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Example determining the tl length
Example determining the tl length
Example determining the tl length
Example determining the tl length
Example determining the tl length
Example determining the tl length
Example determining the tl length
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Example determining the tl length

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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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