So I have completed (a) as this (original on the left):
I have then went onto (b) and I have equated T(s)=Z(s) as follows:
and due to
hence
Does this look correct to you smarter people?
Thanks in advance!! All replies are welcome :)
I am assuming that it would repeat? There is no symmetry that I can see? It would be odd if there was no 'missing' parts of the wave however
Thanks for the reply
Hi HDG,
Did you get the question correct? I got a final answer also of 97.271%
How did you select the fields to create the scattergraph?
Thanks for any help!
ANY AND ALL HELP IS GREATLY APPRECIATED :smile:
I have found old posts for this question however after reading through them several times I am having a hard time knowing where to start.
I am happy with the sketch that the function is correctly drawn and is neither odd nor even. It's title is...
Hi again Rudeman,
Was looking over my old working and found that in the final line of working is incorrect, unless I'm just being an idiot!
= √ (31491.630+j9511.802)
= 179.427+j26.506Ω
Just to update this post.
@FactChecker I think you give the materials too much credit, time and again I find inconsistencies or incorrect parts. Thanks for your help, it is much appreciated!
@FactChecker Yeah in the materials the first graph is even (FIG.1) and the second graph (FIG.2) which is FIG.1 'shifted' is now odd, however I would say that it is neither odd nor even, unless as you say it is shifted down the 'Y' axis. There is however no mention of this Y-axis shift in these...
Hi @bobob thanks for the help that really clears things up! I have tried bulking out the answer for each as follows, does that look correct to you? ( I realise this just builds on what you wrote )
a) Function is neither odd nor even.
a0= V/2, an=0
b) Function is even hence will only...
I thought it wasn't odd, the learning materials seem to be incorrect, it is implying that the diagram shown is odd - seems to be a recurring theme with this learning materials!
@FactChecker Thanks a lot for the insight, not sure how I messed up the A0 figure. I have one query regarding a similar waveform in the learning materials that has been 'shifted odd' and I am unsure how this wave is 'odd' and not neither? wondering if you can give some insight. Thanks again.
Ah I see what you mean I think, so I could maybe say the following:
- The wave is neither odd nor even
- a0 =0 & an = 0
Surely that can't be all there is?
@Connorm1
Hi all, having a look at this now and I am on the same boat as you were - I don't know how much detail they want and where to start with part (b). I see that you were trying to move the wave down the Y-axis to make it Odd, why not shift it to the right by T/4 so that it creates an...
Hi Babadag,
I would've likely said the same however in my learning materials the below is stipulated. I then equated that each as appropriate. Any ideas?
Homework Statement
Figure shows a 50 Hz, high-voltage, transmission line. The relationships between the sending and receiving end voltages and currents are given by the complex ABCD equations:
where 'S' stands for sending-end and 'R' stands for receiving-end
(a) Given the parameter values...
I'm trying to do this but can't seem to work it without a value for the voltage source.
So i made the three nodes at the top V1, V2 & V3 and made the following equations
V2/13 - V3/13 - V3/100 = 0
V2/13 - V1/13 - V1/75 = 0
V1/13 + V3/13 - V2/213 - 2*V2/13 = 0
Struggling to move on from...
Hi Berkeman, I'm trying to do the KCL using loop analysis to check but as I don't have values for voltage or current I can't seem to be able to work it out? It's the value of V I am struggling to cancel out.
Homework Statement
Calculate the insertion loss of the T-network given the values:
Rs = 75Ω
Rl = 100Ω
Ra = 13Ω
Rb = 13Ω
Rc = 213Ω
The transmission matrix and Network are below:
2. Homework Equations
The Attempt at a Solution
Constructing the matrix:
A = 1+Ra/Rc = 1+13/213 = 1.061
B...
Homework Statement
(a) State what is meant by a ‘distortionless’ and a ‘lossless’ transmission line.
(b) A transmission line has the primary coefficients as given below. Determine the line’s secondary coefficients Zo, α and β at a frequency of 1 GHz.
R = 2 Ω/m
L = 8 nH/m
G=0.5 mS/m
C=0.23...
Yeah I was using V1 = z11*I1+z12*I2 with I2=0 so that V1 = z11*I1 which I rearranged to z11 = V1/I1
Just wondering if you could explain in algebraic terms what the original post of:
z21 = V2 / I1 = V1 x (R6 / (R8+R6) ) x ( (R8+R6) / V1) = R6 = 6ohms;
is?
Okay so Z11 = V1 /I1 with V1 volts would mean that current = I1. I think I maybe understand now.
So Z11 = V1 / I1, voltage will run from top terminal 1 to bottom terminal 1 so will. = 14 ohms
However for Z21 = V2 / I1, V2 contacts will be in parallel with the 6 ohm resistor, so Z21 =...
Then Z12 = V1/ I2 , the current would act as in the first drawing so V1 = I2 * Z
So:
Z12 = (Z*I2)/ I2 = (6 * I2) / I2 = 6 ohms
If this is the case then I now don't understand why Z11 is 14 ohms rather than the same as Z21 with 6, unless it is because from the perspective from Z21 the 8 ohm...
Sorry, I assume you mean the red path in the second drawing? If so current would flow from top terminal 1 to bottom terminal one, so with both terminal 2s open the output voltage would equal the voltage across the 6 ohm resistor?
Wait, so would:
Z21 = V2 / I1 = (6 * I1) / I1 = 6ohms as I1...
To try and understand the current flow path, I think I understand why Z12 = 6ohms as it passes only through the 6ohm resistor ( first crude drawing )
I would've though that Z21 = 14ohms as well as it would take the red path ( second drawing), can't quite understand why it is 6ohms unless it...
Hi, sorry to be this a pain as I know this thread is a couple of years old but I'm now working through the same question, was just wondering if anyone could explain why for z21 V2 was changed to V1 x (R6 / (R8+R6) ) x ( (R8+R6) / V1) and how that equals 6 ohms? I'm looking to understand this as...
Sorry got a bit lost there, I follow you now and have gotten 7.5Ω.
One other query regarding part (d), I have calculated out the same procedure as mentioned earlier in this thread using P = 3VI for total power consumed, which gives 11481.667-j11481.667. Could I assume that the 'imaginary'...