SHOCK AND VIBRATION RESPONSE SPECTRA COURSE
Unit 7E. dB/octave Slopes
By Tom Irvine
_____________________________________________________________________
Introduction
NAVMAT P-9492 gives the power spectral density specification shown in Figure 1.
0.1
Overall Level = 6.0 grms
2
0.04 g / Hz
-3 dB / octave
2
PSD ( g / Hz )
+3 dB / octave
0.01
0.001
20
80
350
2000
FREQUENCY (Hz)
Figure 1.
The task is to determine the coordinates of the endpoints.
Derivation
2
Assume that a1 and a 2 each has an amplitude in G /Hz. The difference in dB between
a1 and a 2 is
a 
∆dB = 10 log  2 
a 1 
(1)
1
Furthermore,
a 2 = a1  10 ∆dB / 10 




(2)
Additional equations are needed.
The slope N between two coordinates (f1, a1 )and (f 2 , a 2 )in a log-log plot is
a 
log  2 
 a1 
N=
f 
log  2 
f1 
(3)
Solve for a 2 .
f 
a 
N log  2  = log  2 
f1 
 a1 
(4)
N

a 
f 2  
log    = log  2 
f1  
 a1 



(5)
N
f 2 
a 2 
  = 
f1 
 a1 
(6)
N
a 2  f 2 
 =  
a1  f1 
(7)
N
f 2 
a 2 = a1  
f1 
(8)
Take the anti-log.
Thus,
2
Now consider a one-octave frequency separation.
f 2 = 2 f1
(9)
Substitute equation (9) into (3).
a 
log  2 
 a1 
N=
log[]
2
(10)
Substitute equation (1) into (10).
N=
∆dB / 10
log[]
2
(11)
Note that ∆dB represents the dB/octave slope in equation (11). Again, equations (10) and
(11) assume a one-octave frequency separation.
Now substitute equation (11) into (8).
∆dB / 10 

f 
2 
a 2 = a1  2  log[]
f1 
(12)
Example
Calculate the amplitude at 2000 Hz for the power spectral density in Figure 1.
Note
f1 = 350 Hz
f 2 = 2000 Hz
a1 = 0.04 G 2 / Hz
− 3 dB / 10 




2000
Hz
2 
 log[]
a 2 = 0.04 G 2 / Hz 

 350 Hz 
3
(13)
a 2 = 0.007 G 2 / Hz
(14)
Homework
1. Calculate the amplitude at 20 Hz for the power spectral density curve in Figure 1.
2. Calculate the overall GRMS value for the power spectral density curve in Figure 1.
Use program psdint.exe. Give the answer to two decimal places.
4