Figure 2.26 (a) VCO with $1/f$$1/f$1//f1 / f noise on its supply, and (b) resulting phase noise spectrum.
图 2.26 (a) 具有 $1/f$$1/f$1//f1 / f 噪声的 VCO 电源，以及(b) 由此产生的相位噪声谱。
we have  我们有

We say flicker-noise-induced phase noise follows a $1/f^3$$1/f^3$1//f^(3)1 / f^{3} profile. Illustrated in Fig. 2.26(b), this behavior manifests itself as a $-30-\mathrm{dB}/\mathrm{dec}$$-30-\mathrm{dB}/\mathrm{dec}$-30-dB//dec-30-\mathrm{dB} / \mathrm{dec} slope on a $\log -\log$$\log -\log$log-log\log -\log scale, ${}^5$${}^5$^(5){ }^{5} followed by a $-20\mathrm{dB}/\mathrm{dec}$$-20\mathrm{dB}/\mathrm{dec}$-20dB//dec-20 \mathrm{~dB} / \mathrm{dec} slope corresponding to white-noise-induced phase noise. In this case, too, the phase fluctuations at low offset frequencies (e.g., below 10 Hz ) become very large, violating the narrowband FM approximation and yielding a Lorentzian shape, but we are typically interested in much higher offsets and can assume $S_{\text{out}}$$S_{\text{out }}$S_("out ")S_{\text {out }} (the output voltage spectrum) is a shifted copy of $S_{\varphi n}$$S_{\varphi n}$S_(phi n)S_{\phi n}.
我们说闪烁噪声引起的相位噪声遵循 $1/f^3$$1/f^3$1//f^(3)1 / f^{3} 轮廓。如图 2.26(b) 所示，这种行为表现为 $-30-\mathrm{dB}/\mathrm{dec}$$-30-\mathrm{dB}/\mathrm{dec}$-30-dB//dec-30-\mathrm{dB} / \mathrm{dec} 斜率在 $\log -\log$$\log -\log$log-log\log -\log 规模上， ${}^5$${}^5$^(5){ }^{5} 后面是对应于白噪声引起的相位噪声的 $-20\mathrm{dB}/\mathrm{dec}$$-20\mathrm{dB}/\mathrm{dec}$-20dB//dec-20 \mathrm{~dB} / \mathrm{dec} 斜率。在这种情况下，低偏移频率（例如，低于 10 Hz）的相位波动也变得非常大，违反了窄带 FM 近似，并产生洛伦兹形状，但我们通常对更高的偏移感兴趣，可以假设 $S_{\text{out}}$$S_{\text{out }}$S_("out ")S_{\text {out }} （输出电压谱）是 $S_{\varphi n}$$S_{\varphi n}$S_(phi n)S_{\phi n} 的一个平移副本。

We have seen that jitter in the time domain can arise from (deterministic) sidebands or (random) phase noise. We learned in Section 2.2.1 that the peak-to-peak deterministic jitter in radians is equal to 4 times the normalized magnitude of the output sidebands. In this section, we derive the relation between random jitter and the phase noise spectrum. The randomness suggests that we cannot specify a peak-to-peak jitter and must seek an rms value. ${}^6$${}^6$^(6){ }^{6}
我们已经看到，时域中的抖动可以来自（确定性）边带或（随机）相位噪声。我们在第 2.2.1 节中了解到，单位弧度的峰峰值确定性抖动等于输出边带的归一化幅度的 4 倍。在本节中，我们推导随机抖动与相位噪声谱之间的关系。随机性表明我们无法指定峰峰值抖动，必须寻求均方根值。 ${}^6$${}^6$^(6){ }^{6}

