--- title: "Double-stub microstrip network and its S-matrix" subject: "RF and Microwave Engineering" chapter: 2 key: "semester_7/rf_and_microwave_engineering:double-stub-microstrip-smatrix" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/double-stub-microstrip-smatrix site: BCT Engineering --- # Double-stub microstrip network and its S-matrix RF and Microwave Engineering · Chapter 2: RF and Microwave Transmission Lines - Asked in: 2077 Chaitra, 2+8 marks - Video: [Double shunt stub matching network problem solving using smith chart by Dr. Niraj Kumar VIT Chennai](https://www.youtube.com/watch?v=AuO61UXmMlw) — RF Design Basics - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/double-stub-microstrip-smatrix - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-2 ## Question Sketch a double-stub perfectly matched network using microstrip and prepare its s-matrix. ## Short answer Microstrip double-stub tuner uses two shunt stubs spaced 3λ/8; shorted stubs end in via holes to ground, open stubs simply end. From the cascade ABCD of shunt stub, line, shunt stub, convert to S; perfect match requires S22=ΓL*, giving S11=0 and |S21|=1. Referenced to the load, S=[[0,e^-jφ],[e^-jφ,0]]. ## Answer #### Sketch: double-stub matching network in microstrip Two shunt stubs are etched in the same microstrip as the main line. They are spaced d = 3λ/8 apart (λ is the guided wavelength in the microstrip). A shorted stub ends in a via hole to the ground plane, and an open stub simply ends. ```text Top view (microstrip, ground plane below) stub 2 (l2) stub 1 (l1) || || || || ●via (short) ●via (short) || || =====##======== d =======##=====[ ZL ] port 1 (Z0) 3λ/8 d1 to load ``` ```text Electrical equivalent Z0 o---+-------- d --------+-- d1 --[ZL] | | jb2 (stub 2) jb1 (stub 1) ``` #### Preparing its S-matrix For a two-port with both ports referenced to Z0, the S-parameters follow from the normalised ABCD matrix: ```text S11 = (A + B − C − D)/Δ S12 = 2(AD − BC)/Δ S21 = 2/Δ S22 = (−A + B − C + D)/Δ Δ = A + B + C + D ``` A double-stub tuner is a cascade of shunt stub 2, a line of length d, and shunt stub 1: ```text [ABCD] = |1 0| |cos βd j sin βd| |1 0| |jb2 1| |j sin βd cos βd | |jb1 1| ``` Building blocks: ```text Shunt jb: S = 1/(2+jb) · | −jb 2 | | 2 −jb | Line βd: S = | 0 e^(−jβd) | | e^(−jβd) 0 | ``` **Condition for a perfect match:** with the load ΓL on port 2, ```text Γin = S11 + S12 S21 ΓL/(1 − S22 ΓL) = 0 ``` For a lossless, reciprocal tuner this holds when **S22 = ΓL\*** (conjugate match at the load side). Unitarity then gives |S21|² = 1 − |ΓL|², and S12 = S21. If port 2 is referenced to the load itself (the reference impedance is ZL), the matched network becomes an ideal matched, lossless, reciprocal two-port: ```text S = | 0 e^(−jφ) | | e^(−jφ) 0 | ``` Here φ is the total phase delay. S11 = S22 = 0 (no reflection) and |S21| = 1 (all power reaches the load). **Numerical example** (ZL = 110 + j110 Ω on a 100 Ω line, first stub at the load, d = 3λ/8, b1 = −1.384, b2 = −2.844, stub lengths 0.0996λ and 0.0538λ): ```text S11 = 0.466∠145.5° S12 = 0.885∠−45.8° S21 = 0.885∠−45.8° S22 = 0.466∠−57.2° ΓL = 0.466∠+57.2° → S22 = ΓL* ✓ |S11|² + |S21|² = 1 (lossless) ✓ Γin = 0 (VSWR = 1) ✓ ``` Before matching, the line sees S11 = ΓL = 0.466∠57.2° (VSWR = 2.75). After matching, S11 at the input = 0. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is a short-circuited shunt stub realised in microstrip? A. A strip ending in a via hole to the ground plane B. A strip with an open end C. A strip with a series resistor D. A coaxial T-junction with a sliding short Answer: A. A shorted stub ends in a via to the ground plane; an open stub simply ends. (Key point: Microstrip shorted stub) 2. How is the double-stub tuner's ABCD matrix formed? A. Sum of the two stub matrices B. Shunt stub 2, line of length d, shunt stub 1 multiplied in cascade C. Only the line matrix, since stubs add nothing D. Two series reactances and a line Answer: B. The tuner is a cascade [1 0; jb2 1][cos βd, j sin βd; j sin βd, cos βd][1 0; jb1 1]. (Key point: ABCD cascade) 3. Which expression converts normalised ABCD to S21, with Δ = A + B + C + D? A. S21 = (A + B - C - D)/Δ B. S21 = Δ/2 C. S21 = 2/Δ D. S21 = 2(AD - BC) Answer: C. S21 = 2/Δ, S11 = (A + B - C - D)/Δ, S22 = (-A + B - C + D)/Δ, S12 = 2(AD - BC)/Δ. (Key point: ABCD to S) 4. In the example, S11 = S22 magnitude 0.466 and |S21| = 0.885. What does this confirm? A. The network is lossy B. The line is matched at port 1 of the network C. The stubs are open-circuited D. The network is lossless, since |S11|^2 + |S21|^2 = 1 Answer: D. 0.466^2 + 0.885^2 = 1, confirming a lossless network; S22 = ΓL* gives Γin = 0 when loaded. 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