Pulse Modulation

Linear prediction estimates the present sample of a signal as a linear combination of its past samples. It is the basis of DPCM, ADPCM and speech coders (LPC), because only the small prediction error needs to be transmitted.

Predictor. With $p$ past samples:

$$\hat{x}[n] = \sum_{k=1}^{p} w_k\, x[n-k]$$

The prediction error is

$$e[n] = x[n] - \hat{x}[n]$$

x[n] --+------------------>( + )--> e[n]
       |                     ^ -
       v                     |
   [z^-1]->[z^-1]-> ... ->[z^-1]   (tapped delay line)
     |w1     |w2            |wp
     +-------+---- ... -----+--> sum = x^[n]

Optimum coefficients. The weights are chosen to minimise the mean-square error

$$J = E\{e^2[n]\} = E\left\{\left(x[n] - \sum_{k=1}^{p} w_k x[n-k]\right)^2\right\}$$

Setting $\partial J/\partial w_k = 0$ gives the normal (Wiener–Hopf) equations:

$$\sum_{j=1}^{p} w_j\, R_x(k-j) = R_x(k), \qquad k = 1, 2, \ldots, p$$

or in matrix form $\mathbf{R}\mathbf{w} = \mathbf{r}$, so $\mathbf{w}_o = \mathbf{R}^{-1}\mathbf{r}$, where $R_x(k)$ is the autocorrelation of the input. The minimum error is

$$J_{min} = R_x(0) - \mathbf{r}^T\mathbf{w}_o$$

Example: first-order predictor ($p = 1$): $w_1 = R_x(1)/R_x(0) = \rho$, and $J_{min} = R_x(0)(1-\rho^2)$. For highly correlated speech ($\rho \approx 0.9$), the error variance is only 19% of the signal variance.

Key points:

• Prediction gain $G_p = \sigma_x^2/\sigma_e^2$ gives an SQNR improvement in DPCM of $10\log G_p$ dB (typically 4–11 dB for speech).
• If the statistics change with time, an adaptive predictor updates the weights (e.g. by the LMS algorithm), as in ADPCM.
• In DPCM the predictor works on quantised samples, so transmitter and receiver predictions stay identical.

Need for non-uniform quantization

In uniform quantization the step size $\Delta$ is fixed, so the quantization noise power $\Delta^2/12$ is the same for all signal levels. Hence:

• Weak signals (low amplitude) get a poor SQNR, while loud signals get a high SQNR.
• Speech has a wide dynamic range (about 40 dB) and small amplitudes occur much more often than large ones.
• To keep weak signals acceptable with uniform steps, many more bits would be needed.

Non-uniform quantization uses small steps for small amplitudes and large steps for large amplitudes. This gives nearly constant SQNR over a wide range of input levels with fewer bits (8 bits instead of about 12 for telephony). It is implemented by companding ($\mu$-law, A-law).

SQNR of delta modulation (no slope overload)

Let the message be a single tone $m(t) = A_m\cos 2\pi f_m t$, step size $\Delta$, sampling rate $f_s = 1/T_s$, and the receiver LPF bandwidth $W$.

1. Condition for no slope overload. The staircase can rise by at most $\Delta$ per $T_s$:

$$\left|\frac{dm}{dt}\right|_{max} = 2\pi f_m A_m \le \frac{\Delta}{T_s} \;\Rightarrow\; A_{m,max} = \frac{\Delta f_s}{2\pi f_m}$$

2. Signal power (with maximum amplitude):

$$S = \frac{A_m^2}{2} = \frac{\Delta^2 f_s^2}{8\pi^2 f_m^2}$$

3. Granular noise power. With no slope overload, the error $q(t)$ lies within $\pm\Delta$ and is uniformly distributed:

$$E[q^2] = \frac{1}{2\Delta}\int_{-\Delta}^{\Delta} q^2\,dq = \frac{\Delta^2}{3}$$

This power is spread uniformly over $0$ to $f_s$. The output LPF passes only $W$, so

$$N_q = \frac{\Delta^2}{3}\cdot\frac{W}{f_s}$$

4. SQNR:

$$\text{SQNR} = \frac{S}{N_q} = \frac{\Delta^2 f_s^2/(8\pi^2 f_m^2)}{\Delta^2 W/(3f_s)} = \frac{3 f_s^3}{8\pi^2 f_m^2 W}$$

In dB:

$$\text{SQNR}_{dB} = 30\log_{10} f_s - 20\log_{10} f_m - 10\log_{10} W - 14.2$$

So SQNR improves by 9 dB each time $f_s$ is doubled. Example: $f_s = 64$ kHz, $f_m = 1$ kHz, $W = 4$ kHz gives SQNR $= 2490 \approx 33.96$ dB.

Companding (compressing + expanding) is a way to obtain non-uniform quantization: the signal is compressed by a non-linear amplifier before a uniform quantizer at the transmitter, and expanded by the inverse characteristic at the receiver.

m(t)->[Compressor]->[Uniform Q]->channel->[Expander]->m(t)
       y = c(x)                          x = c^-1(y)

Why it is necessary

• With uniform quantization, noise $\Delta^2/12$ is fixed, so weak signals have poor SQNR.
• Speech has a large dynamic range and small amplitudes are most frequent.
• The compressor boosts small amplitudes and reduces large ones, so effectively small inputs see fine steps and large inputs coarse steps. The SQNR stays nearly constant (within a few dB) over about 40 dB of input range with only 8 bits.

