Smith Chart (introduction)

Chapter 5 Impedance matching and tuning 5.1 Matching with lumped elements L-section matching networks using Smith chart 5.2 Single-stub tuning shunt stub, series stub 5.3 Double-stub tuning forbidden region 5.4 The quarter-wave transformer frequency response 5.5 The theory of small reflections single-section transformer, multi-section transformer 5.6 Binomial multisection matching transformers 5.7 Chebyshev multisection matching transformers 5.8 Taper lines exponential taper, triangular taper 5.9 The Bode-Fano criterion Γ-Bandwidth 微波電路講義 5-1

• Impedance matching concept given ZL, design a matching network to have Γin=0 or selected value

Zo

Γin

matching network

ΓL

ZL

ZL

Zin (=Zo)

Discussion 1. Matching network usually uses lossless components: L, C, transmission line and transformer. 2. There are ∞ possible solutions for the matching circuit. 3. Properly use Smith chart to find the optimal design. 4. Factors for selecting matching circuit are complexity, bandwidth, implementation and adjustability. 5-2

微波電路講義

5.1 Matching with lumped elements (2-element L-network) • Smith chart solution L constant G-circle

constant R-circle

C Z-plane CW → add series L (reduce series C) CCW → add series C (reduce series L)

Y-plane CW → add shunt C (reduce shunt L) CCW → add shunt L (reduce shunt C) 5-3

微波電路講義

(explanation)

constant R-circle → L or C in series

j1 j2

L

j0.5

L B

(1)CW A → B :1 + j 0.5 + jx = 1 + j 2 → jx = j1.5 = jωL : add an L in series (2)CCW B → A :1 + j 2 + jx = 1 + j 0.5 → jx = − j1.5 =

A

1 j ωC

: add a C in series, or reduce extra L

C C -j0.5

(3)CCW C → D :1 − j 0.5 + jx = 1 − j 2 → jx = − j1.5 =

D C -j1

-jj2

1 j ωC

: add a C in series (4)CW D → C :1 − j 2 + jx = 1 − j 0.5 → jx = j1.5 = jωL : add an L in series or reduce extra C

in Z-plane CW → add a series L (or reduce series C) CCW → add a series C (or reduce series L) 5-4

微波電路講義

constant G-circle → L or C in shunt -j1

(1)CW A → B :1 − j 2 + jb = 1 − j 0.5 → jb = j1.5 = jωC -j0.5 : add a C in shunt, or reduce shunt L

L -j2

L A

(2)CCW B → A :1 − j 0.5 + jb = 1 − j 2 → jb = − j1.5 =

B

1 j ωL

: add an L in shunt

D

C C j2

C j1

j0.5

(3)CCW C → D :1 + j 2 + jb = 1 + j 0.5 → jb = − j1.5 =

1 j ωL

: add an L in shunt, or reduce shunt C (4)CW D → C :1 + j 0.5 + jb = 1 + j 2 → jb = j1.5 = jωC : add a C in shunt

in Y-plane CW → add a shunt C (or reduce shunt L) CCW → add a shunt L (or reduce shunt C) 5-5

微波電路講義

Discussion 1. ZL inside 1+jx circle, two possible solutions Smith chart solution (shunt-series elements) 1+jb circle

1+jx circle

B

A: Zo

ZL

series-shunt elements? Zo “N” Z o = jX + analytical solution

ZL

A

B:

jB +

jX jB

1 1 R L + jX L

⇒ B > 0 → C ,B < 0 → L

ZL

X > 0 → L, X < 0 → C 5-6

微波電路講義

2. ZL outside 1+jx circle, two possible solutions Smith chart solution (series-shunt elements) 1+jb circle

B

A

1+jx circle

A Zo

ZL

B Zo

ZL shunt-series elements? “Y” analytical solution 1 1 = jB + Zo R L + j( X + X L ) jX ⇒ B > 0 → C ,B < 0 → L jB ZL X > 0 → L, X < 0 → C 5-7

