}\cdot \delta(t - \nu \cdot T_{\rm A}) \hspace{0.2cm}\circ\!\!\!\!\!\!\!\!\bullet\, \hspace{0.2cm} P_{\delta}(f)= \sum_{\mu = -\infty}^{+\infty} \delta(f - \mu \cdot f_{\rm A}) \hspace{0.05cm}.$$

The spectrum $Q_{\rm A}(f)$ of the sampled signal is obtained from the Convolution Theorem, where $Q(f)$ denotes the continuous spectrum of the analog signal $q(t)$ :

$$Q_{\rm A}(f) = Q(f) \star P_{\delta}(f)= \sum_{\mu = -\infty}^{+\infty} Q(f - \mu \cdot f_{\rm A}) \hspace{0.05cm}.$$

We refer you here to the second part of the tutorial video Pulse Code Modulation which explains sampling and signal reconstruction in terms of system theory.

$\text{Example 1:}$ The top graph schematically shows the spectrum $Q(f)$ of an analog source signal $q(t)$ with frequencies up to $f_{\rm N, \ max} = 5 \ \rm kHz$.

Periodic continuation of the spectrum by sampling

If one samples $q(t)$ with the sampling rate $f_{\rm A} = 20 \ \rm kHz$ $($so at the respective distance $T_{\rm A} = 50 \ \rm µ s)$ , one obtains the periodic spectrum sketched in green $Q_{\rm A}(f)$.
Since the Dirac functions are infinitely narrow, $q_{\rm A}(t)$ also contains arbitrary high frequency components and accordingly $Q_{\rm A}(f)$ is extended to infinity (middle graph).
Drawn below (in red) is the spectrum $Q_{\rm A}(f)$ for the sampling parameters $T_{\rm A} = 100 \ \rm µ s$ ⇒ $f_{\rm A} = 10 \ \rm kHz$.

$\text{Conclusion:}$ From this example, the following important lessons can be learned regarding sampling:

If $Q(f)$ contains frequencies up to $f_\text{N, max}$, then according to the Sampling theorem the sampling rate $f_{\rm A} ≥ 2 - f_\text{N, max}$ should be chosen. At smaller sampling rate $f_{\rm A}$ $($thus larger spacing $T_{\rm A})$ overlaps of the periodized spectra occur, i.e. irreversible distortions.

If exactly $f_{\rm A} = 2 - f_\text{N, max}$ as in the lower graph of $\text{Example 1}$, then $Q(f)$ can be calculated from $Q_{\rm A}(f)$ - resp. in PCM system $V(f)$ from $V_{\rm Q}(f)$ - can be completely reconstructed by an ideal rectangular low-pass filter $H(f)$ with cutoff frequency $f_{\rm G} = f_{\rm A}/2$ .

On the other hand, if sampling is performed with $f_{\rm A} > 2 - f_\text{N, max}$ as in the middle graph of the example, a low-pass filter $H(f)$ with a smaller slope can also be used on the receiver side for signal reconstruction, as long as the following condition is met:

$$H(f) = \left\{ \begin{array}{l} 1 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{5}c}{\rm{for} } \\{\rm{for} } \\ \end{array}\begin{array}{*{10}c} {\hspace{0.04cm}\left \vert \hspace{0.005cm} f\hspace{0.05cm} \right \vert \le f_{\rm N, \hspace{0.05cm}max},} \\ {\hspace{0.04cm}\left \vert\hspace{0.005cm} f \hspace{0.05cm} \right \vert \ge f_{\rm A}- f_{\rm N, \hspace{0.05cm}max}.} \\ \end{array}$$

