Transmission Lines D.E

Thread Starter

bumclouds

Joined May 18, 2008
81
Hey guys,

For one of my lab reports on transmission lines, im required to derive the equation for the reflected waveform when there is a capacitive termination.



So what is in the above figure, except a capacitor instead of a resistor.

The question asks to show that the reflected voltage at the load end is:


Here is my attempt at the solution:
Firstly:


Then the equation for a capacitor can be sub'ed in:


And then given that I+=\[\frac{V^{+}}{Z_{0}}\] and I-=\[\frac{-V^{-}}{Z_{0}}\]
it becomes:


then:
\[V^{+}-V^{-}\] =
 

Thread Starter

bumclouds

Joined May 18, 2008
81
And then the general solution form (i think) is this:

To solve for B:
At t = ∞, the capacitor will look like an open circuit and hence:

And we can substitute that into
to yield:


Now solving for C:
At t=0 the capacitor has a voltage of 0. Sub'ing this into our general soln:

And now putting C back into the general soln:

 

Thread Starter

bumclouds

Joined May 18, 2008
81
But if we substitute t=∞, it yields \[V_{L}\]=-2V, so the gen soln needs to be negated so that:


Now if we substitute this general solution for \[V_{L}\] into an equation we did earlier we get:


but FAIL!

k becomes = 0!!

therefore it can't be rearranged to prove that:
.


So where have I gone wrong here??
 

Mark44

Joined Nov 26, 2007
626
Hey guys,

For one of my lab reports on transmission lines, im required to derive the equation for the reflected waveform when there is a capacitive termination.



So what is in the above figure, except a capacitor instead of a resistor.

The question asks to show that the reflected voltage at the load end is:


Here is my attempt at the solution:
Firstly:


Then the equation for a capacitor can be sub'ed in:


And then given that I+=\[\frac{V^{+}}{Z_{0}}\] and I-=\[\frac{-V^{-}}{Z_{0}}\]
it becomes:


then:
\[V^{+}-V^{-}\] =
Caveat: My expertise is more on the mathematics side, not the electrical or electronics side.

I think you might have a mistake in the equation above, namely
V+ - V- = C*Z0 * d/dt(V+ - V-)

If you solve this differential equation, you get
V+ - V- = (e^K) * (e^(t/(C*Z0)), which isn't exponential decay as long as C >= 0 and Z0 >= 0. Since these are physical quantities, it seems reasonable to assume they are positive. Now, since you are given the answer you should arrive at, and you got something quite different, what you started with in your derivation is suspect.
Mark
 

Thread Starter

bumclouds

Joined May 18, 2008
81
I think you might have a mistake in the equation above, namely
V+ - V- = C*Z0 * d/dt(V+ - V-)

If you solve this differential equation, you get
V+ - V- = (e^K) * (e^(t/(C*Z0))

My maths is a bit weak.

I don't understand how you got from:
\[V^{+} - V^{-} = CZ_{0}\frac{d}{dt}(V^{+}+V^{-})\]
to
\[V^{+} - V^{-} = e^{k}e^{\frac{t}{CZ_{0}}\]
 

Thread Starter

bumclouds

Joined May 18, 2008
81
Now i'm looking at the way my lab tutor has solved it. He does this >>






\[V^{+} - V^{-} = CZ_{0}\frac{d}{dt}(V^{+}+V^{-})\]
then: \[V^{+}-V^{-}\] = \[CZ_{0}\frac{dV^{-}}{dt}\]

What the..? I dont understand how that last step works - but when I follow this step - I arrive at the correct solution.
 

Mark44

Joined Nov 26, 2007
626
My maths is a bit weak.

I don't understand how you got from:
\[V^{+} - V^{-} = CZ_{0}\frac{d}{dt}(V^{+}+V^{-})\]
to
\[V^{+} - V^{-} = e^{k}e^{\frac{t}{CZ_{0}}\]
Actually, I misread a sign in your differential equation. What I started with was \[V^{+} - V^{-} = CZ_{0}\frac{d}{dt}(V^{+} - V^{-})\]

In other words, I had the difference of the two voltages on both sides, rather than what you had, which was a difference of voltages on the left side of the equation and a sum of voltages on the right.


If you're interested in what I did after that, here is how I got what I did.
This is a slight variant of an elementary differential equation (or DE). To simplify things a bit, let y = V+ - V-


So, your DE now looks like this:
y = C*Z0 * dy/dt

Moving factors around a bit yields this:
1/(C*Z0) * dt = dy/y

The DE is now separated, so you can integrate each side with respect to t, getting:
1/(C*Z0) * t + k = ln(y)

The natural log term should be ln |y|, but I am assuming that y >= 0, which might not be a valid assumption. If not, let me know.

Taking each side of the previous equation as the exponent on e, we get:
e^[(1/C*Z0) t + k] = y

But e^[(1/C*Z0) t + K] = e^(1/C*Z0)t * e^k = K*e^(1/C*Z0)t, where K = e^k

Since y = V+ - V-, then we end up with this:
V+ - V- = K*e^(1/C*Z0)t

Again, this result is erroneous, since I didn't start with the same equation you were working with.
 

Mark44

Joined Nov 26, 2007
626
Now i'm looking at the way my lab tutor has solved it. He does this >>






\[V^{+} - V^{-} = CZ_{0}\frac{d}{dt}(V^{+}+V^{-})\]
then: \[V^{+}-V^{-}\] = \[CZ_{0}\frac{dV^{-}}{dt}\]

What the..? I dont understand how that last step works - but when I follow this step - I arrive at the correct solution.
It all hinges on V+. I don't see in this thread where you have defined what V+ and V- represent, which would be helpful information.

Is V+ constant? if so, dV+/dt = 0, which I'm guessing is how your tutor got rid of it.

Mark
 
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