A.H. Systems SAS-512-2, Руководство по эксплуатации

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A.H. Systems SAS-512-2 Log Periodic Antenna
TYPICAL CONVERSION FORMULAS
LOG -> LINEAR VOLTAGE FIELD STRENGTH & POWER DENSITY
dB

V to Volts V = 10 ((dB

V – 120) / 20) dB

V/m to V/m V/m = 10 (((dB

V/m) -120) / 20)
Volts to dB

V dB

V = 20 log(V) + 120 V/m to dB

V/m dB

V/m = 20 log(V/m) + 120
dBV to Volts V = 10 (dBV / 20)
dB

V/m to dBmW/m 2 dBmW/m 2 = dB

V/m – 115.8
Volts to dBV dBV = 20log(V)
dBmW/m 2 to dB

V/m dB

V/m = dBmW/m 2 + 115.8
dBV to dB

V dB

V = dBV +120 dB

V/m to dB

A/m dB

A/m = dB

V/m – 51.5
dB

V to dBV dBV = dB

V - 120 dB

A/m to dB

V/m dB

V/m = dB

A + 51.5
LOG -> LINEAR CURRENT
dB

A/m to dBpT DBpT = dB

A/m + 2
dB

A to uA

A = 10 (dB

A / 20) dBpT to dB

A/m dB

A/m = dBpT – 2

A to dB

A dB

A = 20 log(

A)
W/m 2 to V/m V/m = SQRT(W/m 2 * 377)
dBA to A A = 10 (dBA / 20) V/m to W/m 2 W/m 2 = (V/m) 2 / 377
A to dBA dBA = 20log(A)

T to A/m A/m =

T / 1.25
dBA to dB

A dB

A = dBA + 120 A/m to

T

T = 1.25 * A/m
dB

A to dBA dBA = dB

A -120
E-FIELD ANTENNAS
LOG -> LINEAR POWER Correction Factor
dB

V/m = dB

V + AF
dBm to Watts W = 10 ((dBm – 30)/10) Field Strength V/m = 30 * watts * Gain
numeric
meters
Watts to dBm dBm = 10log(W) + 30 Required Power Watts = (V/m * meters) 2
30 * Gain
numeric
dBW to Watts W = 10 (dBW / 10) LOOP ANTENNAS
Watts to dBW dBW = 10log(W) Correction Factors
dB

A/m = dB

V + AF
dBW to dBm dBm = dBW + 30 Assumed E-field for
shielded loops
dB

V/m = dB

A/m + 51.5
dBm to dBW dBW = dBm - 30
dBpT = dB

V + dBpT/

V
TERM CONVERSIONS
dBm to dB

V
dB

V =
dBm + 107
(50

)
dB

V = dBm + 10log(Z) + 90
CURRENT PROBES
dB

V to dBm dBm = dB

V – 107
(50

)
dBm = dB

V – 10log(Z) – 90
Correction Factor
dB

A = dB

V – dB
(ohm)
dBm to dB

A dB

A = dBm – 73
(50

)
dB

A = dBm – 10log(Z) + 90
Power needed for injection probe given voltage(V) into
50

load and Probe Insertion Loss (I
L
)
dB

A to dBm dBm = dB

A + 73
(50

)
dBm = dB

A + 10log(Z) – 90
Watts = 10
((I
L
+ 10log(V 2 /50))/10)
dB

A to dB

V dB

V = dB

A + 34
(50

)
dB

V = dB

A + 20log(Z)
dB

V to dB

A dB

A = dB

V – 34
(50

)
dB

A = dB

V – 20log(Z)

A.H. Systems inc. – May 2014
REV C
12