Natural and discrete sampling

Multiplication by the Dirac pulse provides only an idealized description of the sampling, since a Dirac function $($duration $T_{\rm R} → 0$, height $1/T_{\rm R} → ∞)$ is not realizable. In practice, the Dirac pulse $p_δ(t)$ must be replaced, for example, by a square pulse

$$p_{\rm R}(t)= \sum_{\nu = -\infty}^{+\infty}g_{\rm R}(t - \nu \cdot T_{\rm A})\hspace{0.3cm} {\rm mit}\hspace{0.3cm} g_{\rm R}(t) = \left\{ \begin{array}{l} 1 \\ 1/2 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{5}c}{\rm{f\ddot{u}r}}\\{\rm{f\ddot{u}r}} \\{\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{10}c}{\hspace{0.04cm}\left|\hspace{0.06cm} t \hspace{0.05cm} \right|} < T_{\rm R}/2\hspace{0.05cm}, \\{\hspace{0.04cm}\left|\hspace{0.06cm} t \hspace{0.05cm} \right|} = T_{\rm R}/2\hspace{0.05cm}, \\ {\hspace{0.005cm}\left|\hspace{0.06cm} t \hspace{0.05cm} \right|} > T_{\rm R}/2\hspace{0.05cm} \\ \end{array}$$

where the rectangular pulse duration $T_{\rm R}$ should be significantly smaller than the sampling distance $T_{\rm A}$ .

The graph above shows the square pulse $p_{\rm R}(t)$. Below are two different sampling methods using this square pulse:

square pulse (top) and natural and discrete sampling

In natural sampling the sampled signal $q_{\rm A}(t)$ is obtained by multiplying $q(t)$ by $p_{\rm R}(t)$. In the ranges $p_{\rm R}(t) = 1$ thus $q_{\rm A}(t)$ has the same progression as $q(t)$.

In discrete sampling the signal $q(t)$ is - at least mentally - first multiplied by the Dirac pulse $p_δ(t)$ Then each Dirac pulse $T_{\rm A} - δ(t - ν - T_{\rm A})$ is replaced by a square pulse $g_{\rm R}(t - ν - T_{\rm A})$ .

Here and in the following frequency domain consideration, an acausal description form is chosen for simplicity. For a (causal) realization, $g_{\rm R}(t) = 1$ would have to hold in the range from $0$ to $T_{\rm R}$ and not as here for $ \ -T_{\rm R}/2 < t < T_{\rm R}/2.$

Frequency domain view of natural sampling

$\text{Definition:}$ The natural sampling can be represented by the convolution theorem in the spectral domain as follows:

$$q_{\rm A}(t) = p_{\rm R}(t) \cdot q(t) = \left [ \frac{1}{T_{\rm A} } \cdot p_{\rm \delta}(t) \star g_{\rm R}(t)\right ]\cdot q(t) \hspace{0.3cm} \Rightarrow \hspace{0.3cm}Q_{\rm A}(f) = \left [ P_{\rm \delta}(f) \cdot \frac{1}{T_{\rm A} } \cdot G_{\rm R}(f) \right ] \star Q(f) = P_{\rm R}(f) \star Q(f)\hspace{0.05cm}.$$

The graph shows the result for

an (unrealistic) rectangular spectrum $Q(f) = Q_0$ limited to the range $|f| ≤ 4 \ \rm kHz$ ,
the sampling rate $f_{\rm A} = 10 \ \rm kHz$ ⇒ $T_{\rm A} = 100 \ \rm µ s$, and.
the square pulse duration $T_{\rm R} = 25 \ \rm µ s$ ⇒ $T_{\rm R}/T_{\rm A} = 0.25$.

Spectrum in natural sampling with a square pulse

One can see from this plot:

The spectrum $P_{\rm R}(f)$ in natural sampling, in contrast to $P_δ(f)$ is not a Dirac pulse $($all weights equal $1)$, but the weights here are related to the function $G_{\rm R}(f)/T_{\rm A} = T_{\rm R}/T_{\rm A} - {\rm si}(πfT_{\rm R})$ evaluated
Because of the zero of the $\rm si$-function, the diraclines vanish here at $±4f_{\rm A}$.

The spectrum $Q_{\rm A}(f)$ results from the convolution with $Q(f)$. The rectangle around $f = 0$ has height $T_{\rm R}/T_{\rm A} - Q_0$, the proportions around $\mu - f_{\rm A} \ (\mu ≠ 0)$ are less high.

