Page 1 IIT Bombay Course Code : EE 611 Department: Electrical Engineering Instructor Name: Jayanta Mukherjee Email: jayanta@ee.iitb.ac.in Lecture 8 EE 611 Lecture 8 Jayanta Mukherjee
Page 2 IIT Bombay Topics Covered • Impedance Matrix • Properties of Loss Less and Reciprocal Devices • Scattering Parameters EE 611 Lecture 8 Jayanta Mukherjee
IIT Bombay Page 3 Impedance Matrix N Port Device Z N I N Z 1 Z 2 + - V N I 2 I 1 + - - V 1 + V 2 V Z Z Z I N 1 11 12 1 1 V Z Z Z I 2 21 22 2 N 2 = = = V Z ZI 3 N Z Z Z − − − ( N 1 ) 1 ( N 1 ) 2 ( N 1 ) N V Z Z Z I N N 1 N 2 NN N EE 611 Lecture 8 Jayanta Mukherjee
IIT Bombay Page 4 Admittance Matrix N Port Device Z N I N Z 1 Z 2 + - V N I 2 I 1 + - - V 1 + V 2 I Y Y Y V N 1 11 12 1 1 I Y Y Y V N 2 21 22 2 2 = = = I Y YV N 3 Y Y Y − − − N N N N ( 1 ) 1 ( 1 ) 2 ( 1 ) I Y Y Y V N N N NN N 1 2 = - 1 Property Y Z EE 611 Lecture 8 Jayanta Mukherjee
IIT Bombay Page 5 Reciprocal and Non Reciprocal Materials • For non-reciprocal medium - semiconductor + power supply - plasma - Ferrite + magnetic field The dielectric constant ε and permeability μ are non- reciprocal matrix and we have Z ij ≠ Z ji • For reciprocal medium we have Z ij = Z ji and [Z] t = [Z] EE 611 Lecture 8 Jayanta Mukherjee
IIT Bombay Page 6 Reciprocal 2 Port network • Circuit representation of a 2 port reciprocal network Z 11 -Z 12 Z 22 -Z 12 Z Z 11 12 Z 12 Z Z 12 22 EE 611 Lecture 8 Jayanta Mukherjee
IIT Bombay Page 7 Lossless Networks • If a network is lossless (i.e. no lossy lines, no lumped resistors, no radiation loss etc.) then • The total complex power P tot dissipated by the N-port device is given by N 1 1 1 1 ∑ = + + ⋅ ⋅ ⋅ + = * * * * P V I V I V I V I tot 1 1 2 2 N N n n 2 2 2 2 = n 1 • For lossless devices Re {P tot }=0 • Plugging in P tot in the impedance matrix for V n gives N N 1 ∑ ∑ = * P Z I I tot nm m n 2 = = 1 1 n m • The real part of the above has to be zero regardless of the values of I n which excite the network EE 611 Lecture 8 Jayanta Mukherjee
IIT Bombay Page 8 Properties of a Lossless reciprocal impedance matrix • Consider that only one of the port currents is not zero. Say that port is called n’ ; our previous equation is then { } { } 1 1 = 2 = * Re Z I I I Re Z 0 n ' n ' n ' n ' n ' n ' n ' 2 2 so the diagonal entries of the matrix must be purely imaginary • Now consider that only two currents are non zero. Call these I n’ and I n” . We then get EE 611 Lecture 8 Jayanta Mukherjee
IIT Bombay Page 9 Properties of a Lossless reciprocal impedance matrix { } 1 + + + = * * * * Re 0 Z I I Z I I Z I I Z I I n ' n ' n ' n ' n " n " n " n " n ' n " n ' n " n " n ' n " n ' 2 { } 1 ⇒ + = * * Re Z I I Z I I 0 n ' n " n ' n " n " n ' n " n ' 2 { ( ) } 1 ⇒ + = * * Re Z I I I I 0 n ' n " n ' n " n " n ' 2 showing again that the Z must be purely imaginary nm since the term in parenthesi s is purely real = Therefore for lossless devices we have Re{Z} 0 EE 611 Lecture 8 Jayanta Mukherjee
IIT Bombay Page 10 Summary of Impedance matrix properties • If a network is reciprocal, the Z and Y matrices are symmetric • If a network is lossless, the diagonal impedance or admittance matrix elements are purely imaginary • If a network is lossless and reciprocal, all impedance or admittance matrix elements are purely imaginary EE 611 Lecture 8 Jayanta Mukherjee
