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- 1. 2/16/2010 Example Determining the tl length.doc 1/7 Example: Determining Transmission Line Length A load terminating at transmission line has a normalized impedance z L′ = 2.0 + j 2.0 . What should the length of transmission line be in order for its input impedance to be: a) purely real (i.e., xin = 0 )? b) have a real (resistive) part equal to one (i.e., rin = 1.0 )? Solution: a) Find z L′ = 2.0 + j 2.0 on your Smith Chart, and then rotate clockwise until you “bump into” the contour x = 0 (recall this is contour lies on the Γr axis!). When you reach the x = 0 contour—stop! Lift your pencil and note that the impedance value of this location is purely real (after all, x = 0 !). Now, measure the rotation angle that was required to move clockwise from z L′ = 2.0 + j 2.0 to an impedance on the x = 0 contour—this angle is equal to 2 β ! You can now solve for , or alternatively use the electrical length scale surrounding the Smith Chart. Jim Stiles The Univ. of Kansas Dept. of EECS
- 2. 2/16/2010 Example Determining the tl length.doc 2/7 One more important point—there are two possible solutions! Solution 1: 2β = 30 = 0.042λ z L′ = 2 + j 2 Γ (z ) ′ zin = 4.2 + j 0 x =0 Jim Stiles The Univ. of Kansas Dept. of EECS
- 3. 2/16/2010 Example Determining the tl length.doc 3/7 Solution 2: z L′ = 2 + j 2 ′ zin = 0.24 + j 0 x =0 Γ (z ) 2β = 210 = 0.292λ Jim Stiles The Univ. of Kansas Dept. of EECS
- 4. 2/16/2010 Example Determining the tl length.doc 4/7 b) Find z L′ = 2.0 + j 2.0 on your Smith Chart, and then rotate clockwise until you “bump into” the circle r = 1 (recall this circle intersects the center point or the Smith Chart!). When you reach the r = 1 circle—stop! Lift your pencil and note that the impedance value of this location has a real value equal to one (after all, r = 1 !). Now, measure the rotation angle that was required to move clockwise from z L′ = 2.0 + j 2.0 to an impedance on the r = 1 circle—this angle is equal to 2β ! You can now solve for , or alternatively use the electrical length scale surrounding the Smith Chart. Again, we find that there are two solutions! Jim Stiles The Univ. of Kansas Dept. of EECS
- 5. 2/16/2010 Example Determining the tl length.doc 5/7 Solution 1: z L′ = 2 + j 2 Γ (z ) r =1 ′ zin = 1.0 − j 1.6 2β = 82 = 0.114λ Jim Stiles The Univ. of Kansas Dept. of EECS
- 6. 2/16/2010 Example Determining the tl length.doc 6/7 Solution 2: ′ zin = 1.0 + j 1.6 Γ (z ) z L′ = 2 + j 2 r =1 2β = 339 = 0.471λ Jim Stiles The Univ. of Kansas Dept. of EECS
- 7. 2/16/2010 Example Determining the tl length.doc 7/7 ′ Q: Hey! For part b), the solutions resulted in zin = 1 − j 1.6 and ′ zin = 1 + j 1.6 --the imaginary parts are equal but opposite! Is this just a coincidence? A: Hardly! Remember, the two impedance solutions must result in the same magnitude for Γ --for this example we find Γ ( z ) = 0.625 . Thus, for impedances where r =1 (i.e., z ′ = 1 + j x ): Γ= jx z ′ − 1 (1 + jx ) − 1 = = z ′ + 1 (1 + jx ) + 1 2 + j x and therefore: 2 Γ = jx 2 2+ j x 2 x2 = 4 +x2 Meaning: x = 2 4 Γ 2 1− Γ 2 of which there are two equal by opposite solutions! x = ± 2 Γ 1− Γ 2 Which for this example gives us our solutions x = ±1.6 . Jim Stiles The Univ. of Kansas Dept. of EECS

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