Lecture 2 Transmission lines 1.Transmission line parameters, equations 2.Wave propagations 3.Lossless line, standing wave and reflection coefficient 4.Input impedence 5.Special cases of lossless line 6.Power flow 7.Smith chart 8.Impedence matching 9.Transients on transmission lines

Lecture 2 1.Transmission line parameters, equations Vg(t) V BB’ (t) V AA’ (t) A A’ B’ B L V AA’ (t) = Vg(t) = V0cos(  t), V BB’ (t) = V AA’ (t-t d ) = V AA’ (t-L/c) = V0cos(  (t-L/c)), V BB’ (t) = V AA’ (t) Low frequency circuits: Approximate result V BB’ (t) = V AA’ (t)

Lecture 2 1.Transmission line parameters, equations Vg(t) V BB’ (t) V AA’ (t) A A’ B’ B L V BB’ (t) = V AA’ (t-t d ) = V AA’ (t-L/c) = V0cos(  (t-L/c)) = V0cos(  t- 2  L/ ), Recall:  =c, and  = 2  If >>L, V BB’ (t)  V0cos(  t) = V AA’ (t), If <= L, V BB’ (t)  V AA’ (t), the circuit theory has to be replaced.

Lecture 2 1.Transmission line parameters, equations Vg(t) V BB’ (t) V AA’ (t) A A’ B’ B L  = 2  f  t = 0.06  e. g:  = 1GHz, L = 1cm Time delay  t = L/c = 1cm /3x10 10 cm/s = 30 ps Phase shift V BB’ (t) = V AA’ (t)  = 2  f  t = 0.6   = 10GHz, L = 1cm Time delay  t = L/c = 1cm /3x10 10 cm/s = 30 ps Phase shift V BB’ (t)  V AA’ (t)

Lecture 2 Transmission line parameters Vg(t) V BB’ (t) V AA’ (t) A A’ B’ B L time delay V BB’ (t) = V AA’ (t-t d ) = V AA’ (t-L/v p ), Reflection: the voltage has to be treat as wave, some bounce back power loss: due to reflection and some other loss mechanism, Dispersion: in material, V p could be different for different wavelength

Lecture 2 Types of transmission lines Transverse electromagnetic (TEM) transmission lines B E E B a) Coaxial lineb) Two-wire linec) Parallel-plate line d) Strip linee) Microstrip line

Lecture 2 Types of transmission lines Higher-order transmission lines a) Optical fiber b) Rectangular waveguidec) Coplanar waveguide

Lecture 2 Lumped-element Model Represent transmission lines as parallel-wire configuration Vg(t) V BB’ (t) V AA’ (t) A A’ B’ B zz zz zz Vg(t) R’  z L’  z G’  z C’  z R’  z L’  z G’  z R’  z C’  z L’  z G’  z C’  z

Lecture 2 Transmission line equations Represent transmission lines as parallel-wire configuration V(z,t) R’  z L’  z G’  z C’  z V(z+  z,t) V(z,t) = R’  z i (z,t) + L’  z  i (z,t)/  t + V(z+  z,t), (1) i (z,t) i (z+  z,t) i (z,t) = G’  z V(z+  z,t) + C’  z  V(z+  z,t)/  t + i (z+  z,t), (2)

Lecture 2 Transmission line equations V(z,t) = R’  z i (z,t) + L’  z  i (z,t)/  t + V(z+  z,t), (1) V(z,t) R’  z L’  z G’  z C’  z V(z+  z,t) i (z,t) i (z+  z,t) -V(z+  z,t) + V(z,t) = R’  z i (z,t) + L’  z  i (z,t)/  t -  V(z,t)/  z = R’ i (z,t) + L’  i (z,t)/  t, (3) Rewrite V(z,t) and i (z,t) as phasors, for sinusoidal V(z,t) and i (z,t) : V(z,t) = Re( V(z) e jtjt ), i (z,t) = Re( i (z) e jtjt ),