μ-law companding (North America, Japan)

$$y = \frac{\ln(1 + \mu|x|)}{\ln(1+\mu)}\,\text{sgn}(x), \qquad 0 \le |x| \le 1$$

• $\mu = 0$ gives uniform quantization; the standard value is $\mu = 255$ (8-bit PCM).
• For small $x$ it is nearly linear ($y \approx \mu x/\ln(1+\mu)$); for large $x$ it is logarithmic.
• Example with $\mu = 255$: $x = 0.01 \to y = 0.228$; $x = 0.1 \to y = 0.591$; $x = 0.5 \to y = 0.876$.

A-law companding (Europe, India, Nepal – ITU-T G.711)

$$y = \begin{cases} \dfrac{A|x|}{1+\ln A}\,\text{sgn}(x), & 0 \le |x| \le \dfrac{1}{A} \\[2mm] \dfrac{1+\ln(A|x|)}{1+\ln A}\,\text{sgn}(x), & \dfrac{1}{A} \le |x| \le 1 \end{cases}$$

• Standard value $A = 87.6$. It is linear for small signals and logarithmic for large ones.
• In practice both laws are realised by a 13-segment (A-law) or 15-segment (μ-law) piecewise-linear approximation.

 y (out)
 1 |            . . . .   <- compressed
   |      . .
   |   .          ....  <- uniform (y = x)
   | .       ....
   |.   ....
   |. ..
   +-------------------- x (in)
   0                  1

Why DPCM is preferred over PCM

Speech and video samples taken at or above the Nyquist rate are highly correlated: adjacent samples differ only slightly. PCM codes every sample fully, so it sends much redundant information.

DPCM predicts each sample from past samples and quantizes only the difference (prediction error). Since the error has a much smaller range than the signal:

• For the same number of bits, the step size is smaller, so SQNR improves by the prediction gain $G_p = \sigma_m^2/\sigma_e^2$ (typically 4–11 dB for speech).
• For the same SQNR, fewer bits per sample are needed, so bit rate and bandwidth are lower (e.g. 32 kbps ADPCM vs 64 kbps PCM).

DPCM transmitter

m[n]-->(+)--e[n]-->[Quantizer]--+--eq[n]->[Encoder]->
        ^ -                     |
        |                       v
      m^[n]                    (+)<------+
        |                       | mq[n]  |
        +----[Predictor]<-------+        |
                  |                      |
                  +----------------------+

1. Prediction error: $e[n] = m[n] - \hat{m}[n]$.
2. Quantized error: $e_q[n] = e[n] + q[n]$, which is encoded and transmitted.
3. The predictor input is the reconstructed sample $m_q[n] = \hat{m}[n] + e_q[n]$.
4. Then $m_q[n] = \hat{m}[n] + e[n] + q[n] = m[n] + q[n]$: the quantized sample differs from the original only by the quantization error of $e[n]$, so errors do not accumulate.
5. The predictor (a linear FIR filter, $\hat{m}[n] = \sum w_k m_q[n-k]$) works on $m_q$, the same signal available at the receiver.

DPCM receiver

in->[Decoder]--eq[n]-->(+)--+--> mq[n] -->[LPF]--> m(t)
                       ^    |
                  m^[n]|    |
                       +--[Predictor]

The decoder recovers $e_q[n]$; the same predictor as in the transmitter gives $\hat{m}[n]$; their sum gives $m_q[n] = m[n] + q[n]$. A low-pass filter reconstructs the analog signal.

Note: delta modulation is a special case of DPCM with a 1-bit quantizer and a one-sample delay as the predictor.

SQNR is the ratio of signal power to quantization noise power at the output of a quantizer.

Step 1: Step size

Let the message $m(t)$ lie in $[-m_p, m_p]$ and be quantized uniformly into $L$ levels. Then the step size is

$$\Delta = \frac{2m_p}{L}$$

Step 2: Quantization noise power

The quantization error $q = m - m_q$ lies in $[-\Delta/2, \Delta/2]$. For a fine quantizer it is uniformly distributed, with pdf $f(q) = 1/\Delta$:

$$\begin{aligned} N_q = E[q^2] &= \int_{-\Delta/2}^{\Delta/2} q^2\cdot\frac{1}{\Delta}\,dq = \frac{1}{\Delta}\left[\frac{q^3}{3}\right]_{-\Delta/2}^{\Delta/2} \\ &= \frac{\Delta^2}{12} = \frac{(2m_p/L)^2}{12} = \frac{m_p^2}{3L^2} \end{aligned}$$

Step 3: SQNR in terms of levels

With message power $P = \overline{m^2(t)}$:

$$\text{SQNR} = \frac{P}{N_q} = \frac{3L^2\, P}{m_p^2}$$

Step 4: SQNR in terms of bits

With binary coding of $n$ bits per sample, $L = 2^n$:

$$\text{SQNR} = \frac{3P}{m_p^2}\,2^{2n} = \frac{3P}{m_p^2}\,4^n$$ $$\text{SQNR}_{dB} = 10\log_{10}\!\left(\frac{3P}{m_p^2}\right) + 6.02\,n\ \text{dB}$$

Special case: full-scale sinusoid

For $m(t) = m_p\cos\omega_m t$, $P = m_p^2/2$:

$$\text{SQNR} = \frac{3L^2}{2} = 1.5L^2 = 1.5\times 2^{2n}$$ $$\text{SQNR}_{dB} = 1.76 + 6.02\,n\ \text{dB}$$

Example: $n = 8$ bits gives SQNR $= 1.76 + 48.16 = 49.92$ dB.

