# dividing ratio problem

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

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

#### wwfeldman #### yefj 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.

#### andre_luis

##### Super Moderator
Staff member 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.

#### yefj 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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