EE335- Double Stub Matching This is not in the book but will be on the midterm. 11 Lecture A single stub must be repositioned to match for different loads. Sometimes this is inconvenient. The double stub method fixes the location of the stubs and varies the lengths. Usually the stubs are separated by standard distances: λ/4, λ/8, or 3λ/4.
Y2 Y2' d21=d2-d1
d2 Y1 Y1'
d1
ZL
YL jb2 L2 jb1 L1
The circuit is to be matched at the beginning of the matching network. This implies that Y2 = 1+0j =Y2’+jb2 Y’2 = Y2 –jb2 = 1-jb2 The admittance just after the second stub must fall on the unit circle of conductance (g = 1) Y1 is Y’2 delayed by the distance d21: WTL (CCW) a distance d21 this gives the auxiliary circle (g =1 circle rotated by the distance d21) Y1 = Y’1 +jb1 Y’1 is the load admittance rotated by the distance from the load.
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Goal is to get to origin of Smith Chart where r =1, g =1, and Γ = 0 Method: 1 2 3 4 5 6 7 8 Normalize load and plot (A) Draw SWR circle Reflect through origin to get admittance (B) Rotate WTG to position of first stub Draw Auxiliary Circle by rotating g=1 circle by d21 (CCW) Shift to Aux. Circle by adding stub1 (C&C’) Rotate d21 to second position of second stub (new SWR circle) (D&D’) Add stub to cancel imaginary part (to origin)
A draw back of double stub tuning is that a certain range of loads cannot be matched once the stub locations are fixed. 3 stubs are necessary to guarantee that a match is always possible.
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Z 0 = 50 EXAMPLE: Z L = 30 − 40 j d1 = 0 d 2 = .375λ
B
D
C’ D’
A C
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1 2 3 4 5 6
7 8
Z L 30 − 40 j = = 0.6 − 0.8 j (A) Z0 50 SWR Circle yL = 0.6 + 0.8 j (B) Rotate WTG to position of first stub (d1 = 0) no rotation Draw Aux. Circle , rotate g = 1 circle d21 = .375λ y1 = yL +jb1 y1-yL=jb1 (C) = 0.6-1.9j, (C’) = 0.6-.09j C: 0.6-1.9j-(0.6+0.8j) = -2.7j .306λ-.25λ=.056λ C’:0.6-0.09j-(0.6+0.8j) = -0.89j .384λ-.25λ = .134λ D: 1+2.5j D’: 1-0.53j .31λ-0.25λ=.06λ b2j = -2.5j b2’j = 0.53j .08λ+.25λ=.33λ zL =
Try these problems and see if you get these answers: Z 0 = 50 Z L = 60 − 80 j d 1 = 0.1λ d 2 = .225λ L1 = .34λ L2 = .43λ L1′ = .10λ ′ L2 = .22λ Z 0 = 50 Z L = 80 + 60 j d1 = 0.2λ d 2 = .45λ L1 = .172λ L2 = .323λ L1′ = .103λ ′ L2 = .177λ
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