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Unformatted text preview: 1 Unlocking the Phase Lock Loop  Part 2 Intuitive Guide to Principles of Communications
www.complextoreal.com Unlocking the Phase Lock Loop  Part 2
Tracking and Acquisition behavior
A PLL produces a carrier that synchronizes in phase and frequency with an independent
external signal. The PLL controls the phase of its internal signal such that the error  the
difference between its phase and that of the input signal  is kept at a minimum while the
frequency of its signal is kept the same as the input signal.
PLL does this by continuously adjusting its frequency although what we track is phase. In
Part I we looked at how the PLL using just three functional blocks, a phase detector, a
low pass filter and a VCO accomplishes this.
A PLL has two main behavior modes;
Tracking Mode  The PLL is already locked on to the signal. What the PLL does to keep
itself aligned with the incoming signal in light of frequency and phase disturbances, is
called Tracking. This behavior is analyzed assuming linear behavior.
Acquisition Mode – The PLL is either out of lock or is just starting up to lock on to a
signal. This mode is nonlinear, difficult to analyze and understand.
Tracking
Tracking assumes that PLL has already acquired the signal and is now trying to keep up
with it. It is kind of like what a bicycle rider does to stay on the bike. The tracking
behavior has a steadystate and a transient mode of response. When PLL is locked up and
nothing is changing, the PLL shows a steady state behavior. When things are changing,
we see transient behavior. Transients also occur at acquisition and occur again as soon as
there is a change in the incoming signals frequency or phase. However as long as the
changes are small, the PLL recovers from these disturbances and gets back to steady
state.
For example, take the bicycle rider who runs into a little bump in the road, the bicycle
lurches to the side but the rider recovers (does not fall down) and gets back to his steady
speed. This immediate response to a changed condition comes under the tracking mode.
Whereas getting on the bike and starting up would be considered an acquisition mode.
During the transient, whether during acquisition or tracking, the error signal generated by
the PLL oscillates in response to large differences between the VCO and the incoming
signal phase. (Its always phase!) The transient part is very important because we can
www.complextoreal.com 2 Unlocking the Phase Lock Loop  Part 2 intuitively see that if a transient is large, the system will be unable to return to steady
state and the PLL will lose lock.
How a PLL behaves during the transient phase of tracking is a function of the loop
natural frequency, its damping factor and the loop gains.
The phase error signal transfer equation of a 2nd order PLL was developed in Part I, (Eq.
16) and is given by
θ e ( s) = s θi ( s)
s + K v K d F ( s) (1) Here θ e (s ) is the error produced by the PLL, the θi (s ) is the phase of the input signal. Kv
is the gain of the VCO, Kd the gain of the phase detector and F(s) the transfer function of
the filter and may include a gain.
You recall that the error signal is a measure of phase difference between the input signal
and the VCO signal. We will see how this error signal changes when it is subjected to
disturbances.
I wanted very much to explain the behavior of PLL without ever referring to the Laplace
transform, but as much as I would like, this just cannot be done. In Part I, I gave some
important Laplace relationships that are needed for understanding the PLL. Here is one
more property that we need to explain the transient behavior. This property is called the
Final Value Theorem and goes like this, lim θ ( t ) = lim s θ ( s ) t→ ∞ s→ 0 (2) It states that to find the steadystate behavior of a signal, we take the limit of s times its
Laplace Transform. This relationship helps us determine the steady state response of a
system.
Quite a few things can wrong for a PLL that is already in lock. We can classify these
disturbances in four broad categories.
1. There is a sudden phase shift at the input.
2. The input frequency does a jump.
3. The input frequency begins to change slowly such as due to Doppler.
4. There is noise at the input or inside the PLL. www.complextoreal.com 3 Unlocking the Phase Lock Loop  Part 2 Case 1: There is a sudden phase change at the input.
What happens to the steady state error in response to a phase step and how does the PLL
which was locked react to this? How big of phase step can a PLL recover from? We will
now answer these questions.
The change in the phase of the incoming signal is represented as a
θ i (t ) = u (t )∆θ (3) u(t) is the unit step function and ∆θ is the size of the phase change. The instantaneous
phase has just done a jump from its old value to a new value by ∆θ .
