4.3 FEEDBACK DEMODULATORS: THE PHASE-LOCKED LOOP
4.3.2 Phase-Locked Loop Operation in the Tracking Mode
The Linear Model
As we have seen, in the tracking mode the phase error is small, and linear analysis can be used to define PLL operation. Considerable insight into PLL operation can be gained by investigating the steady-state errors for first-order, second-order, and third-order PLLs with a variety of input signals.
The Loop Transfer Function and Steady-State Errors
The frequency-domain equivalent of Figure 4.20 is illustrated in Figure 4.21. It follows from Figure 4.21 and (4.104) that
Θ(𝑠) = 𝐾𝑡[Φ(𝑠) − Θ(𝑠)]𝐹 (𝑠)
𝑠 (4.106)
from which the transfer function relating the VCO phase to the input phase is 𝐻(𝑠) = Θ(𝑠)
Φ(𝑠) = 𝐾𝑡𝐹 (𝑠)
𝑠 + 𝐾𝑡𝐹 (𝑠) (4.107)
immediately follows. The Laplace transform of the phase error is
Ψ(𝑠) = Φ(𝑠) − Θ(𝑠) (4.108)
Loop f ilter F(s) Loop
gainKt
+ −
Φ(s) Σ
Demodulated output Ψ(s)
Θ(s)
VCO 1/s
Figure 4.21
Linear PLL model in the frequency domain.
Therefore, we can write the transfer function relating the phase error to the input phase as 𝐺(𝑠) = Ψ(𝑠)
Φ(𝑠) = Φ(𝑠) − Θ(𝑠)
Φ(𝑠) = 1 − 𝐻(𝑠) (4.109)
so that
𝐺(𝑠) = 𝑠
𝑠 + 𝐾𝑡𝐹 (𝑠) (4.110)
The steady-state error can be determined through the final value theorem from Laplace transform theory. The final value theorem states that thelim𝑡→∞𝑎(𝑡)is given bylim𝑠→0𝑠𝐴(𝑠), where𝑎(𝑡)and𝐴(𝑠)are a Laplace transform pair.
In order to determine the steady-state errors for various loop orders, we assume that the phase deviation has the general form
𝜙(𝑡) = 𝜋𝑅𝑡2+ 2𝜋𝑓Δ𝑡 + 𝜃0, 𝑡 > 0 (4.111) The corresponding frequency deviation is
1 2𝜋
𝑑𝜙
𝑑𝑡 = 𝑅𝑡 + 𝑓Δ, 𝑡 > 0 (4.112)
We see that the frequency deviation is the sum of a frequency ramp,𝑅Hz/s, and a frequency step𝑓Δ. The Laplace transform of𝜙(𝑡)is
Φ(𝑠) = 2𝜋𝑅
𝑠3 + 2𝜋𝑓Δ
𝑠2 + 𝜃0
𝑠 (4.113)
Thus, the steady-state phase error is given by 𝜓𝑠𝑠= lim
𝑠→0𝑠 [2𝜋𝑅
𝑠3 +2𝜋𝑓Δ
𝑠2 +𝜃0
𝑠 ]
𝐺(𝑠) (4.114)
where𝐺(𝑠)is given by (4.110).
In order to generalize, consider the third-order filter transfer function defined in Table 4.4:
𝐹 (𝑠) = 1
𝑠2(𝑠2+ 𝑎𝑠 + 𝑏) (4.115)
If𝑎 = 0and𝑏 = 0,𝐹 (𝑠) = 1, which is the loop filter transfer function for a first-order PLL. If 𝑎≠0, and𝑏 = 0,𝐹 (𝑠) = (𝑠 + 𝑎)∕𝑠, which defines the loop filter for second-order PLL. With
Table 4.4 Steady-state Errors
𝜽𝟎≠𝟎 𝜽𝟎≠𝟎 𝜽𝟎≠𝟎
𝒇𝚫= 𝟎 𝒇𝚫≠𝟎 𝒇𝚫≠𝟎
PLL order 𝑹 = 𝟎 𝑹 = 𝟎 𝑹≠𝟎
1(𝑎 = 0,𝑏 = 0) 0 2𝜋𝑓Δ∕𝐾𝑡 ∞
2(𝑎≠0,𝑏 = 0) 0 0 2𝜋𝑅∕𝐾𝑡
3(𝑎≠0,𝑏≠0) 0 0 0
𝑎≠0and𝑏≠0we have a third-order PLL. We can therefore use𝐹 (𝑠), as defined by (4.115) with𝑎and𝑏taking on appropriate values, to analyze first-order, second-order, and third-order PLLs.
