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# dividing ratio problem

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#### yefj

##### Advanced Member level 4
Hello,the output of the power dividing network is shown by the table bellow. The network is shown by the diagram bellow. I want to take 1W input and follow the diving ratio and get the outputs. for example if we have 1W input and i want to see the #5 output we go thre K1 and K3 junctions,how do i divide the 1W input,i am suppose to get 0.7817 output. If i have 1:1 dividing ratio then power is divided 0.5 0.5 So if we try and follw the same logic how does the power being devided if we follow 1:k1 and 1:k3 ratios and get 0.7817(5th output)?

what did you do?
what did you get?
did you normalized output so that 4# = 1?

i have tried to go from leaves to root, but the math is not summing to 1 :-(

show your work

Hello , my work is as follows:
1.if we have 1:1 ratio this means that we have 0.5 0.5 power dividing.
so assuming 1W input, we start between #6 #7 0.32W and 0.3201W which is K6=0.9998
so the total power in the branch is 0.64W and we have 1:K5(0.72) ratio between 0.64 and #8 0.32W
so its 1:2 ratio not 1:0.72
i have mathematical logic error.
Where did i go wrong?
Thanks.

One would expect to see you tabulate te calculation for a set of samples. BTW, the last value in the lower center of the figure is cut off, that makes all the difference, whether 50 or 60 ohms.

Hello Andre the input impedance is 50ohms the missing part.
I have tried to derive #8 power out of #6 #7 branch as shown bellow.
Where did i go wrong?
Thanks.

"assuming 1W input, we start between #6 #7 0.32W and 0.3201W which is K6=0.9998
so the total power in the branch is 0.64W and we have 1:K5(0.72) ratio between 0.64 and #8 0.32W
so its 1:2 ratio not 1:0.72
i have mathematical logic error.
Where did i go wrong?"

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