The Laplace transform of a unit step function with a phase change of ∆θ is θi ( s ) = ∆θ
s (4) This is Laplace transform of the phase step equation (3). Apply this value of θi (s) to (1),
multiply it by s and then take limit per Final Value Theorem of Eq. 2, we get
lim θe (t ) = lim
t →∞ s→0 s2
∆θ
s + K o K d F ( s) s (5) =0
Setting s = 0 in this limit, the expression evaluates to 0. This says that the error signal in
time will go to zero in response to a phase step at the input. (assuming that F(s) is not = 0
which is not possible anyway.)
In other words, no matter, how big the phase step, the loop will eventually track it out
and there will be no remaining steadystate error from a phase step. Imagine a swing, no matter high we pull the swing to start it; it will eventually come to
rest. Same situation here. The loop tracks out the initial phase push. Theoretically
speaking a PLL can recover from any phase step but we don’t mean any phase step, it is
limited generally to 180o or less as is obvious from the swing problem. But we can
actually recover from phases larger than 180o because mathematically the swing would
then flip over the top and once again will return to steady state.
Second question is; how long will it take the loop to track out this phase error? This is an
important question and helps determine what the loop parameters need to be so that
tracking time is not excessive and most importantly is within a bit or two of the signal
symbol rate. www.complextoreal.com 4 Unlocking the Phase Lock Loop  Part 2 The equation that describes this behavior is given below (From Reference 1, 2 and 3.) It
is developed by substituting the filter parameters in (1) and then applying the Final value
Theorem. (You will not find this equation explicitly derived in any of the PLL books.
Most books just tabulate these equations, including books by Gardener and Best.) θ e (t ) = ∆θ cos 1 − ζ 2 ωnt − (valid for ζ < 1) sin 1 − ζ 2 ωn t e −ζωn t 1− ζ 2 ζ (6) A little bit about the damping factor, ζ. The larger the damping factor the looser the
system and more violent the oscillations. (Most PLL use a damping factor of .707, which
is about optimum for signal processing applications.)
Figures 1 shows how the loop behaves in time domain to a phase step error.
1.2 Error/phase step 1
0.8
0.6
0.4
0.2
0
0.2 0 2 4 0.4 6 8 10 12 wn t 0.6 zeta = .3 zeta = .5 zeta = .707 zeta = .9 Figure 1 – Normalized transient response to a phase step In this graph both the x and the y axis have been normalized. The yaxis is error signal
magnitude divided by the phase step. It starts at 1.0, because initially the error signal is
exactly equal to the phase step. Then the error signal level begins to drop and at (wnt) >
10 it has completely tracked out the phase step and the steady state error has returned to
zero.
The xaxis is also normalized by the natural frequency. The time to reach steady state is
inversely proportional to wn. A large wn means a small settling time, t.
Now we plot the same curves but remove the normalization so we can see how the
magnitude of the error signal varies as the phase step is changed. www.complextoreal.com 5 Unlocking the Phase Lock Loop  Part 2 0.2 Error/phase step 0.15
0.1
0.05
0
0.05 0 2 4 6 8 10 12 wn t 0.1
zeta = .5 zeta = .707 Figure 2 – Transient response, wn = 1, phase step = 10o or .1745 rad which is seen as
the maximum yaxis value. Error/phase step 0.2
0.15
0.1
0.05
0
0.05 0 2 4 6 8 10 12 wn t  0.1
zeta = .5 zeta = .707 Figure 3 – Transient response, wn = 2, phase step = 10 o In this pair of graphs we see that when the natural frequency is increased, the settling
time decreases while the magnitude of the error does not change. In the following two
figures, phase shift has been increased to 45 o. Again we see the settling time decreases as
wn increases. www.complextoreal.com 6 Unlocking the Phase Lock Loop  Part 2 1 Error/phase step 0.8
0.6
0.4
0.2
0
0.2 0 2 4 6 8 10 12 wn t 0.4 zeta = .5 zeta = .707 Figure 4  Transient response, wn = 1, phase step = 45 o or .785 rad.
Note that y axis starts at .785 rads.
