IIT Bombay Course Code : EE 611 Department: Electrical Engineering - PowerPoint PPT Presentation

Page 1 IIT Bombay Course Code : EE 611 Department: Electrical Engineering Instructor Name: Jayanta Mukherjee Email: jayanta@ee.iitb.ac.in Lecture 4 EE 611 Lecture 4 Jayanta Mukherjee

IIT Bombay Page 2 Impedance of Loaded Transmission Lines V + γ= j β+α V - Z 0 Z L Z(d) x 0 l d l 0 The impedance along a transmission line at position x is given by V ( x ) = where the complex voltage V(x) and current I(x) are : Z ( x ) , I ( x ) + - V V + γ − γ γx γx = + = − - x - x , V(x) V e V e I x e e ( ) Z Z 0 0 + = - The reference plane for V and V is located at x 0 EE 611 EE 611 EE 611 Lecture 4 Jayanta Mukherjee Lecture 1 Jayanta Mukherjee Lecture 1

IIT Bombay Page 3 Impedance Calculation = The impedance at the position x l is the load impedance Z L V l V ( ) = = = L Z ( l ) Z L I ( l ) I L Now from the voltage and current wave solutions we have + − γ − γ = = = + l l (2) V V ( l ) Z I V e V e L L L + − V V − γ γ = = − l l (3) I I ( l ) e e L Z Z 0 0 + - Solving for the incident wave amplitudes V and V we obtain ( ) 1 + γ = + l V Z Z I e L 0 L 2 ( ) 1 − γ − = − l V Z Z I e L 0 L 2 + - Substituti ng the incident wave V and V amplitudes in Eqn 2 and 3 = + γ V(x 0 ) Z Z tan( l ) − = = L 0 Z( l) Z = 0 + γ I(x 0 ) Z Z tan( l ) 0 L EE 611 EE 611 EE 611 Lecture 4 Jayanta Mukherjee Lecture 1 Jayanta Mukherjee Lecture 1

IIT Bombay Page 4 Lossless Case For a loss free line we have γ =j β and the impedance reduces to: + β Z jZ d tan( ) = L 0 Z ( d ) Z + β 0 Z jZ tan( d ) L 0 The impedance Z is then periodic function of frequency and position: • In terms of the electrical angle θ = β d the impedance Z repeats every period π • In terms of position d it repeats every half wavelength λ /2 since we have β d=(2 π / λ )d EE 611 EE 611 EE 611 Lecture 4 Jayanta Mukherjee Lecture 1 Jayanta Mukherjee Lecture 1

IIT Bombay Page 5 Impedance of a Shorted Transmission Line V + γ= j β V - Z 0 Z L Z(d) + β Z jZ d tan( ) = L 0 Z ( d ) Z + β 0 Z jZ tan( d ) L 0 x 0 l d l 0 • For a short circuited line, Z L =0 and we have Z(d)=jZ 0 tan( β d) = ∞ Z • For an open circuited line, and we have Z(d)=-jZ 0 cot( β d) L • For a matched load, Z L =Z 0 , and we have Z(d)=Z 0 for all values of d EE 611 EE 611 EE 611 Lecture 4 Jayanta Mukherjee Lecture 1 Jayanta Mukherjee Lecture 1

IIT Bombay Page 6 Impedance of a Shorted Transmission Line Im[Z(d)] Inductance λ/4 3/4λ 0 d λ/2 λ Capacitance • The input impedance alternates between shorts (Z=0) = ∞ Z and opens ( ) • The short is transformed into an open for d= λ /4 EE 611 EE 611 EE 611 Lecture 3 Jayanta Mukherjee Lecture 1 Jayanta Mukherjee Lecture 1

IIT Bombay Page 7 Matched Line Z 0 Z 0 Z L • When Z L =Z 0 , the load is said to be matched EE 611 EE 611 EE 611 Lecture 4 Jayanta Mukherjee Lecture 1 Jayanta Mukherjee Lecture 1

IIT Bombay Page 8 Impedance Matching vs Conjugate Impedance Matching Z 0 Z 0 Z G Z 0 Z L • Impedance matching should be distinguished from conjugate * used for maximum power transfer impedance matching Z G =Z L • Both load matching and conjugate impedance matching can * = Z 0 happen when Z G = Z L EE 611 EE 611 EE 611 Lecture 4 Jayanta Mukherjee Lecture 1 Jayanta Mukherjee Lecture 1

IIT Bombay Page 9 Quarter Wave Transformer for a Resistive Load R L •For a line of one quarter of a wavelength (d= λ /2) we have π λ π 2 β = = π = ∞ and since tan( /2) the line impedance d λ 4 2 is + β 2 R jZ tan( d) Z = = λ = = L 0 0 Z Z(d /4) Z in 0 + β Z jR tan( d) R 0 L L 2 Z = = 0 The line input is then matched to the generator : Z R in G R L = if we use Z R R 0 G L EE 611 EE 611 EE 611 Lecture 4 Jayanta Mukherjee Lecture 1 Jayanta Mukherjee Lecture 1

IIT Bombay Page 10 Quarter Wave Transformer for a Resistive Load R L R G 1/2 R L Ζ 0 = (R G R L ) β π/2 d= λ/4 d= The impedance matching is only realized at the frequency where the transmission line length is quarter wavelength EE 611 EE 611 EE 611 Lecture 4 Jayanta Mukherjee Lecture 1 Jayanta Mukherjee Lecture 1