Consider the waveform $V_0\cos [{\omega }_{0}t+{\varphi }_n(t)]$$V_0\cos \left[{\omega }_{0}t+{\varphi }_n(t)\right]$V_(0)cos[omega_(0)t+phi_(n)(t)]V_{0} \cos \left[\omega_{0} t+\phi_{n}(t)\right]. We know that ${\varphi }_n(t)$${\varphi }_n(t)$phi_(n)(t)\phi_{n}(t) represents the fluctuations in the zero crossings and, as with any other time-domain quantity, exhibits an average “power” given by
考虑波形 $V_0\cos [{\omega }_{0}t+{\varphi }_n(t)]$$V_0\cos \left[{\omega }_{0}t+{\varphi }_n(t)\right]$V_(0)cos[omega_(0)t+phi_(n)(t)]V_{0} \cos \left[\omega_{0} t+\phi_{n}(t)\right] 。我们知道 ${\varphi }_n(t)$${\varphi }_n(t)$phi_(n)(t)\phi_{n}(t) 表示零交叉的波动，并且与任何其他时域量一样，表现出由以下公式给出的平均“功率”

(Recall the definition of average power in Section 2.1.4.) This quantity is difficult to compute directly, but we
（回忆第 2.1.4 节中平均功率的定义。）这个量直接计算是困难的，但我们

also know from Parseval’s theorem that the average power is equal to the area under the spectrum:
也可以从帕尔塞瓦尔定理知道，平均功率等于频谱下的面积：
$$\overline{{\varphi }_n^2(t)}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{S}_{\varphi n}(f)df.$$$$\overline{{\varphi }_n^2(t)}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }} S_{\varphi n}(f)df.$$bar(phi_(n)^(2)(t))=int_(-oo)^(+oo)S_(phi n)(f)df.\overline{\phi_{n}^{2}(t)}=\int_{-\infty}^{+\infty} S_{\phi n}(f) d f .

The rms jitter, $J_{rms}$$J_{rms}$J_(rms)J_{r m s}, is equal to the square root of $\overline{{\varphi }_n^2(t)}$$\overline{{\varphi }_n^2(t)}$bar(phi_(n)^(2)(t))\overline{\phi_{n}^{2}(t)} :
rms 抖动， $J_{rms}$$J_{rms}$J_(rms)J_{r m s} ，等于 $\overline{{\varphi }_n^2(t)}$$\overline{{\varphi }_n^2(t)}$bar(phi_(n)^(2)(t))\overline{\phi_{n}^{2}(t)} 的平方根：

This relationship proves extremely useful in computing the rms (random) jitter from the phase noise profile. Note that $J_{rms}$$J_{rms}$J_(rms)J_{r m s} is in radians and must be divided by $2\pi$$2\pi$2pi2 \pi and multiplied by the period if we wish to express it in seconds.
这个关系在计算相位噪声特征中的均方根（随机）抖动时非常有用。请注意， $J_{rms}$$J_{rms}$J_(rms)J_{r m s} 是以弧度为单位的，如果我们希望以秒为单位表示它，则必须将其除以 $2\pi$$2\pi$2pi2 \pi 并乘以周期。

The output phase noise of a phase-locked loop can be approximated by the profile shown in Fig. 2.27, where $S_{\varphi n}(f)$$S_{\varphi n}(f)$S_(phi n)(f)S_{\phi n}(f) is flat for $|f|<f_1$$|f|<f_1$|f| < f_(1)|f|<f_{1} and falls in proportion to $1/f^2$$1/f^2$1//f^(2)1 / f^{2} for $|f|>f_1$$|f|>f_1$|f| > f_(1)|f|>f_{1}. Compute the rms jitter.
相位锁定环的输出相位噪声可以通过图 2.27 所示的轮廓进行近似，其中 $S_{\varphi n}(f)$$S_{\varphi n}(f)$S_(phi n)(f)S_{\phi n}(f) 在 $|f|<f_1$$|f|<f_1$|f| < f_(1)|f|<f_{1} 时是平坦的，并且在 $|f|>f_1$$|f|>f_1$|f| > f_(1)|f|>f_{1} 时与 $1/f^2$$1/f^2$1//f^(2)1 / f^{2} 成比例下降。计算均方根抖动。

