It was empirically determined that sine transfer characteristics (E/g)
varied in frequency between test units and from day to day test operations.
These variations were approximately, ñ3%(Frequency) and less than ñ3 dB
(Amplitude) as measured at minimum and maximum peaks in the (E/G) curve. In
order to compensate for these variances in the preparation of the synthetic
random tape a method was evolved to minimize the potential for overtesting at
resonances and anti-resonances of the vibration system by reducing the
synthetic random voltage requirements in these frequency bands and also to
prevent under testing at conditions of amplitude variances. The results of
these investigations concluded that no positive compensation factors,
resulting in an increase in voltage requirements were to be used and negative
compensation factors were limited to - 4 dB, to reduce the under test
potential.

The choice of points in the sine transfer characteristics requiring
compensation are chosen at frequency bands in which large abrupt changes or
peaks, either positive or negative, exist in the transfer characteristics.
Calculate the required compensation factors for each of these frequency bands
as follows:

**4-3.2.1 Frequency Variation.** Utilizing the test data
transfer characteristic curve obtained in Chapter
3, determine for each resonance or anti-resonance, two frequency bands (in
accordance with the procedure delineated below) and select the one that
encompasses the fewest number of cycles:

a. Establish the center frequency and calculate a value equal to
±6% of that frequency. (The ±6% includes ±3% to reflect the variance in
unit to unit resonances and ±3% which considers the slopes reflecting the
increase and decrease of the resonant reference).

b. At each center frequency determine the bandwidth at the half level
points (-6 dB), and increase that bandwidth by adding ±3% of the center
frequency to reflect unit to unit resonances.

**4-3.2.2 Amplitude Variation.**
Determine the average amplitude variation in dB for each filter bandwidth
within the frequency variation established in para. 4-3.2.1. (a minimum of
three 25 Hz bands is required). Reduce each of these amplitudes by 0.5 dB to
compensate for resonant and anti-resonant conditions. Then calculate the
overall mean amplitude of the average level of each of these bands. Typically
this would include one or two 25 Hz filters on either side of the maximum or
minimum peak response frequencies.

**4-3.2.3 Compensation Factor.** In each filter
bandwidth determine the compensation factor by subtracting its unmodified
amplitude from the mean amplitude calculated in 4-3.2.2. The positive
differences are discarded and the negative differences are then added to the
filter amplitude to reduce the level to the calculated mean.

A sample curve which illustrates the procedures described in par.
4-3.2.1 and 4-3.2.2 for a typical maximum and minimum peak is shown in Figure
4-5. The calculations utilizing these procedures are presented for two
examples:

Example No. 1. (E/g) *maximum* peak occurs in
filter No. 31 which has an average amplitude of 7.1 dB.

a. Calculate the frequency variance.

(1) Center frequency of filter No. 31 is 662
Hz.

(2) ±6% of 622 Hz = ±39.72 Hz

(3) The half level bandwidth (-6
dB) = 150 Hz.

(4) ±3% of 662 Hz = ±19.86 Hz.

(5) ±19.86 Hz (39.72
Hz) + 150 Hz = 189.72 Hz.

(6) Select 79.44 Hz since this is the lower
of the two values.

This frequency spread encompasses a range of
662 Hz ±39.72 Hz or 622.28 to
701.72 Hz, and includes filter Nos. 30, 31 and 32.

b. Calculate mean amplitude in the compensation
band

Mean = 17.8 dB/3 = 5.9
dB

c. Calculate compensation factor

Filter No. 30: 5.9 dB (mean) - 6.1 dB (A) = - 0.3
dB

Filter No. 32: 5.9 dB (mean) - 7.1 dB (A) = - 1.2 dB

Filter No.
32: 5.9 dB (mean) - 6.2 dB (A) = - 0.3 dB

Example No. 2. (E/g) minimum peak occurs between filter
Nos. 48 and 49, with an average amplitude of 1.8 dB.

a. Calculate the frequency variance

(1) Center frequency between filter No. 48 and 49 is
1100 Hz.

(2) ±6 of 1100 Hz = ±66 Hz or 132 Hz.

(3) The half level
bandwidth ( - 6 dB) = 54 Hz.

(4) ±3% of 1100 Hz = ±33 Hz.

(5) ±33 Hz
(66 Hz) + 54 Hz = 120 Hz.

(6) Select 120 Hz since this is the lower of
the two values.

This Frequency spread encompasses a range of 1100 Hz ±60
Hz to 1160 Hz and includes filter Nos. 47, 48, 49 and 50 (only full bands are
to be considered).

b. Calculate the mean amplitude in the compensation
bands.

c. Calculate compensation factor

Filter No. 47: 3.3 dB (mean) - 5.8 dB (A) = - 2.5
dB

Filter No. 48: 3.3 Db (mean) - 1.8 dB (A) = + 1.5 dB*

Filter No.
49: 3.3 dB (mean) - 1.7 dB (A) = + 1.6 dB*

Filter No. 50: 3.3 dB (mean)
- 6.0 dB (A) = - 2.7 dB

**NOTE**

*Positive compensation
factors are not used in order to minimize over test
potential.

The
calculated compensation factors are tabulated in column F of the sample
Tabulation Sheet, Figure
4-3.

**NOTE**

This method of compensation
has been kept simplistic to provide the convenient application to typical test
articles. Individual test systems may lend themselves to more precise methods
of variance compensations.