Synchronization of Frame, Symbol Timing, and Carrier

Abstract

The literature on synchronization is vast, and yet there seem no clear-cut fundamentals identified. Our approach here is to treat the topic from modern implementation point of view, namely, the most of demodulation functions are implemented digitally, after sampling, rather than by analog signal processing. Another point is a fast synchronization, in fact, aiming as fast as theoretically possible with minimum variance. An extensive development for TED is based on two or four samples per symbol, which works with a large carrier frequency offset.

List of sections in Chap. 7:

• 7.1 Packet Synchronization Examples

• 7.2 Symbol Timing Synchronization

  • Digital TEDs with 2 or 4 samples per symbol

  • Embedding digital TED into timing recovery loop

  • Simulations of Doppler clock frequency shift

• 7.3 Carrier Phase Synchronization

• 7.4 Quadrature Phase Imbalance Correction

• 7.5 References with Comments

Appendix 7

7.1.1 A.1 Raised Cosine Pulse and Its Pre-filtered RC Pulse

A raised cosine pulse is often used in digital data transmission for baseband pulse-shaping. We also consider its pre-filtered pulse, which is motivated by eliminating self-noise or pattern-dependent noise. This becomes serious when the modulation order is high, using a band-limited pulse, less than 1/T with say β << 1, in order to be of high bandwidth efficiency. The impulse response confined in time, e.g., rectangular, may eliminate self-noise at the expense of using more bandwidth.

The pre-filtering is defined as ${h}_{\mathrm{pf}}(t)=k\ {h}_{\mathrm{RC}}(t)\otimes {h}_{\mathrm{RC}}(t)\ \cos \frac{2\pi t}{T}$ where k is a scaling factor and ⊗ denotes convolution in time domain. In other words, in frequency domain, ${H}_{\mathrm{pf}}(f)=\frac{k}{2}{H}_{\mathrm{RC}}(f)\left[{H}_{\mathrm{RC}}\left(f-\frac{1}{T}\right)+{H}_{\mathrm{RC}}\left(f+\frac{1}{T}\right)\right]$.

Raised Cosine Pulse and Its Pre-filtered One

	Raised cosine pulse	Pre-filtered raised cosine pulse
Time domain	${h}_{\mathrm{RC}}(t)=\frac{\sin \left(\frac{\pi t}{T}\right)}{\frac{\pi t}{T}}\frac{\cos \left(\frac{\beta \pi t}{T}\right)}{1-4{\left(\frac{\beta t}{T}\right)}^2}$	${h}_{\mathrm{pf}}(t)=\frac{\sin \left(\frac{\beta \pi t}{T}\right)}{\frac{\pi t}{T}}\frac{\cos \left(\frac{\pi t}{T}\right)}{1-{\left(\frac{\beta t}{T}\right)}^2}$
Frequency domain	${\displaystyle \begin{array}{l}{H}_{\mathrm{RC}}(f)=T\kern6.25em \mid \\ {}=\frac{T}{2}\left\{1+\cos \frac{\pi \mathrm{T}}{\beta}\left(f-\frac{1-\beta }{2T}\right)\right\}\kern1em \\ {}=\frac{T}{2}\left\{1+\cos \frac{\pi \mathrm{T}}{\beta}\left(f+\frac{1-\beta }{2T}\right)\right\}\ \end{array}}$	${\displaystyle \begin{array}{l}{H}_{\mathrm{pf}}(f)=0\kern4.5em \mid f\mid <\frac{1-\beta }{2T}\\ {}={\mathrm{Tcos}}^2\frac{\pi T}{\beta}\left(f-\frac{1}{2T}\right)\kern1em \frac{1-\beta }{2T}<f\\ {}\kern1em <\frac{1+\beta }{2T}\\ {}={\mathrm{Tcos}}^2\frac{\pi T}{\beta}\left(f+\frac{1}{2T}\right)\kern0.5em -\frac{1+\beta }{2T}<f\\ {}\kern1em <-\frac{1-\beta }{2T}\end{array}}$

Exercise

Raised cosine pulse and its pre-filtered RC pulse with β =0.41 are shown below.

Exercise

Sketch the frequency response of RC and pre-filtered RC.

Hint

Use the argument to define pre-filtering.

Note that the pre-filtered pulse is not used for transmission but used for only timing recovery. It can be approximately done by a high-pass filter, and for digital implementation see Reference [5] as an example.