Figure 2.27 Spectrum of phase-locked VCO.
图 2.27 相位锁定压控振荡器的频谱。

We have  我们有

where the factor of 2 accounts for the power symmetrically distributed over positive and negative frequencies. Since $S_{\varphi n}(f)=S_0$$S_{\varphi n}(f)=S_0$S_(phi n)(f)=S_(0)S_{\phi n}(f)=S_{0} at $|f|=f_1$$|f|=f_1$|f|=f_(1)|f|=f_{1}, the profile beyond this frequency is given by $S_0{f}_1^2/f^2$$S_0{f}_1^2/f^2$S_(0)f_(1)^(2)//f^(2)S_{0} f_{1}^{2} / f^{2}. It follows that
其中的 2 因子考虑了在正频率和负频率上对称分布的功率。由于 $S_{\varphi n}(f)=S_0$$S_{\varphi n}(f)=S_0$S_(phi n)(f)=S_(0)S_{\phi n}(f)=S_{0} 在 $|f|=f_1$$|f|=f_1$|f|=f_(1)|f|=f_{1} 处，因此该频率之外的轮廓由 $S_0{f}_1^2/f^2$$S_0{f}_1^2/f^2$S_(0)f_(1)^(2)//f^(2)S_{0} f_{1}^{2} / f^{2} 给出。由此可得

Interestingly, the rms jitter is equal to 4 times the area under the spectrum from 0 to $f=+f_1$$f=+f_1$f=+f_(1)f=+f_{1}. It is useful to memorize this result.
有趣的是，rms 抖动等于从 0 到 $f=+f_1$$f=+f_1$f=+f_(1)f=+f_{1} 的频谱下的面积的 4 倍。记住这个结果是有用的。

The control line of a VCO is driven by a resistor $R_1$$R_1$R_(1)R_{1}. Determine the output rms jitter.
VCO 的控制线由电阻 $R_1$$R_1$R_(1)R_{1} 驱动。确定输出的均方根抖动。

As explained in Example 2.11, in response to $V_{n,R1}^2=2kT{R}_1$$V_{n,R1}^2=2kT{R}_1$V_(n,R1)^(2)=2kTR_(1)V_{n, R 1}^{2}=2 k T R_{1}, the VCO produces a phase noise profile given by $S_{\varphi n}(f)=2kT{R}_1{K}_{VCO}^2/(2\pi f{)}^2$$S_{\varphi n}(f)=2kT{R}_1{K}_{VCO}^2/(2\pi f{)}^2$S_(phi n)(f)=2kTR_(1)K_(VCO)^(2)//(2pi f)^(2)S_{\phi n}(f)=2 k T R_{1} K_{V C O}^{2} /(2 \pi f)^{2}. The area under this profile is infinite, which agrees with our observation that the phase fluctuations can accumulate indefinitely [Fig. 2.24(b)]. A similar situation arises if the VCO is modulated by flicker noise. We say the “absolute” jitter of a stand-alone oscillator is unbounded and its rms value is infinite.
如示例 2.11 所述，作为对 $V_{n,R1}^2=2kT{R}_1$$V_{n,R1}^2=2kT{R}_1$V_(n,R1)^(2)=2kTR_(1)V_{n, R 1}^{2}=2 k T R_{1} 的响应，VCO 产生的相位噪声轮廓由 $S_{\varphi n}(f)=2kT{R}_1{K}_{VCO}^2/(2\pi f{)}^2$$S_{\varphi n}(f)=2kT{R}_1{K}_{VCO}^2/(2\pi f{)}^2$S_(phi n)(f)=2kTR_(1)K_(VCO)^(2)//(2pi f)^(2)S_{\phi n}(f)=2 k T R_{1} K_{V C O}^{2} /(2 \pi f)^{2} 给出。该轮廓下的面积是无限的，这与我们观察到的相位波动可以无限积累的情况一致[图 2.24(b)]。如果 VCO 受到闪烁噪声的调制，也会出现类似的情况。我们说独立振荡器的“绝对”抖动是无界的，其均方根值是无限的。

We have thus far seen two types of jitter: (a) deterministic jitter, resulting from periodic frequency modulation (manifested by sidebands in the spectrum), and (b) absolute jitter, defined as the phase difference between a noisy oscillator and a noiseless oscillator running at the same nominal frequency.
到目前为止，我们已经看到了两种类型的抖动：（a）确定性抖动，由周期性频率调制引起（在频谱中表现为边带），以及（b）绝对抖动，定义为一个有噪声的振荡器与一个在相同名义频率下运行的无噪声振荡器之间的相位差。

