Abstract

The magnitude of the noise present in the signal from a spectroscope depends strongly upon the frequency-band-pass width of the amplifier. It is, therefore, possible to reduce the noise by inserting a low-pass filter. At the same time a distortion is produced in the spectrum recorded. It is shown for two different kinds of filter that this distortion may be divided into two effects: (1) a displacement of the spectrum as a whole, which is easily corrected for, and (2) a real distortion depending upon the width of the spectral lines. It is shown that the only means to limit this distortion is to scan the spectrum at a sufficiently low speed. An expression is given for this speed, which is proportional to the width of the narrowest line (often equal to the effective slit-width), and inversely proportional to the time-constant of the filter and to a constant characterizing the maximum tolerable distortion.

© 1953 Optical Society of America

Full Article  |  PDF Article

References

  • View by:
  • |
  • |
  • |

  1. S. Brodersen, J. Opt. Soc. Am. 43, 877 (1953).
    [Crossref]

1953 (1)

Cited By

OSA participates in Crossref's Cited-By Linking service. Citing articles from OSA journals and other participating publishers are listed here.

Alert me when this article is cited.


Figures (9)

Fig. 1
Fig. 1

Simple RC-network.

Fig. 2
Fig. 2

Effect of a simple RC network on a spectral line.

Fig. 3
Fig. 3

Effect of a critically damped network on a spectral line.

Fig. 4
Fig. 4

Line displacement caused by a simple RC-network.

Fig. 5
Fig. 5

Line displacement caused by a critically damped network.

Fig. 6
Fig. 6

Displacement-corrected distortion caused by a simple RC-network.

Fig. 7
Fig. 7

Displacement-corrected distortion caused by a critically damped network.

Fig. 8
Fig. 8

Experimental curves from a critically damped network. The numerals on the curves indicate Kmin. (Water vapor bands at 1540 cm−1 Beckman IR3, P=2.7 sec, Lmin=seff=4.2 cm−1.)

Fig. 9
Fig. 9

Same as Fig. 8 but with a slight underdamping.

Equations (26)

Equations on this page are rendered with MathJax. Learn more.

i N 2 = 2 I e Δ f ,
e N 2 = 4 k T R Δ f ,
Δ f = 1 / 4 R C ,
y = t 1 = - t f ( t 1 ) × ( 1 / R C ) exp [ - ( t - t 1 ) / R C ] d t 1 .
y = k { 1 - exp [ - ( t - t 0 ) / R C ] } .
f ( ν 1 ) = k × exp [ - ν 1 2 / ( 2 σ 2 ) ] ,
y = k / R C t 1 = - t exp [ - ν 1 2 / ( 2 σ 2 ) ] × exp [ - ( t - t 1 ) / R C ] d t 1 .
ν 1 t 1 = σ K R C / 2 = ν t ( = 8 σ Δ f K ) .
y = k exp [ - ν 2 2 σ 2 ] × K 2 exp [ 1 2 ( ν σ - K 2 ) 2 ] × v = - ( ν / σ ) - ( K / 2 ) exp ( - v 2 2 ) d v .
4 π 2 P 2 ( t - t 1 ) exp [ - 2 π P ( t - t 1 ) ] ,
y = k { 1 - ( 1 + 2 π P ( t - t 0 ) ) exp [ - 2 π P ( t - t 0 ) ] } ,
Δ f = π / ( 4 P ) .
y = k 4 π 2 P 2 t 1 = - t exp [ - ν 1 2 2 σ 2 ] ( t - t 1 ) × exp [ - 2 π P ( t - t 1 ) ] d t 1 .
ν 1 t 1 = σ K P / ( 2 π ) = ν t ( = 8 σ Δ f K ) .
y = k exp [ - ν 2 2 σ 2 ] × K 2 [ 1 + ( ν σ - K ) × exp [ 1 2 ( ν σ - K ) 2 ] × v = - ( ν / σ ) - K exp ( - v 2 2 ) d v ] .
Δ / σ = 2 / K .
T 1 ( sec / cm - 1 ) = K R C / ( 2 σ )
T 1 ( sec / cm - 1 ) = K P / ( 2 π σ ) .
T 1 ( sec / cm - 1 ) = R C / Δ
T 1 ( sec / cm - 1 ) = P / ( π Δ ) ,
Δ d l d ν = R C ( 1 T 1 d l d ν )
Δ d l d ν = P π ( 1 T 1 d l d ν ) ,
T 1 ( sec / cm - 1 ) = 1.18 K min R C / L min
T 1 ( sec / cm - 1 ) = 0.37 K min P / L min ,
1 T 1 d l d ν = 0.85 K min R C ( L min d l d ν )
1 T 1 d l d ν = 2.7 K min P ( L min d l d ν ) ,