- #1
bpetersen
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Homework Statement
I am trying to calculate the power of Rload in the circuit attached (SEE ATTACHED CIRCUIT). The coupler I am trying to model is the Agilent 778D. Can someone verify that I calculate this out correctly?
Homework Equations
y dB = 10 * log10(x)
V = i * R
P = v * i
The Attempt at a Solution
i. Calculating the percent of power the coupler uses:
-0.6dB = 10 * log10( x )
-0.06 = log10( x )
10^( -0.06 ) = 10^( log10( x ) )
x = 87.10%
>> So the Coupler in the diagram consumes 12.90% of power seen at its input <<
Power seen at couplers input = Power_Source - Power_R1
So Power_Coupler = 0.1290 * ( Power_Source - Power_R1)
Power_Source = Power_R1 + Power_Coupler + Power_Load
Power_Source = Power_R1 + 0.1290 * ( Power_Source - Power_R1) + Power_Load
Power_Source = Power_R1 + 0.1290 * Power_Source + 0.1290 * Power_R1 + Power_Load
0.8710 * Power_Source = Power_R1 + 0.1290 * Power_R1 + Power_Load
0.8710 * Power_Sourse = 1.1290 * Power_R1 + Power_Load
0.8710 * V_Source = 1.1290 * V_R1 * i + V_Load * i
0.8710 * V_Source = 1.1290 * V_R1 + V_Load
0.8710 * V_Source = 1.1290 * i * R_1 + i * R_load
0.8710 * V_Source = i * ( 1.1290 * R_1 + R_Load )
V_Source = i * (1.1290 * R_1 + R_Load) / 0.8710
i = (V_Load) / R_Load
V_Source = V_Load / R_Load * (1.1290 * R1 + R_Load ) / 0.8710
V_Source * R_Load * 0.8710 = V_Load * 1.1290 * R1 + R_Load
V_Load = V_Source * (R_Load * 0.8710) / (1.1290 * R1 + R_Load)
V_Load * i = V_Source * i * (R_Load * 0.8710) / (1.1290 * R1 + R_Load)
P_Load = P_Source * (R_Load * 0.8710) / (1.1290 * R1 + R_Load)
So if R1 = R_Load = 50Ohms
P_Load = P_Source * 0.4091
Thusly,
The power of the load = 40.91 % of the Power Source
y dB = 10 * log10 (0.4091) = -3.8817 dB
Conclusion:
The power distribution is as follows:
R1 = 46.19%
Coupler = 12.90%
RLoad = 40.91%
and I should see a -3.8817 dB attenuation of the signal at the load
