Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

dividing ratio problem

Status
Not open for further replies.

yefj

Advanced Member level 2
Joined
Sep 12, 2019
Messages
650
Helped
0
Reputation
0
Reaction score
2
Trophy points
18
Activity points
3,551
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)?

1625850512794.png


1625850543701.png
 

wwfeldman

Advanced Member level 4
Joined
Jan 25, 2019
Messages
1,058
Helped
222
Reputation
443
Reaction score
282
Trophy points
83
Activity points
8,054
what did you do?
what did you get?
did you normalized output so that 4# = 1?
 

yefj

Advanced Member level 2
Joined
Sep 12, 2019
Messages
650
Helped
0
Reputation
0
Reaction score
2
Trophy points
18
Activity points
3,551
i have tried to go from leaves to root, but the math is not summing to 1 :-(
 

yefj

Advanced Member level 2
Joined
Sep 12, 2019
Messages
650
Helped
0
Reputation
0
Reaction score
2
Trophy points
18
Activity points
3,551
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
Joined
Nov 7, 2006
Messages
9,479
Helped
1,183
Reputation
2,385
Reaction score
1,181
Trophy points
1,403
Location
Brazil
Activity points
54,954
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

Advanced Member level 2
Joined
Sep 12, 2019
Messages
650
Helped
0
Reputation
0
Reaction score
2
Trophy points
18
Activity points
3,551
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?"
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top