Conclusions:

• SQNR increases as the square of the number of levels.
• SQNR increases exponentially with bits: each extra bit adds about 6 dB.
• But each extra bit increases the bit rate $n f_s$ and the bandwidth $n f_s/2$, so there is a trade-off between SQNR and bandwidth.

Nyquist sampling theorem

A band-limited signal with no frequency components above $f_m$ Hz is completely described by, and can be exactly recovered from, its samples taken uniformly at a rate

$$f_s \ge 2f_m$$

The minimum rate $f_s = 2f_m$ is the Nyquist rate and $T_s = 1/(2f_m)$ is the Nyquist interval. Recovery is done by passing the samples through an ideal low-pass filter of cut-off $f_m$. If $f_s < 2f_m$, spectral copies overlap (aliasing) and recovery is impossible.

Nyquist rate and interval of the given signal

$$x(t) = 6\cos 50\pi t + 20\sin 300\pi t - 10\cos 100\pi t$$

Frequencies of the components ($\omega = 2\pi f$):

Highest frequency: $f_m = 150$ Hz.

$$\text{Nyquist rate } f_N = 2f_m = 2 \times 150 = 300\ \text{Hz (samples/s)}$$ $$\text{Nyquist interval } T_N = \frac{1}{f_N} = \frac{1}{300} = 3.33\ \text{ms}$$

Effect of quantizing with 512 levels

$$n = \log_2 512 = 9\ \text{bits/sample}$$

At the Nyquist rate the bit rate is

$$R_b = n f_N = 9 \times 300 = 2700\ \text{bits/s}$$

and the minimum transmission bandwidth is $R_b/2 = 1350$ Hz.

Answer: Nyquist rate = 300 Hz, Nyquist interval = 3.33 ms (9 bits/sample, 2.7 kbps).

Delta modulation (DM)

Delta modulation is 1-bit DPCM: the signal is oversampled, and only one bit per sample tells whether the signal is above or below the previous staircase approximation, which then moves up or down by a fixed step $\Delta$.

Equations:

$$\begin{aligned} e[n] &= m[n] - m_q[n-1] \\ e_q[n] &= \Delta\,\text{sgn}(e[n]) \quad (\text{bit 1 for } +\Delta,\ 0 \text{ for } -\Delta) \\ m_q[n] &= m_q[n-1] + e_q[n] = \sum_{i=1}^{n} e_q[i] \end{aligned}$$

So the receiver output is the accumulated (integrated) sum of $\pm\Delta$ steps.

DM transmitter

m(t)->[Sampler]->m[n]->(+)--e[n]-->[1-bit    ]--+--> bits
                        ^ -        [quantizer]  |
                        |                       v
                  mq[n-1]                     (+)
                        |                       |
                        +--[Delay z^-1]<--mq[n]-+
                         (accumulator)

DM receiver

bits->[Decoder +/-D]->(+)--+--> mq[n] --> [LPF] --> m(t)
                       ^   |
                       +-[z^-1]  (accumulator)

Noise in DM:

• Slope overload: if the signal slope exceeds $\Delta/T_s$, the staircase cannot follow. No overload needs $\left|\frac{dm}{dt}\right|_{max} \le \Delta f_s$; for a tone, $A_m \le \frac{\Delta f_s}{2\pi f_m}$.
• Granular noise: when the signal is nearly flat, the staircase hunts by $\pm\Delta$.
• SQNR (no overload, single tone): $\text{SQNR} = \frac{3f_s^3}{8\pi^2 f_m^2 W}$.

Adaptive delta modulation (ADM)

In ADM the step size is varied according to the signal slope: large steps for steep parts (to avoid slope overload) and small steps for flat parts (to reduce granular noise).

A common rule (Jayant / song algorithm):

$$\Delta[n] = \begin{cases} K\,\Delta[n-1], & b[n] = b[n-1] \\ \Delta[n-1]/K, & b[n] \ne b[n-1] \end{cases} \qquad (K \approx 1.5)$$

with limits $\Delta_{min} \le \Delta[n] \le \Delta_{max}$. Consecutive equal bits mean the staircase is lagging (step grows); alternating bits mean granular hunting (step shrinks).

m(t)->(+)->[1-bit Q]--+--------------> bits
       ^-             |
       |        [Step-size logic]
       |              | D[n]
       |              v
       +--[Accum.]<-[x]
Receiver: same step logic + accumulator + LPF

The receiver uses the same step-size logic on the received bits, so no side information is needed.