7.1.2 A.2 Poisson Sum Formula for a Correlated Signal

In this appendix, we like to show the identity, used extensively in the text.

$$ \sum \limits_{n=-\infty}^{n=\infty }h\left(t- nT\right)h\left(t\mp \Delta T- nT\right)=\sum \limits_{m=-\infty}^{m=\infty}\left[\frac{1}{T}{\int}_{v=-\infty}^{v=\infty }H\left(\frac{m}{T}-v\right)H(v){\mathrm{e}}^{\mp j2\pi \Delta Tv} dv\right]{\mathrm{e}}^{j\frac{2\pi }{T} mt} $$

This identity is quickly seen by applying three identities used in Fourier transform. Firstly multiplication in time domain can be expressed by convolution in frequency domain and vice versa, and secondly periodic extension of a signal in time domain, i.e., periodic signal, can be expressed as a sampling in frequency domain. Thirdly a delayed time domain signal transform can be expressed by multiplying a delay factor (e^∓j2πfΔT where ΔT is the delay) to the original transform.

Define Fourier transform pair as

$$ u(t)={\int}_{-\infty}^{\infty }U(f){\mathrm{e}}^{j2\pi ft} df\iff U(f)={\int}_{-\infty}^{\infty }u(t){\mathrm{e}}^{-j2\pi ft} dt. $$

$$ {\displaystyle \begin{array}{l}\mathrm{Multiplication}\kern6em \mathrm{Convolution}\\ {}u(t)=v(t)w(t)\kern2.75em \iff \kern2em U(f)={\int}_{-\infty}^{\infty }V\left(f-v\right)W(v) dv\end{array}} $$

$$ {\displaystyle \begin{array}{l}\mathrm{Periodic}\ \mathrm{extension}\\ {}\sum \limits_{k=-\infty}^{k=\infty }u\left(t- kT\right)=\frac{1}{T}\sum \limits_{m=-\infty}^{m=\infty }U\left(\frac{m}{T}\right){\mathrm{e}}^{j2\pi \frac{m}{T}t}\end{array}} $$

$$ {\displaystyle \begin{array}{l}\mathrm{Delayed}\ \mathrm{signal}\ \mathrm{transform}\\ {}u\left(t\mp \Delta T\right)={\int}_{f=-\infty}^{f=\infty}\left[U(f){\mathrm{e}}^{\mp j2\pi f\Delta T}\right]{\mathrm{e}}^{j2\pi f t} df\end{array}} $$

By choosing v(t) = h(t), and w(t) = h(t ∓ ΔT), and applying the above three identities, the result is obtained.

Exercise

Fill the additional details for the above derivation.

7.1.3 A.3 Review of Phase-Locked Loops

The phase-locked loops (PLL) are extensively used in synchronization subsystems – carrier and timing – and in frequency synthesizers, e.g., frequency variable oscillators, of RF systems. We cover its basics here, to be concrete and simple in describing, starting with one of frequency and phase synchronization of sine wave as shown on the left. The key components of PLL – VCO, phase detector, and loop filter – will be explained, and then the linear model of PLL will be derived below. A thorough understanding of PLL is important and useful as well.

VCO

The model of voltage-controlled oscillator is shown in Fig. 7.60. The characteristics of input control voltage to output frequency are shown in RHS of the figure: control voltage ranges from v1 to v2 and its corresponding output frequency. Its slope is given as a loop parameter K_o. In fact, it is used in Chap. 2 as part of FSK signal generation. See Fig. 2.56 if you are curious.

Fig. 7.60

VCO and its input and output relationship

Phase detector

In Fig. 7.61, its symbol is on LHS, and its actual implementation is on RHS when the input is sinusoids. We use the relationship of $\cos \left(2\pi {f}_ct+{\theta}_i\right)\sin \left(2\pi {f}_ct+{\theta}_o\right)=\frac{1}{2}\sin \left({\theta}_o-{\theta}_i\right)+\frac{1}{2}\sin \left(4\pi {f}_ct+{\theta}_o+{\theta}_i\right),$ and the higher harmonic is low-pass filtered. The gain of phase detector is denoted by K_d . Note that sin(θ_o − θ_i) ≈ θ_o − θ_i when the phase difference is small.

Fig. 7.61

Phase detector symbol (LHS) and its implementation in case of sinusoids

Loop filter

For the second-order loop, we often use a filter with direct path and integration path as shown in Fig. 7.62. When there is no integration path (b = 0), it becomes a first-order loop. A frequency offset, if available, can be inserted as shown in dotted line.

