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W.C.Chew<br />

ECE 350 Lecture Notes<br />

<strong>7.</strong> <strong>The</strong> <strong>Smith</strong> <strong>Chart</strong><br />

We have seen from Equation (6.9) that a generalized impedance can be<br />

dened as<br />

V<br />

Z(z) = ~ (z)<br />

~I(z) = Z e ;jz + v e +jz<br />

0<br />

e ;jz ; v e<br />

+jz:<br />

(1)<br />

<strong>The</strong> above can be written as<br />

1+ v e 2jz<br />

Z(z) =Z 0<br />

1 ; v e = Z 1+;(z)<br />

2jz 0<br />

1 ; ;(z) (2)<br />

where ;(z) is as dened in (6.16). When z = 0, Z(0) = Z L , and ;(0) = v ,<br />

and (2) becomes (6.25). Hence (6.25) is a special case of (2). We can introduce<br />

a normalized generalized impedance to be<br />

Similarly,<br />

Z n (z) = Z(z)<br />

Z 0<br />

= 1+;(z)<br />

1 ; ;(z) : (3)<br />

;(z) = Z n(z) ; 1<br />

Z n (z)+1 : (4)<br />

Given ;(z), we can solve for Z n (z) in (3), and given Z n (z), we can solve for<br />

;(z) in (4). It turns out that the mapping of Z n (z) to;(z) and the mapping<br />

of ;(z) toZ n (z) are one-to-one. We shall next discuss a graphical method to<br />

solve (3) and (4) rapidly using the <strong>Smith</strong> <strong>Chart</strong>.<br />

X n<br />

Z n = R n + jX n<br />

R n = .5 R n = 1 R n = 2<br />

Im Γ<br />

X n = 1<br />

1<br />

X n = 1<br />

R n = 0<br />

R n = 1<br />

R n = .5<br />

R n = 2<br />

R n = 0<br />

0 0.5 1 2<br />

X n = 0<br />

R n<br />

0 0.5 1 2<br />

Re Γ<br />

–1<br />

X n = –1<br />

|Γ| = 1<br />

circle<br />

X n = –1<br />

Z n – plane<br />

Γ–plane<br />

Z n is a complex number and can be represented by a point on the Z n -plane,<br />

and ; is a complex number and can be represented by apoint on the complex<br />

; plane.<br />

1


We noted that from Equation (4) that:<br />

(i) When Z n =0 ;=;1.<br />

(ii) When Z n =1,or R n =1X n =0 ;=0.<br />

(iii) When Z n !1in any direction, ; ! 1.<br />

(iv) When Z n = jX n j;j =1.<br />

(v) When Z n = j, or R n =0X n =1 ;=j.<br />

(vi) When Z n = ;j, or R n =0X n = ;1 ;=;j.<br />

If one works out the mapping from Z n -plane to ;-plane completely, one<br />

nds that the R n = 0 line on Z n -plane maps onto the unit-circle on the ;-<br />

plane. Furthermore, the other R n = constant lines map into circles as shown.<br />

<strong>The</strong> X n = constant lines map into arcs like the X n 1 lines as shown. Hence,<br />

if one puts grids on the ;-plane, one can read o the R n and X n associated<br />

with the corresponding ; immediately, and, given the value of ;, one can<br />

read o the values of R n and X n immediately.<br />

<strong>The</strong> mappings (3) and (4) are known as bilinear transforms. A bilinear<br />

transform always maps a circle onto a circle.<br />

Properties of a <strong>Smith</strong> <strong>Chart</strong><br />

(i) <strong>The</strong> normalized admittance Y n = 1=Z n , or the reciprocal of Z n , can be<br />

found easily from a <strong>Smith</strong> <strong>Chart</strong>, because<br />

Zn<br />

;= Z n ; 1<br />

Z n +1 = 1 ; 1<br />

1+ 1<br />

Zn<br />

= 1 ; Y n<br />

1+Y n<br />

= ; Y n ; 1<br />

Y n +1 : (5)<br />

(ii) <strong>The</strong> change of impedance along the line is obtained by adding or subtracting<br />

phase to ;(z) via the relationship<br />

;(z) = v e 2jz : (6)<br />

(iii)<br />

VSWR = 1+j vj<br />

1; j v j = R nmax (7)<br />

since the <strong>Smith</strong> <strong>Chart</strong> is a graphical tool to solve Equation (7), and j v j<br />

is real, corresponding to a number on the X n = 0 line. Notice that<br />

1 < VSWR < 1 always.<br />

2

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