Merits and demerits

Comparison of PCM, DPCM and DM

SQNR of PCM

Let $m(t)$ have peak value $m_p$ and be quantized uniformly into $L = 2^n$ levels.

Step size: $\Delta = \dfrac{2m_p}{L}$

Noise power: the error $q$ is uniform in $(-\Delta/2, \Delta/2)$:

$$N_q = \int_{-\Delta/2}^{\Delta/2}\frac{q^2}{\Delta}\,dq = \frac{\Delta^2}{12} = \frac{m_p^2}{3L^2}$$

SQNR: with signal power $P$,

$$\text{SQNR} = \frac{P}{N_q} = \frac{3L^2P}{m_p^2} = \frac{3P}{m_p^2}\,2^{2n}$$

For a full-scale sinusoid, $P = m_p^2/2$:

$$\text{SQNR} = 1.5\times 2^{2n} \quad\Rightarrow\quad \text{SQNR}_{dB} = 1.76 + 6.02\,n$$

Each extra bit adds about 6 dB. Example: 8-bit PCM gives $1.76 + 48.16 = 49.92$ dB.

Aperture effect in flat-top sampling

In flat-top sampling each sample is held for a pulse width $T$. The flat-top signal is the ideally sampled signal convolved with a rectangular pulse $h(t)$ of width $T$:

$$S(f) = f_s\sum_k M(f - kf_s)\,H(f), \qquad H(f) = T\,\text{sinc}(fT)\,e^{-j\pi fT}$$

The message spectrum is multiplied by $\text{sinc}(fT)$, so high frequencies are attenuated. This amplitude distortion (plus a delay $T/2$) is the aperture effect. It is reduced by keeping $T \ll T_s$ (duty ratio ≤ 0.1) or by an equalizer after the reconstruction filter with response $1/H(f)$.

DPCM scheme

PCM sends full samples although adjacent samples are highly correlated. DPCM sends only the quantized prediction error:

m[n]->(+)-e[n]->[Quant.]--+--eq[n]-->[Encoder]--> out
       ^-                 |
     m^[n]               (+)<-+
       |                  |   |
       +--[Predictor]<-mq[n]  |
              +---------------+
Receiver: eq[n]->(+)->mq[n]->[LPF]; feedback via
          the same predictor gives m^[n]

$e[n] = m[n] - \hat m[n]$, $m_q[n] = \hat m[n] + e_q[n] = m[n] + q[n]$. Since $e[n]$ is small, fewer bits are needed, or SQNR improves by the prediction gain.

Numerical: maximum amplitude without slope overload

Given: $W = 3$ kHz, $f_s = 5 \times$ Nyquist rate, $f_m = 2$ kHz, $\Delta = 250$ mV.

$$f_s = 5 \times 2W = 5 \times 6\ \text{kHz} = 30\ \text{kHz}$$

No slope overload requires $2\pi f_m A_m \le \Delta f_s$:

$$A_{m,max} = \frac{\Delta f_s}{2\pi f_m} = \frac{0.25 \times 30000}{2\pi \times 2000} = \frac{7500}{12566.4} = 0.597\ \text{V}$$

Answer: maximum amplitude $\approx 0.597$ V (597 mV).

SQNR of uniformly quantized PCM

Let the message $m(t)$ have peak amplitude $m_p$ (range $-m_p$ to $+m_p$) and be quantized uniformly into $L$ levels.

Step size:

$$\Delta = \frac{2m_p}{L}$$

Quantization noise. The error $q$ is uniformly distributed over $(-\Delta/2, \Delta/2)$ with pdf $1/\Delta$:

$$N_q = E[q^2] = \int_{-\Delta/2}^{\Delta/2} q^2\frac{1}{\Delta}\,dq = \frac{\Delta^2}{12} = \frac{m_p^2}{3L^2}$$

Signal-to-quantization noise ratio. With signal power $P = \overline{m^2(t)}$:

$$\text{SQNR} = \frac{P}{N_q} = 3L^2\frac{P}{m_p^2}$$

In terms of bits. For $n$-bit binary coding, $L = 2^n$:

$$\text{SQNR} = \frac{3P}{m_p^2}\,2^{2n}, \qquad \text{SQNR}_{dB} = 10\log_{10}\frac{3P}{m_p^2} + 6.02\,n$$

Sinusoidal message with full-scale amplitude, $P = m_p^2/2$:

$$\text{SQNR} = 1.5\times 2^{2n}, \qquad \text{SQNR}_{dB} = 1.76 + 6.02\,n\ \text{dB}$$

(For a uniformly distributed signal, $P = m_p^2/3$ and SQNR $= 2^{2n}$, i.e. $6.02n$ dB.)

Relation between SQNR and bits

• SQNR grows exponentially with $n$ ($\propto 4^n$); in dB it grows linearly: each added bit improves SQNR by about 6 dB.
• Example: 7 bits → 43.9 dB; 8 bits → 49.9 dB (for a full-scale sine).
• But bandwidth $B_T = nf_s/2$ grows linearly with $n$. So SQNR improves exponentially with bandwidth: a key trade-off in PCM.