If one uses an ideal, rectangular lowpass for signal reconstruction.

$$H(f) = \left\{ \begin{array}{l} T_{\rm A}/T_{\rm R} = 4 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{5}c}{\rm{for}}\\{\rm{for}} \\ \end{array}\begin{array}{*{10}c} {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| < f_{\rm A}/2}\hspace{0.05cm}, \\ {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| > f_{\rm A}/2}\hspace{0.05cm}, \\ \end{array}$$

so for the output spectrum $V(f) = Q(f)$ and accordingly $v(t) = q(t)$.

$\text{Conclusion:}$

For natural sampling, a rectangular–low-pass filter is sufficient for signal reconstruction as for ideal sampling (with Dirac pulse).
However, for amplitude matching in the passband, a gain by the factor $T_{\rm A}/T_{\rm R}$ must be considered.

Frequency domain view of discrete sampling

$\text{Definition:}$ In discrete sampling the multiplication of the Dirac pulse $p_δ(t)$ with the source signal $q(t)$ takes place - at least mentally - first and only afterwards the convolution with the square pulse $g_{\rm R}(t)$:

$$q_{\rm A}(t) = \big [ {1}/{T_{\rm A} } \cdot p_{\rm \delta}(t) \cdot q(t)\big ]\star g_{\rm R}(t) \hspace{0.3cm} \Rightarrow \hspace{0.3cm}Q_{\rm A}(f) = \big [ P_{\rm \delta}(f) \star Q(f) \big ] \cdot G_{\rm R}(f)/{T_{\rm A} } \hspace{0.05cm}.$$

It is irrelevant, but quite convenient, that here the factor $1/T_{\rm A}$ has been added to the valuation function $G_{\rm R}(f)$ .
Thus, $G_{\rm R}(f)/T_{\rm A} = T_{\rm R}/T_{\rm A} - {\rm si}(πfT_{\rm R}).$

The upper graph shows (highlighted in green) the spectral function $P_δ(f) \star Q(f)$ after ideal sampling. In contrast, discrete sampling with a square pulse yields the spectrum $Q_{\rm A}(f)$ corresponding to the lower graph.

Spectrum when discretely sampled with a square pulse

You can see:

Each of the infinitely many partial spectra now has a different shape. Only the middle spectrum around $f = 0$.&nbsp is important;
All other spectral components are removed at the receiver side by the low pass of the signal reconstruction.

If one uses for this low pass again a rectangular filter with the gain around $T_{\rm A}/T_{\rm R}$ in the passband, one obtains for the output spectrum:

$$V(f) = Q(f) \cdot {\rm si}(\pi f T_{\rm R}) \hspace{0.05cm}.$$

$\text{Conclusion:}$ With discrete sampling and rectangular filtering, attenuation distortions gaccording to the weighting function ${\rm si}(πfT_{\rm R})$.

These are the stronger, the larger $T_{\rm R}$ is. Only in the limiting case $T_{\rm R} → 0$ holds ${\rm si}(πfT_{\rm R}) = 1$.

However, ideal equalization can fully compensate for these linear attenuation distortions.
To obtain $V(f) = Q(f)$ respectively, $v(t) = q(t)$ then must hold:

$$H(f) = \left\{ \begin{array}{l} (T_{\rm A}/T_{\rm R})/{\rm si}(\pi f T_{\rm R}) \\ 0 \\ \end{array} \right.\quad\begin{array}{*{5}c}{\rm{for} }\\{\rm{for} } \\ \end{array}\begin{array}{*{10}c} {\hspace{0.04cm}\left \vert \hspace{0.005cm} f\hspace{0.05cm} \right \vert < f_{\rm A}/2}\hspace{0.05cm}, \\ {\hspace{0.04cm}\left \vert \hspace{0.005cm} f\hspace{0.05cm} \right \vert > f_{\rm A}/2} \\ \end{array}$$

Quantization and quantization noise

The second functional unit Quantization of the PCM transmitter is used for value discretization.