微波電路講義

3. Ex. 5.1 ZL=200-j100, Zo=100Ω, f=500MHz

B

3 1 2

A

L C

A:

1. zL=2-j1, yL=0.4+j0.2 Solution A 2. y=0.4+j0.5 → jb=j0.3 → jB=jωC=jb/Zo C=b/Zoω =0.92pF z=1-j1.2 → jx=j1.2 → jX=j ωL=jxZo L=xZo/ω =38.8nH Solution B 3. y=0.4-j0.5 → jb=-j0.7→-jB=1/jωL=-jb/Zo L=-Zo/ωb=46.1nH z=1+j1.2 → jx=-j1.2 → jX=1/jωC=-jxZo C=-1/xZoω =2.61pF frequency response (p.227, Fig.5.3(c))

B:

C

L 5-8

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4. Possible 3-element L-network

1+jx circle

series ? “Y”

Zo

ZL

Zo

ZL

1+jb circle

5-9

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5. Possible 4-element L-network shorter paths for a wider operational bandwidth 1+jx circle Zo

ZL

Zo

ZL

1+jb circle

5-10

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6. Lumped elements (size<λ/10) capacitor: chip capacitor, MIM capacitor (<25pF), interdigital gap capacitor (<0.5pF), open stub(<0.1pF) inductor: chip inductor, loop inductor, spiral inductor (<10nH) resistor: chip resistor, planar resistor All these lumped elements inherently have parasitic elements in the microwave range. (p.228, “point of interest”) Size (mil)

B

C D

0402

0603

0805

1206

A

39

62

78

125

B

24

38

39

62

C

8

12

20

23

D

20

31

49

62

1mil=0.001in=25um=1/40mm A

5-11

微波電路講義

5.2 Single-stub tuning • equivalent microstrip elements a series C --a series L in series with a high impedance microstrip line a shunt C in shunt with an open microstrip line a shunt L in shunt with a short microstrip line an open-circuited microstrip line

Z in =

Zo 1 ≡ j tan β l jω C

a short-circuited microstrip line Z in = jZ o tan β l ≡ j ω L a high impedance microstrip line jZ o tan

βl 2

Z1 Zo, β

≡

jZ o tan

βl

jZ o β l

2

Z2 Y3 j sin β l Zo

l 5-12

≈ 微波電路講義

(derivation of high impedance line) From p.185 Table 4.1

Zo, β

⎡A ⎢C ⎣

l jZ o tan

βl 2

Z1

jZ o tan

βl 2

Z2 Y3 j sin β l Zo

high Zo,Y3→0 β l <<

jZ o sin β l ⎤ cos β l ⎥⎦

Z1 Z 2 ⎤ ⎡ Z1 1 Z Z + + + 1 2 Z3 ⎥ ⎡ A B ⎤ ⎢⎢ Z 3 ⎥ ⎢C D ⎥ = ⎢ 1 Z2 ⎥ ⎣ ⎦ 1 + ⎢ Z ⎥ Z3 ⎣ 3 ⎦ Zo j sin β l 1 Z3 = , Y3 = = jYo sin β l j sin β l Zo 1+

jZ o β l

B ⎤ ⎡ cos β l =⎢ ⎥ D ⎦ ⎣ jYo sin β l

Z1 = cos β l , Z1 = (cos β l − 1) Z 3 Z3

= −2 sin 2

π 2 5-13

βl 2

× − jZ o

βl 1 = jZ o tan βl βl 2 2 sin cos 2 2

微波電路講義

Discussion 1. Shunt stub

Smith chart solution

yin = jb = jωC

d G Zo

Zo Zo

ZL

constant Γ-circle

y in = 1 − jb

l 1

y in = − jb =

jω L

Zo

Zo

Zo l

G

d

ZL

y in = 1 + jb 5-14

微波電路講義

2. Series stub

Smith chart solution

zin = 1 + jx d G

Zo l

ZL

Zo Zo

zin = − jx =

1 jωC

zin = 1 − jx

d

Zo ZL

Zo

l

G

Zo

zin = jx = jωL 5-15

微波電路講義

3. Ex. 5.2 ZL=60-j80, Zo=50Ω, f=2GHz, using a shunt short stub

3 G

1. zL=1.2-j1.6, yL=0.3+j0.4 Solution A 2. y=1+j1.47 → d1=0.11λ 3. y=-j1.47 → l1=0.095λ, short stub

4

l1

B S.C.