1 Error/phase step 0.8
0.6
0.4
0.2
0
0.2 0 2 4 0.4 6
wn t zeta = .5 8 10 12 zeta = .707 Figure 5  Transient response, wn = 2, phase step = 45 o or .785 rad. 1
Error/phase step 0.8
0.6
0.4
0.2
0
0.2 0 2 4 0.4
zeta = .5 6
wn t 8 10 12 zeta = .707 Figure 6  Transient response, wn = 3, phase step = 45o or .785 rad. www.complextoreal.com 7 Unlocking the Phase Lock Loop  Part 2 Error/phase step 2
1.5
1
0.5
0
0.5 0 2 4 6 8 10 12 wn t 1
zeta = .5 zeta = .707 Figure 7  Transient response, wn = 3, phase step = 90o or 1.57 rad. The interesting thing about these curves is that no matter how large the phase error, it can
be tracked out. Only the magnitude of the error signal changes. The tracking time does
not seem to be a function of the phase step but is definitely a function of the natural
frequency of the PLL.
Important result: The settling time of a PLL in response to a phase step is a function of its
natural frequency. A larger natural frequency gives a smaller settling time.
Example: For a system with a natural frequency of 20 rad/sec, the settling time is
ωn t ≈ 6
t= for ζ = .707 6
= .0003 sec
20000 Keep in mind that wn, the natural frequency is different from loop bandwidth by
approximately a factor of two. Most PLLs are specified by a loop bandwidth in
sympathy to specifying a PLL as a filter.
A large natural frequency would then seem to be the ticket to a fastreacting system. But
a large natural frequency also lets in more noise. The noise degrades the performance by
a different mechanism. There is balance somewhere between large and a small wn. Large
wn is good for response performance and bad for noise tolerance.
Case 2: The input frequency does a jump. Now let's assume that instead of phase change, the input signal does an abrupt frequency
shift such that a ∆ω is presented at the input to the PLL.
We can write this as the sum of the original frequency plus a step function times the
frequency shift. ωi (t ) = ω0 + ∆ω u (t ) (7) www.complextoreal.com 8 Unlocking the Phase Lock Loop  Part 2 The frequency ωo has now changed by ∆ ω . The phase change now seen by the PLL
(because phase change is all a PLL is capable of experiencing, so we convert everything
into phase) instead of being a step is a ramp, that is the phase changes continuously over
a certain period of time t, to give us the frequency shift we specify. We define input
phase as a function of time as
θ i (t ) = ∆ω t (8) The Laplace Transform of this function is θi ( s ) = ∆ω
s2 (9) Now apply this to PLL transfer function (1) multiply it by s and then take the limit to find
the time domain response.
lim θe (t ) = lim
t →∞ s →0 ∆ω
s2
s + K v Kd F ( s) s 2 ∆ω
=
Ko K d F ( s) (10) This is steady state error due to a frequency step. This result is different from the one we
obtained for phase step in (3). First of all the steadystate error is not zero like it was for
phase step. This result says that a steady state error will exist and will not be tracked out
when there is a frequency change at the input to the PLL.
The denominator of this equation is the gain of the loop which means we should use an
active filter with a large gain and make sure that all the other components also have large
gains. This will give large loop gain and better tolerance for frequency shifts.
KO = Kv K d F ( s)
The loop gain is equal to K0 which has dimensions of time1 or frequency, Hz.
Now we can rewrite (10) as θe = ∆ω
KO (11) The time domain behavior for this case is given by (Ref. 1, 2) www.complextoreal.com 9 Unlocking the Phase Lock Loop  Part 2 θ e (t ) = ∇ω ωn 1
1− ζ 2 sin 1 − ζ 2 ω n t e −ζω nt (12) The error in this case starts at zero and begins to build up and then damps out according
to the damping factor of the loop. Intuitively this is because the phase step is zero to
begin with but continues building up. Error * Freq step 0.01
0.008
0.006
0.004
0.002
0
0.002
0.004
0 2 4 zeta = .3 6 zeta = .5 wn t8 10 zeta = .707 12 14 zeta = .9 Figure 8 –
Response to a frequency step This error unlike the response to a phase step, never goes zero; see the steady state
section above wn t > 6. Since this steady state error is a function of the loop gain, it can be
made manageably small. We do that by making the loop gain as large as possible.