Lecture 2 Transmission line equations V(z,t) R’  z L’  z G’  z C’  z V(z+  z,t) i (z,t) i (z+  z,t) Recall: di(t)/dt = Re(d i e jtjt )/dt ),= Re(i jtjt e jj -  V(z,t)/  z = R’ i (z,t) + L’  i (z,t)/  t, (3) - d V(z)/ dz = R’ i (z) + j  L’ i (z), (4)

Lecture 2 Transmission line equations Represent transmission lines as parallel-wire configuration V(z,t) R’  z L’  z G’  z C’  z V(z+  z,t) V(z,t) = R’  z i (z,t) + L’  z  i (z,t)/  t + V(z+  z,t), (1) i (z,t) i (z+  z,t) i (z,t) = G’  z V(z+  z,t) + C’  z  V(z+  z,t)/  t + i (z+  z,t), (2)

Lecture 4 Transmission line equations V(z,t) R’  z L’  z G’  z C’  z V(z+  z,t) i (z,t) i (z+  z,t) - i (z+  z,t) + i (z,t) = G’  z V (z +  z,t) + C’  z  V(z +  z,t)/  t -  i(z,t)/  z = G’ V (z,t) + C’  V (z,t)/  t, (5) Rewrite V(z,t) and i (z,t) as phasors, for sinusoidal V(z,t) and i (z,t) : V(z,t) = Re( V(z) e jtjt ), i (z,t) = Re( i (z) e jtjt ), i (z,t) = G’  z V(z+  z,t) + C’  z  V(z+  z,t)/  t + i (z+  z,t), (2)

Lecture 2 Transmission line equations V(z,t) R’  z L’  z G’  z C’  z V(z+  z,t) i (z,t) i (z+  z,t) Recall: dV(t)/dt = Re(d V e jtjt )/dt ),= Re(V jtjt e jj -  i(z,t)/  z = G’ V (z,t) + C’  V (z,t)/  t, (6) - d i(z)/ dz = G’ V (z) + j  C’ V (z), (7)

Lecture 2 V(z,t) R’  z L’  z G’  z C’  z V(z+  z,t) i (z,t) i (z+  z,t) - d i(z)/ dz = G’ V(z) + j  C’ V (z), (7) - d V(z)/ dz = R’ i (z) + j  L’ i (z), (4) Telegrapher’s equation in phasor domain Take d /dz on both sides of eq. (4) - d² V(z)/ dz² = R’ d i (z)/dz + j  L’ d i (z)/dz, (8)

Lecture 2 - d i(z)/ dz = G’ V(z) + j  C’ V (z), (7) - d V(z)/ dz = R’ i (z) + j  L’ i (z), (4) Telegrapher’s equation in phasor domain substitute (7) to (8) d² V(z)/ dz² = ( R’ + j  L’) (G’+ j  C’)V(z), - d² V(z)/ dz² = R’ d i (z)/dz + j  L’ d i (z)/dz, (8) d² V(z)/ dz² - ( R’ + j  L’) (G’+ j  C’)V(z) = 0, (9) or d² V(z)/ dz² -  ² V(z) = 0, (10)  ² = ( R’ + j  L’) (G’+ j  C’), (11)

Lecture 2 V(z,t) R’  z L’  z G’  z C’  z V(z+  z,t) i (z,t) i (z+  z,t) - d i(z)/ dz = G’ V(z) + j  C’ V (z), (7) - d V(z)/ dz = R’ i (z) + j  L’ i (z), (4) Telegrapher’s equation in phasor domain Take d /dz on both sides of eq. (7) - d² i(z)/ dz² = G’ d V (z)/dz + j  C’ d V (z)/dz, (12)

