22.08.2015   Frequency Response Testing. in Nuclear Reactors

Spectrum of a Binary Pulse Chain and Its Bandwidtht

As will be shown in later chapters, two-level (binary) signals are well suited for tests on reactor systems. In this section, the upper bound on the energy in any harmonic of a binary signal will be derived. We will see in Chapter 3 that the harmonic amplitude of several important signals is less than this upper bound by a constant factor at all harmonics. Also, the upper bound indicates the limits on spectral shape that can be achieved in any binary signal.

image096 Подпись: (2.12.1)

Consider a periodic signal u(t) consisting of Z equal-amplitude positive and negative pulses of length At seconds. Each pulse in the chain is called a bit. It should be noted that it is possible to have strings of bits of the same sign, giving a pulse that is several bits long. This is called a run. The Fourier coefficient Ck of the fcth harmonic is

image098 Подпись: (2.12.2)

where cok = 2knjT and T = ZAt. Equation (2.12.1) can be written as

where и,- is the polarity of u(t) over the interval (і — 1) At < t < і At and A is the amplitude of the signal. Performing the integration and rearranging

f See Buckner (5).

Подпись: P k~ Z2 Подпись: for к > 0 (2.12.5)

image102 image103

Since

Подпись: Pk < 2A2 Подпись: sin(nk/Z) nk/Z Подпись: (2.12.7)

we obtain

This gives the upper bound on the power spectrum of a binary pulse chain.

The actual spectrum for a particular pulse chain depends on the actual set of polarities щ in the signal.

The bandwidth of a signal (the frequency range over which the signal power is at least half as large as its greatest value) is useful in estimating the maxi­mum frequency at which results can be obtained. Since the spectra for several important signals have the same shape but a smaller magnitude than the maximum (Eq. 2.12.7), bandwidth relations derived for the spectral shape of the maximum will have general utility. The shape function is

Подпись: 2 (2.12.8) sin(o> At/2)

_ to At/2

The bandwidth obtained by setting

Подпись: 2 = 0.5 (2.12.9) sin(o> At/2)

_ со At/2

is to At/2 = 1.39 or to = 2.78/Дг. Since to is a harmonic frequency for a periodic signal, we may write

2kn/T = 2.78/ Дг

or

T

к = 0.44— = 0.44Z (2.12.10)

At

That is, in a signal with Z bits, all harmonics out to harmonic number 0.44Z have at least half as much power as the largest harmonic.