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CHAPTER 2

THE BASIC LOOP

We begin by studying the basic phase-locked loop (PLL), whose loop filter is repre-

sented by a frequency-independent gain. We will look at its transient response and its

frequency response in the same way that we will later look at those for the more com-plicated cases, but it is important that we first gain a clear understanding by studying

this simpler case.

2.1 STEADY-STATE CONDITIONS

The center frequency of the voltage-controlled oscillator (VCO) fc is the frequency

that occurs when the phase detector is in the center of its output range or the frequency

midway between the frequencies that occur at the edges of the phase detector range.

If the transfer function from the phase detector to the VCO output frequency is linear,

these are the same thing. The output from the phase detector is not necessarily zero

at that point; it depends on the particular phase detector realization. Even if it is zero,

a voltage shift may be implemented between the phase detector and the VCO. In

fact, most VCOs are operated with a tuning voltage that is always positive or always

negative. While these details are of considerable practical concern, they are easily

understood, and our studies will concentrate on changes from some initial or average

condition.

Figure 2.1 is similar to Fig. 1.6 except that the independent variable has been

changed from the tuning voltage u2 to the phase difference . It illustrates the

Phase-Lock Basics, Second Edition. By William F. EganCopyright C 2008 John Wiley & Sons, Inc.

15

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16 2 THE BASIC LOOP

Fig. 2.1 Loop output frequency vs. phase difference at the phase detector with a typical VCO.

The hold-in range, over which the frequency will vary as the phase detector covers its output

range, is indicated.

one-to-one correspondence between the VCOs frequency and its phase (relative to

the reference phase) in steady state. In this simple first-order, loop, the one-to-onerelationship holds not only in steady state but always. If we know one variable, we

know the other because there is no storage, no memory, in the filter. The loops that we

will study later will be different in this regard; the filter will permit the relationship

between the VCOs frequency and phase to change with time. That will make those

loops more versatile, and more challenging.

2.2 CLASSICAL ANALYSIS

Before using Laplace transforms to analyze the loops dynamics, let us first determine

the response by more directly solving Eq. (1.15). We will find that the PLL responds

to changes in phase or frequency in a manner similar to that in which other simple

circuits respond to changes in voltage or current. Thus we find that, whereas, in other

circuits voltage and current are the variables that describe the state of the circuit and

phase and frequency are parameters of those variables, in the PLL frequency and

phase are state variables. They too, nevertheless, have frequencies and phases, which

are their modulation parameters. For example, Eq. (2.1),

out= c + A sin(m t+ 0), (2.1)

describes the state variable out, which represents the output frequency of the loop,

but out can be seen to possess a frequency (its modulation frequency) m and a

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2.2 CLASSICAL ANALYSIS 17

modulation phase,0. These parameters must also be parameters ofu 1 andu 2, other

state variables in the loop, since u 1, andu 2 are proportional to out.

2.2.1 Transient Response

Equation (1.15) can be rewritten as

dout= K d. (2.2)

The time derivative of this equation is

dout

dt = Kd

dt = K(in out), (2.3)

where Eq. (1.9) was used. We rewrite (2.3) as

dout= K(in out) dt. (2.4)

Ifinis constant, it may be subtracted fromouton the left side, since the differential

will not be affected by inclusion of a constant. This gives

d(out in) = K(out in) dt (2.5)

or

d = K dt, (2.6)

where

out

in. (2.7)

We now write Eq. (2.6) as

d

= K dt (2.8)

and integrate both sides to obtain

1d = K dt (2.9)

with the solution

ln = K t+ C. (2.10)

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18 2 THE BASIC LOOP

Taking the exponential of both sides, we obtain

=eK t+C

=eCeK t. (2.11)

We can determine the value ofeC by evaluating this expression at t= 0. When wedo so and substitute the results in Eq. (2.11), we obtain

(t) = (0)eK t. (2.12a)

The output frequency is the input frequency plus ,

out= in +(t) = in +(0)eK t. (2.12b)

Thus we see that this simple loop responds to an initial frequency error (difference

between the output and reference frequencies) (0) by exponentially decreasing

the error with a time constant equal to 1/K. As a result, the step response shown in

Fig. 2.2ais obtained. This is the same response to a frequency step that a single-pole

low-pass circuit has to a voltage step (Fig. 2.2b).

We might intuit this exponential response through the following considerations.

After the input frequency steps, there is a frequency error that causes to

change. This change in (multiplied byK) causes the VCO frequency to change in

such a direction as to decrease the frequency error (Fig. 2.3). This, in turn, decreases

the rate of phase change and thus the rate at which the VCO frequency changes.

As always, when the rate of change of a variable is proportional to the value of the

variable, it changes exponentially. Of course, the exponential shape will be perturbed

to the degree that there is curvature in the gain characteristics over the operating

region.