Lecture 2 - d i(z)/ dz = G’ V(z) + j  C’ V (z), (7) - d V(z)/ dz = R’ i (z) + j  L’ i (z), (4) Telegrapher’s equation in phasor domain substitute (4) to (12) d² i(z)/ dz² = ( R’ + j  L’) (G’+ j  C’)i(z), d² i(z)/ dz² - ( R’ + j  L’) (G’+ j  C’) i(z) = 0, (9) or d² i(z)/ dz² -  ² i(z) = 0, (13)  ² = ( R’ + j  L’) (G’+ j  C’), (11) - d² i(z)/ dz² = G’ d V (z)/dz + j  C’ d V (z)/dz, (12)

Lecture 2 Wave equations d² i(z)/ dz² -  ² i(z) = 0, (13) d² V(z)/ dz² -  ² V(z) = 0, (10)  =  + j ,  = Re ( R’ + j  L’) (G’+ j  C’),  = Im ( R’ + j  L’) (G’+ j  C’),

Lecture 2 Wave equations d² i(z)/ dz² -  ² i(z) = 0, (13) d² V(z)/ dz² -  ² V(z) = 0, (10)  =  + j ,  = Re ( R’ + j  L’) (G’+ j  C’),  = Im ( R’ + j  L’) (G’+ j  C’), V(z) = V 0 (14) Solving the second order differential equation + e -z-z + - V0V0 e zz i(z) = I 0 (15) + e -z-z + - I0I0 e zz

Lecture 2 Wave equations V(z) = V 0 (14) + e -z-z + - V0V0 e zz i(z) = I 0 (15) + e -z-z + - I0I0 e zz where: + V0V0 - V0V0 andare determined by boundary conditions. + I0I0 - I0I0 andare related to - V0V0 + V0V0 andby characteristic impedance Z 0.

Lecture 2 recall: - d V(z)/ dz = R’ i (z) + j  L’ i (z), (4) V(z) = V 0 (14) V0V0 e zz i(z) = I 0 (15) + e -z-z + - I0I0 e zz e -z-z e -z-z V0V0 + e zz V0V0 - - = ( R’ + j  L’) i (z), (16) i (z) = ( R’ + j  L’)  e -z-z (V0(V0 + e zz V0V0 - - ) I0I0 + =  V0V0 + I0I0 - = -- V0V0 - (17) (18) Characteristic impedance Z 0

Lecture 5 Characteristic impedance Z 0 I0I0 + = ( R’ + j  L’)  V0V0 + I0I0 - = -- V0V0 - (17) (18) = ( R’ + j  L’)  + Z 0  Define characteristic impedance Z 0 I0I0 + V0V0 =  = ( R’ + j  L’) (G’+ j  C’) ( R’ + j  L’) (G’+j  C’) recall:

Lecture 5 Summary:  = ( R’ + j  L’) (G’+ j  C’) + Z 0  I0I0 + V0V0 = ( R’ + j  L’) (G’+j  C’) V(z) = V 0 (14) V0V0 e zz i(z) = I 0 (15) + e -z-z + - I0I0 e zz e -z-z (19) (20)

Lecture 5 Example, an air line : solution: R’ = 0 , G’ = 0 / , Z 0 = 50 ,  = 20 rad/m, f = 700 MHz L’ = ? and C’ = ? Z0Z0 = ( R’ + j  L’) (G’+j  C’) = j  L’ j  C’ = 50   = ( R’ + j  L’) (G’+ j  C’) = jj L’C’  =  + j ,  =  L’C’ = 20 rad/m

Lecture 5 lossless transmission line : = ( R’ + j  L’) (G’+ j  C’)  =  + j , If R’<< j  L’ and G’ << j  C’,  = ( R’ + j  L’ ) (G’+ j  C’) = jj L’C’  = 0  =  L’C’ lossless line

Lecture 2 lossless transmission line :  = 0  =  L’C’ lossless line Z0Z0 = ( R’ + j  L’) (G’+j  C’) = j  L’ j  C’ Z0Z0 = L’ C’  = 2  / = 2  /  =  L’C’ 1 Vp =  /  = L’C’ 1

