1. A product emits a radiated electric field level of 36 dBμV/m at 100 MHz when measured at a...

Question

1. A product emits a radiated electric field level of 36 dBμV/m at 100 MHz when measured at a distance of 30 ft from the product. Determine the emission level when measured at the FCC Class B distance of 3 m. [45.68 dBμV/m]

2. An antenna measures the radiated emissions at 220 MHz from a product as shown in Fig. P XXXXXXXXXXIf the receiver measures a level of –93.5 dBm at220 MHz, determine the voltage at the base of the antenna in dBμV. The cable loss at 220 MHz is 8 dB/100 ft. [29.5 dBμV] If the product providing these emissions is located a distance of 20 m and the antenna provides 1.5 V for every V/m of incident electric field at 220 MHz, determine whether the emissions comply with the CISPR 22 Class B and FCC Class B limits and by how much. [Fails CISPR 22 by 2 dB but passes FCC Class B by 3.54 dB]

Baljit · Accepted Answer

1.  Now we know that Electric field emission level is inversely proportional to  distance
    E (R)=k/R       here k=constant ,E=Electric Field Emission Level  ,R=Distance
  
So Electric field at Distance R’ 
                     E(R’ )=k/R’ 
                     E(R)/E (R’ ) =R’ /R
                     E(R’ )=E(R)*(R/R’ ) 
                     E(R’ )dB = E(R)dB + 20log10(R /R’)
Here in our case we have to find  E(R’ )dB 
Given
E(R)dB =36dBuV/m
R=30ft=9.1m
R’ =3m
so 
E(R’)dB =36dBuV/m+20log(9.

1. A product emits a radiated electric field level of 36 dBμV/m at 100 MHz when measured at a distance of 30 ft from the product. Determine the emission level when measured at the FCC Class B distance...

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