Keith Dysart wrote:
On Mar 16, 10:21 am, Cecil Moore wrote:
snip
Would you please explain how energy is conserved in the
following example at the zero-crossing point for Vs?

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| |
Vs 45 degrees | Shorted
100v RMS 50 ohm line | Stub
| |
| |
+--------------+----------------------+
gnd

At the zero-crossing of Vs, Ps(t) = 0, i.e. the source
is supplying zero watts at that time but Prs(t) = 100w.
Where is the 100 watts coming from?


For the first 90 degrees of time, the circuit can be represented as

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| /
Vs \ 50 ohm resistor
100v RMS /
| \
| |
+--------------+----------------------+
gnd

After 90 degrees of time has passed, the circuit can be represented as

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| |
Vs ---
100v RMS --- 50 ohm inductive
| |
| |
+--------------+----------------------+
gnd

The sudden switch in circuit design at time 90 degrees is not unique to
start up, but is true for any adjustment made to Vs and any returning
wave from the shorted stub,. As a result, a true stable circuit will
never be found unless some voltage adjustment is allowed for the 90
timing shift caused by the shorted stub. Keith (in his analysis of the
circuit) recognizes that Vs drives into a reactive circuit.

If we want to understand how constructive and destructive interference
act to cause a 50 ohm resistor to evolve into a 50 ohm capacitor, then
we need to examine how traveling waves might do this. It would be nice
to have a formula or wave sequence that fully addressed this evolution.

From circuit theory, we have the inductive reactance of a
short-circuited line less than 1/4 wavelength long is

XL = Zo * tan (length degrees)

= 50 * tan(45)

= 50 ohms


From traveling wave theory, we would have the applied wave from the
source arriving 90 degrees late to the stub side of resistor Rs. This
ignores the fact that current must already be passing through resistor
Rs because voltage has been applied to Rs from the Vs side 90 degrees
earlier. Whoa! Things are not adding up correctly this way!

We need to treat the wave going down the 50 ohm line as a single wave
front. The wave reverses at the short-circuit, reversing both direction
of travel and sign of voltage. When the wave front reaches the input
end, 90 degrees after entering (for this case), the voltage/current
ratio is identical to the starting ratio (the line was 45 degrees long,
tan(45) = 1), and the returning voltage directly adds to the voltage
applied from Vs. As a result, the current flowing through Rs will
increase, and Vs will see a changed (decreased) impedance.

After 90 degrees of signal application, we should be able to express the
voltage across Rs as

Vrs(t) = Vs(t) - Vg(t) + Vref(t)

Vs(t) = Vs(wt + 90), Vg(t) = Vs(wt + 90)/2, Vref(t) = Vs(wt)/2

Substitute so

Vrs(t) = Vs(wt + 90) - Vs(wt + 90)/2 + Vs(wt)/2

= Vs(wt + 90)/2 + Vs(wt)/2

Allow Vs to be represented by a sine wave, we have

2Vrs(t) = Vs*sin(wt + 90) + Vs*sin(wt)
= 2Vs*(sin(wt + 45)(cos(45))
Vrs(t) = Vs(sin(wt + 45)(cos(45))

Vs is defined as 100v RMS, which equals 100 * 1.414 = 141.4v Peak. The
maximum voltage would occur when the sin term was 90 degrees and equals
1, which would occur at wt = 45 degrees. We would have

Vrs(45) = 141.42 * sin(90)(cos(45)
= 141.42 * 1 * 0.7071
= 100v

Now consider the current. After the same 90 degrees of signal
application, we should be able to express the current through Rs as

Irs(t) = Is(t) + Iref(t)

Is(t) = Is(wt + 90), Iref(t) = Is(wt)

The reflected current has been shifted by 90 degrees due to the
reflection so we must rewrite Iref(t) to read

Iref(t) = Is(wt + 90)

Substitute,

Irs(t) = Is(wt + 90) + Is(wt)

Allow Is to be represented by a sine wave, we have

Irs(t) = Is*sin(wt + 90) + Is*sin(wt)

= 2*Is(sin(wt + 45)(cos(45))

How do we find Is? Is is the initial current found by dividing the
applied voltage at peak (141.42v) by the initial resistance (100 ohms).


Is = 141.42/100 = 1.4142a

The maximum current would occur when the sin term was 90 degrees and
equals 1, which would occur at wt = 45 degrees. We would have

Irs(t) = 2*Is(sin(wt + 45)(cos(45))

= 2 * 1.4142 * 1 * 0.7071

= 2a

These results agree with the results from Keith and from circuit theory.

We have a theory and at least the peaks found from the theory agree with
the results from others. How about Cecil's initial question which is
At the zero-crossing of Vs, Ps(t) = 0, i.e. the source
is supplying zero watts at that time but Prs(t) = 100w.
Where is the 100 watts coming from?

We will use the equation

Vrs(t) = Vs(t) - Vg(t) + Vref(t)

Vs(t) = Vs(wt + 90), Vg(t) = Vs(wt + 90)/2, Vref(t) = Vs(wt)/2

The challenge looks at the time when Vs(t) = Vs*sin(wt + 90) = 0, which
occurs when wt = -90. When wt = -90, all the power to the source
resistor Rs is from the reflected portion described as Vref(t) =
Vs(wt)/2. The voltage across Rs would be

Vrs(-90) = Vref(-90)
= Vs*sin(-90)/2
= 141.4/2
= 70.7v

The power to Rs would be (Vrs^2)/50 = (70.7^2)/50 = 100w, all coming
from the reflection.

In summary, power to the resistor Rs comes via two paths, one longer
than the other by 90 degrees (in this example). The short path is the
series path of two resistors composed of the source resistor Rs and the
input to the 50 ohm transmission line measured by Vg. The long path is
the series path of one resistor Rs and one capacitor composed of the
shorted transmission line. Both paths are available at all times.
Power flows through both paths to Rs at all times, but because of the
time differential in arrival timing, at some point Rs will receive power
only from Vs, and at another point, receive power only from Vref.

73, Roger, W7WKB