Important result: The steadystate error due to a sudden frequency shift of ∆f, can be
made small by using high gain loop and by using active filter rather than a passive filter. www.complextoreal.com 10 Unlocking the Phase Lock Loop  Part 2 Error * Freq step 0.008
0.006
0.004
0.002
0
0.002
0.004
0 2 4 6 zeta = .707 wn t8 10 12 14 zeta = .9 Figure 9 –
Response to a frequency step, .01 Hz, wn = 2 0.012 Error * Freq step 0.01
0.008
0.006
0.004
0.002
0
0.002
0.004
0 2 4 zeta = .707 6 wn t8 10 12 14 zeta = .9 Figure 10 –
Response to a frequency step, .02 Hz, wn = 2 www.complextoreal.com 11 Unlocking the Phase Lock Loop  Part 2 Error * Freq step 0.008
0.006
0.004
0.002
0
0.002
0.004
0 2 4 zeta = .707 6 wn t8 10 12 14 zeta = .9 Figure 11 –
Response to a frequency step, .01 Hz, wn = 3 Just as in response to a phase step, a larger wn gives better performance.
Case 3: The input frequency begins to change slowly In case 1, we changed the phase only. In case 2, we changed the phase slowly over a
period of time so that there was a frequency shift. Now we look at what the loop does, in
. response to a frequency ramp or a slowly changing frequency at the rate of rate of ∇ ω
per second2. This is can happen in situations where the transmitter or the receiver are
moving relative to each other causing a Doppler effect. The PLL sees this as changing
frequency.
. w1 (t ) = ω 0 + ∆ ω t (13) If you remember your dynamics class from college, you see that these equations are same
as displacement, velocity, and acceleration equations. This error is also called
acceleration error, where the previous one is called the velocity error.
The Laplace transform of a frequency ramp is
. ∆ω
θi ( s ) = 3
s (14) If we repeat the same process as above, take the derivative per the Final Value Theorem,
substitute the expression for the phase error and then take the limit, we get www.complextoreal.com 12 Unlocking the Phase Lock Loop  Part 2 lim
t →∞ θ e (t )
dt = lim [ s θ e (s ) ]
s →0 s 3θi ( s )
= lim
s →0 s + K K F ( s )
vd (15) . ∆ω
= lim
s→0 K K F ( s)
vd
. ∆ω
radian / sec
=
Ko The last line is the important result. This is the steady state error that is caused by the
frequency ramp. After time t, the accumulated error becomes
. ∆ω
t.
Ko (16) Note that this is similar to the value derived for the frequency step, Case 2. Only thing is
that this value keeps increasing with time. But if Ko (the total loop gain) is sufficiently
large, then we can reduce this error but only if the situation does not continue
indefinitely. In most real situation, unbounded accelerations do not happen so this error
can be mitigated by keeping the loop gain large. There is one other thing that we can do
to reduce this error even more. We can use a third order loop, this gives us thirdorder
term in the denominator and controls the magnitude of the error effectively.
The transient response for this case is given by (Ref. 1, 2) and is plotted below.
. . . ∆ω t ∆ω ∆ω θ e (t ) =
+ 2 − 2 cos 1 − ζ 2 ωn t −
Kv
ωn ωn sin 1 − ζ 2 ωn t e −ζωnt (17) 1−ζ 2 ζ www.complextoreal.com 13 Unlocking the Phase Lock Loop  Part 2 0.004
0.0035 Error 0.003
0.0025
0.002
0.0015
0.001
0.0005
0
0
z = .30 5 10
z = .50 15 20 25 30 z = .707 wn t = .90
z 35 40 Steadystate Error Figure 12 – Response to a frequency ramp, wn = 1
0.0014
0.0012 Error 0.001
0.0008
0.0006
0.0004
0.0002
0
0 10
z = .707 20
z = .90 30 40 50 60 wn Steadystate Error
t Figure 13 –
Response to a frequency ramp, wn = 3. Note that the tracking time is smaller but the
error still continues to grow. www.complextoreal.com 14 Unlocking the Phase Lock Loop  Part 2 0.0009
0.0008
0.0007 Error 0.0006
0.0005
0.0004
0.0003
0.0002
0.0001
0
0 10 20 z = .707 30
z = .90 40 50 60 70 80 wn Steadystate Error
t Figure 14 –
Response to a frequency ramp, wn = 3. Note that the tracking time is even smaller. Important result: A thirdorder loop can mitigate the frequency ramp errors more
effectively than a secondorder loop. Thirdorder loops have been used in deep space
satellite tracking and are suitable for a situation with Doppler such as cellular phones.
