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Crank Diagram

Module by: Bill Wilson. E-mail the author

We actually still have some options open to us. One of thenicest, at least in terms of getting some insight, is call a crank diagram. Note that this equation is a complex equation, which requires us to take a the ratio of two complex numbers; 1+ Γ ν e(j2βs) 1 Γ ν 2 β s and 1 Γ ν e(j2βs) 1 Γ ν 2 β s .

Let's plot these two quantities on the complex plane, starting at s=0 s 0 (the load end of the line). We can represent Γ ν Γ ν , the reflection coefficient, by its magnitude and its phase, | Γ ν | Γ ν and φ Γ φ Γ . For the numerator we plot a 1, and then add the complex vector ΓΓ which has a length |Γ| Γ and sits at an angle φ Γ φ Γ with respect to the real axis Figure 1. The denominator is just the same thing, except the ΓΓ vector points in the opposite direction Figure 2.

Figure 1: Plotting 1+ Γ ν 1 Γ ν
Plot
Plot (7_08.png)
Figure 2: Plotting 1 Γ ν 1 Γ ν
Another Plot
Another Plot (7_09.png)
The top vector is proportional to Vs V s and the bottom vector is proportional to Is I s Figure 3. Of course, for s=0 s 0 we are at the load so Vs=0= V L V s 0 V L and Is=0= I L I s 0 I L .
Figure 3: Showing that 1+ Γ ν = V L V + 1 Γ ν V L V + and 1 Γ ν = Z 0 I L V + 1 Γ ν Z 0 I L V +
Another Crank Diagram
Another Crank Diagram (7_10.png)
As we move down the line, the two "ΓΓ" vectors rotate around at a rate of -2βs -2 β s Figure 4. As they rotate, one vector gets longer and the other gets shorter, and then the opposite occurs. In any event, to get Zs Z s we have to divide the first vector by the second. In general, this is not easy to do, but there are some places where it is not too bad. One of these is when 2βs= θ Γ 2 β s θ Γ Figure 5.
Figure 4
Rotating the Phasors On the Crank Diagram
Rotating the Phasors On the Crank Diagram (7_11.png)
Figure 5: Rotating to a V max V max
Rotating a Crank Diagram
Rotating a Crank Diagram (7_12.png)
At this point, the voltage vector has rotated around so that it is just lying on the real axis. Obviously its length is now 1+|Γ| 1 Γ . By the same token, the current vector is also lying on the real axis, and has a length 1|Γ| 1 Γ . Dividing one by the other, and multiplying by Z 0 Z 0 gives us Zs Z s at this point.
Zs= Z 0 1+| Γ ν |1| Γ ν | Z s Z 0 1 Γ ν 1 Γ ν
(1)
Where is this point, and does it have any special meaning? For this, we need to go back to our expression for Vs V s in this equation.
Vs= V + ejβs(1+ Γ ν e-2jβs)= V + ejβs(1+| Γ ν |ej( θ Γ 2βs))= V + ejβs(1+| Γ ν |ejφs) V s V + β s 1 Γ ν -2 β s V + β s 1 Γ ν θ Γ 2 β s V + β s 1 Γ ν φ s
(2)
where we have substituted | Γ ν |ejθ Γ ν θ for the phasor Γ ν Γ ν and then defined a new angle φs= θ Γ 2βs φ s θ Γ 2 β s .

Now let's find the magnitude of Vs V s . To do this we need to square the real and imaginary parts, add them, and then take the square root.

|Vs|=| V + |(1+| Γ ν |ejφs)=| V + |1+| Γ ν |cosφs2+| Γ ν |2sin2φs V s V + 1 Γ ν φ s V + 1 Γ ν φ s 2 Γ ν 2 φ s 2
(3)
so,
|Vs|=| V + |1+2| Γ ν |cosφs+| Γ ν |2cos2φs+| Γ ν |2sin2φs V s V + 1 2 Γ ν φ s Γ ν 2 φ s 2 Γ ν 2 φ s 2
(4)
which, since sin2·+cos2·=1 · 2 · 2 1
|Vs|=| V + |1+| Γ ν |2+2| Γ ν |cosφs V s V + 1 Γ ν 2 2 Γ ν φ s
(5)
Remember, φs φ s is an angle which changes with ss. In particular, φs= θ Γ 2βs φ s θ Γ 2 β s . Thus, as we move down the line |Vs| V s will oscillate as cosφs φ s oscillates. A typical plot for Vs V s (for | Γ ν |=0.5 Γ ν 0.5 and θ Γ = 45 ° θ Γ 45 ° ) is shown here.
Figure 6
Standing Wave Pattern
Standing Wave Pattern (7_13.png)

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