Fig. 2.2 Step response of (a) a simple loop compared to the step response of (b) a low-pass

filter. The filter in this loop is a simple amplifier. The frequency of the loop responds like the

voltage of the RCcircuit.

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2.2 CLASSICAL ANALYSIS 19

Fig. 2.3 Output frequency vs. phase for two phase detector characteristics. Since frequency

is the derivative of phase, the direction of change is determined by the sign of the frequency

difference.

Example 2.1 Transient Response When the input frequency to a particular loop

steps to a new frequency, we require that the output achieve that frequency, to within

an accuracy of 1%, in 1 msec. What are the range of Kvalues that allow a simple

loop to comply with this requirement?

Restating the problem, we require

(1 msec)/(0) 0.01.From (2.12a) this requires

(eK103 sec)

0.01

K

ln(0.01)/103 sec=

4.6

103 sec1 .

2.2.2 Modulation Response

Rearranging Eq. (2.3) we can write

1

K

dout

dt+ out= in. (2.13)

If the input is frequency modulated, then we expect the response to be frequency

modulated also, so we assume a response of the form

out(t) = c + A sin(m t), (2.14)

wherec is the center, or carrier, frequency, A is the peak frequency deviation, and

m is the modulation frequency. Putting Eq. (2.14) into (2.13), we obtain

Am

Kcos(m t)+ c + A sin(m t) = in. (2.15)

We rewrite (2.14) and (2.15) in terms of deviations from the mean:

out out(t) c= A sin(m t), (2.16)in in c= A

m

Kcos(m t) + A sin(m t). (2.17)

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20 2 THE BASIC LOOP

The amplitude ofout is Aand that ofin is

|in| = Am

K2 + 1, (2.18)

so the magnitude of the frequency response is

|H(m )| = 1/m

K

2+ 1. (2.19)

Sinceincontains a component equal to outplus one that leads it by 90 and is

larger bym/K, the phase angle ofin relative to out is tan1[m/K] and the

transfer phase is

H(m ) out(m ) in(m ) = tan1[m/K]. (2.20)

As with the time response, the response of this loop to a sinusoidally modulated

frequency is the same as the response of a low-pass filter to a sinusoidal voltage. In

each case the corner (3 dB) frequency is = K.

Example 2.2 Modulation Response A simple loop has a sinusoidal phase de-tector characteristic (Fig. 1.5) and K= 1000 sec1. The input (reference) frequencyis frequency modulated at an unknown rate with a peak deviation of 50 Hz. What

is the phase shift between the VCO modulation and the input modulation when the

deviation at the output is 10 Hz?

We will use the gain to determine the modulation frequency and then use the

modulation frequency to determine the phase shift. The closed-loop gain is

|H(m )| = 10Hz/50Hz = 0.2.

From (2.19), this occurs at

mK

2+ 1 = 5,

from which we obtain

m

K=

24,

giving the modulation frequency as m= 4899 rad/sec. Insertingm/K into (2.20),we obtain the phase shift as H(m ) = tan1(4.9) = 78.5. Thus the outputmodulation lags the input modulation by 78.5.

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2.3 MATHEMATICAL BLOCK DIAGRAM 21

The usual input and output state variables (y for voltage or current) are shown below

the plot of instantaneous frequency. The input happens to be a sinusoid while the

output is a square wave. These are intended to illustrate the meaning of the frequency

deviations plotted above. Note that the magnitude of the output frequency deviation is

smaller than that of the input deviation and how the times of occurrence of maximum

frequency (minimum period) correspond to the peaks in the frequency plots.

2.3 MATHEMATICAL BLOCK DIAGRAM

The mathematical relationships in the loop can be shown by means of a block diagram

such as Fig. 2.4. Here the input variable is the reference, or input, instantaneous phase

in, which is, in general, a function of time, perhaps a sinusoidal function such as

in(t)

=i

+Aisin (m t

+i ) (2.21)

with average value i , peak deviation Ai , modulation frequency m , and phase i .

Generally, as in any control system, responses are given in terms of the deviation from

Fig. 2.4 Mathematical block diagram of the simple loop with phase variables as input and out-

put. The frequency state variables are shown to indicate how they relate to the phase variables.

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22 2 THE BASIC LOOP

the average or initial or steady-state value, and i is only of interest in establishing

the steady-state operating point. A block is shown precedinginin order to establish

the integral relationship (1/s) betweenin(s) andin(s). The output from the summer

(a subtracter in this case) is the phase errore(s),1 the difference between the inputphase and the VCO phase, out. The phase error is converted to voltage in the phase

detector, which is represented by the gain Kp= u1(s)/e(s). For this simple case theloop filter is merely an amplifier with gain KLF;Kv represents the tuning sensitivity

out(s)/u2(s) of the VCO. The output from this block is the output frequency out.

