--- title: "G21(s) and Y21(s) of ladder with bridging capacitor" subject: "Electrical Circuit Theory" chapter: 5 key: "semester_3/electrical_circuit_theory:g21-y21-two-port-1ohm-1h" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/g21-y21-two-port-1ohm-1h site: BCT Engineering --- # G21(s) and Y21(s) of ladder with bridging capacitor Electrical Circuit Theory · Chapter 5: Frequency Response of Network - Asked in: 2074 Chaitra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/g21-y21-two-port-1ohm-1h - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-5 ## Question Find the forward voltage ratio transfer function G21(s) and forward transfer admittance Y21(s) in the following circuit. [Figure: port 1 (V1, I1) followed by a 1 Ω series resistor to node A. From A to node B: a 10 Ω resistor in series with a 1 H inductor. From B to the port 2 terminal: a 10 Ω resistor in series with a 1 H inductor. A 1/2 F capacitor in series with a 10 Ω resistor runs from node A to the port 2 terminal (as drawn, passing across node B). Shunt elements: a 1/2 F capacitor at node A and a 1 H inductor at node B to the bottom wire. Port 2 has V2, I2.] ## Short answer Use nodal analysis with port 2 open for G21 and shorted for Y21. Results: G21(s)=2s(3s²+50s+102)/(3s⁴+66s³+328s²+488s+44); Y21(s)=−s(3s²+50s+102)/(15s⁴+236s³+1006s²+1384s+220). ## Answer G21 = V2/V1 with port 2 open (I2 = 0). Y21 = I2/V1 with port 2 short-circuited (V2 = 0), I2 being the current entering port 2. **Reading of the figure (assumption).** Name the node after the 1 Ω port resistor P, the node at the 1 H shunt N and the port 2 top terminal O. The elements are: P–N: 10 Ω + 1 H in series; N–O: 10 Ω + 1 H in series; shunt at P: ½ F; shunt at N: 1 H; and the ½ F capacitor in series with the 10 Ω resistor runs from P to O. This lower branch only crosses the vertical wire of N (no dot at the crossing), so it is not connected to N. #### Branch admittances ```text yPN = yNO = 1/(s + 10) (10 Ω + 1 H) yPO = 1/(10 + 2/s) = s/(10s + 2) (10 Ω + ½ F) shunt at P: s/2 shunt at N: 1/s ``` #### Node equations (V1 applied through the 1 Ω resistor) ```text P: (VP − V1)·1 + (s/2)VP + (VP − VN)yPN + (VP − VO)yPO = 0 N: (VN − VP)yPN + VN/s + (VN − VO)yNO = 0 O: (VO − VN)yNO + (VO − VP)yPO = I2 ``` #### G21 (I2 = 0) Solving the three equations (Cramer's rule, checked in Python): ```text G21 = 2s(3s² + 50s + 102) / (3s⁴ + 66s³ + 328s² + 488s + 44) ``` #### Y21 (VO = 0) Put VO = 0, solve P and N for VP and VN, then I2 = −[VN·yNO + VP·yPO] (current flowing into the shorted port, so it is negative of the current leaving it): ```text Y21 = I2/V1 = − s(3s² + 50s + 102) / (15s⁴ + 236s³ + 1006s² + 1384s + 220) ``` **Answer: G21(s) = 2s(3s² + 50s + 102)/(3s⁴ + 66s³ + 328s² + 488s + 44); Y21(s) = −s(3s² + 50s + 102)/(15s⁴ + 236s³ + 1006s² + 1384s + 220).** Both have a zero at s = 0 (the inductors short d.c. at the output side) and share the same numerator polynomial. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are G21 and Y21 defined? A. G21 = V2/V1 with I2 = 0; Y21 = I2/V1 with V2 = 0 B. G21 = V2/V1 with V2 = 0; Y21 = I2/V1 with I2 = 0 C. G21 = I2/V1 with I2 = 0; Y21 = V2/V1 D. Both defined with port 2 open Answer: A. G21 needs port 2 open, Y21 needs port 2 shorted. (Key point: Definitions) 2. What is the admittance of a 10 ohm resistor in series with 1 H, and of 10 ohm in series with 1/2 F? A. 1/(10s+1) and s/(s+10) B. 1/(s+10) and s/(10s+2) C. s+10 and 10s+2 D. 1/(s+10) and 1/(10s+2) Answer: B. Series impedances 10 + s and 10 + 2/s are inverted. (Key point: Branch admittances) 3. How is the lower branch (1/2 F + 10 ohm) connected? A. From N to O B. From P to ground C. From P to O, not connected to N D. From N to ground Answer: C. It crosses the wire at N without a dot, so it joins P and port 2. (Key point: Figure reading) 4. Why does Y21 come out negative? A. Because of the inductors B. Because V1 is negative C. Because the capacitor is reversed D. I2 is the current entering port 2, opposite to the current leaving the shorted port Answer: D. The shorted port current flows out, so I2 = -(branch currents). 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