Chapter2_Notes

Do we get the current i1 vg rg z0 209 v1 v g

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Unformatted text preview: /2). V (z, 5T 5T ) = (1 + ΓL + ΓL Γg )V1+ , V (z, ) = (1 + ΓL )V1+ 2 2 0≤z&lt;l/2 l/2≤z ≤l (213) • Clearly this will go on for a while . . . • What about the current? The procedure the same, but the reﬂection coeﬃcients come with a negative sign: − + + − + I1 = −ΓL I1 , I2 = −Γg I1 = Γg ΓL I1 (214) • Is there a limit to this? In eq. 213 we keep adding up components that are products of Γn Γn,n−1 . This inﬁnite series has a Lg limit 1 1 + x + x2 + . . . = for |x| &lt; 1. (215) 1+x Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 121 • Net result: we can calculate lim∞ V V∞ = V1+ 1 + ΓL 1 − Γg ΓL (216) • After using expressions for V1+ (eq. 209), ΓL (eq. 211), and Γg (eq. 212) ⇒ V g ZL (217) V∞ = Rg + ZL What is this? (It’s called the steady-state voltage) • For the steady-state current I∞ = V∞ Vg = ZL Rg + ZL Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (218) Electromagnetics I: Transmission lines 122 • Bounce diagrams How do we keep track of all these waves bouncing back and forth? We use a graphical method involving bounce diagrams, as shown in Fig. 33. Note the coordinates: • The horizontal axis is the position along transmission line; starts at the generator, at z = 0, and ends at the end of transmission line z = l. • The vertical axis is time. • The reﬂection coeﬃcients Γg and ΓL at each end are indicated. Note the diﬀerence between voltages and currents: currents have negative Γ-s. So, how do we use the bounce diagram? • Select a point on a transmission line • Draw a line parallel to time axis through this point Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines Γ = Γg z=0 t=0 l/4 l/2 3l/4 123 Γ = ΓL Γ = -Γg z=0 t=0 z=l + V1 T ΓLV1+ z=l T ΓgΓLI1+ -ΓgΓL2I1+ 4T 3T 2 Γg ΓL2I1+ I(l/4, 4T) 5T t Γ = -ΓL + 3T 22+ Γg ΓL V1 V(l/4, 4T) 3l/4 I1 2T 2+ ΓgΓL V1 4T l/2 -ΓLI1+ ΓgΓLV1+ 2T l/4 t 5T t (a) Voltage bounce diagram t (b) Current bounce diagram (1+ΓL+ΓgΓL+ΓgΓL2+Γg2ΓL2)V1+ V(l/4, t) (1+ΓL+ΓgΓL+ΓgΓL2)V1+ (1+ΓL)V1+ (1+ΓL+ΓgΓL)V1+ V1+ V1+ t T 4 T 7T 2T 9T 4 4 3T 15T 4T 17T 4 4 5T (c) Voltage versus time at z = l/4 Figure 2-35 Figure 33: Bounce diagrams and voltage variation at a point. Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 124 • Construct a zig-zag curve starting from the generator; note that it takes time t = T to go from one end of the line to another, • On each section of the curve indicate the product of the reﬂection coeﬃcients up to that point (whenever the curve “hits” one of the ends, the wave is multiplied by reﬂection coeﬃcient) • For your chosen point on transmission line look for its line’s intersection with the zig-zag curve; note the times of the intersection • Between two intersections, the voltage (or current) stays constant • On a V vs. t diagram indicate the times at which the two lines intersect • Starting from the ﬁrst intersection, keep adding up terms (as indicated on the bounce diagram) but remember that between two times voltage is constant. Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 125 • For currents, we have to be careful about the signs - best to indicate them on the bounce diagram itself and then just add them up. Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 126 V(0, t) 6V 3V t 12 µs 0 (a) Observed voltage at the sending end t=0 Rg = Z0 Z0 + Vg z=0 Rf Z0 ZL = Z0 z=d (b) The fault at z = d is represented by a fault resistance Rf Figure Time-domain reﬂectometer for ex. 2–13. Figure 34:2-36 Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU....
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