However, out is needed to complete the loop so out(s) is integrated (multiplied

by 1/s) to produce it. Then the loop is completed by subtracting out(s) fromin(s)

in the summer. Note that the minus sign at the summer represents180 of phaseshift around the loop that does not appear in the transfer functions of the individual

blocks.The generic control system block diagram is shown in Fig. 2.5. The well-known

equations describing its transfer function are the response of the controlled variable

Cto the reference R,

C

R= G F

1+ G FGR, (2.22)

where G F and GR are forward and reverse transfer functions, respectively, and the

response of the errorEto the reference2

E

R= 1

1+ G FGR. (2.23)

The correspondence between the usual designations for loop signals, C,R , andE,and

state variables of the phase-locked loop depends on the input and output of interest.

Most often, in this text, we will be interested in the configuration shown in Fig. 2.4

so we identifyCasout(s) and Rasin(s). Because the feedback path has unity gain,

we have alsoG R= 1 and G Fis the entire open-loop transfer function G(s),G(s) = Kp KLF K/s K/s. (2.24)

Fig. 2.5 Generic control system diagram.

1 With reference to Eq. (1.9),e(s) = (s) except thate(s) is used to refer to a change without referenceto the steady-state phase difference.2 Note:G FGR is the open-loop gain andC/Rand E/R are closed-loop gains.

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2.3 MATHEMATICAL BLOCK DIAGRAM 23

Thus from Eq. (2.22) we obtain

out(s)

in(s) =G(s)

1+ G(s) (2.25a)

= K/s1+ K/s =

1

1+ s/K, (2.25b)

while from Eq. (2.23) we obtain

e(s)

in(s)= 1

1+ G(s) (2.26a)

= 11+ K/s =

s

s + K. (2.26b)

When we are interested in inputs or outputs at points other than those used above, we

can rewrite the equations or, what is often simpler, relate the desired input or output

to that shown in Fig. 2.4 by the transfer function of the segment that connects them.

Equation (2.25) represents a low-pass characteristic with a cutoff (3 dB) fre-quency of = K. This is analogous to the low-pass filter of Fig. 2.6a in which thevoltages have been given names corresponding to the phases. Equation (2.26) says

that the error phase has a high-pass characteristic, analogous to Fig. 2.6b. It is gen-erally true, even in more complex loops, that the output has a low-pass relationship

to the input while the error has a high-pass relationship. This reflects the fact that the

error responds immediately to a change in the input while, and because, the output

response is delayed.

Figure 2.7 is essentially the same as Fig. 2.4 except that the two 1/s blocks have

been moved forward through the summer; since both integration and summing are

linear operations, either can be done first. The input and output are now frequencies

rather than phases, illustrating that the response of output frequency to input frequency

is the same as the response of output phase to input phase. We can show the same

Fig. 2.6 Electrical analogs to the PLL. Voltages have names of analogous loop variables.

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24 2 THE BASIC LOOP

Fig. 2.7 Mathematical block diagram of the simple loop with frequency variables as input

and output.

results as follows:

out(s)

in(s)= sout(s)

sin(s)= out(s)

in(s). (2.27)

A similar relationship holds for error responses.

It is apparent from Figs. 2.4 and 2.7 that

in= out + e (2.28a)

and

in

=out

+e (2.28b)

and thus

1 = outin

+ ein

= outin

+ ein

(2.28c)

so that the output can easily be obtained from error and vice versa in most cases.

Example 2.3 Modulation Response Use (2.25b) to obtain the results of Exam-

ple 2.2. Ats= j 4899/sec, we haveout

in= 1

1+ j 4899/1000 = 0.2 78.5

as before.

2.4 BODE PLOT

One of the most important tools with which we will work is a logarithmic plot of open-

loop gain |G FGR | and phase shiftG FGR versus modulation frequency, s m .This plot, called the Bode plot, is shown for this simple loop in Fig. 2.8. Sometimes

we will forgo the phase plot since the phase shift is implied by the slope of the gain.

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2.4 BODE PLOT 25

Fig. 2.8 Bode plot for the simple loop. Note that the abscissa axis is marked with the values

ofm (e.g., K), rather than the actual distance, log(m sec/rad). Similarly, the ordinate axis is

marked with gain rather than its logarithm or value in dB.