A

1

Solution B 4. y=1-j1.47 → d2=0.26λ 5. y=j1.47 → l2=0.405λ, short stub

2 5

l2 d

l

frequency response (p.231, Fig.5.5(c)) Solution A has a wider bandwidth. ZL 5-16

微波電路講義

4. Ex. 5.3 ZL=100+j80, Zo=50Ω, f=2GHz, using series open stub 3 4

G

B

1

O.C.

A

l1

l2

2 5

1. zL=2+j1.6 Solution A 2. z=1-j1.33 → d1=0.12λ 3. z=j1.33 → l1=0.397λ, open stub Solution B 4. z=1+j1.33 → d2=0.463λ 5. z=-j1.33 → l2=0.103λ, open stub frequency response (p.234, Fig.5.6(c)) It can not be implemented in microstrip lines.

5-17

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5. Analytical solution for shunt stub Y in = − jB

d

ZL

Zo l ⎧⎪⎛ Z L + jZ o tan β d ⎞ −1 ⎫⎪ Yo = Re ⎨⎜ Z o ⎟ ⎬ Z jZ tan d + β o L ⎠ ⎪⎭ ⎩⎪⎝

Y in = Y o + jB

→ d

⎧ ⎛ Z o ⎞ −1 open stub ⎪ ⎧⎪⎛ Z L + jZ o tan β l ⎞ −1 ⎫⎪ ⎪ ⎝⎜ tan β l ⎠⎟ − B = − Im ⎨⎜ Z o → l ⎟ ⎬=⎨ −1 Z jZ tan l + β o L ⎠ ⎪⎭ ⎪⎛ ⎞ ⎪⎩⎝ 1 − ⎪⎜ Z o tan β l ⎟ short stub ⎠ ⎩⎝ 5-18

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6. Analytical solution for series stub Z in = Z o + jX

d ZL

Zo

Z in = − jX

l ⎧ Z L + jZ o tan β d ⎫ Z o = Re ⎨ Z o ⎬ + β Z jZ tan d o L ⎩ ⎭

→ d

Zo ⎧ open stub ⎧ Z L + jZ o tan β l ⎫ ⎪ − X = − Im ⎨ Z o ⎬ = ⎨ tan β l ⎩ Z o + jZ L tan β l ⎭ ⎪ Z o tan β l short stub ⎩ 5-19

→ l

微波電路講義

5.3 Double-stub tuning 4, 5 1±jb

2, 3

1

λ/8

G

Zo

1

5

ZL 2

l2

l1

4 forbidden region

3

Discussion 1. There exists a forbidden region for ZL. It can be tuned out by adding a certain length of line. 5-20

微波電路講義

2. Ex. 5.4 ZL=60-j80, Zo=50Ω, f=2GHz, using double-shuntopen-stubs 1. zL=1.2-j1.6, yL=0.3+j0.4 l1’ 7 Solution A G 6 2. y=0.3+j 0.286 → 6. b1’ = - 0.114 l2’ → l1’ = 0.482λ 5 4. y=1+j1.38 → 7. b2’ = - 1.38 → l2’ =0.35λ O.C. 2 B A Solution B 4 3. y=0.3+j1.714 → 8. b1 =1.314 1 3 → l1 =0.146λ 9 l1 5. y=1-j3.38 → 9. b2 = 3.38 l2 8 → l2 =0.204λ λ/8

l2

l1

frequency response (p.239, Fig.5.9(c)) ZL 5-21

微波電路講義

3. Analytical solution Y2 = Yo - jB2 Y1

d

Zo

ZL

Zo l2

l1

jB2

jB1

⎛ Z + jZ o tan β d Y1 = YL + jB1 , Y2 = Yo − jB 2 = ⎜⎜ Z o 1 Z o + jZ 1 tan β d ⎝ Re {Y2 } = Yo → B1 → l1

