Example Time & Frequency Seminar Questions

Below are a few examples of the types of questions that we will answer for you at the NIST Time and Frequency Seminar. More detailed answers are provided at the seminar.

QUESTION: What is frequency stability?

ANSWER: An oscillating signal with good frequency stability produces a sine wave at a desired frequency. If the signal frequency deviates over any time interval, this is a result of something which is undesirable. Undesirable noise mechanisms cause random or systematic processes to exist on the sine wave signal of an oscillator. To account for the noise components at the output of an oscillator, we have:

V(t) = V₀[1+a(t)]sin(2πν₀t+φ(t))

where V₀  = nominal peak voltage amplitude, a(t) = deviation of amplitude from nominal, i.e. ΔV/V₀, ν₀ = nominal fundamental frequency, φ(t) = deviation of phase from desired phase. Ideally "a(t)" and "φ(t)" should equal zero for all time, but in actuality they are measurable consequences of various types of noise.
 

QUESTION: Of the five noisy signals below (amplitude vs. running time is shown), which of these have you encountered?


ANSWER: You have probably encountered all five. They are (in descending order) white, flicker, random-walk, flicker-walk, and random-run noise. Each has a specific cause.

QUESTION: Of the two plots of phase-noise below, do you know which would be characteristic of an amplifier and which would be characteristic of an oscillator?

ANSWER: The blue plot, showing flicker PM and white PM noises, is characteristic of the residual noise introduced by an amplifier. The red plot, showing flicker FM and white PM noises, is typical of a high-quality oscillator.

QUESTION: Why are the Total and Allan deviations (shown on the top plot below) recommended for characterization of fractional frequency fluctuations of an oscillator (shown on the bottom plot below)?   ANSWER: The standard deviation applied to the measurement of the frequency of an oscillator implies a false assumption that there exists a true mean frequency. The Total and Allan deviations can estimate frequency stability even with non-white noise where the mean is changing, such as the frequency step that is shown in the figure.

QUESTION: What is an optical frequency divider and how does it work?   ANSWER: An optical frequency divider connects radio frequencies (RF) at 25 MHz – 1 GHz to optical frequencies of 200 – 500 THz with a special device called a mode-locked laser (MLL). The MLL is a pulsed laser that uses non–linear distortion in certain optical materials to produce a 'comb' of frequencies. Locking circuits ensure that the frequencies in the laser light (200 – 500 THz) are exact multiples (harmonics) of the pulse repetition rate (100 MHz – 1 GHz). The MLL shines on a photodetector producing RF in which optical phase noise is reduced by N2. The noise reduction over existing microwave RF technology is over 1000x!

ANSWER: First, RF spectrum analyzers cannot distinguish phase-modulation (PM) noise from amplitude-modulation (AM) noise. Second, they have insufficient dynamic range in most cases. Third, inaccuracies can be high, especially around steep parts of the displayed function.   DISCUSSION: RF spectrum analyzers are basically very sensitive receivers. The received frequency spectrum is swept through a range of pre-selected frequencies, converting a received frequency to a DC-level that is displayed as signal strength vs. received frequency.

QUESTION: What is the two-sample Allan variance, &sigmay2(&tau), of the following sequence of fractional frequency fluctuation values yk, each value averaged over one second? y1 = 4.36 × 10-5      y5 = 4.47 × 10-5 y2 = 4.61 × 10-5      y6 = 3.96 × 10-5 y3 = 3.19 × 10-5      y7 = 4.10 × 10-5 y4 = 4.21 × 10-5      y8 = 3.08 × 10-5 (Assume no dead time between averages) ANSWER:Since each average of the fractional frequency fluctuation values is for one second, then the first variance calculations will be at &tau = 1s. We are given M = 8 (eight values); therefore, the number of pairs in sequence is M-1 = 7. We have: Therefore, and Using the same data, one can calculate the variance for &tau=2s by averaging pairs of adjacent values and using these new averages as the data values for the same procedure as above. For three second averages (&tau=3s) take adjacent threesomes, find their averages, and proceed in a similar manner. More data must be acquired for longer averaging times. One sees that with large numbers of data values it is helpful to use a computer or programmable calculator. The confidence of the estimate on &sigmay(&tau) improves nominally as the square root of the number of data values used. In this example there are seven terms to average, and the confidence can be approximately expressed as being no better than 1/&radic7 × 100% = 38%. This represents a 1-sigma or 68% confidence interval in the estimate of 5.6 × 10-6 &tau=1s average. Attend the NIST Metrology Seminar to learn exact methods of computing and improving confidence.