Page 1 IIT Bombay Course Code : EE 611 Department: Electrical Engineering Instructor Name: Jayanta Mukherjee Email: jayanta@ee.iitb.ac.in Lecture 9 EE 611 Lecture 9 Jayanta Mukherjee
Page 2 IIT Bombay Topics Covered • Properties of S parameters (Contd..) • Properties of 2 port networks • Signal flow graphs EE 611 Lecture 9 Jayanta Mukherjee
IIT Bombay Page 3 S Matrix for a Passive Reciprocal Lossless 2-port Junction • The S matrix of a general 2 port device is S S = 11 12 S S S 21 22 • = If the device is reciprocal , S S since S must be symmetric. 12 21 This leaves three complex va lues to describe the device. • If the device is lossless, then S must be unitary [ ] [ ] [ ] ⋅ = H S S U 2 2 + + * * * * S S S S S S S S 1 0 S S ⋅ 11 12 = = 11 12 11 12 12 22 11 12 2 2 * * + + * * S S 0 1 S S S S S S S S 21 22 12 22 12 11 22 12 22 12 EE 611 Lecture 9 Jayanta Mukherjee
IIT Bombay Page 4 S Matrix for a Passive Reciprocal Lossless 2-port Junction 2 2 2 2 • + = + = From the first 2 equations : S S S S 1 we get : 11 12 22 12 2 = = − S S and S 1 S 11 22 12 11 • To establish the phase relationsh ips, we define θ θ θ = = = j j j e S e e S S and S and S S 1 2 12 11 11 22 11 12 12 • + = * * S S S S From the third equation 0 we get : 12 11 22 12 − θ θ − θ θ + = j j j j S e S e S e S e 0 12 1 2 12 12 11 22 12 ( ) ( ) θ − θ θ - θ • = − j j Simplifyin g this gives e e , or 1 12 12 2 ( ) + θ θ j θ = − j 2 e e 1 2 12 θ + θ π ( ) θ = − + 1 2 n so that 2 1 and there are only three parameters 12 2 2 ( ) θ θ S , and which characteri ze the system at a given frequency 11 1 2 EE 611 Lecture 9 Jayanta Mukherjee
IIT Bombay Page 5 Example of direct calculation of S Matrices • Consider the S matrix of the network shown below Port 1 Port 2 Port 1 Port 2 jX jX a 1 a 2 b 1 b 2 Z 2 Z 1 Z 2 Z 2 Z 1 • To measure S 11 and S 21 we must terminate port 2 in a matched load of impedance Z 2 • To find S 11 , terminate port 2 with Z 2 . Then we have S 11 = (b 1 /a 1 ) EE 611 Lecture 9 Jayanta Mukherjee
IIT Bombay Page 6 Example of direct calculation of S Matrices (2) + • - V /V Since b and a are at the same port, this is identical to so 1 1 1 1 − + − b V Z jX Z = = = Γ = 1 1 2 1 S + in 11 + + a Z jX Z V 1 2 1 1 + − Z jX Z • = 1 2 We can repeat the same process to find S . We can verify 22 + + Z jX Z 1 2 = that S S as it should for a lossless, reciprocal , 2 - port circuit 11 22 EE 611 Lecture 9 Jayanta Mukherjee
IIT Bombay Page 7 Derivation of S 21 ( ) b • = = 2 We calculate now S S port 2 remains terminate d by Z ohms 21 12 2 a 1 = a 0 2 • S ince the current entering ports 1 and 2 must be equal but opposite we get ( ) ( ) I I 1 1 = − = − ⇒ − = − − 1 2 I I or a b a b 1 2 1 1 2 2 Z Z Z Z 1 2 1 2 b • = 1 Now using the fact that S when port two is matched, along 11 a 1 = with the fact that a 0 we have : 2 Z Z 2 ( ) ( ) 1 b b Z − = ⇒ = = − = 1 2 2 2 2 a S S S 1 1 + + 1 11 21 11 a Z Z Z jX Z Z 1 1 2 1 1 2 EE 611 Lecture 9 Jayanta Mukherjee
IIT Bombay Page 8 Results Summary + Z - Z jX • = 2 1 In the previous slides we derived : S + + 11 Z Z jX 2 1 − + Z Z jX • = 1 2 By symmetry S + + 22 Z Z jX Port 1 Port 2 2 1 jX 2 Z Z • = 1 2 We also derived S + + 21 Z 1 Z Z jX Z 2 1 2 2 Z Z • = = 1 2 By symmetry S S + + 12 21 Z Z jX 1 2 • We can confirm that 4 Z Z 2 = 2 = − 2 = 1 2 S 1 S S ( ) + + 21 12 11 2 2 Z Z X 1 2 • We have to be careful about the port termi nations EE 611 Lecture 9 Jayanta Mukherjee
IIT Bombay Page 9 Input reflection coefficient of a 2 port network a 2 b 2 b 1 a 1 2 Port Network Z 1 Z 2 Z L • Since port 2 is terminate d in Z , the relationsh ip between L Z - Z = Γ Γ = L 2 a and b is fixed at a b with + 2 2 2 L 2 L Z Z L 2 • Our generalize d S matrix tel ls us that = + = + Γ b S a S a S a S b 1 11 1 12 2 11 1 12 L 2 = + = + Γ b S a S a S a S b 2 21 1 22 2 21 1 22 L 2 EE 611 Lecture 9 Jayanta Mukherjee