Fig. 7.62

Loop filter in circuits (LHS) and its flow diagram (RHS); direct path with gain a and integration path with gain b. Dotted line indicates the addition of frequency offset possible

Linear model of PLL is useful to see its behavior when it is in lock or tracking mode, i.e., sin (θ_o − θ_i) ≈ θ_o − θ_i being small or the approximation works well. Its model is in Fig. 7.60 and its transfer function H(s) = $\frac{\theta_o(s)}{\theta_i(s)}$ is shown as well. Note that VCO is modeled as an integrator, ${K}_o\frac{1}{s}$ .

Exercise

Confirm the transfer function H(s) using the flow graph representation in Fig. 7.63.

Hint

${\theta}_o(s)={K}_o{K}_d\ a\left\{{\theta}_i(s)-{\theta}_o(s)\right\}\frac{1}{s}+{K}_o{K}_d\ b\left\{{\theta}_i(s)-{\theta}_o(s)\right\}\ \frac{1}{s^2}$ . And H(s) = $\frac{\theta_o(s)}{\theta_i(s)}=\frac{sK_o{K}_da+{K}_o{K}_db}{s^2+{sK}_o{K}_da+{K}_o{K}_db}$.

Exercise

Show by letting K_oK_d a = A, K_oK_d b = B, H(s) = $\frac{\theta_o(s)}{\theta_i(s)}$=$\frac{sA+B}{s^2+ sA+B}$.When b = 0, H(s) =$\frac{A}{s+A}$.

In the literature H(s) is often expressed in terms of ζ and ω_n.

Exercise

The transfer function is often expressed as $(s)=\frac{s2{\zeta \omega}_n+{\omega_n}^2}{s^2+s2{\zeta \omega}_n+{\omega_n}^2}=\frac{sA+B}{s^2+ sA+B}$ . Show ${\omega}_n=\sqrt{B}$, $\zeta =\frac{A}{2\sqrt{B}}$, called natural (radian) frequency and damping factor, respectively.

Answer

B = ω_n², A = 2ζω_n; then ${\omega}_n=\sqrt{B}$, $\zeta =\frac{A}{2\sqrt{B}}$.

Exercise

When the damping factor$\zeta =\frac{1}{\sqrt{2}}$, express B in terms of A.

Answer

$B=\frac{1}{2}\ {A}^2.$ Try other cases of $\zeta =\frac{1}{2}$ and ζ = 1; B = A² and $B=\frac{1}{4}\ {A}^2$.

Fig. 7.63

Linearized model of PLL for the behavior in lock mode (tracking mode) and its transfer function is shown as well

Now we try to understand the behavior of Fig. 7.63, linearized model of PLL through examples and exercises. We state two theorems from Laplace transform.

The final value theorem of Laplace transform states that $\underset{t\to \infty }{\lim }y(t)=\underset{s\to 0}{\ \lim } sY(s)$.

Hint

Search Internet, like Wikipedia, to get the proof and the conditions under which it is valid. At least understand it so that one can use it.

Similarly the initial value theorem of Laplace transform states that $\underset{t\to {0}^{+}}{\lim }y(t)=\underset{s\to \infty }{\ \lim } sY(s)$.

We define the phase error as θ_e(s) = θ_i(s) − θ_o(s)=$\frac{s^2}{s^2+ As+B}{\theta}_i(s)$ in Laplace transform domain. Using the final value theorem we obtain the following:

Input	Error; θe(t)= θi(t) − θo(t)
	$\underset{t\to \infty }{\lim }{\theta}_e(t)$= $\underset{s\to 0}{\ \lim }\ s\ \left(\frac{s^2}{s^2+ As+B}{\theta}_i(s)\right)$ = 0; i.e., tracks out the input phase jump where θi(s) = $\frac{\Delta \theta }{s}$ (step).
	$\underset{t\to \infty }{\lim }{\theta}_e(t)=\underset{s\to 0}{\ \lim }\ s\ \left(\frac{s^2}{s^2+ As+B}{\theta}_i(s)\right)=0$; i.e., tracks out the input frequency jump where θi(s) =$\frac{\Delta \upomega}{s^2}$.

Exercise

For the first-order loop (i.e., b = 0), $\underset{t\to \infty }{\lim }{\theta}_e(t)=\underset{s\to 0}{\ \lim }\ s\ \left(\frac{s}{s+A}{\theta}_i(s)\right)=0$ when there is a phase jump, i.e., θ_i(s) = $\frac{\Delta \theta }{s}$. On the other hand, if there is a frequency jump (i.e., phase ramp), i.e., θ_i(s) = $\frac{\Delta \omega }{s^2}$, then $\underset{t\to \infty }{\lim }{\theta}_e(t)$=$\frac{1}{A}$ . This is the reason the second-order loop is often used in practice.