Lecture 2 For TEM transmission line : Vp = L’C’ 1 L’C’ =  =  1  =  L’C’ =   = rrrr c Z0Z0 = L’ C’ summary : V(z) = V V0V0 e jzjz i(z) = I 0 + e -j  z + - I0I0 e jzjz e = rrrr c Vp = L’C’ 1  =  =  

Lecture 5 Voltage reflection coefficient : Vg(t) VLVL A z = 0 B l ZLZL z = - l Z0Z0 ViVi i(z) = V(z) = V V0V0 e jzjz - e -j  z e + V0V0 Z0Z0 e jzjz - V0V0 Z0Z0 V L = + = - V0V0 + V0V0 V(z) z = V0V0 Z0Z0 - V0V0 Z0Z0 i(z) z = 0 i L = = ZLZL = VLVL iLiL = + - V0V0 + V0V0 - + V0V0 Z0Z0 - V0V0 Z0Z0 + V0V0 - V0V0 = ZLZL Z0Z0 - ZLZL Z0Z0 +

Lecture 5 Voltage reflection coefficient : - V0V0 + V0V0 = ZLZL Z0Z0 - ZLZL Z0Z0 +    Current reflection coefficient : - i0i0 + i0i0 = -  i  - V0V0 + V0V0 = -  Notes : 1.|  |  1, how to prove it? 2.If Z L = Z 0,  = 0. Impedance match, no reflection from the load Z L.

Lecture 2 An example : A’ z = 0 A Z 0 = 100  R L = 50  C L = 10pF f = 100MHz Z L = R L + j/  C L = 50 –j159 = ZLZL Z0Z0 - ZLZL Z0Z0 + 

Lecture 2 Standing wave Input impedance i(z) = V(z) = V V0V0 e jzjz - e -j  z e + V0V0 Z0Z0 e jzjz - V0V0 Z0Z0 - V0V0 + V0V0 with  = i(z) = V(z) = V 0 ( ) + +  e jzjz - e -j  z (e(e + V0V0 Z0Z0 e jzjz  ) |V(z)| = | V 0 | | | + e -j  z |||| e jzjz + e jrjr = | V 0 | [ 1+ |  |² + 2|  |cos(2  z +  r )] + 1/2

Lecture 2 Standing wave i(z) = V(z) = V V0V0 e jzjz - e -j  z e + V0V0 Z0Z0 e jzjz - V0V0 Z0Z0 - V0V0 + V0V0 with  = i(z) = V(z) = V 0 ( ) + +  e jzjz - e -j  z (e(e + V0V0 Z0Z0 e jzjz  ) |V(z)| = | V 0 | | | + e -j  z |||| e jzjz + e jrjr = | V 0 | [ 1+ |  |² + 2|  |cos(2  z +  r )] + 1/2

Lecture 2 Standing wave i(z) = V(z) = V 0 ( )  e jzjz e -j  z (e(e + V0V0 Z0Z0 e jzjz  ) |i(z)| = | V 0 | /|Z0|| | + e -j  z |||| e jzjz - e jrjr = | V 0 |/|Z 0 | [ 1+ |  |² - 2|  |cos(2  z +  r )] + 1/2 = | V 0 | [ 1+ |  |² + 2|  |cos(2  z +  r )] + 1/2 |V(z)|

Lecture 2 Special cases = | V 0 | [ 1+ |  |² + 2|  |cos(2  z +  r )] + 1/2 |V(z)| 1.Z L = Z 0,  = 0 = | V 0 | + |V(z)| 2. Z L = 0, short circuit,  = -1 |V(z)| |V0| /4- /2- /4 = | V 0 | [ 2 + 2cos(2  z +  )] + 1/2 |V(z)| 2|V0| /4- /2- /4

Lecture 2 Special cases = | V 0 | [ 1+ |  |² + 2|  |cos(2  z +  r )] + 1/2 |V(z)| 3. Z L = , open circuit,  = 1 = | V 0 | [ 2 + 2cos(2  z )] + 1/2 |V(z)| 2|V0| /4- /2- /4

