Engineering Find V(L) for Lumped vs distributive circuits

  • Thread starter Thread starter nagr
  • Start date Start date
  • Tags Tags
    Circuits
Click For Summary
The discussion focuses on finding the load voltage V(L) across a load in both lumped and distributed circuit scenarios. In the first problem, the user grapples with reflections due to mismatched impedances, while attempting to graph V(L) as a function of length and wavelength without given frequency values. The second problem emphasizes that lumped circuits inherently do not involve reflections, and the user seeks to relate load voltage to source voltage using specific equations and parameters. Key equations include the reflection coefficient and the relationship between input impedance and voltage. The user expresses confusion over missing values and assumptions regarding relative permittivity, highlighting the complexity of the problems.
nagr
Messages
5
Reaction score
0

Homework Statement


Find V(L) voltage across load
29384ki.png



Homework Equations


Vg=2cos(wt)
Zg=Zl=200
Zo=50
\Gamma=\frac{Zl-Zo}{Zl+Zo}
L'R'=με (all TEM lines)
μ0=4∏E-7
ε0=8.854E-12
11aycee.png


The Attempt at a Solution


Basically I think that with the 1st one, i don't have to deal with reflection or whatever. i could be wrong, since Zo does not equal Zl in both problems, which means there has to be reflection. but the equation I used for the 1st problem was in my book, and its similar to the voltage source equation. 2cos(wt-θ) where θ=wl/c. Since we have to graph V(L) for both as a function of length which is in terms of wavelength, this is also listed in book as θ=2∏*(l/λ). The only other thing that confuses me is that frequency is not given, so i can't find w. How can I vary both length (in terms of wavelength) and frequency, which is nowhere mentioned in the problem?? maybe i am doing this wrong.

for the 2nd problem. i think this is basically the same as the 1st one, except now we deal with reflection. since we also deal with R' L' G' C' (with R'=G'=0). the main breakthrough I've had with this problem is finding \Gamma=0.6 which was easy. But i have no idea what to do from here. if I ASSUME that the relative permittivity is the same as in in air (that is εr=1) THEN i calculate R' L' using the equations Zo=50=sqrt(L'/C') and β=w*sqrt(L'C')=w*sqrt(με), and if i do that I get C'=6.671E-11 and L'=1.668E-7 but, i don't want to assume anything. can someone help me? i have the equations but there's no problem like this in my book, plus I am lacking values like frequency and εr, so I am not sure what exactly to do.
 
Last edited:
Physics news on Phys.org
1st problem: there is definitely reflection, not that that's important here.
What equation relates load voltage to source voltage with the given parameters w, L', C', l, RL, RG, Z0?

2nd problem: this is a lumped circuit. There are by definition no reflections with lumped circuits.
 
the only equation i see in my book is V(z)=V0+(e-jβz)+V0-(ejβz)
at the load voltage VL is at z=0 and given by V0+ + V0-
V0+/V0- are related to reflection coefficient by: V0- = \GammaV0+

to find V0+ i would have to find Zin. where Zin=Z0[(zL+jtanβl)/(1+j(zLtanβl)] where zL=ZL/Z0=4 and β=w*sqrt(L'C')=w*sqrt(με). So I have Zin=50[(4+jtanβl)/(1+j(4tanβl)] but i have trouble finding β without L' and C'. my book also has another equation β=2∏/λ but I am not sure if i should use that.
 
Last edited:
Your answer will include ω, L' and C', also l plus the given parameters Zg, ZL. You can't get numerical answers without those but you can compare V(0) for a transmission line vs. the lumped circuit.

You also have the relations in that post, plus β = ω√(L'C') and Z0 = √(L'/C').

Start with your equation from post 3 for V(z), solve for V(0) as function of V(l).

You then have to compute Zin since that forms a voltage divider with Zg. You have the formula for that too I see.

So V(l) = Vs[Zin/(Zin + Zg).

Then relate V(l) to V(0).
 
Last edited:

Similar threads

Replies
19
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 28 ·
Replies
28
Views
3K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
16
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K