Mix of disturbances The following Figure shows how the PLL might react to a sequence of frequency and
phase shifts. At time t = 0, the PLL sees a frequency shift, which results in a steady state
error of approximately .03 radians, then a 80o phase shift is introduced, this is tracked out
and PLL returns to the previous steady state error. Figure 15  f = 5.1c Hz, then 80o phase shift at t = 128, wn = 2 www.complextoreal.com 15 Unlocking the Phase Lock Loop  Part 2 Figure 18  phase shift is 30o at t = 128 sec. Figure 19  phase shift is 110o at t = 128 sec. In Fig.18 and 19, you see response to two different phase shifts, each resulting in zero
steady state error as was predicted. www.complextoreal.com 16 Unlocking the Phase Lock Loop  Part 2 Case 4: Response of PLL to noise coming with the input signal and generated
within the PLL We can describe a generic modulated carrier as
s (t ) = A(t )sin(φ (t )) (18) where A(t) is the amplitude, and φ (t ) the changing center frequency and the phase. We
can write the phase term as
φ (t ) = ω0 t + θ (t ) (19) θ (t ) the instantaneous phase term is assumed to be sinusoidal, and we write it as
θ (t ) = m sin(ωm t ) (20) where ωm is the modulating frequency. Noise is not really sinusoidal but this is a
reasonable assumption for now. Now we can write the equation as
s (t ) = A(t )sin(ω0 t + m sin(ωm t )) (21) We will see what happens to the carrier as a random phase fluctuation is introduced via
the m sin(ωm t ) phase term. When the phase fluctuates, it introduces an unintended from of
FM modulation in to the carrier. (Notice that equation (21) is very much like the FM
modulation equation.) So whatever the spectrum of the signal, this additional phase noise
acts like an unintended FM modulation and results in a superimposed noise.
The FM modulation resulting from the phase noise, or jitter has Bessel function
components just like a FM signal but they are small and incidental. In general we assume
that the noise level is small compared to the signal power, so m, the noise mod index is
assumed to be << 1. Because of this, the noise spectrum has basically just two sidebands.
(This was covered in narrowband FM. A narrowband FM has just a few sidebands.) The
power in these sideband is related to the carrier power by m2 / 2 .
This allows us to make the following important statement about the phase noise process.
The power in phase noise, θ n2 is equal to onehalf the ratio between the signal power and
the noise power which can then be written as SNR as follows. www.complextoreal.com 17 Unlocking the Phase Lock Loop  Part 2 P /P
m2
=ns
2
2
1
=
P
2s
Pn θ n2 = = 1
2 SNR (22) rad 2 m2 is the ratio between signal and noise power and by manipulation, we get that the
power in components generated by the phase jitter is equal to the inverse of the SNR.
This seems intuitive. Better SNR means lower the jitter due to random phase noise and
conversely, large phase noise means lower SNR.
θ n2 = = 1
2 SNR (23) rad 2 Example: a signal has a SNR of 6 dB. The equivalent phase noise rms value can be
calculated by
= 1
2 ×4 = 1
rad 2
8
1
= .353 rad = 190
8 θ RMS = θ n2 = Example: A signal has rms phase noise of 10o. What is the SNR?
SNR = 1
2(rms phase noise)
1 = 10 2 57.3 = 12.1 dB 2 Noise bandwidth The noise bandwidth of any system with a transfer function given by H( ω) can be written
as
BL = 1
2π ∫ ∞ 0 2 H ( jω ) dω www.complextoreal.com 18 Unlocking the Phase Lock Loop  Part 2 This scary looking equation says that the noise bandwidth of a system is the area under its
spectral density.
If we plug in the transfer function of the PLL (active or passive filter), and solve it, we
get
BL = ωn 1
ζ + 2
4ζ (24) for ζ = .7, we get
BL ≅ .5ωn
Loop bandwith vs. Damping factor
1.4 1.2 1 0.8 0.6 0.4
0 0.5 1 1.5 ζ Figure 20 – Loop bandwidth vs. the damping factor.