For this purpose the whole value range of the analog source signal $($for example the range $± q_{\rm max})$ is divided into $M$ intervals.
Each sample $q_{\rm A}(ν ⋅ T_{\rm A})$ is then assigned a representative $q_{\rm Q}(ν ⋅ T_{\rm A})$ of the associated interval (for example, the interval center) .

$\text{Example 2:}$ The graph illustrates quantization using the quantization step number as an example $M = 8$.

To illustrate quantization with $M = 8$ steps

In fact, a power of two is always chosen for $M$ in practice because of the subsequent binary coding.
Each of the samples marked by circles $q_{\rm A}(ν - T_{\rm A})$ is replaced by the corresponding quantized value $q_{\rm Q}(ν - T_{\rm A})$ The quantized values are entered as crosses.
However, this process of value discretization is associated with an irreversible falsification.
The falsification $ε_ν = q_{\rm Q}(ν - T_{\rm A}) \ - \ q_{\rm A}(ν - T_{\rm A})$ depends on the quantization level number $M$ The following bound applies:

$$\vert \varepsilon_{\nu} \vert < {1}/{2} \cdot2/M \cdot q_{\rm max}= {q_{\rm max} }/{M}\hspace{0.05cm}.$$

$\text{Definition:}$ One refers to the root mean square error magnitude $ε_ν$ as quantization noise power:

$$P_{\rm Q} = \frac{1}{2N+1 } \cdot\sum_{\nu = -N}^{+N}\varepsilon_{\nu}^2 \approx \frac{1}{N \cdot T_{\rm A} } \cdot \int_{0}^{N \cdot T_{\rm A} }\varepsilon(t)^2 \hspace{0.05cm}{\rm d}t \hspace{0.3cm} {\rm with}\hspace{0.3cm}\varepsilon(t) = q_{\rm Q}(t) - q(t) \hspace{0.05cm}.$$

Notes:

For calculating the quantization noise power $P_{\rm Q}$ the given approximation of "spontaneous quantization" is usually used.
Here, one ignores sampling and forms the error signal from the continuous-time signals $q_{\rm Q}(t)$ and $q(t)$.
$P_{\rm Q}$ also depends on the source signal $q(t)$ . Assuming that $q(t)$ takes all values between $±q_{\rm max}$ with equal probability and the quantizer is designed exactly for this range, we get accordingly Exercise 4.4:

$$P_{\rm Q} = \frac{q_{\rm max}^2}{3 \cdot M^2 } \hspace{0.05cm}.$$

In a speech or music signal, arbitrarily large amplitude values can occur - even if only very rarely. In this case, for $q_{\rm max}$ usually that amplitude value is used which is exceeded only at $1\%$ all times (in amplitude).

PCM encoding and decoding

The block PCM coding is used to convert the discrete-time (after sampling) and discrete-value (after quantization with $M$ steps) signal values $q_{\rm Q}(ν - T_{\rm A})$ into a sequence of $N = {\rm log_2}(M)$ binary values. Logarithm to base 2 ⇒ Binary logarithm.

{{GraueBox|TEXT= $\text{Example 3:}$ Each binary value ⇒ bit is represented by a rectangle of duration $T_{\rm B} = T_{\rm A}/N$ resulting in the signal $q_{\rm C}(t)$ .

[[File: Mod_T_4_1_S5a_vers2.png|center|frame | PCM coding with the dual code $(M = 8,\ N = 3)$]

You can see:

The dual code is used here. This means that the quantization intervals $\mu$ are numbered consecutively from $0$ to $M-1$ and then written in simple binary. With $M = 8$ for example $\mu = 6$ ⇔ 110.
The three binary symbols of the coded signal $q_{\rm C}(t)$ are obtained by replacing 0 by L ("Low") and 1 by H ("High"). In the example, this gives: HHL HHL LLH LHL HLH LHH.
The bit duration $T_{\rm B}$ is here shorter than the sampling distance by a factor $N = {\rm log_2}(M) = 3$ $T_{\rm A} = 1/f_{\rm A}$, and the bit rate is $R_{\rm B} = {\rm log_2}(M) - f_{\rm A}$.
If one uses the same mapping in decoding $(v_{\rm C} ⇒ v_{\rm Q})$ as in coding $(q_{\rm Q} ⇒ q_{\rm C})$, then, if there are no transmission errors: $v_{\rm Q}(ν - T_{\rm A}) = q_{\rm Q}(ν - T_{\rm A}). $
An alternative to dual code is Gray code, where adjacent binary values differ only in one bit. For $N = 3$:

$\mu = 0$: LLL, $\mu = 1$: LLH, $\mu = 2$: LHH, $\mu = 3$: LHL, $\mu = 4$: HHL, $\mu = 5$: HHH, $\mu =6$: HLH, $\mu = 7$: HLL. }}

Signal-to-noise power ratio

The digital pulse code modulation $\rm (PCM)$ is now compared to the analog modulation methods $\rm (AM, \ FM)$ regarding the achievable sink SNR $ρ_v = P_q/P_ε$ with AWGN noise.

Sink SNR at AM, FM, PCM 30/32

As denoted in previous chapters (for example) $ξ = {α_{\rm K}}^2 - P_{\rm S}/(N_0 - B_{\rm NF})$ the power parameter. This summarizes different influences:

the channel transmission factor $α_{\rm K}$ (quadratic),
the transmit power $P_{\rm S}$,
the AWGN noise power density $N_0$ (reciprocal), and.
the signal bandwidth $B_{\rm NF}$ (also reciprocal);
for a harmonic oscillation: Frequency $f_{\rm N}$ instead of $B_{\rm NF}$.

The two comparison curves for amplitude modulation (AM) and for frequency modulation (FM) can be described as follows:

Two-sideband FM without carrier:

$$ρ_v = ξ \ ⇒ \ 10 · \lg ρ_v = 10 · \lg \ ξ,$$

Frequency modulation with $η = 3$:

$$ρ_υ = 3/2 \cdot η^2 - ξ = 13.5 - ξ \ ⇒ \ 10 · \lg \ ρ_v = 10 · \lg \ ξ + 11.3 \ \rm dB.$$

The curve for the PCM 30/32 system should be interpreted as follows:

If the power parameter $ξ$ is sufficiently large, then no transmission errors occur. The error signal $ε(t) = v(t) \ - \ q(t)$ is then due to quantization alone $(P_ε = P_{\rm Q})$.
With the quantization step number $M = 2^N$ holds approximately in this case:

$$\rho_{v} = \frac{P_q}{P_\varepsilon}= M^2 = 2^{2N} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{v}=20 \cdot {\rm lg}\hspace{0.1cm}M = N \cdot 6.02\,{\rm dB}$$

$$ \Rightarrow \hspace{0.3cm} N = 8, \hspace{0.05cm} M =256\text{:}\hspace{0.2cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{v}=48.16\,{\rm dB}\hspace{0.05cm}.$$

Note that the given equation is exactly valid only for a sawtooth shaped source signal. However, for cosine shaped source signal the deviation from this is not very large.
As $ξ$ (smaller transmit power or larger noise power density) decreases, the transmission errors increase. Thus $P_ε > P_{\rm Q}$ and the sink-to-noise ratio becomes smaller.
The PCM $($with $M = 256)$ is superior to the analog methods $($AM and FM$)$ only in the lower and middle $ξ$-range. But if transmission errors do not play a role anymore, no improvement can be achieved by a larger $ξ$ (horizontal curve section with yellow background).
An improvement is only achieved by increasing $N$ (number of bits per sample) ⇒ larger $M = 2^N$ (number of quantization steps). For example, for a Compact Disc (CD) with parameter $N = 16$ ⇒ $M = 65536$ the value

$$10 · \lg \ ρ_v = 96.32 \ \rm dB.$$

$\text{Example 4:}$ The following graph shows the limiting influence of quantization:

White dotted is the source signal $q(t)$, green dotted is the sink signal $v(t)$ after PCM with $N = 4$ ⇒ $M = 16$.
Sampling times are marked by crosses.
Transfer errors are excluded for the time being. Sampling and signal reconstruction are best fit to $q(t)$ .