IIT Bombay Page 11 Scattering Parameters • S-parameters can be measured with a network analyzer • They have a natural relation with the flow of power • S-parameters are readily represented by flow-graph • The measurement of S-parameters relies on 50 ohm resistive terminations: usually active devices do not oscillate for such terminations • Devices are measured in the medium in which they are used EE 611 Lecture 8 Jayanta Mukherjee
IIT Bombay Page 12 Normalized waves, voltages and currents N Port Device Z N I N Z 1 Z 2 + - V N I 2 I 1 + - - V 1 + V 2 EE 611 Lecture 8 Jayanta Mukherjee
IIT Bombay Page 13 Normalized waves, voltages and currents Normalized incident a and reflected waves b n n + − V V = = n n a b n n Z Z n n Normalized voltage : ( ) V + − = + = + ⇒ = = + n V V V Z a b V a b n n n n n n n n n Z n Normalized current : + − ( ) V V 1 I = − = − = ⇒ = = − n n n I a b I Z I a b n n n n n n n n Z Z Z Z n n n n Normalizat ion is needed to handle different characteri stic impedance Z n EE 611 Lecture 8 Jayanta Mukherjee
IIT Bombay Page 14 Normalized Z matrix 1 1 0 0 0 0 Z Z 1 1 1 1 0 0 0 0 = Z Z Z Z 2 2 0 0 0 0 1 1 0 0 Z Z 1 1 or equivalent ly Z = ij Z ij Z Z i j Property : If a N port network is reciprocal both Z znd Z are symmetrica l EE 611 Lecture 8 Jayanta Mukherjee
IIT Bombay Page 15 Definition of Scattering Parameters b S S S a 1 11 12 1 N 1 b S S S a 2 21 22 2 N 2 = = = S b Sa N 3 S ( ) S ( ) S ( ) − − − N 1 1 N 1 2 N 1 N b S S S a N N 1 N 2 NN N Therefore to measure S all ports must be matched except port j ij b = i S ij a j = ≠ a 0 for k j k Alternativ ely we can write : − V Z i j = S ij + V Z j i + = ≠ V for k j 0 k EE 611 Lecture 8 Jayanta Mukherjee
IIT Bombay Page 16 2 port scattering parameters For a linear device (small - signal) we have a linear relationsh ip between and . a's b's = + b S a S a 1 11 1 12 2 = + b S a S a 2 21 1 22 2 In matrix form b S S a a = = 1 11 12 1 1 S b S S a a 2 21 22 2 2 b = 1 S Input reflection coefficien t with output matched 11 a = 1 a 0 2 b = 2 S Forward transmiss ion coefficien t with output matched 21 a = 1 a 0 2 b = 2 S Output reflection coefficien t with input matched 22 a = 2 a 0 1 b = 1 S Reverse transmiss ion coefficien t with input matched 12 a = 2 0 a 1 EE 611 Lecture 8 Jayanta Mukherjee
IIT Bombay Page 17 S Matrix properties • Reciprocal devices have symmetric impedance matrices, and that lossless reciprocal devices have purely imaginary impedance matrices • Let’s find similar identities for S matrix • Our normalized impedance matrix equation is [V] = [Z] [I] • Now since [V] = [a] + [b] and since [I] = [a] - [b] we get [U] ([a] + [b]) = [Z] ([a] - [b]) Where [U] is the identity matrix EE 611 Lecture 8 Jayanta Mukherjee
IIT Bombay Page 18 S Matrix properties (2) • Re-write as { [Z] + [U] } [b] = { [Z] – [U] } [a] • Now since we define our S-matrix through [b] = [S] [a], we find [S] = { [Z] + [U] } -1 { [Z] - [U] } (1) which is the relation between the normalized Z and S matrices • We can also get another similar relationship using [a] = (1/2) { [V] + [I] } = (1/2) { [Z] + [U] } [I] [b] = (1/2) { [V] – [I] } = (1/2) { [Z] – [U] } [I] EE 611 Lecture 8 Jayanta Mukherjee
IIT Bombay Page 19 S Matrix properties (3) • Solving the above equations for [b] in terms of [a], we find [b] = { [Z] - [U] } { [Z] + [U] } -1 [a] So another formula for the S matrix is [S] = { [Z] – [U] } { [Z] + [U] } -1 (2) • Taking the transpose of (1) we get [S] t = { [Z] – [U] } t { [Z] + [U] } -1 t but [U] and [Z] are symmetric if the medium is reciprocal so [S] t = { [Z] – [U] } { [Z] + [U] } -1 EE 611 Lecture 8 Jayanta Mukherjee
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