Substituting (4.115) into (4.110) yields
𝐺(𝑠) = 𝑠3
𝑠3+ 𝐾𝑡𝑠2+ 𝐾𝑡𝑎𝑠 + 𝐾𝑡𝑏 (4.116) Using the expression for𝐺(𝑠)in (4.114) gives the steady-state phase error expression
𝜓𝑠𝑠= lim
𝑠→0
𝑠(𝜃0𝑠2+ 2𝜋𝑓Δ𝑠 + 2𝜋𝑅)
𝑠3+ 𝐾𝑡𝑠2+ 𝐾𝑡𝑎𝑠 + 𝐾𝑡𝑏 (4.117) We now consider the steady-state phase errors for first-order, second-order, and third-order PLLs. For various input signal conditions, defined by 𝜃0, 𝑓Δ, and 𝑅 and the loop filter parameters𝑎and𝑏, the steady-state errors given in Table 4.4 can be determined. Note that a first-order PLL can track a phase step with a zero steady-state error. A second-order PLL can track a frequency step with zero steady-state error, and a third-order PLL can track a frequency ramp with zero steady-state error.
Note that for the cases given in Table 4.4 for which the steady-state error is nonzero and finite, the steady-state error can be made as small as desired by increasing the loop gain 𝐾𝑡. However, increasing the loop gain increases the loop bandwidth. When we consider the effects of noise in Chapter 8, we will see that increasing the loop bandwidth makes the PLL performance more sensitive to the presence of noise. We therefore see a trade-off between steady-state error and loop performance in the presence of noise.
EXAMPLE 4.6
We now consider a first-order PLL, which from (4.110) and (4.115), with𝑎 = 0and 𝑏 = 0, has the transfer function
𝐻(𝑠) = Θ(𝑠) Φ(𝑠)= 𝐾𝑡
𝑠 + 𝐾𝑡 (4.118)
The loop impulse response is therefore
ℎ(𝑡) = 𝐾𝑡𝑒−𝐾𝑡𝑡𝑢(𝑡) (4.119)
The limit of ℎ(𝑡)as the loop gain 𝐾𝑡 tends to infinity satisfies all properties of the delta function.
Therefore,
𝐾lim𝑡→∞𝐾𝑡𝑒−𝐾𝑡𝑡𝑢(𝑡) = 𝛿(𝑡) (4.120)
which illustrates that for large loop gain𝜃(𝑡) ≈ 𝜙(𝑡). This also illustrates, as we previously discussed, that the PLL serves as a demodulator for angle-modulated signals. Used as an FM demodulator, the VCO input is the demodulated output since the VCO input signal is proportional to the frequency deviation of the PLL input signal. For PM the VCO input is simply integrated to form the demodulated output, since phase deviation is the integral of frequency deviation.
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EXAMPLE 4.7
As an extension of the preceding example, assume that the input to an FM modulator is𝑚(𝑡) = 𝐴𝑢(𝑡). The resulting modulated carrier
𝑥𝑐(𝑡) = 𝐴𝑐cos [
2𝜋𝑓𝑐𝑡 + 𝑘𝑓𝐴∫
𝑡𝑢(𝛼)𝑑𝛼 ]
(4.121) is to be demodulated using a first-order PLL. The demodulated output is to be determined.
This problem will be solved using linear analysis and the Laplace transform. The loop transfer function (4.118) is
Θ(𝑠) Φ(𝑠) = 𝐾𝑡
𝑠 + 𝐾𝑡 (4.122)
The phase deviation of the PLL input𝜙(𝑡)is 𝜙(𝑡) = 𝐴𝑘𝑓∫
𝑡𝑢(𝛼)𝑑𝛼 (4.123)
The Laplace transform of𝜙(𝑡)is
Φ(𝑠) =𝐴𝑘𝑓
𝑠2 (4.124)
which gives
Θ(𝑠) =𝐴𝐾𝑓 𝑠2
𝐾𝑡
𝑠 + 𝐾𝑡 (4.125)
The Laplace transform of the defining equation of the VCO, (4.99), yields
𝐸𝑣(𝑠) = 𝑠𝐾𝑣Θ(𝑠) (4.126)
so that
𝐸𝑣(𝑠) =𝐴𝐾𝑓 𝐾𝑣
𝐾𝑡
𝑠(𝑠 + 𝐾𝑡) (4.127)
Partial fraction expansion gives
𝐸𝑣(𝑠) = 𝐴𝐾𝑓 𝐾𝑣
(1 𝑠− 1
𝑠 + 𝐾𝑡 )
(4.128) Thus, the demodulated output is given by
𝑒𝑣(𝑡) =𝐴𝐾𝑓
𝐾𝑣 (1 − 𝑒−𝐾𝑡𝑡)𝑢(𝑡) (4.129)
Note that for𝑡 ≫ 1∕𝐾𝑡 and𝐾𝑓= 𝐾𝑣 we have, as desired,𝑒𝑣(𝑡) = 𝐴𝑢(𝑡)as the demodulated output.