Lecture 2 Voltage maximum = | V 0 | [ 1+ |  |² + 2|  |cos(2  z +  r )] + 1/2 |V(z)| | V 0 | [ 1+ |  |], + |V(z)| max = when 2  z +  r = 2n . –z =  r /4  + n /2 n = 1, 2, 3, …, if  r <0 n = 0, 1, 2, 3, …, if  r >= 0

Lecture 2 Voltage minimum = | V 0 | [ 1+ |  |² + 2|  |cos(2  z +  r )] + 1/2 |V(z)| | V 0 | [ 1 - |  |], + |V(z)| min = when 2  z +  r = (2n+1) . –z =  r /4  + n /2 + /4 Note: voltage minimums occur /4 away from voltage maximum, because of the 2  z, the special frequency doubled.

Lecture 2 Voltage standing-wave ratio VSWR or SWR 1 - |  | |V(z)| min S  |V(z)| max = 1 + |  | S = 1, when  = 0, S = , when |  | = 1,

Lecture 2 An example Vg(t) VLVL A z = 0 B l ZLZL z = - l Z0Z0 ViVi Voltage probe S = 3, Z 0 = 50 ,  l min = 30cm, l min = 12cm, Z L =?  l min = 30cm,  = 0.6m, S = 3,  |  | = 0.5, Solution: -2  l min +  r = - ,   r = -36º,  , and Z L.

Lecture 2 Input impudence Vg(t) VLVL A z = 0 B l ZLZL z = - l Z0Z0 ViVi Zg IiIi Z in (z) = V(z) I(z) = + +  e jzjz e -j  z V0V0 ( ) + -  e jzjz e V0V0 ( ) Z0Z0 = + (1  e j2  z ) - (1  e j2  z ) Z0Z0 + (1  e -j2  l ) - (1  e -j2  l ) Z0Z0 Z in (-l) =

Lecture 2 An example A 1.05-GHz generator circuit with series impedance Zg = 10-  and voltage source given by Vg(t) = 10 sin(  t +30º) is connected to a load ZL = 100 +j5-  through a 50- , 67-cm long lossless transmission line. The phase velocity is 0.7c. Find V(z,t) and i(z,t) on the line. Solution: Since, Vp = ƒ, = Vp/f = 0.7c/1.05GHz = 0.2m.  = 2  /,  = 10 .  = (Z L -Z 0 )/(Z L +Z 0 ),  = 0.45exp(j26.6º) + (1  e -j2  l ) - (1  e -j2  l ) Z0Z0 Z in (-l) = = j17.4  V 0 [ exp(-j  l)+  exp(j  l) ] + = Z in (-l) + Zg Z in (-l) Vg

Lecture 2 short circuit line Z L = 0,  = -1, S =  Vg(t) VLVL A z = 0 B l Z L = 0 z = - l Z0Z0 Zg IiIi Z in sc i(z) = V(z) = V 0 ) - e jzjz + (e(e -j  z (e(e + V0V0 Z0Z0 e jzjz ) = -2jV 0 sin(  z) + = 2V 0 cos(  z)/Z 0 + Z in = V(-l) i(-l) = jZ 0 tan(  l)

Lecture 2 short circuit line Z in = V(-l) i(-l) = jZ 0 tan(  l) If tan(  l) >= 0, the line appears inductive,j  L eq = jZ 0 tan(  l), If tan(  l) <= 0, the line appears capacitive,1/j  C eq = jZ 0 tan(  l), l = 1/  [  - tan (1/  C eq Z 0 )], The minimum length results in transmission line as a capacitor:

Lecture 2 An example: tan (  l) = - 1/  C eq Z 0 = , Choose the length of a shorted 50-  lossless line such that its input impedance at 2.25 GHz is equivalent to the reactance of a capacitor with capacitance Ceq = 4pF. The wave phase velocity on the line is 0.75c. Solution: Vp = ƒ,   = 2  / = 2  ƒ/Vp = 62.8 (rad/m)  l = tan (-0.354) + n , = n ,

