Find V(L) for Lumped vs distributive circuits

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  • Thread starter nagr
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In summary, the two problems have the same equations, but different constants. The first problem has reflection, while the second does not. The second problem has a voltage divider, so you need to find Zin first. Once you have Zin, you can find V(0) using the equation V(0) = Vs[Zin/(Zin + Zg)].
  • #1
nagr
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Homework Statement


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



Homework Equations


Vg=2cos(wt)
Zg=Zl=200
Zo=50
[itex]\Gamma[/itex]=[itex]\frac{Zl-Zo}{Zl+Zo}[/itex]
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 [itex]\Gamma[/itex]=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.
 
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  • #2
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.
 
  • #3
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- = [itex]\Gamma[/itex]V0+

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.
 
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  • #4
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).
 
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  • #5


I would suggest approaching this problem by first identifying the key variables and equations involved. In this case, the variables are V(L), Vg, Zg, Zl, Zo, Γ, L', R', μ, ε, μ0, and ε0. The equations involved are Vg=2cos(wt), Zg=Zl=200, Zo=50, Γ=(Zl-Zo)/(Zl+Zo), L'R'=με, μ0=4πE-7, and ε0=8.854E-12.

Next, I would suggest considering the physical principles behind these equations and how they relate to the problem at hand. For example, Vg represents the voltage of the source, Zg represents the impedance of the source, and Zo represents the characteristic impedance of the transmission line. Γ represents the reflection coefficient, which is a measure of how much of the signal is reflected back from the load. L' and R' represent the inductance and resistance per unit length of the transmission line, respectively, and μ and ε represent the permeability and permittivity of the medium through which the signal is traveling.

Based on this understanding, I would then suggest approaching the problem by first considering the lumped circuit case, where the transmission line is short enough that it can be treated as a lumped circuit rather than a distributed one. In this case, we can use the equation V(L)=Vg*Γ, where Vg is the voltage of the source and Γ is the reflection coefficient. We can then solve for V(L) by plugging in the given values for Vg and Γ.

For the distributed circuit case, we must consider the effects of the transmission line's length and the frequency of the signal. In this case, we can use the equation V(L)=Vg*e^(-jβl), where Vg is the voltage of the source and β is the propagation constant, which is equal to w*sqrt(με). We can then solve for V(L) by plugging in the given values for Vg, β, and l. However, as you mentioned, the problem does not specify the frequency or relative permittivity, so you may need to make some assumptions or seek clarification from your instructor.

Overall, the key is to carefully consider the problem and the equations involved, and to use your understanding of the physical principles to guide your approach
 

1. What is the difference between lumped and distributive circuits?

The main difference between lumped and distributive circuits is in how the elements of the circuit are placed. In a lumped circuit, all the components are placed close together, while in a distributive circuit, the components are spread out along the length of the circuit. This affects how the circuit behaves and is analyzed.

2. How does the distance between components affect the behavior of a circuit?

The distance between components in a circuit can affect the behavior in terms of signal transmission, impedance, and attenuation. In lumped circuits, the components are close together, allowing for quick signal transmission and low impedance. In distributive circuits, the components are spread out, leading to slower signal transmission and higher impedance.

3. How does the type of circuit affect the calculation of V(L)?

The type of circuit, whether it is lumped or distributive, affects the calculation of V(L) because it determines the method used to analyze the circuit. In lumped circuits, V(L) can be calculated using Kirchhoff's laws and Ohm's law, while in distributive circuits, more complex methods such as the transmission line theory may be needed.

4. Can lumped and distributive circuits be used interchangeably?

No, lumped and distributive circuits cannot be used interchangeably. The type of circuit used depends on the specific application and the desired behavior of the circuit. For example, lumped circuits are commonly used in low-frequency applications, while distributive circuits are more suitable for high-frequency applications.

5. Are there any advantages of using a lumped circuit over a distributive circuit?

One advantage of using a lumped circuit is that it is easier to analyze and design, as the calculations involved are simpler. It is also more cost-effective, as fewer components are needed. However, distributive circuits have the advantage of better signal transmission at high frequencies, making them more suitable for certain applications.

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