⎞ ⎟⎟ ⎠

−1

Y1 → Im {Y2 } = − B 2 → l 2

5-22

微波電路講義

5.4 The quarter-wave transformer • frequency response

|Γ|

ZL/ Z0=10 Δθ

| Γm |

l

ZL/ Z0=2 Zo

Z1 =

Γ

Z1

ZL (real) θm θo π-θm θ Γm: max. tolerated Γover the bandwidth

ZoZL

Γ (θ ) ≈

ZL − Zo 2 ZoZ L

cos θ , for θ near θ o =

π , θ = βl 2

2 ZoZ L Γm 4 Δf ), Z L → Z o , Δ f increases = 2 − cos − 1 ( 2 fo π 1− Γm ZL − Zo 5-23

微波電路講義

(derivation of Γ (θ) ≈

Γ (θ ) =

Z 12 = Z o Z L

=

θ→ θo =

≈

π 2

Z in ( θ ) − Z o Z in ( θ ) + Z o

ZL + Z1 + = Z + Z1 L Z1 + Z1

cos θ )

2 ZoZL

jZ 1 tan θ − Zo Z 1 Z L + jZ 12 tan θ − Z o Z 1 − jZ L Z o tan θ jZ L tan θ = jZ 1 tan θ Z 1 Z L + jZ 12 tan θ + Z o Z 1 + jZ L Z o tan θ + Zo jZ L tan θ

Z1 (Z L − Z o ) ZL − Zo 1 = = Z 1 ( Z L + Z o ) + j 2 Z 12 tan θ Z L + Z o + j 2 Z o Z L tan θ j2 ZoZ L ZL + Zo + tan θ ZL − Zo ZL − Zo ZL − Zo j2 ZoZ L

cos θ

(derivation of Γ (θ ) =

ZL − Zo

2 ZoZL Γm Δf 4 = 2 − cos − 1 ( )) 2 fo π 1− Γm ZL − Zo 1

2 ZoZL Z + Zo 2 [( L ) +( tan θ ) 2 ]1 / 2 ZL − Zo ZL − Zo

1 + tan 2 θ = sec 2 θ

=

1 (Z L − Z o )2 + 4Z L Z o 4ZoZ L [ tan 2 θ ]1 / 2 + 2 2 (Z L − Z o ) (Z L − Z o )

1

=

[1 +

4ZoZ L sec 2 θ ]1 / 2 2 (Z L − Z o )

5-24

微波電路講義

1

Γ (θ ) = [1 + Γ 2m =

4ZoZ L sec 2 θ ]1 / 2 2 (Z L − Z o )

2 ZoZ L 2 1 1 1 ) → 2 −1 = ( cos 2 θ m ZL − Zo Γm 2 ZoZ L 1 2 1+ ( ) Z L − Z o cos θ m

→ cos θ m =

Γm

1 − Γ 2m Z L − Z o

T E M line: θ = β l = f fo

=

2 ZoZL πf m 2θ m f πf 2 πf λ o 2 πf v p ( fo ) = = → θm = → m = π 2 fo 2 fo vp ( f ) 4 v p ( f ) 4 fo fo

2( f o − f m ) 2 fm 4 = 2− = 2 − cos − 1 π fo fo

5-25

Γm

2 ZoZ L

1 − Γ m2 Z L − Z o

微波電路講義

5.5 The theory of small reflections • single-section transformer

Z1

Γ1

Γ2

T12

e

Γ3 Γ in

Γ1 e − jθ

Γ2

Z2

Γ3

ZL (real)