Lecture 2 open circuit line Z L = 0,  = 1, S =  Vg(t) VLVL A z = 0 B l Z L =  z = - l Z0Z0 Zg IiIi Z in oc i(z) = V(z) = V 0 ) + e jzjz - (e(e -j  z (e(e + V0V0 Z0Z0 e jzjz ) = 2V 0 cos(  z) + = 2jV 0 sin(  z)/Z 0 + Z in = V(-l) i(-l) = -jZ 0 cot(  l) oc

Lecture 2 Application for short-circuit and open-circuit Network analyzer Measure Z in oc Z in sc and Calculate Z 0 Measure S paremeters Z in = -jZ 0 cot(  l) oc Z in = jZ 0tan (  l) sc =Z0Z0 Z in sc Z in oc Calculate  l = -jZ0Z0 Z in sc Z in oc

Lecture 2 Line of length l = n /2 = Z L tan(  l) = tan((2  / )(n /2)) = 0, + (1  e -j2  l ) - (1  e -j2  l ) Z0Z0 Z in (-l) = Any multiple of half-wavelength line doesn’t modify the load impedance.

Lecture 2 Quarter-wave transformer l = /4 + n /2 = Z 0 ²/Z L  l = (2  / )( /4 + n /2) =  /2, + (1  e -j2  l ) - (1  e -j2  l ) Z0Z0 Z in (-l) = + (1  e -j  ) - (1  e -j  ) Z0Z0 = (1 +  ) Z0Z0 (1 -  ) =

Lecture 2 An example: A 50-  lossless tarnsmission is to be matched to a resistive load impedance with Z L = 100  via a quarter-wave section, thereby eliminating reflections along the feed line. Find the characteristic impedance of the quarter-wave tarnsformer. Z L = 100  /4 Z 01 = 50  Z in = Z 0 ²/Z L = 50  Z 0 = (Z in Z L ) = (50*100) ½ ½

Lecture 2 Matched transmission line: 1.Z L = Z 0 2.  = 0 3.All incident power is delivered to the load.

Lecture 8 Instantaneous power Time-average power i(z) = V(z) = V 0 ( ) + +  e jzjz - e -j  z (e(e + V0V0 Z0Z0 e jzjz  ) At load z = 0, the incident and reflected voltages and currents: V = V 0 + i = i + V0V0 Z0Z0 i V = V 0 - r i = - V0V0 Z0Z0 r

Lecture 2 Instantaneous power + i P(t) = v(t) i(t) = Re[V exp(j  t)] Re[ i exp(j  t)] i i = Re[|V 0 |exp(j  )exp(j  t)] Re[|V 0 |/Z 0 exp(j  )exp(j  t)] ++ + = (|V 0 |²/Z 0 ) cos²(  t +  ) r P(t) = v(t) i(t) = Re[V exp(j  t)] Re[ i exp(j  t)] r r = Re[|V 0 |exp(j  )exp(j  t)] Re[|V 0 |/Z 0 exp(j  )exp(j  t)] +- + = - |  |²(|V 0 |²/Z 0 ) cos²(  t +  +  r ) + +

Lecture 2 Time-average (|V 0 |²/Z 0 ) cos²(  t +  )dt + + Time-domain approach: P av = i T 1 0 T P (t)dt i = 22  0 T = (|V 0 |²/2Z 0 ) + P av r = -|  |² (|V 0 |²/2Z 0 ) + = P av i P av + P av r = (1-|  |²) (|V 0 |²/2Z 0 ) + Net average power:

Lecture 2 Time-average Phasor-domain approach P av r = -|  |² (|V 0 |²/2Z 0 ) + = (1-|  |²) (|V 0 |²/2Z 0 ) + = (½)Re[V i*] P av P av = (1/2) Re[V 0 V 0 * /Z 0 ] i + + = (|V 0 |²/2Z 0 ) P av