That the absolute jitter is unbounded presents a quandary. How do we specify the jitter of a stand-alone oscillator? Fortunately, we can consider the “cycle-to-cycle” jitter, $J_{cc}$$J_{cc}$J_(cc)J_{c c}, i.e., the difference between consecutive periods, $T_k-T_{k-1}$$T_k-T_{k-1}$T_(k)-T_(k-1)T_{k}-T_{k-1}. Of course, this measurement must be performed for many cycles, with the rms value defined as
绝对抖动是无界的，这带来了一个难题。我们如何指定独立振荡器的抖动？幸运的是，我们可以考虑“周期到周期”的抖动， $J_{cc}$$J_{cc}$J_(cc)J_{c c} ，即连续周期之间的差异， $T_k-T_{k-1}$$T_k-T_{k-1}$T_(k)-T_(k-1)T_{k}-T_{k-1} 。当然，这项测量必须在多个周期内进行，均方根值定义为

We expect the cycle-to-cycle jitter to be very small (but not negligible!) because phase noise components do not have much time to accumulate from one output period to the next (Section 2.1.6). As an example, consider the white-noise-induced phase noise profile depicted in Fig. 2.28(a). We assume $S_{\varphi n}(f)=\alpha /f^2$$S_{\varphi n}(f)=\alpha /f^2$S_(phi n)(f)=alpha//f^(2)S_{\phi n}(f)=\alpha / f^{2}, where $\alpha$$\alpha$alpha\alpha is a constant. This profile contains all noise frequencies but with declining power levels as $|f|$$|f|$|f||f| increases. Now,
我们预计周期间抖动会非常小（但不可忽视！），因为相位噪声成分在一个输出周期到下一个输出周期之间没有太多时间积累（第 2.1.6 节）。作为一个例子，考虑图 2.28(a)中描绘的白噪声引起的相位噪声特征。我们假设 $S_{\varphi n}(f)=\alpha /f^2$$S_{\varphi n}(f)=\alpha /f^2$S_(phi n)(f)=alpha//f^(2)S_{\phi n}(f)=\alpha / f^{2} ，其中 $\alpha$$\alpha$alpha\alpha 是一个常数。该特征包含所有噪声频率，但随着 $|f|$$|f|$|f||f| 的增加，功率水平逐渐下降。现在，

Figure 2.28 (a) Phase noise resulting from white noise, and (b) a 1-GHz oscillator output waveform.
图 2.28 (a) 由白噪声引起的相位噪声，以及 (b) 1 GHz 振荡器输出波形。
suppose this phase noise appears at the output of a $1-\mathrm{GHz}$$1-\mathrm{GHz}$1-GHz1-\mathrm{GHz} oscillator [Fig. 2.28(b)]. Since each cycle is 1 ns long, we surmise that only high-frequency phase noise components can affect the zero crossings 1 ns later. For example, a phase noise component around 1 MHz can be approximated by ${\varphi }_0\cos [2\pi (1\mathrm{MHz})t]$${\varphi }_0\cos [2\pi (1\mathrm{MHz})t]$phi_(0)cos[2pi(1MHz)t]\phi_{0} \cos [2 \pi(1 \mathrm{MHz}) t], which changes negligibly in 1 ns .
假设这个相位噪声出现在 $1-\mathrm{GHz}$$1-\mathrm{GHz}$1-GHz1-\mathrm{GHz} 振荡器的输出端[图 2.28(b)]。由于每个周期长达 1 纳秒，我们推测只有高频相位噪声成分会在 1 纳秒后影响零交叉。例如，约 1 MHz 的相位噪声成分可以用 ${\varphi }_0\cos [2\pi (1\mathrm{MHz})t]$${\varphi }_0\cos [2\pi (1\mathrm{MHz})t]$phi_(0)cos[2pi(1MHz)t]\phi_{0} \cos [2 \pi(1 \mathrm{MHz}) t] 来近似，在 1 纳秒内变化微乎其微。