The transient time is set by the total loop gain𝐾𝑡, and𝐾𝑓∕𝐾𝑣is simply an amplitude scaling of the demodulated output signal.
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As previously mentioned, very large values of loop gain cannot be used in practical applications without difficulty. However, the use of appropriate loop filters allows good performance to be achieved with reasonable values of loop gain and bandwidth. These filters make the analysis more complicated than our simple example, as we shall soon see.
Even though the first-order PLL can be used for demodulation of angle-modulated signals and for synchronization, the first-order PLL has a number of drawbacks that limit its use for most applications. Among these drawbacks are the limited lock range and the nonzero steady- state phase error to a step-frequency input. Both these problems can be solved by using a second-order PLL, which is obtained by using a loop filter of the form
𝐹 (𝑠) = 𝑠+ 𝑎
𝑠 = 1 + 𝑎
𝑠 (4.130)
This choice of loop filter results in what is generally referred to as aperfect second-order PLL.
Note that the loop filter defined by (4.130) can be implemented using a single integrator, as will be demonstrated in a Computer Example 4.4 to follow.
The Second-Order PLL: Loop Natural Frequency and Damping Factor With𝐹 (𝑠)given by (4.130), the transfer function (4.107) becomes
𝐻(𝑠) = Θ(𝑠)
Φ(𝑠) = 𝐾𝑡(𝑠 + 𝑎)
𝑠2+ 𝐾𝑡𝑠 + 𝐾𝑡𝑎 (4.131)
We also can write the relationship between the phase errorΨ(𝑠)and the input phaseΦ(𝑠).
From Figure 4.21 or (4.110), we have 𝐺(𝑠) = Ψ(𝑠)
Φ(𝑠)= 𝑠2
𝑠2+ 𝐾𝑡𝑎𝑠 + 𝐾𝑡𝑎 (4.132) Since the performance of a linear second-order system is typically parameterized in terms of the natural frequency and damping factor, we now place the transfer function in the standard form for a second-order system. The result is
Ψ(𝑠)
Φ(𝑠) = 𝑠2
𝑠2+ 2𝜁𝜔𝑛𝑠 + 𝜔2𝑛 (4.133)
in which𝜁is the damping factor and𝜔𝑛is the natural frequency. It follows from the preceding expression that the natural frequency is
𝜔𝑛=√
𝐾𝑡𝑎 (4.134)
and that the damping factor is
𝜁 = 1 2
√𝐾𝑡
𝑎 (4.135)
A typical value of the damping factor is 1∕√
2 = 0.707. Note that this choice of damping factor gives a second-order Butterworth response.
In simulating a second-order PLL, one usually specifies the loop natural frequency and the damping factor and determines loop performance as a function of these two fundamental parameters. The PLL simulation model, however, is a function of the physical parameters𝐾𝑡
and𝑎. Equations (4.134) and (4.135) allow𝐾𝑡and𝑎to be written in terms of𝜔𝑛and𝜁. The results are
𝑎 = 𝜔𝑛 2𝜁 = 𝜋𝑓𝑛
𝜁 (4.136)
and
𝐾𝑡= 4𝜋𝜁𝑓𝑛 (4.137)
where2𝜋𝑓𝑛= 𝜔𝑛. These last two expressions will be used to develop the simulation program for the second-order PLL that is given in Computer Example 4.4.
EXAMPLE 4.8
We now work a simple second-order example. Assume that the input signal to the PLL experiences a small step change in frequency. (The step in frequency must be small to ensure that the linear model is applicable. We will consider the result of large step changes in PLL input frequency when we consider operation in the acquisition mode.) Since instantaneous phase is the integral of instantaneous frequency and integration is equivalent to division by𝑠, the input phase due to a step in frequency of magnitude Δ𝑓is
Φ(𝑠) =2𝜋Δ𝑓
𝑠2 (4.138)
From (4.133) we see that the Laplace transform of the phase error𝜓(𝑡)is
Ψ(𝑠) = Δ𝜔
𝑠2+ 2𝜁𝜔𝑛𝑠 + 𝜔2𝑛 (4.139)
Inverse transforming and replacing𝜔𝑛by2𝜋𝑓𝑛yields, for𝜁 < 1, 𝜓(𝑡) = Δ𝑓
𝑓𝑛√
1 − 𝜁2𝑒−2𝜋𝜁𝑓𝑛𝑡[sin(2𝜋𝑓𝑛√
1 − 𝜁2𝑡)]𝑢(𝑡) (4.140)
and we see that𝜓(𝑡)→0as𝑡→∞. Note that the steady-state phase error is zero, which is consistent with the values shown in Table 4.4.
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