A.H. Systems SAS-551 운영 매뉴얼 - 페이지 12

{카테고리_이름} A.H. Systems SAS-551에 대한 운영 매뉴얼을 온라인으로 검색하거나 PDF를 다운로드하세요. A.H. Systems SAS-551 14 페이지. Passive monopole antenna
A.H. Systems SAS-551에 대해서도 마찬가지입니다: 운영 매뉴얼 (12 페이지)

페이지 인쇄
A.H. Systems SAS-551 운영 매뉴얼
A.H. Systems Model SAS-551 Passive Monopole Antenna
TYPICAL CONVERSION FORMULAS
LOG -> LINEAR VOLTAGE
dBV to Volts
V = 10
Volts to dBV
dBV = 20 log(V) + 120
dBV to Volts
V = 10
Volts to dBV
dBV = 20log(V)
dBV to dBV
dBV = dBV +120
dBV to dBV
dBV = dBV - 120
LOG -> LINEAR CURRENT
A = 10
dBA to uA
A to dBA
dBA = 20 log(A)
dBA to A
A = 10
A to dBA
dBA = 20log(A)
dBA to dBA
dBA = dBA + 120
dBA to dBA
dBA = dBA -120
LOG -> LINEAR POWER
dBm to Watts
W = 10
Watts to dBm
dBm = 10log(W) + 30
dBW to Watts
W = 10
Watts to dBW
dBW = 10log(W)
dBW to dBm
dBm = dBW + 30
dBm to dBW
dBW = dBm - 30
TERM CONVERSIONS
dBm to dBV
dB
dB
dBm = dBV – 107
dBV to dBm
dBm = dB
dBA = dBm – 73
dBm to dBA
dB
dBA to dBm
dBm = dBA + 73
dBm = dB
dBA to dBV
dBV = dBA + 34
dB
dBA = dBV – 34
dBV to dBA
dB
© A.H. Systems inc. – REV E
((dB  V – 120) / 20)
(dBV / 20)
(dB  A / 20)
(dBA / 20)
((dBm – 30)/10)
(dBW / 10)
V =
dBm + 107
(50)
V = dBm + 10log(Z) + 90
(50)
V – 10log(Z) – 90
(50)
A = dBm – 10log(Z) + 90
(50)
A + 10log(Z) – 90
(50)
V = dB
A + 20log(Z)
(50)
V – 20log(Z)
A = dB
FIELD STRENGTH & POWER DENSITY
dBV/m to V/m
V/m = 10 (((dBV/m) -120) / 20)
V/m to dBV/m
dBV/m = 20 log(V/m) + 120
2
dBV/m to dBmW/m
dBmW/m
2
dBmW/m
to dBV/m dBV/m = dBmW/m
dBA/m = dBV/m – 51.5
dBV/m to dBA/m
dBA/m to dBV/m
dBV/m = dBA + 51.5
dBA/m to dBpT
DBpT = dBA/m + 2
dBA/m = dBpT – 2
dBpT to dBA/m
2
W/m
to V/m
V/m = SQRT(W/m
2
V/m to W/m
W/m
T to A/m
A/m = T / 1.25
A/m to T
T = 1.25 * A/m
E-FIELD ANTENNAS
Correction Factor
dBV/m = dB
Field Strength
V/m =
Required Power
Watts = (V/m * meters)
LOOP ANTENNAS
Correction Factors
dBA/m = dBV + AF
Assumed E-field for
dBV/m = dBA/m + 51.5
shielded loops
dBpT = dBV + dBpT/V
CURRENT PROBES
dBA = dBV – dB
Correction Factor
Power needed for injection probe given voltage(V) into
50 load and Probe Insertion Loss (I
Watts = 10
= dBV/m – 115.8
2
2
+ 115.8
2
* 377)
2
2
= (V/m)
/ 377
V + AF
30 * watts * Gain
numeric
meters
2
30 * Gain
numeric
(ohm)
)
L
2
((I
+ 10log(V
/50))/10)
L
12