A linear increase in gain with m originates in a term such as s= jand thereforeimplies both a gain that is increasing proportionally with frequency and a 90 phaseshift. In general the segments of the Bode plot correspond to approximations of the

form

F() = C(j)n = Cnn(90). (2.29)

Thus, for example, when n= 2, the gain drops as the square of the frequency andthe phase is 180. The slope is often expressed logarithmically. The magnitude isdescribed by

|F()

| 20 dB log10(C

n )

=20ndB log10()

+C. (2.30)

Note that we precede log10 by 20 dB rather than 10 dB because G FGR is a voltage

gain. We can break the loop at a point where the variable is voltage and G FGR will

describe the open-loop gain there.

In each decade of frequency change the function changes by

|F(10)/F()| 20ndB log10(10) = 20ndB (2.31)

so the slope is 20n dB per decade. Similarly, a two-to-one change in frequency

corresponds to

|F(2)/F()| 20ndB log10(2) = 6ndB (2.32)

or 6n dB per octave (an octave is a two-to-one change).

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26 2 THE BASIC LOOP

Figure 2.8 also illustrates thatK is both the gain at = 1 and the radian frequencyat which the open-loop gain is unity. As has been shown, this is the frequency at which

the closed-loop gain is

3 dB. At frequencies well below

=K, Eq. (2.25) can be

approximated as

out(s)

in(s)= 1, (2.33)

whereas at frequencies well above = Kthe equation is approximately

out(s)

in(s) =

K

s = j

K

. (2.34)

This equals the open-loop transfer function, Eq. (2.24). Thus, at low frequencies,

where the gain is high, the output follows the input faithfully, whereas, at high fre-

quencies, where the gain becomes low, the loop is essentially open and the response

is as if there were no loop (except that the low-frequency gain keeps it locked

otherwise none of these equations would be valid). At the loop corner frequency,

= K, it is easy to show from Eq. (2.25b) that the closed-loop gain is 1/2, or3 dB, and the phase shift is 45, just as in the case of the low-pass filter.

Similarly, the high-pass characteristic of Eq. (2.26) approaches unity at high fre-

quencies and zero at low frequencies and, at = K, also has a gain of3 dB, butthe phase shift is +45.

The more complex loops that we will study also have these general tendencies,

but we will be able to shape their characteristics more exactly to our needs because

of the increased number of parameters at our disposal.

2.5 NOTE ON PHASE REVERSALS

It is not uncommon that the transfer functions of the various blocks in Fig. 2.4 have

negative signs. It is only necessary that the total phase shift around the loop not be

altered. The negation of an even number of transfer functions has no important effect.

In fact, when a balanced mixer is used, it will automatically operate on the proper

slope to produce a correct number of phase reversals.

2.6 SUMMARY OF TRANSIENT RESPONSES OF

THE FIRST-ORDER LOOP

Here we will summarize the transient responses of the first-order loop based on

previous material and simple extensions of that material.

The unit step response is given in Eq. (2.12b) and shown in Fig. 2.2. Since a

unit ramp t is the time integral of a unit step, we can obtain the time response to

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2.6 SUMMARY OF TRANSIENT RESPONSES OFTHE FIRST-ORDER LOOP 27

TABLE 2.1 Summary of First-Order Loop Transient Responses

Inpout Response (sec not shown)

f(t) F(s) Error Output

Step 11

seK t 1-eK t

Rampt

sec

1

s21

K(1 eK t) t 1

K(1 eK t)

Parabolat2

2sec21

s31

K[t+ 1

k(ekt1)]

t2

2 1

K[t+ 1

K(ekt1])

t by integrating the time response to the unit step. Similarly, the integral of t is a

parabola t2/2, and we can obtain the response to this by integrating the ramp response.

Alternately, we can take the inverse Laplace transform of the transform obtained by

multiplying the transfer function of Eq. (2.25b) by 1/s, 1/s2, and 1/s3 to obtain step,

ramp, and parabola responses. (The parabolic input is important because it represents

the phase corresponding to a frequency ramp.)

The error response can be obtained by following the above process but starting

with equations for the error response, Eq. (2.12a) or the Laplace transform of the error

response, Eq. (2.26b).

Either the output response or the error response can be obtained from the other by

subtraction according to Eq. (2.28c).

Table 2.1 summarizes these responses. If an input is a 5-Hz step, since that is 5 Hz

times the given unit input, the output or error will be 5 Hz times the given output or

error. If the input is a 2-rad or 2 step, then the output will be 2 rad or 2 times thegiven response. A 10-Hz step is also a 10-cycle/sec ramp (10 cycles t/sec), 10 cyclestimes the given ramp input, so the output phase will be 10 cycles times the given ramp

response. If the input is a frequency ramp of 100 rad/sec/sec [(100 rad/sec) t/sec],the output frequency will be obtained by multiplying the ramp output response by

100 rad/sec. However, we can also obtain the output phase by considering the inputas a parabolic phase [100 rad t2/(2 sec2)] and obtaining the phase of the responseby multiplying the given response by 100 radians.