T12T21 Γ 3 e − j 2 θ = Γ1 + 1 − Γ 2 Γ 3e − j 2θ

Γ 1 + Γ 12 Γ 3 e − j 2 θ + (1 − Γ 1 )(1 + Γ 1 ) Γ 3 e − j 2 θ = 1 + Γ1 Γ 3e − j 2θ

− jθ

e − jθ

≈ Γin

T21 T12

Γ 2 = −Γ1 , T21 = 1 + Γ1 , T12 = 1 + Γ 2

e − jθ

T 21 Γin

Γ1

θ

Γ3

Γ1 + Γ 3e − j 2θ = 1 + Γ1 Γ 3e − j 2θ ≈ Γ 1 + Γ 3 e − j 2 θ if Z 1 ≈ Z 2 ≈ Z L 5-26

微波電路講義

• multisection transformer

Zo

Γ

Γo

θ

θ

θ

Z1

Z2

ZN

Γ1

Γ2

ΓN

ZL (real)

Γ (θ ) = Γ o + Γ1e− j 2θ + Γ 2e− j 4θ + ......Γ N e− j 2 Nθ , if Γ o = Γ N , Γ1 = Γ N −1...... ⎧ e− jNθ [ Γ o (e jNθ + e− jNθ ) + Γ1 (e j ( N −2)θ + e− j ( N −2)θ ) + ......Γ ( N −1) / 2 (e jθ + e− jθ )] N odd = ⎨ − jNθ jNθ − jNθ j ( N − 2) θ + e− j ( N −2)θ ) + ......Γ N / 2 ] N even ⎩e [ Γ o (e + e ) + Γ1 (e 1 ⎧ − jNθ θ θ θ + − + − + 2 e [ Γ cos N Γ cos( N 2) .... Γ cos( N 2 n ) .... Γ( N −1) / 2 cosθ ] N odd o n 1 ⎪⎪ 2 =⎨ ⎪2e− jNθ [ Γ cos Nθ + Γ cos( N − 2)θ + ....Γ cos( N − 2n)θ + .... 1 Γ ] N even o n N /2 1 ⎪⎩ 2 given Γ (θ ), design Z1 , Z 2 ,....Z n 5-27

微波電路講義

5.6 Binomial multisection matching transformer • maximal flatness response for Γ (θ) Γ (θ) = A(1 + e

N

) = A∑ CnN e − j 2 nθ = Γ o + Γ1e − j 2θ + Γ 2 e − j 4θ + ......Γ N e − j 2 Nθ

− j 2θ N

n =0

⇒ Γ n = ACnN

Discussion 1. Maximal flatness response,

d N −1 Γ(θ ) dθ N −1

2. Γ (0) = A 2 N = Z L − Z o → A = 2− N Z L − Z o Z L + Zo

3. Γ n = → ln

Z n +1 − Z n 1 Z n +1 ≈ ln Z n +1 + Z n 2 Zn

=0 θ=

π 2

or l =

λ 4

Z L + Zo

(ln x ≈ 2

x −1 ) x +1

Z n +1 Z − Zo N Z ≈ 2 Γ n = 2 ACnN = 2 × 2− N L Cn ≈ 2− N CnN ln L Zn Z L + Zo Zo

(p.249, Table 5.1 for Zn values) 5-28

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Γ m N1 4 Δf −1 1 = 2 − cos [ ( ) ], N ↑ , Δ f ↑ , Γ m = 2 N A con N θ m fo π 2 A

4.

