8. Examples on Using the Smith Chart

W.C.ChewECE 350 Lecture Notes<strong>8.</strong> <strong>Examples</strong> on Using the Smith Chart(a) Find the voltages at A on the transmission line.20 ΩZ 0 = 50 W, v = 1.5 × 10 8 ms –1V s =10 sin ωtvoltsZ sABZ L(30+j25) Ωz = –l = –1 m25 MHzz = 0The voltage source sets up a forward going and a backward going waveon the transmission lines. Hence,The corresponding current isIn impedance at position Z isV (z) =V 0 e ;jz + v V 0 e jz : (1)I(z) = V 0Z 0e ;jz ; vV 0Z 0e jz : (2)Z(z) = V (z)I(z) = Z e ;jz + v e jz0e ;jz ; v e = Z 1+;(z)jz 01 ; ;(z) (3)where;(z) = v e 2jz v = Z L ; Z 0: (4)Z l + Z 0We can use the Smith Chart to nd Z(;l). To use the Smith Chart, we haveto normalize all the impedances with respect to the characteristic impedanceof the line. Hence,Z nL = Z LZ 0=30 + j2550=0:6+j0:5: (5)We can locate Z nL on the Smith Chart which is the complex ; plane. ;(0)or v can also be deduced from the Smith Chart. Since ;(z) is given by (4),at z = ;l, we have;(;l) = v e ;2jl : (6)1

At f = 25MHz, and with v = 1:5 10 8 ms ;1 , = v=f = 6m. Thenl = 2l = l. Therefore, 3;(;l) = v e ;j 2 3 l : (7)At z = ;l = ;1m ;(;1) = v e ;j 2 3 . From the Smith Chart, we can readZ n (;1)=2:15 ; j0:3 or Z(;1) = 107:5 ; j15 : (8)So, an equivalent circuit for the point A is:20 ΩAV s = 10 sin ωtZsZ(–1)(107.5–j15) ΩSinceIn phasor representation, V S =10e ;j 2Z(;1)V A = V SZ S + Z(;1)= ;j10. Hence,= ;j10107:5; j15127:5 ; j15 = 108:54e;j7:9128:38e ;j6:7 e;j90 10V=8:5e ;j91:2 V: (9)V A = V (;1) = V 0 e j [1 + ;(;1)] (10)we can nd V o from the above. Once V o is found, we can nd V B fromV B = V (0) = V o [1 + v ]: (11)j1j0.5j2j0.2z Lρ V20120 o Γ(-l)Z(-l)0.2 0.5 1 2−j0.2−j0.5−j2−j1

(b) Find Z L from VSWR and d min using a Smith ChartThe voltage on the transmission line isV (z) =V o (e ;jz + v e +jz )=V o e ;jz [1 + ;(z)]: (12)If V (z) =jV (z)je j(z) , the real time voltage can be written asV (z t) =

j1j0.5j2j0.2ρVZ=-d Rn=2.5=VSWRmin00.2 0.5 1 25λ16−j0.2−j0.5−j2−j14