IIT Bombay Page 11 Reflection Coefficient Γ L Γ( x) V + (x) load V - (x) 0 x=-d d We define the reflection coefficien t at a position x as the ratio of the reflected wave to the incident wave : − − − − γ x V e V V V γ γ − − γ − γ Γ = = = = = Γ x l d d d 0 2 0 2 ( ) L 2 2 (x) e e e e + − γ + L x + + V e V V V L 0 0 Γ = Γ = where (x l) L EE 611 EE 611 EE 611 Lecture 4 Jayanta Mukherjee Lecture 1 Jayanta Mukherjee Lecture 1

IIT Bombay Page 12 Reflection Coefficient Along a Line For a loss less line the reflection coefficien t can be written as : β α − β α Γ = Γ = Γ Γ = Γ - 2j d j d j ( 2 ) (d) e if we define e e L L L L Im[ Γ] Toward the load Γ L | e j α Γ L | = Toward the generator α Re[ Γ] −2β d Γ( d) for d>0 Γ plane EE 611 EE 611 EE 611 Lecture 4 Jayanta Mukherjee Lecture 1 Jayanta Mukherjee Lecture 1

IIT Bombay Page 14 Reflection Coefficient Along a Line • As we move along the line, Γ (d) moves along a circle of radius | Γ L | • The reflection coefficient Γ (d) rotates clock wise as d increases and we move towards the generator EE 611 EE 611 EE 611 Lecture 4 Jayanta Mukherjee Lecture 1 Jayanta Mukherjee Lecture 1

IIT Bombay Page 15 Power Dissipated at the Load Power dissipated by the load : Γ { } { } L 1 = = * * P VI VI Re Re Z L L 2 rms amplitude   ( )  + −    V V   + − = + −   V V Re     γ= j β Z Z   Z 0   0 0 2 2 + − V V = − Z Z 0 0 2 2 + − V V = − Z Z 0 0 + − + 2 = − = − Γ (incident minus reflected power) P P P ( 1 ) L EE 611 EE 611 EE 611 Lecture 4 Jayanta Mukherjee Lecture 1 Jayanta Mukherjee Lecture 1

IIT Bombay Page 16 Relation between impedance and Reflection Coefficient Z(x) Z L Γ( x) Γ L V + (x) load γ= j β Z 0 V - (x) 0 x=-d d + − γ − γ + + Γ x x V ( x ) V e V e 1 ( x ) = = = Z ( x ) Z + − 0 − Γ I ( x ) 1 ( x ) V V − γ γ − x x e e Z Z 0 0 Inverting : − Z(x) - Z Z Z Γ = Γ = 0 L 0 (x) and particular ly at the load : L + + Z(x) Z Z Z 0 L 0 = Γ = Note : for a matched load Z Z and we have 0 (no reflection ) L 0 L EE 611 EE 611 EE 611 Lecture 4 Jayanta Mukherjee Lecture 1 Jayanta Mukherjee Lecture 1

IIT Bombay Page 17 The Smith Chart Bilateral Transform connecting the impedance Z and the Reflection coefficient Γ . The smith chart maps the x-plane on the Γ plane Z − 1 − − Z Z Z Z z 1 Γ = = = = = + 0 0 with z r jx + + Z Z Z Z z 1 + 0 0 1 Z 0 x j 1 0.5 1 0.5 -1 0 1 0 1 0 Open 0.5 r 0 0.5 1 Short 0 -0.5 -0.5 -1 -j -1 EE 611 EE 611 EE 611 Lecture 4 Jayanta Mukherjee Lecture 1 Jayanta Mukherjee Lecture 1

IIT Bombay Page 18 Extended Smith Chart For negative resistance r<0 we have | Γ |>1 Z − 1 − Z Z Z Γ = = 0 0 + Z Z Z + 0 1 x Z j -0.5 0 1 − z 1 0.5 1 = 0.5 + z -1 0 1 1 0 1 0 0.5 Open r -0.5 0 0.5 1 Short Z 0 = = + -0.5 with z -0.5 r jx -1 Z 0 -j -1 For r=-1 (Re{Z}=-50 ohms) we have Γ =infinity EE 611 EE 611 EE 611 Lecture 4 Jayanta Mukherjee Lecture 1 Jayanta Mukherjee Lecture 1

IIT Bombay Page 19 Active and Passive Load Im[ Γ L ] | Γ L | >1 Active Devices Re[ Γ L ] | Γ L | <1 Passive Devices EE 611 EE 611 EE 611 Lecture 4 Jayanta Mukherjee Lecture 1 Jayanta Mukherjee Lecture 1

IIT Bombay Page 20 Inductance and Capacitance on Smith Chart Locus of the reflection coefficient for an inductor and a capacitor In a Z Smith chart ω= infinity ω=0 ω=0 Short Short Open Open ω= infinity Γ plane Z smith chart EE 611 EE 611 EE 611 Lecture 4 Jayanta Mukherjee Lecture 1 Jayanta Mukherjee Lecture 1

IIT Bombay Page 21 Y Smith Chart Γ The Y - smith chart can be obtained by expressing in terms of Y : Z Y 1 − − − L L 1 1 1 − − Z z y Y 1 y 1 = = = Γ = − = − 0 0 L + + Z 1 Y z 1 y 1 + + + L L 1 1 1 Z y Y 0 0 Open Short Open Short Γ plane - Γ plane Normal Rotated EE 611 EE 611 EE 611 Lecture 4 Jayanta Mukherjee Lecture 1 Jayanta Mukherjee Lecture 1

IIT Bombay Page 22 Y Smith Chart • The Y Smith Chart is obtained by inverting the Z smith chart • In the rotated Y-Smith Chart the short and open are exchanged EE 611 EE 611 EE 611 Lecture 4 Jayanta Mukherjee Lecture 1 Jayanta Mukherjee Lecture 1