IIT Bombay Page 10 Input reflection coefficient a 2 b 2 b 1 a 1 2 Port Network Z 1 Z 2 Z L • We can solve the second equation t o get S = 21 b a Γ 2 1 1 - S 22 L • Now work on the first equation t o get b in terms of a : 1 1 S = + Γ 21 b a S S Γ 1 1 11 12 L 1 - S 22 L • The input reflection coefficien t of the loaded two port is now : Γ b S S Γ = = + 1 12 21 L S − Γ in 11 a 1 S 1 22 L EE 611 Lecture 9 Jayanta Mukherjee
IIT Bombay Page 11 Attributes of Input Reflection Coefficient for Loaded 2-port Network Γ in Input Reflection Coefficien t : Γ S S S S 1 Γ = + 2 11 12 Z L 12 L 21 S − Γ in 11 S S 1 S 21 22 22 L Properties : • Γ = Γ = S for 0 (matched load) in 11 L • Γ = = S for S 0 (unilatera l device) in 11 12 Γ out S S Output Reflection Coefficien t 11 12 Z S 1 2 Γ S S S S 21 22 Γ = + 21 S 12 S Γ out 22 1 - S 11 S EE 611 Lecture 9 Jayanta Mukherjee
IIT Bombay Page 12 Application: Matching using an attenuator Z S 0 S + 12 E S Z 1 Z 2 S 0 - 21 Γ s Γ out S ≠ Consider a generator with a poor matching (Z 50 ohm). An attenuator (device reducing the power) can provide a broadband match at the cost of reduced output power. The S parameter matrix for an ideal attenuator (Chapter 5) is given by : EE 611 Lecture 9 Jayanta Mukherjee
IIT Bombay Page 13 Application: Matching using an attenuator 0 S = 12 S att S 0 12 An attenuator is matched and provides an attenuatio n of 2 = + L -10log S (dB). The match provided by the generator 10 21 attenuator circuit is : Γ S S Γ = + = Γ 21 S 12 S S S S out 22 Γ 21 12 1 - S 11 S 2 2 ⇒ Γ = Γ ≈ Γ << S 0 for S 1 out 21 S 21 S EE 611 Lecture 9 Jayanta Mukherjee
IIT Bombay Page 14 Signal Flow Graphs • S-parameters were introduced in the previous lectures to represent linear N-port networks • As we design microwave circuits we will need to analyze bigger circuits realized with multiple building blocks Flow graph techniques will provide us: • an analysis technique applicable to S parameters • the means to visualize the power flow in a circuit EE 611 Lecture 9 Jayanta Mukherjee
IIT Bombay Page 15 Signal Flow Graphs • Each variable is designated as a node • The S parameters and reflection coefficients are represented by branches • Branches enter dependent variable nodes and emanate from independent variable nodes. The independent/dependent variable nodes are the incident/reflected waves respectively. • A node is equal to the sum of the branches entering it S 21 a 1 b 2 S S 11 12 Z 1 Z 2 S 11 S 22 S S S 12 21 22 b 1 a 2 a 1 a 2 b 1 b 2 EE 611 Lecture 9 Jayanta Mukherjee
IIT Bombay Page 16 Signal Flow Graph Examples • A length of transmission line of length l satisfies b 2 =a 1 e -j β l , b 1 =a 2 e -j β l so the corresponding signal flow graph is e -j β l a 1 b 2 a 1 a 2 Z 0 b 1 b 2 e -j β l a 2 b 1 l • Suppose we have a load with reflection coefficient Γ L connected to port 2. Then we know a 2 = Γ L b 2 and we can include this as a “branch” in a signal flow graph. EE 611 Lecture 9 Jayanta Mukherjee
IIT Bombay Page 17 Loaded Port Network driven by a Generator • A basic circuit we need to be analyzed is given below. Γ L Γ S Z S S S 11 12 Z L Z 0 Z 0 S S 21 22 EE 611 Lecture 9 Jayanta Mukherjee
IIT Bombay Page 18 Flow Graph for a Generator • A basic circuit we need to be analyzed is given below. = Starting from : V E - I Z and divide by Z g S 1 S 0 Z S V Z I E + = − g S S 0 1 S + Z 11 12 S V g Z 0 Z 0 Z Z Z E S Z L S S 0 0 0 21 22 - - ( ) E Z a 1 + = − − a 2 S S a b a b b 1 b 2 1 1 1 1 Z Z 0 0 a 1 b 2 S 21 1 + − 1 Z E Z b S = + b 2 S S S a 1 b 1 1 1 Z Z Z S 11 0 0 Γ S S 22 0 Γ L − Z Z Z = + = + Γ 0 S 0 S 12 a 2 a E b b b b 1 + + 1 S 1 S 1 S Z Z Z Z 0 S S 0 Γ b S S EE 611 Lecture 9 Jayanta Mukherjee
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