Practical considerations in sampling

1. Real signals are not strictly band-limited. An anti-aliasing low-pass filter is used before sampling to cut off components above $f_s/2$.
2. Sample above the Nyquist rate to provide a guard band, because practical reconstruction filters cannot have a sharp cut-off. Example: speech band-limited to 3.4 kHz is sampled at 8 kHz, not 6.8 kHz.
3. Finite pulse width: practical samples are pulses of width $\tau$ (natural or flat-top), not impulses. Flat-top samples cause the aperture effect, which needs equalization.
4. Non-ideal reconstruction filter: a practical LPF has a transition band, which needs the guard band above.
5. Timing jitter and sample-and-hold accuracy in the ADC must be small.

Natural sampling

In natural sampling the message is multiplied by a periodic train of rectangular pulses $c(t)$ of width $\tau$ and period $T_s$; the top of each pulse follows the message.

m(t) ---->(x)----> s(t) = m(t) c(t)
           ^
           |  c(t): pulses, width tau, period Ts
   _   _   _   _
  | | | | | | | |      (tops follow m(t))
 _| |_| |_| |_| |_

Derivation. $c(t)$ is periodic, so it has the Fourier series

$$c(t) = \sum_{n=-\infty}^{\infty} C_n\, e^{j2\pi n f_s t}, \qquad C_n = \frac{\tau}{T_s}\,\text{sinc}(n f_s\tau)$$

The sampled signal is

$$s(t) = m(t)\,c(t) = \sum_{n=-\infty}^{\infty} C_n\, m(t)\,e^{j2\pi n f_s t}$$

Using the frequency-shift property, $m(t)e^{j2\pi nf_st} \leftrightarrow M(f - nf_s)$:

$$S(f) = \sum_{n=-\infty}^{\infty} C_n\, M(f - n f_s) = \frac{\tau}{T_s}\sum_{n} \text{sinc}(n f_s \tau)\, M(f - n f_s)$$

Interpretation:

• The spectrum consists of copies of $M(f)$ centred at $nf_s$, each weighted by a constant $C_n$ that depends only on $n$.
• Each copy keeps the original shape of $M(f)$ (no distortion), unlike flat-top sampling where the shape is multiplied by $\text{sinc}(fT)$.
• The baseband term ($n = 0$) is $\frac{\tau}{T_s}M(f)$. If $f_s \ge 2W$, the copies do not overlap, and an LPF of cut-off $W$ recovers $m(t)$ exactly (scaled by $\tau/T_s$).

S(f)
 |   /\         /\          /\
 |  /  \  ...  /  \  ...   /  \   (heights follow C_n)
 +-/----\-----/----\------/----\---> f
   -W 0 W   fs-W fs fs+W  2fs

PAM vs PPM

Generation of PWM

Pulse width modulation (PWM) keeps amplitude and position of the leading edge fixed and varies the width of each pulse in proportion to the message sample.

             +-----------+
m(t) ------->|  (+)      |       +------------+
             |   sum     |------>| Comparator |--> PWM
saw-tooth -->|           |       | (vs. Vref) |
(fs)         +-----------+       +------------+

saw  /|  /|  /|      m(t) + saw compared with Vref
    / | / | / |
   /  |/  |/  |
PWM  ___    _____    __
    |   |  |     |  |  |      width ~ m(nTs)
 ___|   |__|     |__|  |__

Working (comparator method):

1. A sawtooth (ramp) generator runs at the sampling frequency $f_s$.
2. The message $m(t)$ is added to the ramp (or compared directly with the ramp).
3. A comparator outputs HIGH while the ramp is below the message (or the sum exceeds the reference) and LOW otherwise.
4. The time at which the crossing occurs depends on the message value, so the trailing edge moves and the pulse width is proportional to $m(t)$.

Alternative: a monostable multivibrator triggered at $f_s$ whose timing is controlled by the message voltage (e.g. 555 timer with the message on the control pin).

PPM from PWM: differentiate the PWM and use the trailing-edge spikes to trigger a fixed-width monostable; pulse positions then follow the message.

Non-uniform quantization using a uniform quantizer

A non-uniform quantizer is equivalent to a compressor, followed by a uniform quantizer, with an expander at the receiver. This is called companding.

x(t)->[Compressor]--y-->[Uniform  ]-->[Encoder]-->channel
       y = c(x)          quantizer
channel->[Decoder]-->[Expander x = c^-1(y)]-->x^(t)

Technique:

1. The input $x$ (normalised to $|x| \le 1$) is passed through a non-linear compressor $y = c(x)$ with large slope near zero and small slope near full scale.
2. A uniform quantizer with step $\Delta$ operates on $y$. Step $\Delta$ in $y$ corresponds to a step in $x$ of

$$\Delta_x \approx \frac{\Delta}{c'(x)}$$

so where $c'(x)$ is large (small $x$) the effective step is fine, and where $c'(x)$ is small (large $x$) the step is coarse. 3. At the receiver, the expander $c^{-1}(y)$ undoes the compression. Compressor + expander together have a linear overall response. 4. If $c(x)$ is logarithmic, $\Delta_x \propto |x|$, so the noise is proportional to the signal and SQNR is nearly constant for all levels.

In practice $c(x)$ is implemented digitally as a piecewise-linear segmented characteristic, so standard uniform ADC/DACs can be used.

