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No capacitors are involved in the design.the capacitor resistor combination may make a filter
How about applying ohms law and calculating the output voltages?can anyone derive this?
Here are my analysis [Producing results which DO NOT match the values given in the table ]
Lets assume the loss from port 1 to port 2 is 1.72db . Since its a loss , I take it as -1.72db .
Using (db)=20 log V2/V1 we get V2/V1 =0.82 or V2=0.82V1
Now , this implies for a 2 way unequal power divider , V3= (1-0.82) V1 i.e. V3= 0.18V1
Converting this 0.18 into db gives -14db
Thus for a 2 port unequal resistive power divider , if S21= -1.72 db , then S31 = -14db
***The paper says S31=-20db
and also the table values do not match
HaiHere is a paper by G Adams on resistive power dividers. Can anyone please explain to me table 1 ?
The paper says if u get 1.7 db loss from port one to port 2 , u'll get 20 db loss from port 1 to port 3
can anyone derive this?
The obviously erronous assumption is a lossless power splitter. But by using resistors, a power divider must have losses. The amount of dissipated input power depends on the particular circuit and can't be derived from the attenuation numbers. Thus all calculations based on it are void.Assuming input power =1
1-0.67=0.33 and hence P3=0.33*P1
Hai
Based on equations in article i get the following values of resistors:
Rs = 0.1 Zo
Rp = 5 Zo
Rt = 4.45 Zo
Ru = 1.25 Zo
So S21 would be 20log10(0.82) = -1.73 dB, as it should be,
S31 = 20log10(0.9*0.55/5) = -20 dB
So everything is correct.