Although damping factor .5 gives the smallest bandwidth, generally the .7 is used
instead since it gives better settling time.
Result: The optimum loop bandwidth is approximately onehalf the natural
frequency. Now remember that the noise bandwidth is kind of like a gate. It limits the amount of
noise that can pass though the system. So if a PLL has a noise bandwidth, then it makes
sense that it can reduce phase jitter that comes out of it.
Another important relationship;
If H(f) is the response of a system then the output spectral energy is given by
2 2 Y( f ) = H( f ) X ( f ) 2 (25) www.complextoreal.com 19 Unlocking the Phase Lock Loop  Part 2 We can represent the phase noise as a spectrum which extends to Bi/2 and has a value
of θ n2 . Its power spectral density is this
2 X(f) = θ n2 (26) Bi / 2 Recall that Power Spectral Density is a measure of power per frequency which is what
we see in Eq above. Now we have the input power spectral density, the response of the
system H(f), we can calculate the output spectral density by relationship.
This gives us after some manipulation of terms the following for the output phase jitter.
θ n2_ out = Pn BL
Ps Bi (27) Generalize Eq. 8 to say that this is also equal to
θ n2_ out = 1
2SNRL rad 2 (28) which says that the output phase noise (that which sneaked through the PLL) is inversely
proportional to the PLL bandwidth. This noise is also called untracked phase noise.
Equating to Eq (9) and Eq. (10), we get
SNRL = SNRi Bi
2 Bl (29) The SNR out of the loop is improved by the ratio of input signal noise bandwidth and the
loop bandwidth. Bi is generally the bandwidth of the incoming signal. It can also be set
equal to the prefilter or the symbol rate.
1. The SNR needs to be larger than 4 otherwise the RMS phase noise is too large.
2. The SNR out of the loop is improved by the PLL. So the PLL actually gives us
a signal gain.
Since noise can cause phase shift to go above 90o, unlocking can take place and will do
so. Viterbi has derived a formula for predicting how often the PLL will unlock. This is
given in several books such as Ref 1 (Page 54) and Ref 2 and will not be elaborated here. www.complextoreal.com 20 Unlocking the Phase Lock Loop  Part 2 Acquisition In all of the above cases we have assumed that the signal has already been acquired and
the loop is just responding to error conditions. In all of these cases, it is assumed that the
error is small and the behavior can be assumed linear.
What made it linear? In part I, we assumed that sin ( θ ) ≅ θ , when we developed the
transfer equations for the PLL. This requirement that the phase change be small makes
the analysis linear.
If the phase or the frequency of the input signal changes to such a large extent that the
this assumption is not valid, then we enter a nonlinear range of operation. Analysis of
such behavior is difficult, so most books resort to a limiting type of analysis which
identifies limits of operations rather than closed loop time domain equations.
PLL acquisition properties are specified by many different frequencies. All these are
limiting type of analyses.
The Hold Range:
The frequency over which the lock can be maintained successfully. In the linear assumption (while a lock is maintained) we set the Laplace transform of the
phase error due to a frequency offset as θe = ∆ω
KO but the real equation is
sin (θ e ) = ∆ω
KO Which says that as long as this ratio is less than 1, then there is lock, otherwise no. From
this we define a hold range of a loop by the condition
∆ω = ± K O (30) Which says that as long as the frequency step is less than the loop gain, the signal will be
held by the loop. Conversely, to make a loop better respond to large frequency steps, we
need to make the loop gain large.
The Lock Range
The lock range is the frequency range over which acquisition can be made within a single
beat note. A beat note is the difference between the VCO and the input signal frequencies
and accompanies with it sidebands at dc. The dc term is used to control the PLL and if www.complextoreal.com 21 Unlocking the Phase Lock Loop  Part 2 this can be done in just one loop around then the frequency of the input signal can be
considered to be within the lock range.