μ-law

$$y = \frac{\ln(1+\mu|x|)}{\ln(1+\mu)}\,\text{sgn}(x), \qquad 0 \le |x| \le 1$$

• $\mu = 0$: uniform (linear). Standard $\mu = 255$ for 8-bit PCM (North America, Japan).
• Small $|x|$ ($\mu|x| \ll 1$): $y \approx \frac{\mu x}{\ln(1+\mu)}$, linear with large gain (46 for $\mu = 255$).
• Large $|x|$ ($\mu|x| \gg 1$): $y \approx \frac{\ln(\mu|x|)}{\ln(1+\mu)}$, logarithmic.

Example ($\mu = 255$): $x = 0.01 \to 0.228$, $x = 0.1 \to 0.591$, $x = 0.5 \to 0.876$, $x = 1 \to 1$.

 y
1.0|              ____----   mu = 255
   |        __----      __-- mu = 5
   |    _-'      ___---
   |  /    __---    .  mu = 0 (y = x)
   | / _--     .
   |/-    .
   +---------------------> x
   0                   1.0

Larger $\mu$ gives more compression and a wider dynamic range with nearly flat SQNR (about 38 dB over 40 dB input range for 8-bit $\mu = 255$).

Flat-top sampling produces pulses whose amplitude equals the instantaneous sample value $m(nT_s)$ and stays constant (flat) for the pulse duration $T$. It is produced by a sample-and-hold circuit.

Mathematical model

The flat-top signal is

$$s(t) = \sum_{n=-\infty}^{\infty} m(nT_s)\,h(t - nT_s), \qquad h(t) = \begin{cases} 1, & 0 < t < T \\ 0, & \text{otherwise} \end{cases}$$

This equals the ideally sampled signal convolved with $h(t)$:

$$s(t) = \left[m(t)\sum_n \delta(t - nT_s)\right] * h(t) = m_\delta(t) * h(t)$$

Taking the Fourier transform:

$$S(f) = M_\delta(f)\,H(f) = f_s\sum_{k=-\infty}^{\infty} M(f - kf_s)\;H(f)$$

where

$$H(f) = T\,\text{sinc}(fT)\,e^{-j\pi fT}$$

 |H(f)|
  T |\___
    |    \__              sinc envelope
    |       \___          attenuates higher
    +-----------\----->f  message frequencies
    0   W        1/T

Merits

1. Easy to generate with a sample-and-hold circuit; the constant level gives the ADC time to quantize, so it is the usual practical method before PCM encoding.
2. Pulse tops are flat, so noise on the top is easier to handle and amplitude detection is simple.
3. Since the amplitude is constant during the pulse, the energy per sample is well defined, giving a better SNR than very narrow pulses.

Demerits

1. Aperture effect: each spectral copy is multiplied by $\text{sinc}(fT)$, so higher message frequencies are attenuated: amplitude distortion. Example: at $f = W$ with $T = T_s$ and $f_s = 2W$, $\text{sinc}(0.5) = 0.637$, i.e. a 3.9 dB loss.
2. Delay distortion: the term $e^{-j\pi fT}$ introduces a delay of $T/2$.
3. An equalizer with response $\frac{1}{H(f)} = \frac{\pi f}{\sin(\pi fT)}e^{j\pi fT}$ (within the message band) is needed after the reconstruction LPF. The distortion is negligible if $T/T_s \le 0.1$.
4. Wider pulses need more bandwidth than ideal impulses and increase crosstalk in TDM.

Compared with natural sampling (where each copy of $M(f)$ keeps its shape), flat-top sampling distorts the spectrum, but it is preferred in practice because of the simple hold-circuit implementation.

Pulse code modulation (PCM)

PCM is a method of converting an analog signal into a digital (binary) signal by three steps: sampling at or above the Nyquist rate, quantizing each sample to one of $L$ levels, and encoding each level into an $n$-bit binary code word ($L = 2^n$). The binary pulses are then transmitted.

Numerical

Given: $W = 3$ kHz, sampling at twice the Nyquist rate, step size $\Delta = 0.5\%$ of $m_p$, binary coding.

Step 1: Sampling rate

$$f_N = 2W = 6\ \text{kHz}, \qquad f_s = 2 \times 6 = 12\ \text{kHz}$$

Step 2: Number of levels. The signal swings from $-m_p$ to $+m_p$, so

$$\Delta = \frac{2m_p}{L} = 0.005\,m_p \;\Rightarrow\; L = \frac{2}{0.005} = 400$$

Step 3: Bits per sample. $n$ must be an integer with $2^n \ge 400$:

$$n = \lceil\log_2 400\rceil = \lceil 8.64\rceil = 9 \quad (2^9 = 512)$$

Step 4: Bit rate

$$R_b = n f_s = 9 \times 12000 = 108\ \text{kbps}$$

Step 5: Minimum channel bandwidth. A binary signal of rate $R_b$ needs at least $R_b/2$ Hz (Nyquist bandwidth for ISI-free transmission):

$$B_{min} = \frac{R_b}{2} = \frac{108}{2} = 54\ \text{kHz}$$

Answer: minimum channel bandwidth = 54 kHz (9 bits/sample, 108 kbps).