Influence of quantization with $N = 4$ and $N = 8$

This image can be interpreted as follows:

With $N = 8$ ⇒ $M = 256$ the sink signal $v(t)$ is indistinguishable from the source signal $q(t)$ with the naked eye. The white dotted signal curve applies approximately to both.
From the signal-to-noise ratio $10 · \lg \ ρ_v = 47.8 \ \rm dB$ however, it can be seen that the quantization noise (power $P_\varepsilon$ of the error signal) is only reduced by a factor $1. 6 - 10^{-5}$ smaller than the power $P_q$ of the source signal. This SNR would already be clearly audible with a speech or music signal.
Although the source signal considered here is neither sawtooth nor cosine shaped, but is composed of several frequency components, the given approximation $ρ_v ≈ M^2$ ⇒ $10 · \lg \ ρ_υ = 48.16 \ \rm dB$ deviates only insignificantly from the actual value.
In contrast, for $N = 4$ ⇒ $M = 16$ deviations between the sink signal (marked in green) and the source signal (marked in white) can already be seen in the image, which is also quantitatively expressed by the very small signal-to-noise ratio $10 · \lg \ ρ_υ = 28.2 \ \rm dB$

.

Influence of transmission errors

Starting from the same analog signal $q(t)$ as in the last section and a linear quantization with $N = 8$ bits ⇒ $M = 256$ the effects of transmission errors are now illustrated using the respective sink signal $v(t)$ .

Influence of a transmission error concerning Bit 5' at the dual code

The white dots again mark the source signal $q(t)$. Without transmission error the sink signal $v(t)$ has the same course when neglecting quantization.
Now, exactly one bit of the fifth sample at a time $q(5 - T_{\rm A}) = -0.715$ is corrupted, where this sample has been coded as LLHL LHLL' . This graph is based on dual code, meaning that the lowest quantization interval $(\mu = 0)$ is represented with LLLL LLLL' and the highest interval $(\mu = 255)$ is represented with HHHH HHHH .

Table showing the results of the bit error analysis

The table shows the results of this analysis:

The specified signal-to-noise ratio $10 · \lg \ ρ_v$ was calculated from the presented (very short) signal section of duration $10 - T_{\rm A}$ .

For each transmission error of $10 - 8 = 80$ bits, this corresponds to a bit error rate of $1.25\%$.

The results of this error analysis shown in the graph and table can be summarized as follows:

If only the last bit of the binary word is corrupted $($LSB: Least Significant Bit, LLHL LHLL ⇒ LLHL LHLH$)$, then no difference from error-free transmission is visible to the naked eye $($white curve$)$. Nevertheless, the signal-to-noise ratio is reduced by $3.5 \ \rm dB$ .
A transmission error of the fourth last bit $($green curve, LLHLLHLL ⇒ LLHLHHLL$)$ already leads to a clearly detectable distortion by eight quantization intervals. That is, $v(5T_{\rm A}) \ - \ q(5T_{\rm A}) = 8/256 - 2 = 0.0625$ and the signal-to-noise ratio drops to $10 · \lg \ ρ_υ = 28.2 \ \rm dB$.
Finally, the red curve shows the case where the MSB (Most Significant Bit) is corrupted: LLHLLHLL ⇒ HLHLL. This leads to distortion $v(5T_{\rm A}) \ - \ q(5T_{\rm A}) = 1$ (corresponding to half the modulation range). The signal-to-noise ratio is now only about $4 \ \rm dB$.
At all sampling times except $5T_{\rm A}$ matches $v(t)$ exactly except for the quantization error with $q(t)$ . Outside these time points marked by yellow crosses, however, the single error at $5T_{\rm A}$ leads to strong deviations in an extended range, which is due to the interpolation with the $\rm si$-shaped impulse response of the reconstruction low-pass $H(f)$ .