The foregoing thoughts are similar to our study of noise accumulation in Section 2.1.6, suggesting that phase noise experiences high-pass filtering if observed for a limited time, i.e., for one period. It can be shown that for white-noise-induced phase noise, the cycle-to-cycle jitter of an oscillator running at ${\omega }_0$${\omega }_0$omega_(0)\omega_{0} is equal to [3]
上述思考与我们在第 2.1.6 节中对噪声积累的研究相似，表明相位噪声在有限时间内观察时会经历高通滤波，即在一个周期内。可以证明，对于白噪声引起的相位噪声，运行在 ${\omega }_0$${\omega }_0$omega_(0)\omega_{0} 的振荡器的周期间抖动等于[3]。

as if $S_{\varphi n}(\omega )$$S_{\varphi n}(\omega )$S_(phi n)(omega)S_{\phi n}(\omega) has traveled through a differentiator. If $S_{\varphi n}(\omega )$$S_{\varphi n}(\omega )$S_(phi n)(omega)S_{\phi n}(\omega) is of the form $\alpha /f^2$$\alpha /f^2$alpha//f^(2)\alpha / f^{2}, we have
好像 $S_{\varphi n}(\omega )$$S_{\varphi n}(\omega )$S_(phi n)(omega)S_{\phi n}(\omega) 经过了一个微分器。如果 $S_{\varphi n}(\omega )$$S_{\varphi n}(\omega )$S_(phi n)(omega)S_{\phi n}(\omega) 的形式是 $\alpha /f^2$$\alpha /f^2$alpha//f^(2)\alpha / f^{2} ，我们有

Normalized to the oscillation period, $T_0=1/f_0$$T_0=1/f_0$T_(0)=1//f_(0)T_{0}=1 / f_{0}, this result emerges as
归一化到振荡周期 $T_0=1/f_0$$T_0=1/f_0$T_(0)=1//f_(0)T_{0}=1 / f_{0} ，这个结果出现为

A $1-\mathrm{GHz}$$1-\mathrm{GHz}$1-GHz1-\mathrm{GHz} oscillator exhibits a phase noise of $-90\mathrm{dBc}/\mathrm{Hz}$$-90\mathrm{dBc}/\mathrm{Hz}$-90dBc//Hz-90 \mathrm{dBc} / \mathrm{Hz} at $100-\mathrm{kHz}$$100-\mathrm{kHz}$100-kHz100-\mathrm{kHz} offset. (a) Determine the cycle-to-cycle jitter if only white noise sources are considered. (b) How much jitter do we expect after 50 cycles?
一个 $1-\mathrm{GHz}$$1-\mathrm{GHz}$1-GHz1-\mathrm{GHz} 振荡器在 $100-\mathrm{kHz}$$100-\mathrm{kHz}$100-kHz100-\mathrm{kHz} 偏移处表现出 $-90\mathrm{dBc}/\mathrm{Hz}$$-90\mathrm{dBc}/\mathrm{Hz}$-90dBc//Hz-90 \mathrm{dBc} / \mathrm{Hz} 的相位噪声。(a) 如果仅考虑白噪声源，确定周期间抖动。(b) 我们预计在 50 个周期后会有多少抖动？

(a) From the phase noise value, we can compute $\alpha$$\alpha$alpha\alpha :
(a) 从相位噪声值，我们可以计算 $\alpha$$\alpha$alpha\alpha :

and, thus,  因此，

It follows that $\alpha =10\mathrm{Hz}$$\alpha =10\mathrm{Hz}$alpha=10Hz\alpha=10 \mathrm{~Hz} and $J_{cc,rms}=\sqrt{20/(1\mathrm{GHz}{)}^3}=0.14\mathrm{ps}$$J_{cc,rms}=\sqrt{20/(1\mathrm{GHz}{)}^3}=0.14\mathrm{ps}$J_(cc,rms)=sqrt(20//(1GHz)^(3))=0.14psJ_{c c, r m s}=\sqrt{20 /(1 \mathrm{GHz})^{3}}=0.14 \mathrm{ps}.
因此 $\alpha =10\mathrm{Hz}$$\alpha =10\mathrm{Hz}$alpha=10Hz\alpha=10 \mathrm{~Hz} 和 $J_{cc,rms}=\sqrt{20/(1\mathrm{GHz}{)}^3}=0.14\mathrm{ps}$$J_{cc,rms}=\sqrt{20/(1\mathrm{GHz}{)}^3}=0.14\mathrm{ps}$J_(cc,rms)=sqrt(20//(1GHz)^(3))=0.14psJ_{c c, r m s}=\sqrt{20 /(1 \mathrm{GHz})^{3}}=0.14 \mathrm{ps} 。
(b) The jitter accumulates randomly with time. If the cycle-to-cycle jitters of consecutive cycles are uncorrelated, we can add their squares:
(b) 抖动随着时间随机累积。如果连续周期的周期间抖动不相关，我们可以将它们的平方相加：

