1/28/25 The Terminated ossless Transmission.doc 1/8 The Terminated, ossless Transmission ine Now let s attach something to our transmission line. Consider a lossless line, length, terminated with a load. I(z) I V (z), β z z z z V z Q: What is the current and voltage at each and every point on the transmission line (i.e., what is I ( z ) and V ( z ) for all points z where z z z?)? A: To find out, we must apply boundary conditions! In other words, at the end of the transmission line ( z z) where the load is attached we have many requirements that all must be satisfied!
1/28/25 The Terminated ossless Transmission.doc 2/8 1. To begin with, the voltage and current ( I ( z z ) and V ( z z )) must be consistent with a valid transmission line solution: ( ) ( ) ( ) V z z V z z V z z V e V e jβz jβz ( ) ( ) V z z V z z I ( z z ) V jβz V jβz e e 2. ikewise, the load voltage and current must be related by Ohm s law: V I 3. Most importantly, we recognize that the values I ( z z ), V ( z z ) and I, V are not independent, but in fact are strictly related by Kirchoff s aws! I(zz ) I, β V (zz ) V z z z z
1/28/25 The Terminated ossless Transmission.doc 3/8 From KV and KC we find these requirements: V z z V I z z I These are the boundary conditions for this particular problem. Careful! Different transmission line problems lead to different boundary conditions you must access each problem individually and independently! Combining these equations and boundary conditions, we find that: V I ( ) ( ) V z z I z z V z z V z z V z z V z z ( ) ( ) ( ) ( ) Rearranging, we can conclude: ( ) V z z V z z
1/28/25 The Terminated ossless Transmission.doc 4/8 Q: Hey wait as second! We earlier defined V ( z ) V ( z ) as reflection coefficient Γ ( z ). How does this relate to the expression above? A: Recall that Γ ( z ) is a function of transmission line position z. The value V ( z z) V ( z z) is simply the value of function ( z ) z z (i.e., evaluated at the end of the line): Γ evaluated at ( ) V z z Γ V z z ( z z ) This value is of fundamental importance for the terminated transmission line problem, so we provide it with its own special symbol ( Γ )! Γ Γ ( z z ) Q: Wait! We earlier determined that: so it would seem that: ( z ) Γ ( z ) ( z ) ( z z ) Γ Γ ( ) z z z z Which expression is correct??
1/28/25 The Terminated ossless Transmission.doc 5/8 A: They both are! It is evident that the two expressions: Γ and Γ ( ) z z z z are equal if: z z And since we know that from Ohm s aw: and from Kirchoff s aws: V I ( ) ( ) V V z z I I z z and that line impedance is: ( ) ( z ) V z z I z z z we find it apparent that the line impedance at the end of the transmission line is equal to the load impedance: z z The above expression is essentially another expression of the boundary condition applied at the end of the transmission line.
1/28/25 The Terminated ossless Transmission.doc 6/8 Q: I m confused! Just what are were we trying to accomplish in this handout? A: We are trying to find V(z) and I(z) when a lossless transmission line is terminated by a load! We can now determine the value of V in terms of V. Since: We find: ( ) ( ) V z z V e Γ j βz j βz V z z V e V e Γ V 2j βz And therefore we find: 2 β ( Γ ) V z e V e j z jβz 2 ( Γ ) V z V e e e jβz jβz jβz where: V 2 I ( z ) e ( e Γ ) e j βz jβz jβz Γ
1/28/25 The Terminated ossless Transmission.doc 7/8 z Now, we can further simplify our analysis by arbitrarily assigning the end point z a zero value (i.e., z ): I(z) I V (z), β z z V z If the load is located at z (i.e., if z ), we find that: ( ) ( ) ( ) V z V z V z jβ V e V e V jβ V I ( z ) ( ) ( ) V z V z V V e V jβ V jβ e V V z V V
1/28/25 The Terminated ossless Transmission.doc 8/8 ikewise, it is apparent that if z, Γ and Γ are the same: ( ) ( ) V z V Γ Γ ( z z) Γ V z V Therefore: Γ Γ Thus, we can write the line current and voltage simply as: β V ( z) V e Γ e V β I ( z) e Γ e j z jβz j z jβz for z Q: But, how do we determine V?? A: We require a second boundary condition to determine V. The only boundary left is at the other end of the transmission line. Typically, a source of some sort is located there. This makes physical sense, as something must generate the incident wave!