The equation for the locks range is obtained by
∆ω L = K d K v F ( ∆ω L ) (31) The underlined term is the filter response and for an active filter is given by
F ( ∆ω L ) = K l τ2
τ1 (32) The lock range is a function of the loop filter parameters as given for an active lag filter
here. The time to acquire is given by
T= 2π (33)
ωn
So once again, the time to acquire is a function of the natural frequency. Pullin, pullout and everything in between There is a frequency step above which the PLL does not lock quickly. This frequency is
called the pullout frequency. Viterbi did a lot of simulations and has proposed an
empirical equation to identify this frequency as a function of the damping factor.
∆ω PO = 1.8ωn ( ζ + 1) (34) For zeta = .707, the most common figure, the pullout frequency is app. three times the
natural frequency.
So if natural frequency of a loop is 10 kHz, then the frequency step can be as large as 30
kHz and the PLL will still maintain lock, i.e. no data would be lost.
As long as the input frequency is within the lock range, the PLL can lock on to the
incoming frequency fairly quickly. But there are cases, where even if the frequency is
fairly far away the PLL may still be able to “walk up” to the incoming frequency and
start tracking it. It takes longer but can be done.
Let’s take a look at how this may be done. In Fig 21(a) below, we show a case, where
PLL signal is moving towards the input signal. The signal is assumed to be out of the
PLL range. This configuration gives rise to a FM modulation between the PLL and input www.complextoreal.com 22 Unlocking the Phase Lock Loop  Part 2 signal and generates the sidebands you see with center frequency of these, a beat
frequency.
Beat frequency is the difference between two frequencies. There is also a dc term that
goes with it. The PLL tries to use the dc term to move closer to the signal. But it actually
needs the beat frequency voltage which is out of its range. It tries to lock based on the dc
term. In the Fig. 22, we see it tries again, and now the delta frequency is less so the beat
frequencies are moving closer together but there is still a dc term. The PLL tries again
that keeps trying until this dc term is made zero as we can see in these figures. Figure 21 – PLL frequency = 36, Input signal = 44 PLL reacts to the dc terms but
really should react to the terms at the beat frequency. Because the dc term is small,
the correction the PLL makes is too small. www.complextoreal.com Unlocking the Phase Lock Loop  Part 2 23 Figure 22 – Now the PLL is moving closer to the signal, the beat note is smaller and
so is the dc term. In the last case, the signal has moved to the opposite side. (Note that what you see here is a demonstration. The PLL is fixed and I have moved the
input signal. In a real case, it is the PLL that moves. Also nowhere in a PLL will you see
all these tones on one spectrum.)
In time domain, let’s take a look at what the error signal is doing. www.complextoreal.com 24 Unlocking the Phase Lock Loop  Part 2 Figure 23 – Error signal as the input signal is moving away from the VCO. www.complextoreal.com 25 Unlocking the Phase Lock Loop  Part 2 The above figures show the response of a system with natural frequency of 1.0. The PLL
is at 5.0 Hz. Figure 23 (a) shows the response when the input signal is 5.7 Hz. The error
signal indicates that PLL is able to lock quickly. Figure 23 (b) shows response to an input
signal at 5.75 Hz. Now the transient is a bit longer but lock is still obtained. In the figure
(c), the input signal is at 5.8 Hz, now we see that that lock does not take place and the
PLL goes through a second locking loop and it is taking longer to lock. This pattern
continues as input frequency is moving away from PLL. At 5.9 Hz, the lock never takes
place.
This is called a pullin process and although of interest academically, is not one we want
to encounter in signal processing. We want to able to lock immediately and so should
design a system that has a lockin frequency well within what might be encountered.
References:
1. Phaselock Techniques, Floyd Gardener, John Wiley & Sons, 2nd Edition
2. PhaseLocked Loop Circuit Design, Dan H. Wolaver, Prentice Hall, Ist Edition
3. PhaseLocked Loops: Design, Simulation, and Applications, Ronald E. Best, McGraw
Hill, 4th Edition
4. PhaseLock Basics, William Eagan, Wiley Interscience, July 1998
______________________________________________________ Copyright 1998, 2002 Charan Langton
I can be contacted at
[email protected]
Other tutorials at
www.complextoreal.com www.complextoreal.com ...
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This note was uploaded on 02/07/2011 for the course EEE EE567 taught by Professor Tutorials during the Spring '11 term at Birla Institute of Technology & Science, Pilani  Hyderabad.
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