∵ Γ (θ) = A(1 + e − j 2θ ) N → Γ (θ) = A 2 N cos N θ 1 Δf 4 4 −1 1 Γ m N = 2 − θm = 2 − cos [ ( ) ] fo π π 2 A

5. Ex.5.6 ZL=50Ω, Zo=100Ω, N=3, Γm=0.05 N = 3, A = 2 − N

Z L − Zo Z 1 = N +1 ln L = − 0.0433, 2 Z L + Zo Zo

Δf = 70% fo

ln

Z n +1 Z Z Z = 2 − N C nN ln L ⇒ ln 1 = 2 −3 C 03 ln L → Z1 = 91.7 Ω Zn Zo Zo Zo

ln

Z Z2 Z Z = 2 −3 C13 ln L → Z 2 = 70.7 Ω , ln 3 = 2 −3 C 23 ln L → Z 3 = 54.5Ω Z1 Zo Z2 Zo

(p.250, Fig.5.15 for frequency response of |Γ|) 5-29

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5.7 Chebyshev multisection matching transformers • Equal ripple response for Γ (θ): optimal design Γ (θ ) = 2 e − jNθ [ Γ o cos N θ + Γ 1 cos( N − 2)θ + .... Γ n cos( N − 2 n )θ + ....] = Ae − jNθ TN (sec θ m cos θ )

⇒ Γ n (p.254, Table 5.2)

T1 ( x ) = x , T2 ( x ) = 2 x 2 − 1,

T3 ( x ) = 4 x 3 − 3 x ,

Tn ( x ) = 2 xTn -1 ( x ) − Tn- 2 ( x ), x ≡

T4 ( x ) = 8 x 4 − 8 x 2 + 1,

cos θ , x ≤1 cos θ m

-1

1

π-θm

θm

x

Discussion 1. Γ(0) = ATN (sec θm ) = Z L − Z o → Γ m = A = Z L − Z o Z L + Zo

2. TN (sec θm ) = 1 Z L − Z o

Γm Z L + Zo

→ θm

1 Z L + Z o TN (sec θm ) 4θ Δf (5.63) → = 2− m fo π

3. Optimal design: given Γm, maximal Δf given Δf , minimal Γm. 5-30

微波電路講義

4. Ex.5.7 ZL=100Ω, Zo=50Ω, N=3, Γm=0.05 N = 3, Γ (θ ) = 2 e − j 3θ ( Γ o cos 3θ + Γ1 cos θ ) = Ae − j 3θ T3 (sec θ m cos θ ) = Ae − j 3θ (4 sec 3 θ m cos 3 θ − 3 sec θ m cos θ ) = Ae − j 3θ [sec 3 θ m (cos 3θ + 3 cos θ ) − 3 sec θ m cos θ ] A=

Z L − Zo 1 Δf = Γ m → sec θ m =1.408 → θ m = 44.7 o → = 101% Z L + Z o T3 (sec θ m ) fo

⇒ 2 Γ o = A sec 3 θ m → Γ o = 0.0 698 = Γ 3 2 Γ1 = A(3sec 3 θ m − 3sec θ m ) → Γ1 = 0.1037 = Γ 2 Γo =

Z1 − Z o Z1 + Z o

→ Z1 = 57.5Ω

Γ1 =

Z 2 − Z1 Z 2 + Z1

→ Z 2 = 70.7 Ω

Γo =

Z3 − Z2 Z3 + Z2

→ Z 3 = 87 Ω

frequency response (p.255, Fig.5.17) 5-31

微波電路講義

5.8 Tapered lines • Frequency response

Z+ΔZ Z(z) ZL ΔΓ

Γ

Zo

0

z z+Δz

L

z

Z + ΔZ − Z ΔZ ≈ Z + ΔZ + Z 2Z 1 dZ 1 d ln Z Z o → dΓ = = dz dz 2 Z 2 d Z 1 L → Γ (θ) = ∫ e − j 2βz (ln ) dz 0 dz Zo 2 ΔΓ =

5-32

微波電路講義

Discussion 1. Exponential taper Z ( z ) = Z o e az

0< z< L

Z ( L ) = Z L = Z o e aL → a = → Γ (θ ) =

1 2

∫

L 0

e− j2β z

Z 1 ln L L Zo

Z 1 sin β L d (ln e az ) dz = ln L e − j β L βL dz 2 Zo

L ↑ , Γ (θ ) ↓ (p. 257, Fig.5.19)