In the above we have shown that a “small” input jump of both phase and frequency can be tracked out when the loop is in lock (a linear model assumes it). Next two exercises extend it further to see the time behavior of a linearized model of PLL. This may be skipped in the first reading.

Exercise

In Fig. 7.63, there is a “small” input phase jump, K_i, at time 0; the jump is “small” enough that the linear model is valid. Find the initial output phase and final phase: θ_o(t = 0⁺) and θ_o(t =  + ∞). And find the initial VCO frequency and final VCO frequency: ${\left.\frac{d}{dt}{\theta}_o(t)\right|}_{t={0}^{+}}$ and ${\left.\frac{d}{dt}{\theta}_o(t)\right|}_{t=+\infty }$.

Answer

In terms of s, the output phase is represented by

$$ {\theta}_o(s)=\frac{sK_o{K}_da+{K}_o{K}_db}{s^2+{sK}_o{K}_da+{K}_o{K}_db}{\theta}_i(s)=\frac{sK_o{K}_da+{K}_o{K}_db}{s^2+{sK}_o{K}_da+{K}_o{K}_db}{K}_i\frac{1}{s} $$

$$ \mathrm{VCO}\ \mathrm{frequency}\frac{d}{dt}{\theta}_o(t)\leftarrow \mathrm{Laplace}\ \mathrm{Transform}\to \frac{sK_o{K}_da+{K}_o{K}_db}{s^2+{sK}_o{K}_da+{K}_o{K}_db}{K}_i $$

We use that the Laplace transform of differentiation is $\frac{dx(t)}{dt}$← →s X(s). Using the initial and final value theorems in Laplace transform, we obtain

$$ {\theta}_0\left(t={0}^{+}\right)=\underset{\mathrm{s}\to \infty }{\lim }s{\theta}_0(s)=0,\mathrm{and}\;{\theta}_0\left(t\to \infty \right)=\underset{\mathrm{s}\to 0}{\lim }s{\theta}_0(s)={K}_i $$

$$ \frac{\mathrm{d}}{\mathrm{d}\mathrm{t}}{\theta}_0\left(t={0}^{+}\right)=\underset{\mathrm{s}\to \infty }{\lim }{s}^2{\theta}_0(s)={K}_o{K}_d{aK}_i,\mathrm{and} $$

$$ \frac{\mathrm{d}}{\mathrm{d}\mathrm{t}}{\theta}_0\left(t\to \infty \right)=\underset{\mathrm{s}\to 0}{\lim }{s}^2{\theta}_0(s)=0 $$

As part of intuitive understanding PLL, in particular a linear model of it, we try to see the above result without resorting to the final and initial value theorem of Laplace transform. Under lock condition of PLL where θ₀(t) ≈ 0, if θ_i(t) = K_iu(t) is applied where u(t) is a unit step, what will happen in a linear model of PLL shown in Fig. 7.63

θ₀(t = 0⁺) cannot change instantaneously due to the presence of integrator and thus must be the previous value, i.e., zero. The input to VCO will change instantaneously, through the direct path, with the gain of K_daK_i, and thus VCO frequency will be K_o times gain (K_daK_i). When PLL is settled, the phase error must be zero θ_e(t) = θ_i(t) − θ_o(t), and thus θ₀(t → ∞) = θ_i(t → ∞) = K_i.

Exercise

In the previous exercise, we looked for the initial and final output phase and VCO frequency. In this exercise we try to obtain both phase change in time and VCO frequency change in time.

Answer:

$$ {\displaystyle \begin{array}{ll}& \frac{sK_o{K}_da+{K}_o{K}_db}{s^2+{sK}_o{K}_da+{K}_o{K}_db}{K}_i\ \frac{1}{s}\\ {}& =\frac{sA+B}{s^2+ sA+B}{K}_i\ \frac{1}{s}\leftarrow \mathrm{Laplace}\ \mathrm{transform}\to \mathrm{output}\ \mathrm{phase}\ \mathrm{in}\ \mathrm{time}\end{array}} $$

$$ \frac{sK_o{K}_da+{K}_o{K}_db}{s^2+{sK}_o{K}_da+{K}_o{K}_db}{K}_i=\frac{sA+B}{s^2+ sA+B}{K}_i\kern1em \leftarrow \mathrm{Laplace}\ \mathrm{transform}\to \mathrm{VCO}\ \mathrm{frequency}\ \mathrm{in}\ \mathrm{time} $$

For actual inverse Laplace transform of the above, we need to classify them into three categories – over damped ζ > 1.0, critically damped (ζ = 1.0), and under damped ζ < 1.0 cases with A = 2ζω_nB = ω_n². We leave it as a small project below.

