Page 164 - Phase-Locked Loops Design, Simulation, and Applications
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PLL PERFORMANCE IN THE PRESENCE OF NOISE   Ronald E. Best                              101
                 Hence, the rms value         becomes






                 Since the rms value of phase jitter is about 20°, the value of 180° (limit of dynamic
               stability) is rarely exceeded. Consequently, the PLL does not unlock frequently. At (SNR) =
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               1, however, the rms value of output phase jitter would be as large as 40°, and the dynamic
               limit of stability would be exceeded on every major noise peak, thus making stable operation
               impossible.
                 As a rule of thumb,



                                                                                           (4.15)

               is a convenient design goal.

                 Note: The SNR of a signal can be specified either numerically or in decibels. The numerical
               value SNR is computed from








               whereas the (SNR)  is calculated from
                                 dB






               where U  (rms) and U  (rms) are the rms values of signal and noise, respectively.
                       s            n
                 The designer of practical PLL circuits is vitally interested in how often, on the average, a
               system will temporarily unlock. The probability of unlocking  is decreased with increasing
               (SNR) . We now define  T  to be the average time interval between two lock-outs. For
                     L                    av
               example, if T  = 100 ms, the PLL unlocks (on average) 10 times per second.
                            av
                                                                                                     1
                 For second-order PLLs, T  has been found experimentally as a function of (SNR) . The
                                           av
                                                                                                   L
               resulting curve is plotted in Fig. 4.6.
                 To illustrate the theory, let’s calculate a numerical example.


                    Numerical Example A second-order PLL is assumed to have the following specifications:









                    From Fig. 4.6, we read ω T  = 200. Consequently, T  = 3 ms. This means the PLL unlocks
                                          n av
                                                                   av
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