2. Triangular taper ⎧⎪ Z o e 2 (z/L ) ln Z L Z o 0< z< L 2 Z (z) = ⎨ ( 4 z/L − 2 z 2 /L2 − 1 ) 2 ln Z L Z o Z e L 2< z< L ⎪⎩ o Z L − j β L sin( β L 2) 2 1 Γ (θ ) = ln e [ ] Zo βL 2 2 2

first null at 2 π (p. 258, Fig.5.20) 5-33

微波電路講義

3. Klopfenstein taper Z ( z ) (5.74), (5.75), Γ ( θ ) (5.76), optimal taper

4. Ex.5.8 ZL=100Ω, Zo=50Ω, Γm=0.02 exponential taper: Z ( z ) = Z o eaz , a =

1 ZL = 0.693 L ln L Zo

1 Z sin β L Γ (θ ) = ln L 2 Zo β L 1 Z sin β L 2 2 triangular taper: Γ (θ ) = ln L [ ] 2 Zo β L 2 cos ( β L)2 − A2 Klopfenstein taper: Γ (θ ) = Γ o , A = 3.543, Γ o = 0.346 cosh A frequency response (p. 260, Fig.5.21) 5-34

微波電路講義

5.9 The Bode-Fano criterion

Γ(ω)

Zo

lossless matching network

C

Discussion 1. |Γ | 1 Γm

R

∫

∞ 0

ln

1 π dω ≤ RC Γ (ω)

ln1/ |Γ | ln1/Γm Δω

ω

5-35

Δω

ω

微波電路講義

∫

∞

0

ln

1 dω = Γ ( ω)

= Δω ln

∫

Δω

ln

1 dω Γm

π 1 ≤ : constant Γ m RC

(1) given RC → Δω↑ → Γm↑ (2) Γm≠ 0, unless Δω=0 i.e., Γm=0 only at a finite number of frequencies (3) R and/or C ↑ → Δω ↓ and/or Γm↑ ⇒ high Q load is harder to match

Zo

Γ(ω)

lossless matching network

C

R

parallel resonator Q= ωoRC 5-36

微波電路講義

Solved problems: Prob.5.7 find Z1 and l Zo=40Ω

Z1, l

zL

1.95+j0.98 4+j2

200+j100

zin 2.05 r’max 5.25

r’min 0.8 40 = Z1

200 + j100 + jZ1 tan β l Z1 + j (200 + j100) tan β l

(Z1)

→ 40 Z1 + j8000 tan βl − 4000 tan β l = 200 Z1 + j100 Z1 + jZ12 tan β l → → →

40 Z1 − 4000 tan β l = 200 Z1 → Z1 = −25 tan β l j8000 tan β l = j100 Z1 + jZ12 tan β l j8000 = − j 2500 + j 625 tan 2 βl → tan β l = −4.1 Z1 = 102.5

k=

Z1 kr 'min = 0.8 , 50 kr 'max = 5.25

→ k r 'min r 'max 2

r 'min r 'max =1

=

k 2 = 4.2

→ k = 2.05 Z1 = 50 × 2.05 = 102.5

l = 0.288λ 5-37

微波電路講義

Prob.5.25 find the best RL over operating range of 3.1~10.6GHz 0.6pF Zo

∫

∞ 0

ln

Γ(ω)

UWB network

75Ω

1 π dω ≤ Γ (ω) RC

→ ln

1 π ≤ Γm 2 π (10.6 − 3.1) × 10 9 × 75 × 0.6 × 10 − 12

→ ln

1 ≤ 1.48 → Γ m ≥ 0.228, RL ≤ 6.4 dB Γm

Suggested homework (due 2 weeks): 5, 11, 13, 22 ADS examples: Ch5_prj 5-38

微波電路講義