where $J_N^2$$J_N^2$J_(N)^(2)J_{N}^{2} denotes the jitter (with respect to a noiseless oscillator) after $N$$N$NN cycles, and $J_{m,n}$$J_{m,n}$J_(m,n)J_{m, n} is the jitter between the $m$$m$mm th cycle and the $n$$n$nnth cycle. We then have
其中 $J_N^2$$J_N^2$J_(N)^(2)J_{N}^{2} 表示在 $N$$N$NN 个周期后（相对于无噪声振荡器）的抖动， $J_{m,n}$$J_{m,n}$J_(m,n)J_{m, n} 是第 $m$$m$mm 个周期和第 $n$$n$nn 个周期之间的抖动。我们接下来有

In this example, $J_{50}\approx 1\mathrm{ps}$$J_{50}\approx 1\mathrm{ps}$J_(50)~~1psJ_{50} \approx 1 \mathrm{ps}. It is interesting to note that it takes roughly 50 million cycles ( 50 ms ) for the jitter to accumulate to one period ( 1 ns ).
在这个例子中， $J_{50}\approx 1\mathrm{ps}$$J_{50}\approx 1\mathrm{ps}$J_(50)~~1psJ_{50} \approx 1 \mathrm{ps} 。有趣的是，抖动积累到一个周期（1 ns）大约需要 50 百万个周期（50 ms）。

Another type of jitter that remains bounded is the “period jitter” (also called the “cycle jitter”). Representing the random departures of the period from its average value, $T_{\text{avg}}$$T_{\text{avg }}$T_("avg ")T_{\text {avg }}, this type is expressed as
另一种保持有界的抖动是“周期抖动”（也称为“循环抖动”）。它表示周期相对于其平均值 $T_{\text{avg}}$$T_{\text{avg }}$T_("avg ")T_{\text {avg }} 的随机偏离，这种类型表示为

It can be shown that $J_c=J_{cc}/\sqrt{}2$$J_c=J_{cc}/\sqrt{}2$J_(c)=J_(cc)//sqrt()2J_{c}=J_{c c} / \sqrt{ } 2 for white-noise-induced jitter [3].
可以证明，对于白噪声引起的抖动 $J_c=J_{cc}/\sqrt{}2$$J_c=J_{cc}/\sqrt{}2$J_(c)=J_(cc)//sqrt()2J_{c}=J_{c c} / \sqrt{ } 2 [3]。
Table 2.1 summarizes the three common types of jitter and their characteristics. The deterministic jitter can be obtained from the magnitude of the output FM sidebands and the absolute jitter is unbounded for a stand-alone oscillator. The rms cycle-to-cycle jitter resulting from white noise sources is computed from Eq. (2.40).
表 2.1 总结了三种常见的抖动及其特征。确定性抖动可以从输出 FM 边带的幅度中获得，而绝对抖动对于独立振荡器是无界的。由白噪声源引起的均方根周期间抖动是根据公式(2.40)计算的。

In this section, we show that (random) phase noise directly trades with power dissipation; hence, in most cases, the power drawn by an oscillator is dictated by the phase noise specification.
在本节中，我们展示了（随机）相位噪声与功耗之间的直接权衡；因此，在大多数情况下，振荡器所消耗的功率由相位噪声规范决定。

Let us take $N$$N$NN nominally identical oscillators and add their output voltages (Fig. 2.29). Using Eq. (2.23),
让我们取 $N$$N$NN 个名义上相同的振荡器并将它们的输出电压相加（图 2.29）。使用公式（2.23），