Estimation of SNR degradation due to transmission errors

Now we will try to determine the SNR curve of the PCM system taking into account bit errors, at least approximately. We start from the following block diagram and further assume:

Each sample $q_{\rm A}(νT)$ is quantized by $M$ stages and quantized by $N = {\rm log_2} (M)$ binary sign (bit). In the example $M = 8$ ⇒ $N = 3$.
The binary representation of $q_{\rm Q}(νT)$ yields the amplitude coefficients $a_k\, (k = 1, \text{...} \hspace{0.08cm}, N),$ which can be corrupted by bit errors in the coefficients $b_k$ .
Both $a_k$ and $b_k$ are $±1$, respectively.
A bit error $(b_k ≠ a_k)$ occurs with probability $p_{\rm B}$ .
Each bit is equally likely to be corrupted and in each PCM word there is at most one error ⇒ only one of the $N$ bits can be wrong.

For calculating PCM SNR with bit errors taken into account

From the diagram given in the graph, it can be seen for $N = 3$ and natural binary coding (dual code):

A corruption of $a_1$ changes the quantized value $q_{\rm Q}(νT)$ by $±A$.
A corruption of $a_2$ changes the quantized value $q_{\rm Q}(νT)$ by $±A/2.$.
A corruption of $a_3$ changes the quantized value value $q_{\rm Q}(νT)$ by $±A/4$.

By generalization, we obtain for the deviation $ε_k = υ_{\rm Q}(νT) \ - \ q_{\rm Q}(νT)$ for the case when the amplitude coefficient $a_k$ was transferred incorrectly:

$$\varepsilon_k = - a_k \cdot A \cdot 2^{-k +1} \hspace{0.05cm}.$$

For the error noise power after averaging over all corruption values $ε_k$ (with $1 ≤ k ≤ N)$ taking into account the bit error probability $p_{\rm B}$:

$$P_{\rm F}= {\rm E}\big[\varepsilon_k^2 \big] = \sum\limits^{N}_{k = 1} p_{\rm B} \cdot \left ( - a_k \cdot A \cdot 2^{-k +1} \right )^2 =\ p_{\rm B} \cdot A^2 \cdot \sum\limits^{N-1}_{k = 0} 2^{-2k } = p_{\rm B} \cdot A^2 \cdot \frac{1- 2^{-2N }}{1- 2^{-2 }} \approx {4}/{3} \cdot p_{\rm B} \cdot A^2 \hspace{0.05cm}.$$

Here the summation formula of the geometric series and the approximation $1 - 2^{-2N } ≈ 1$ are used.
For $N = 8$ ⇒ $M = 256$ the associated relative error is, for example, about $\rm 10^{-5}$.

Excluding transmission errors, the signal-to-noise power ratio $ρ_v = P_{\rm S}/P_{\rm Q}$ has been found, where for a uniformly distributed source signal (for example, sawtooth-shaped) the signal power and quantization noise power are to be calculated as follows:

Sink SNR for PCM considering bit errors

$$P_{\rm S}={A^2}/{3}\hspace{0.05cm},\hspace{0.3cm}P_{\rm Q}= {A^2}/{3} \cdot 2^{-2N } \hspace{0.05cm}.$$

Taking into account the transfer errors, the above result gives:

$$\rho_{\upsilon}= \frac{P_{\rm S}}{P_{\rm Q}+P_{\rm F}} = \frac{A^2/3}{A^2/3 \cdot 2^{-2N } + A^2/3 \cdot 4 \cdot p_{\rm B}} = \frac{1}{ 2^{-2N } + 4 \cdot p_{\rm B}} \hspace{0.05cm}.$$

The graph shows $10 - \lg ρ_v$ as a function of the (logarithmized) power parameter $ξ = P_{\rm S}/(N_0 - B_{\rm NF})$, where $B_{\rm NF}$ indicates the signal bandwidth. Let the constant channel transmission factor be ideally $α_{\rm K} = 1$.