Setup. Delta modulation uses step size $\Delta$, sampling rate $f_s = 1/T_s$ and a reconstruction LPF of bandwidth $W$ at the receiver. The message is a single tone:

$$m(t) = A_m\cos 2\pi f_m t$$

Step 1: Maximum amplitude without slope overload

The staircase can change by at most $\Delta$ in $T_s$, so the maximum slope it can follow is $\Delta/T_s = \Delta f_s$. The maximum slope of the tone is

$$\left|\frac{dm}{dt}\right|_{max} = 2\pi f_m A_m$$

No slope overload requires $2\pi f_m A_m \le \Delta f_s$, so

$$A_{m,max} = \frac{\Delta f_s}{2\pi f_m}$$

Step 2: Signal power

$$S = \frac{A_{m,max}^2}{2} = \frac{\Delta^2 f_s^2}{8\pi^2 f_m^2}$$

Step 3: Granular (quantization) noise power

Without slope overload, the error $q(t) = m(t) - m_q(t)$ stays within $\pm\Delta$. Assuming it is uniformly distributed in $(-\Delta, \Delta)$:

$$E[q^2] = \int_{-\Delta}^{\Delta} q^2\frac{1}{2\Delta}\,dq = \frac{1}{2\Delta}\cdot\frac{2\Delta^3}{3} = \frac{\Delta^2}{3}$$

This noise power is spread approximately uniformly over $0$ to $f_s$. The LPF passes only $0$ to $W$:

$$N_q = \frac{\Delta^2}{3}\cdot\frac{W}{f_s}$$

Step 4: SQNR

$$\begin{aligned} \text{SQNR} &= \frac{S}{N_q} = \frac{\Delta^2 f_s^2}{8\pi^2 f_m^2}\cdot\frac{3 f_s}{\Delta^2 W} \\ &= \frac{3 f_s^3}{8\pi^2 f_m^2 W} \end{aligned}$$

Step 5: In dB

Since $10\log_{10}\frac{3}{8\pi^2} = -14.2$ dB,

$$\text{SQNR}_{dB} = 30\log_{10} f_s - 20\log_{10} f_m - 10\log_{10} W - 14.2\ \text{dB}$$

Observations:

• SQNR $\propto f_s^3$: doubling the sampling rate improves SQNR by 9 dB (compared with 6 dB per extra bit in PCM).
• SQNR does not depend on $\Delta$ (at the maximum amplitude), because the allowed amplitude and noise both scale with $\Delta$.

Example: $f_s = 64$ kHz, $f_m = 1$ kHz, $W = 4$ kHz:

$$\text{SQNR} = \frac{3(64000)^3}{8\pi^2(1000)^2(4000)} = 2490 \approx 33.96\ \text{dB}$$

Statement

A band-limited signal $m(t)$ with no frequency components above $W$ Hz ($M(f) = 0$ for $|f| > W$) is uniquely determined by its samples taken at uniform intervals $T_s \le \frac{1}{2W}$, i.e. at a rate $f_s \ge 2W$. It can be recovered exactly from the samples by an ideal low-pass filter of bandwidth $W$.

The minimum rate $2W$ is the Nyquist rate; $1/(2W)$ is the Nyquist interval.

Proof

1. Ideal sampling. Multiply $m(t)$ by a periodic impulse train:

$$m_\delta(t) = m(t)\sum_{n=-\infty}^{\infty}\delta(t - nT_s) = \sum_{n=-\infty}^{\infty} m(nT_s)\,\delta(t - nT_s)$$

2. Spectrum of the impulse train. It is periodic, with Fourier series coefficients $1/T_s$:

$$\sum_n \delta(t - nT_s) = \frac{1}{T_s}\sum_{k=-\infty}^{\infty} e^{j2\pi k f_s t} \;\longleftrightarrow\; f_s\sum_{k}\delta(f - kf_s)$$

3. Spectrum of the sampled signal. Multiplication in time is convolution in frequency:

$$M_\delta(f) = M(f) * f_s\sum_k\delta(f - kf_s) = f_s\sum_{k=-\infty}^{\infty} M(f - kf_s)$$

So $M_\delta(f)$ is $M(f)$ repeated every $f_s$ Hz, scaled by $f_s$.

f_s >= 2W (no overlap):
     /\        /\        /\
    /  \      /  \      /  \
 --/----\----/----\----/----\---> f
  -W  0  W fs-W fs   2fs
f_s < 2W: copies overlap -> aliasing

4. Condition for no overlap. The copy at $f_s$ occupies $f_s - W$ to $f_s + W$. It does not overlap the baseband copy ($-W$ to $W$) if

$$f_s - W \ge W \;\Rightarrow\; f_s \ge 2W$$

5. Recovery. If $f_s \ge 2W$, pass $m_\delta(t)$ through an ideal LPF with $H(f) = T_s$ for $|f| < W$ (any cut-off between $W$ and $f_s - W$):

$$M_\delta(f)\,H(f) = M(f)$$

so $m(t)$ is recovered exactly.