Exercise*(project)

Find the impulse response for three categories above depending on damping factor and then the unit step responses.

Hint

Look for a good textbook on signals and systems such as Oppenheim (Oppenheim, Alan V., et al.,“Systems and Signals”, 2nd ed. 1996, Prentice Hall) or on Laplace transform.

Noise bandwidth of the second order loop

Noise bandwidth of PLL is defined by ${B}_n=\frac{1}{\max_f{\left|H(f)\right|}^2}{\int}_0^{+\infty }{\left|H(f)\right|}^2 df$.

When it is applied to H(s) of the second-order loop where s=j2πf, we obtain ${B}_n=\frac{\omega_n}{2}\left(\frac{1}{4\zeta }+\zeta \right)=\frac{1}{4}\left(A+\frac{B}{A}\right)$.

Exercise* (Project)

Show the above statement, i.e., noise bandwidth is expressed as ${B}_n=\frac{\omega_n}{2}\left(\frac{1}{4\zeta }+\zeta \right)=\frac{1}{4}\left(A+\frac{B}{A}\right)$.

Hint

Note that $\underset{f}{\ \max }{\left|H(f)\right|}^2=1.0$. $2{B}_n=\underset{-\infty }{\overset{+\infty }{\int }}{\left|H(f)\right|}^2 df=\underset{-\infty }{\overset{+\infty }{\int }}{\left|h(t)\right|}^2 dt.$

Using the above result, we show the minimum, ${B}_n=\frac{\omega_n}{2}$, is achieved when ζ = 0.5. This is plotted in Fig. 7.64.

Fig. 7.64

Loop noise bandwidth for second-order loop

Table 10

7.1.4 A.4 Polynomial Interpolation: Farrow Structure

A linear interpolator between two points is a straight line between them. Its impulse response representation is shown on the left, symmetrical around zero and triangular. For |t| < 1, the triangle is represented by y(t) =  −  ∣ t ∣  + 1. The coefficients of its sampled impulse response, with delay μ, h_o, h₁, are shown as well.

An impulse response representation of a polynomial interpolator with p = 2, 3, and 4 can be obtained in addition to linear case of p = 1. We explain it with p = 3 case below.

Fig. 7.65

Getting an impulse response of p = 3 spline interpolator, we can find polynomial coefficients such that it should pass through at time 0, 1, 2, and 3 with 1.0, 0, 0 and 0

P = 1, 2, 3, and 4 polynomial interpolators are summarized in the table below:

The filter tap coefficients are the evaluation of a polynomial, y(t), at t, 1−t, 2−t, 3−t, and 4−t for h₀, h₁, h₂, h₃, and h₄, e.g., for the fourth-order interpolation, as shown in the table.

Exercise

Consider fourth-order interpolation (4th row in the above table). In order to get the coefficients of the interpolating polynomial, show that we need to solve a simultaneous equation as below. It is clear that it can be extended to higher-order polynomial, say, 5th with six taps, as well.

$$ \left[\begin{array}{cccc}1& 1& 1& 1\\ {}16& 8& 4& 2\\ {}81& 27& 9& 3\\ {}256& 64& 16& 4\end{array}\right]\left[\begin{array}{c}{c}_4\\ {}{c}_3\\ {}{c}_2\\ {}{c}_1\end{array}\right]=\left[\begin{array}{c}-1\\ {}-1\\ {}-1\\ {}-1\end{array}\right]\to \left[\begin{array}{c}{c}_4\\ {}{c}_3\\ {}{c}_2\\ {}{c}_1\end{array}\right]=\left[\begin{array}{c}\frac{1}{24}\\ {}\frac{-5}{12}\\ {}\frac{35}{24}\\ {}\frac{-25}{12}\end{array}\right]\mathrm{with}\ {c}_0=1.0 $$

For given samples of x(n) and delay, μ, a resampled one can be computed as shown Fig. 7.66, which is not Farrow structure but a generic one.

Fig. 7.66

Computation of resampled x(n + μ) with μ using four-tap polynomial interpolator; this is not a Farrow structure but another useful view of how to obtain x(n + μ)

In fact, a block of computing tap coefficient given μ (shaded in Fig. 7.66) can be a lookup table. Furthermore it can be any interpolating impulse response spanning six sample periods; see Fig. 7.65. In passing we comment that so-called Farrow structure, not shown here, takes well advantage of polynomial evaluation (iterative) to facilitate implementation in hardware.