6

All TVs are designed with a standard input impedance of 75 ohms. No need to measure. The cable is also 75 ohms, the length doesn't matter. Except the signal will get weaker if it has to travel further. As @Hearth points out older TVs may also have a 300 Ohm balanced feeder antenna connection. Very rare these days though, especially on a cable-ready set.


4

Nice work. Yes, what you're trying makes sense. The diagonal hatched ground plane will work fine. The other configurations are interesting academically, and mathematically sound, but difficult enough to do in practic that the easier diagonal hatched ground plane is by far the more practical approach. Not to worry about the imbalance because of imperfect ...


4

Here a solution with the Extra Element Theorem (EET). with the dependant current source as extra element (See Vorperian's book). To get the resistance a voltage vx is injected and the current ix is determined. The other way to inject a current and get the voltage is possible too. But with them I get a complicate sub circuit. Ro is removed first. Rx is the ...


3

Your mistake was in the original expression for the voltage gain (which you called \$A\$). The voltage gain in Miller's Theorem is the voltage gain of the circuit including the effects of the feedback impedance. Miller's Theorem is valid when the output voltage is proportional to the input voltage with some constant of proportionality that I will call \$K\$ ...


2

From Wikipedia article (https://en.wikipedia.org/wiki/Miller_theorem#Explanation): The Miller theorem implies that an impedance element is supplied by two arbitrary (not necessarily dependent) voltage sources that are connected in series through the common ground. (The emphasis is mine). As there is a current source in the drain branch of your circuit and ...


2

By attaching the antenna to a AC current source..... and I have plotted V vs Freq. How do i now plot Impedance vs Freq using Circuitlab? So, set the current source to be 1 amp RMS and the resulting plot of voltage is transferrable 1:1 to impedance.


2

The transmission line appears initially to be a lumped element of its specified impedance. Only after the signal travels at the speed of light to the open end, and the return signal travels back to z=0, does the transmission line at z=0 appear to be open. Instantaneous response would violate relativity.


2

If in a circuit there is real power positive and reactive power negative, what does mean in terms of power flow? If you have positive real power it means that some component is consuming it and converting it into work, be it mechanical work or heat or both. If you have negative reactive power it means that some component is changing the phase of the voltage ...


2

Real power flow is controlled primarily by angular differences, flowing from leading angles toward lagging. Reactive power is controlled by voltage magnitudes, flowing from higher voltage toward lower. This assumes a system where all bus voltages are reasonably near 1 p.u. and relatively small angles. This is the basis of a decoupled load flow by the way. ...


2

Yes, once you have calculated the impedance matrix for the line, \$Z_{abc}\$, you just need to pre-multiply it with \$A^{-1}\$ and post-multiply it with \$A\$. $$Z_{012}=\begin{pmatrix}Z_{00} & Z_{01} & Z_{02} \\\ Z_{10} & Z_{11} & Z_{12} \\\ Z_{20} & Z_{21} & Z_{22} \end{pmatrix} =A^{-1}Z_{abc}A$$ Turan Gönen called this ...


2

Results from a trace width calculator I use: Using Er = 4.8 (I normally use 4.2 since we use Isola 370HR) Gap between trace and upper ground plane = 6 mils, as used in Figure 5.1 Z = 50 ohms Trace width = 30 mils (36 mils for Er = 4.2) which closely matches with Figure 5.1. With a gap between trace and upper ground plane = 60 mils, the trace width is 100 ...


1

First, I will assume the angular frequency is \$10^5\text{ rad/s}\$ as you state and that the \$105\$ in the problem statement is wrong. This is a steady-state circuit so you can just use phasor math. You know the following, $$2\pi f = 10^5 \text{ rad/s} $$ $$X_L=2\pi f L=j4.6 \Omega$$ Letting our reference phasor be \$sin(105t)\$ then the current down into ...


1

You have used an impedance triangle (and pythagoras) to calculate the total impedance. That technique would be used for a series RL circuit but not for a parallel combination. To calculate the total (parallel) impedance use product/sum or 1/RT = 1/R + 1/XL Using this method I get 3.385 angle +48degrees for the total impedance.


1

To answer your explicit question: you start from the false assumption that any equivalent resistance can be expressed as a combination of parallel or series resistances. There are even simple passive resistive circuits that cannot. For example, try to calculate the equivalent resistance of this one: simulate this circuit – Schematic created using ...


1

MappleSoft Licensee . see https://www.maplesoft.com/products/Maple/students/ Ok. Complicated to see the result as shown. More understandable by following the "classic" procedure ... in the theoretical framework :) But hey, it's only the result that counts. On the other hand, when the diagram is more complicated, it suffices to write "all"...


1

I don't think the method of "looking into the terminals" to find the equivalent resistance works in this case, due to the dependent current source. Look at this excerpt from Electrical Engineering Principles and Applications by Allan Hambley:- Although I am starting to doubt this conclusion when I look at the comments of the question.


1

You ignore half of the wave which starts to propagate along the wires to the right as soon as the battery is connected. The ignored half is the current component of the wave. In the linked article there's clearly stated that the input voltage isn't constant DC which has stayed as is from t=minus eternity, the system is switched ON at t=0, so a wave starts to ...


1

If you are on USB v2, there are no worries as the line frequency is not so critical, as long as you don't go overboard. In USB v3, it gets more critical.


1

How to plot it using Circuitlab, I do not know But I do know how to model it mathematically: $$\underline{\text{Z}}_{\space\text{in}}=\frac{\frac{1}{\text{j}\omega\text{C}}\cdot\left(\text{R}+\text{j}\omega\text{L}\right)}{\frac{1}{\text{j}\omega\text{C}}+\text{R}+\text{j}\omega\text{L}}=\frac{\text{R}+\text{j}\omega\text{L}}{1-\omega^2\text{CL}+\text{R}\...


1

Here, the way you are doing it is correct. In \$A = -g_m r_o\$, \$r_o\$ is the impedance seen at the collector. It is the impedance of the load in parallel with the output impedance of the transistor. In this case it is the output impedance of the transistor \$r_o\$ (you get confused when you use the same letter for two things), \$Z_f\$, and the impedance ...


1

The limit will be the loading by the cable capacitances. You can amass 10s of nF of load capacitance if you use many long cables. This would be like a load of only a few 100 Ohm at 20kHz and the opamp will eventually not be able to drive this due to the lack of output current capabilities, producing distortion.


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