But for the optimal binary system and AWGN noise, the power parameter is also $ξ = E_{\rm B}/N_0$ (energy per bit related to noise power density).
The bit error probability is then given by the Gaussian error function ${\rm Q}(x)$ as follows:

$$p_{\rm B}= {\rm Q} \left ( \sqrt{{2E_{\rm B}}/{N_0} }\right ) \hspace{0.05cm}.$$

For $N = 8$ ⇒ $ 2^{-2{\it N} } = 1.5 - 10^{-5}$ and $10 - \lg \ ξ = 6 \ \rm dB$ ⇒ $p_{\rm B} = 0.0024$ (point marked in red) results:

$$\rho_{\upsilon}= \frac{1}{ 1.5 \cdot 10^{-5} + 4 \cdot 0.0024} \approx 100 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}10 \cdot {\rm lg} \hspace{0.15cm}\rho_{\upsilon}\approx 20\,{\rm dB} \hspace{0.05cm}.$$

This small $ρ_v$ value goes back to the term $4 · 0.0024$ in the denominator (influence of the transmission error) while in the horizontal section of the curve for each $N$ (number of bits per sample) the term $\rm 2^{-2{\it N} }$ dominates - i.e. the quantization noise.

Nonlinear quantization

Often the quantization intervals are not chosen equally large, but one uses a finer quantization for the inner amplitude range than for large amplitudes. There are several reasons for this:

In audio signals, distortions of the quiet signal components (i.e. values near the zero line) are subjectively perceived as more disturbing than an impairment of large amplitude values.
Such an uneven quantization also leads to a larger sink-interval for such a music or speech signal, because here the signal amplitude is not evenly distributed.

The graph shows a speech signal $q(t)$ and its amplitude distribution $f_q(q)$ ⇒ Probability density function.

Uniform quantization of a speech signal

This is the Laplace distribution, which can be approximated as follows:

by a continuous two-sided exponential distribution, and.
by a Dirac function $δ(q)$ to account for the speech pauses (magenta colored).

In the graph, nonlinear quantization is only implied, for example, by means of the 13-segment characteristic, which is described in more detail in the Exercise 4.5 :

The quantization intervals here become wider and wider towards the edges section by section.
The more frequent small amplitudes, on the other hand, are quantized very finely.

Compression and expansion

Non-uniform quantization can be realized, for example, by.

the sampled values $q_{\rm A}(ν - T_{\rm A})$ are first deformed by a nonlinear characteristic $q_{\rm K}(q_{\rm A})$ and
subsequently, the resulting output values $q_{\rm K}(ν · T_{\rm A})$ are uniformly quantized.

Realization of a non-uniform quantization

This results in the signal chain sketched opposite.

$\text{Conclusion:}$ Such non-uniform quantization means:

Through the nonlinear characteristic $q_{\rm K}(q_{\rm A})$ small signal values are amplified and large values are attenuated ⇒ Compression.
This deliberate signal distortion is undone at the receiver by the inverse function $v_{\rm E}(υ_{\rm Q})$ ⇒ expansion.
The total process of transmit-side compression and receiver-side expansion is also called 'companding.

For the PCM system 30/32, the Comité Consultatif International des Télégraphique et Téléphonique (CCITT) recommended the so-called A characteristic:

$$y(x) = \left\{ \begin{array}{l} \frac{1 + {\rm ln}(A \cdot x)}{1 + {\rm ln}(A)} \\ \frac{A \cdot x}{1 + {\rm ln}(A)} \ - \frac{1 + {\rm ln}( - A \cdot x)}{1 + {\rm ln}(A)} \end{array} \right.\quad\begin{array}{*{5}c}{\rm{for}}\\{\rm{for}}\\{\rm{for}} \end{array}\begin{array}{*{10}c}1/A \le x \le 1\hspace{0.05cm}, \ - 1/A \le x \le 1/A\hspace{0.05cm}, \ - 1 \le x \le - 1/A\hspace{0.05cm}. \end{array}$$

Here, for abbreviation $x = q_{\rm A}(ν - T_{\rm A})$ and $y = q_{\rm K}(ν - T_{\rm A})$ is used.
This characteristic curve with the value $A = 87.56$ introduced in practice has a constantly changing slope.
For more details on this type of non-uniform quantization, see the Exercise 4.5.

Note: In the third part of the tutorial video Pulse code modulation are covered: