8/29/2007 Microwave Sources 1/1 Jim Stiles The Univ. of Kansas Dept. of EECS C. Microwave Sources Q: A passive load ZL specifies ()Zzand ()zΓ, but we still don’t explicitly know ()()()() or Vz,Iz V z,V z+−. How are these functions determined? A: All of these quantities are zero, unless a source (generator) is applied to trans. line. The boundary condition enforced by the generator will then explicitly determine these functions! HO: A Transmission Line Connecting Source and Load Q: OK, we can finally ask the question that we have been concerned with since the very beginning: How much power is delivered to the load by the source? A: HO: Delivered Power Q: So the power transferred depends on the source, the transmission line, and the load. What combination of these devices will result in maximum power transfer? A: HO: Special Cases of Source and Input Impedances8/29/2007 A Transmission Line Connecting Source 1/5 Jim Stiles The Univ. of Kansas Dept. of EECS A Transmission Line Connecting Source & Load We can think of a transmission line as a conduit that allows power to flow from an output of one device/network to an input of another. To simplify our analysis, we can model the input of the device receiving the power with it input impedance (e.g., ZL), while we can model the device output delivering the power with its Thevenin’s or Norton’s equivalent circuit. + - Vg Zg + Vi - Ii Ig Zg + Vi - Ii gigiVVZI=+ igigVIIZ=+8/29/2007 A Transmission Line Connecting Source 2/5 Jim Stiles The Univ. of Kansas Dept. of EECS Typically, the power source is modeled with its Thevenin’s equivalent; however, we will find that the Norton’s equivalent circuit is useful if we express the remainder of the circuit in terms of its admittance values (e.g., 0,,()LYYYz). Recall from the telegrapher’s equations that the current and voltage along the transmission line are: 000000jzjzjzjzV(z) V e V eVVI(z) e eZZββββ+−−++−−+=+=− At 0z=, we enforced the boundary condition resulting from Ohm’s Law: ()00000000()()LLLVVVVzZIIzVVZZ+−+−+=== ==⎛⎞−⎜⎟⎝⎠ + - Vg + Vi - Ii LZ z=−A 0z= 0Z gZ8/29/2007 A Transmission Line Connecting Source 3/5 Jim Stiles The Univ. of Kansas Dept. of EECS Which resulted in: 0000LLLZZVVZZ−+−=+Γ So therefore: 000jz jzLjzjzLV(z) V e eVI(z) e eZββββ+− ++−+⎡⎤=+Γ⎣⎦⎡⎤=−Γ⎣⎦ We are left with the question: just what is the value of complex constant 0V+?!? This constant depends on the signal source! To determine its exact value, we must now apply boundary conditions at z=−A . We know that at the beginning of the transmission line: 000jjLjjLV(z ) V e eVI(z ) e eZAAAAAAββββ++ −++−⎡⎤=− = + Γ⎣⎦⎡⎤=− = − Γ⎣⎦ Likewise, we know that the source must satisfy: gigiVVZI=+8/29/2007 A Transmission Line Connecting Source 4/5 Jim Stiles The Univ. of Kansas Dept. of EECS To relate these three expressions, we need to apply boundary conditions at z=−A : From KVL we find: ()iVVz==−A And from KCL: ()iIIz==−A Combining these equations, we find: 000gi gijj jjgLgLVVZIVVVe e Z e eZββ ββ++− +−+=+⎡⎤⎡⎤=+Γ+ −Γ⎣⎦⎣⎦AA AA One equation Æ one unknown (0V+)!! Solving, we find the value of 0V+: + - Vg iV+− Ii LZ z=−A 0z= 0Z gZ ()Vz+=−−A ()Iz=−A8/29/2007 A Transmission Line Connecting Source 5/5 Jim Stiles The Univ. of Kansas Dept. of EECS ()()00011jgin g inZVVeZZβ−+=+Γ + −ΓA where: ()2jin Lzeβ−Γ=Γ =− =ΓAA There is one very important point that must be made about the result: ()()00011jgin g inZVVeZZβ−+=+Γ + −ΓA And that is—the wave ()0Vz+ incident on the load ZL is actually dependent on the value of load ZL !!!!! Remember: ()2jin Lzeβ−Γ=Γ =− =ΓAA We tend to think of the incident wave ()0Vz+ being “caused” by the source, and it is certainly true that ()0Vz+ depends on the source—after all, ()00Vz+= if 0gV=. However, we find from the equation above that it likewise depends on the value of the load! Thus we cannot—in general—consider the incident wave to be the “cause” and the reflected wave the “effect”. Instead, each wave must obtain the proper amplitude (e.g., 00,VV+−) so that the boundary conditions are satisfied at both the beginning and end of the transmission line.8/29/2007 Delivered Power 1/3 Jim Stiles The Univ. of Kansas Dept. of EECS Delivered Power Q: If the purpose of a transmission line is to transfer power from a source to a load, then exactly how much power is delivered to ZL for the circuit shown below ?? A: We of course could determine 00 and VV+−, and then determine the power absorbed by the load (Pabs) as: ()(){}1Re 0 02absPVzIz∗=== However, if the transmission line is lossless, then we know that the power delivered to the load must be equal to the power “delivered” to the input (Pin) of the transmission line: ()(){}1Re2inabsPP Vz Iz∗= = =− =−AA + - Vg ()Vz+− ()Iz LZ z=−A 0z= 0Z gZ Zin8/29/2007 Delivered Power 2/3 Jim Stiles The Univ. of Kansas Dept. of EECS However, we can determine this power without having to solve for 00 and VV+− (i.e., ()() and VzIz). We can simply use our knowledge of circuit theory! We can transform load ZL to the beginning of the transmission line, so that we can replace the transmission line with its input impedance Zin : Note by voltage division we can determine: ()ingginZVzVZZ=− =+A And from Ohm’s Law we conclude: ()gginVIzZZ=− =+A And thus, the power Pin delivered to Zin (and thus the power Pabs delivered to the load ZL) is: + - Vg ()Vz+=−−A ()Iz=−A ()inZZz==−A gZ8/29/2007 Delivered Power 3/3 Jim Stiles The Univ. of Kansas Dept. of EECS ()(){}(){}{}222221Re21Re21Re21Re2inabsginggingingingininginginPP Vz IzVZVZZZZVZZZZVYZZ∗∗∗= = =− =−⎧⎫⎪⎪=⎨⎬++⎪⎪⎩⎭=+=+AA Note that we could also determine Pabs from our earlier expression: ()220012abs LVPZ+=−Γ But we would of course have to first determine 0V+(! ): ()()00011jgin g inZVVeZZβ−+=+Γ + −ΓA8/29/2007 Special Cases of Source and Load Impedance 1/10 Jim Stiles The Univ. of Kansas Dept. of EECS Special Cases of Source and Load Impedance Let’s look at specific cases of Zg and ZL, and determine how they affect 0V+ and Pabs. 0gZZ= For this