6. Interpolation formula. With $f_s = 2W$, the LPF impulse response is $h(t) = \text{sinc}(2Wt)$, so

$$m(t) = \sum_{n=-\infty}^{\infty} m\!\left(\frac{n}{2W}\right)\text{sinc}(2Wt - n)$$

Each sample is replaced by a sinc pulse and the sum rebuilds $m(t)$ at all instants. This proves the theorem. If $f_s < 2W$, the copies overlap (aliasing) and $m(t)$ cannot be recovered; an anti-aliasing filter is therefore used before sampling.

Quantization

Quantization is the process of approximating each continuous-amplitude sample of a signal by the nearest of a finite set of allowed levels. With $L$ levels spaced $\Delta$ apart it turns a sampled (discrete-time, continuous-amplitude) signal into a discrete-amplitude signal that can be coded with $n = \log_2 L$ bits. The difference between the true sample and the quantized value is the quantization error (quantization noise).

SQNR for linear (uniform) quantization

Let the message $m(t)$ lie in the range $-m_p$ to $+m_p$ and let the quantizer have $L$ equally spaced levels.

1. Step size

$$\Delta = \frac{2m_p}{L}$$

2. Error range: in a mid-rise quantizer the sample is replaced by the centre of its step, so the error $q$ lies between $-\Delta/2$ and $+\Delta/2$.
3. Error pdf: for a busy signal with many levels, $q$ is assumed uniformly distributed, $f_Q(q) = 1/\Delta$ for $|q| \le \Delta/2$.
4. Quantization noise power (mean square error):

$$\begin{aligned} N_q = \overline{q^2} &= \int_{-\Delta/2}^{\Delta/2} q^2 \frac{1}{\Delta}\,dq = \frac{1}{\Delta}\left[\frac{q^3}{3}\right]_{-\Delta/2}^{\Delta/2} \\ &= \frac{\Delta^2}{12} = \frac{(2m_p/L)^2}{12} = \frac{m_p^2}{3L^2} \end{aligned}$$

5. SQNR: with signal power $S = \overline{m^2(t)}$,

$$\text{SQNR} = \frac{S}{N_q} = 3L^2\,\frac{\overline{m^2(t)}}{m_p^2}$$

6. Sinusoidal message $m(t) = m_p\cos\omega_m t$ (full load): $S = m_p^2/2$, so

$$\begin{aligned} \text{SQNR} &= \frac{m_p^2/2}{m_p^2/(3L^2)} = \frac{3}{2}L^2 = 1.5 \times 2^{2n} \\ \text{SQNR}_{dB} &= 10\log_{10}1.5 + 20n\log_{10}2 \approx 1.76 + 6.02n\ \text{dB} \end{aligned}$$

Key results

• SQNR rises as $L^2$, i.e. by about 6 dB for each extra bit.
• For a signal whose power equals $m_p^2$ (e.g. a square-like signal) the result is $3L^2$, i.e. $4.77 + 6.02n$ dB.
• Example: 8-bit PCM with a full-scale sine gives $1.76 + 48.16 \approx 49.9$ dB.
• Weak signals give a much lower SQNR because $N_q$ is fixed while $S$ falls; this is why non-uniform quantization (companding) is used for speech.

Nyquist sampling theorem

A signal band-limited to $W$ Hz (no components above $W$) is completely described by, and can be exactly recovered from, its samples taken uniformly at a rate

$$f_s \ge 2W$$

The minimum rate $f_s = 2W$ is the Nyquist rate and $T_s = 1/(2W)$ is the Nyquist interval.

Explanation: ideal sampling multiplies $g(t)$ by an impulse train $\delta_{T_s}(t)$. In the frequency domain this repeats the spectrum every $f_s$:

$$G_\delta(f) = f_s\sum_{n=-\infty}^{\infty} G(f - nf_s)$$

If $f_s \ge 2W$ the copies do not overlap, so a low-pass filter of cut-off $W \le f_c \le f_s - W$ recovers $G(f)$ exactly. If $f_s < 2W$ the copies overlap (aliasing) and recovery is impossible.

Flat-top sampling

In flat-top sampling each sample is held constant for a pulse width $\tau$ (sample-and-hold), giving a PAM wave with flat tops. It is modelled as ideal sampling followed by convolution with a rectangular pulse $h(t)$ of width $\tau$:

$$s(t) = \sum_n g(nT_s)\,h(t - nT_s) = \big[g(t)\,\delta_{T_s}(t)\big] * h(t)$$

Taking the Fourier transform (convolution becomes multiplication):

$$\begin{aligned} S(f) &= \Big[f_s\sum_n G(f - nf_s)\Big] H(f) \\ H(f) &= \tau\,\text{sinc}(f\tau)\,e^{-j\pi f\tau} \end{aligned}$$

 g(t)--(x)-->ideal samples-->[h(t): hold tau]-->s(t)
        |
   impulse train

Meaning of the result

• The spectrum is still repeated every $f_s$, but each copy is multiplied by $H(f)$, a sinc shape.
• So the high frequencies of the message are attenuated: this amplitude distortion is the aperture effect.
• Distortion is small when $\tau \ll T_s$.
• At the receiver it is corrected with an equalizer of response $1/H(f)$ (in the band $|f| \le W$) after the reconstruction low-pass filter.

Flat-top sampling is the one used in practice (sample-and-hold before the ADC) because a constant amplitude during $\tau$ is easy to quantize and generate.