6.302 Feedback Systems
Recitation : Phase-locked Loops
Prof. Joel L. Dawson
Phase-locked loops are a foundational building block for analog circuit design, particularly
for communications circuits. They provide a good example system for this class because they
are an excellent exercise in physical modeling. In these systems, the key variable is the phase
of a sinusoid. As a first step, then we must be precise about what we mean by the phase of a
sinusoid. Consider:
v(t) = cos [φ(t)]
We define the frequency of a sinusoid as the instantaneous rate of change of its phase. That
is:
ω≡
EXAMPLE:
dφ
dt
v(t) = cos (ω0t + φ0)
PHASE = ω0t + φ0 = φ(t)
FREQUENCY =
dφ(t)
= ω0
dt
To be consistent, we write the phase in terms of the frequency:
t
φ = ∫ -∞
ω(t) dt
So to understand phase-locked loops (PLLs) we must make the following conceptual jump...
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6.302 Feedback Systems
Recitation : Phase-locked Loops
Prof. Joel L. Dawson
VOLTAGE PICTURE:
SIGNAL
GENERATOR
cos(ω0t + φ0)
PLL PICTURE:
SIGNAL
GENERATOR
ω0t + φ0
Now, the anatomy of a PLL:
VOLTAGE PICTURE
cos(ω0t + φ0)
PHASE
DETECTOR
Ve
Vc
LOOP
FILTER
VCO
cos(ωt + φ)
VCO = Voltage Controlled Oscillator
ω = k0Vc
Ve = kd (φ0 - φ)
[kd] = V/RAD
Notice, if Ve is constant, φ - φ0 is constant => ω0 = ω
A PLL locks the output of a VCO in frequency and phase to an incoming periodic signal.
PLL PICTURE
φIN
Σ
-
φe
kd
Ve
F(s)
Vc
k0
s
φOUT
VCO is an integrator. Its output frequency is
dφOUT
t
= k0Vc => φOUT = ∫-∞
k0Vcdt
dt
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Cite as: Joel Dawson, course materials for 6.302 Feedback Systems, Spring 2007.
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6.302 Feedback Systems
Recitation : Phase-locked Loops
Prof. Joel L. Dawson
Now, let’s look at how we put together and use PLLs. To start, how does one build a phase
detector?
) ANALOG MULTIPLIER:
VINcos(ωt+φ)
V0
X
LPF
V02
VOUTcos(ω2t+φ2)
V0
= VINV
OUT cos(ωt+φ) cos(ω2t+φ2)
= ½ V
INVOUT [cos(ωt + φ + ω2t + φ2) + cos(ωt + φ - ω2t - φ2)]
IF ω
= ω2 = ω
> V0 = ½ VINVOUT [cos(2ωt + φ + φ2) + cos(φ - φ2)]
After LPF, we lose high-frequency component:
V02 = ½ VINVOUT [cos(φ - φ2)]
So we get zero out of the phase detector when φ - φ2 = ± π/2 .
Linearizing about this condition, we would say:
∆v02 = ±
}
VINVOUT
∆φ
2
k0
(Notice that the constant k0 depends on the amplitude of the sinusoids.)
2) DIGITAL XOR GATE
VIN
0
0
V
IN
V
OUT
VOUT
0
0
VXOR
0
0
Page 3
Cite as: Joel Dawson, course materials for 6.302 Feedback Systems, Spring 2007.
MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.302 Feedback Systems
Recitation : Phase-locked Loops
Prof. Joel L. Dawson
Easiest to analyze in time domain. (Here, assume square wave inputs.)
VIN
Т/2
t
VOUT
t
VXOR
∆t
t
For our phase detector output, we’ll use the average (DC) value of VXOR:
VXOR =
T/2
[·∆t + 0·(Τ/2 - ∆t)] = 2 ·
∆t
T
Now, how do we relate this to phase? Recall that for a sinusoid:
cos(ωt - φ) = cos (2πft - φ)
2π
T
= cos (
THUS: VXOR =
t - φ)
= cos
2π
T
(t -
φ
2π
= cos
2π
T
(t - ∆t) => ∆t =
∙ T)
φ
2π
∙T
φ
2 φe
(
∙ T) = e
2π
T 2π
VXOR
π
φe
Page Cite as: Joel Dawson, course materials for 6.302 Feedback Systems, Spring 2007.
MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.302 Feedback Systems
Recitation : Phase-locked Loops
Prof. Joel L. Dawson
There are many other phase detectors, each with their own strengths and weaknesses. More
on these later...
Application to stabilization of the frequency of a laser
φIN
Σ
C(s)
kd
Reference
Signal
Controller
P
Laser
Cavity
Dynamics
Laser Pump
Power
ωL
s
φOUT
Frequency
of Laser
Light
Locks frequency of laser light to a stable reference.
Typical laser cavity dynamics:
G(s) = e-ST
delay
Typical choices for a controller:
ωn2
s2 + 2ζωns + ωn2
second order system
C(s) = D0s
D0s + P0
I0 s
kk
Returning to a general case, we have L(s) = 0s d F(s), where as a designer you usually have
kk
some control over the form of F(s). Suppose we choose F(s) = , so that L(s) is just 0s d .
What is the steady-state error in response to a constant-frequency input?
cos(ω0t)
ramp in phase
ω0
s2
Page Cite as: Joel Dawson, course materials for 6.302 Feedback Systems, Spring 2007.
MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.302 Feedback Systems
Recitation : Phase-locked Loops
Prof. Joel L. Dawson
Steady-state error, then, is
lim
lim
t—>∞ φe(t) = s —> 0 s
lim
ω
= s —> 0 s0
ω0
=
k0kd
ω0
2 ·
s
+C(s)
+ k sk
0 d
=> Large k0kd for small phase error. But according to root locus,
jω
σ
×
Large k0kd also means large bandwidth. If we have a noisy reference, large bandwidth is not
a good thing.
We can improve things by being more sophisticated in our choice of F(s):
F(s) =
Steady state error is still
τs + => L(s) =
k
s(sτ + )
ω0
k , but bandwidth is reduced:
log |L|
ω
0
/τ
ωC
∡L(s)
ωC'
-90˚
-80˚
˚
Page 6
Cite as: Joel Dawson, course materials for 6.302 Feedback Systems, Spring 2007.
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6.302 Feedback Systems
Recitation : Phase-locked Loops
Prof. Joel L. Dawson
Now we’ve got our improved noise performance, but increasing k will lower our damping ratio:
jω
×
×
σ
Put another way, increasing k will lower our phase margin.
=> we must decide what stability margins are acceptable in our application.
Suppose we decide that a 2% overshoot in the step response is acceptable. Using our chart of 2ns order parameters, we discover that this corresponds to ζ = 0. and Mp = .. This means we should design for a phase margin of
Mp ≈
sinφm
φn ≈ sin- ( m ) ≈ ˚
p
We arrange for this by ensuring that |L| = at the frequency for which ∡L(s) = -3˚. Looking
at our Bode Plot, we see that this frequency is just ω = /τ. On the asymptotic magnitude plot,
kτ
|L(s)| at this frequency is /k τ =kτ. The actual magnitude is √2
.
We therefore choose k using
kτ
√2
= => k =
√2
τ
Page Cite as: Joel Dawson, course materials for 6.302 Feedback Systems, Spring 2007.
MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].