--- title: "Significance of reflection coefficient; R1 = 1 Ω, R2 = 4 Ω" subject: "Filter Design" chapter: 5 key: "semester_5/filter_design:reflection-significance-unequal-termination" source: https://www.bctengineer.com/q/semester_5/filter_design/reflection-significance-unequal-termination site: BCT Engineering --- # Significance of reflection coefficient; R1 = 1 Ω, R2 = 4 Ω Filter Design · Chapter 5: Design of Resistively-Terminated Lossless Filters - Asked in: 2080 Bhadra, 2+5 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/reflection-significance-unequal-termination - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-5 ## Question Describe the significance of reflection coefficient. Derive the 3rd order Butterworth low pass filter resistively-terminated lossless network with unequal termination of R1 = 1Ω and R2 = 4Ω. ## Short answer The reflection coefficient indicates impedance matching (ρ=0 perfect match), stopband attenuation (|ρ|=1 total reflection), and through |ρ|²=1−|t|² gives the magnitude response; it also yields Z1 for synthesis. For R1=1Ω, R2=4Ω, the 3rd-order Butterworth ladder gives L1=6.387 H, C2=0.3608 F, L3=2.170 H, and R2=4Ω. ## Answer #### Significance of reflection coefficient The **reflection coefficient** $\rho(s)$ of a resistively terminated lossless ladder compares the power reflected back towards the source with the maximum power the source can give. Looking into the ladder (input impedance $Z_1$) from a source of resistance $R_1$: $$ \rho(s) = \frac{R_1 - Z_1(s)}{R_1 + Z_1(s)}, \qquad |\rho(j\omega)|^2 = \frac{P_{\text{reflected}}}{P_{\max}} $$ - **Matching:** $\rho = 0$ means $Z_1 = R_1$ (perfect match, maximum power to the load); $|\rho| = 1$ means total reflection (stopband or transmission zero). - **Response:** for a lossless ladder $|\rho|^2 = 1 - |t|^2$, so $\rho$ carries the same information as the magnitude response; small $|\rho|$ in the passband means low loss. - **Synthesis:** it is the bridge from the transfer function to the circuit, because $Z_1 = R_1\dfrac{1\pm\rho}{1\mp\rho}$ is the impedance that is expanded into the LC ladder. - **Termination limits:** with unequal terminations $|\rho(0)| \neq 0$, which shows that the maximum available power cannot be fully delivered at dc. #### Derivation of the 3rd order Butterworth ladder ($R_1 = 1\ \Omega$, $R_2 = 4\ \Omega$) **Step 1: Transmission coefficient.** Third order Butterworth (Table 1): $B(s) = (s+1)(s^2+s+1) = s^3+2s^2+2s+1$. At dc the inductors are shorts and capacitors open, so the ladder reduces to a divider: $V_2/V_1 = \frac{R_2}{R_1+R_2} = \frac{4}{5} = 0.8$. Hence the largest possible dc transmission is $$ |t(0)|^2 = \frac{4R_1R_2}{(R_1+R_2)^2} = \frac{4(1)(4)}{(1+4)^2} = \frac{16}{25} = 0.64 $$ so $|t(j\omega)|^2 = \dfrac{0.64}{1+\omega^6}$, i.e. $t(s) = \dfrac{0.8}{s^3+2s^2+2s+1}$. **Step 2: Reflection coefficient.** $$ |\rho(j\omega)|^2 = 1 - \frac{0.64}{1+\omega^6} = \frac{0.36 + \omega^6}{1+\omega^6} \;\Rightarrow\; \rho(s)\rho(-s) = \frac{0.36 - s^6}{1 - s^6} $$ The zeros satisfy $s^6 = 0.36$, i.e. they lie on a circle of radius $a = (0.36)^{1/6} = 0.8434$ at the Butterworth angles. Choosing the left-half-plane zeros (like the Butterworth poles, scaled by $a$): $$ \rho(s) = \frac{(s+a)(s^2+as+a^2)}{B(s)} = \frac{s^3 + 1.6869 s^2 + 1.4228 s + 0.6}{s^3+2s^2+2s+1} $$ **Step 3: Input impedance** ($R_1 = 1$): $$ Z_1(s) = R_1\frac{1+\rho(s)}{1-\rho(s)} = \frac{2s^3+3.6869s^2+3.4228s+1.6}{0.3131s^2+0.5772s+0.4} $$ **Step 4: Continued fraction expansion** (computed by repeated long division): $$ Z_1(s) = 6.387\,s + \cfrac{1}{0.3608\,s + \cfrac{1}{2.170\,s + 4}} $$ So series $L_1 = 6.387$ H, shunt $C_2 = 0.3608$ F, series $L_3 = 2.170$ H, and the remainder is the load $R_2 = 4\ \Omega$, as required (this choice of $\rho$ sign gives the correct $R_2$; the other sign ends in $1/4\ \Omega$ and is rejected). **Step 5: Circuit.** ```text R1=1 L1=6.387 H L3=2.170 H o-/\/\---UUU----+----UUU---+ | | | (~) V1 C2=0.3608 F R2=4 | | | o---------------+----------+ ``` **Check:** analysing this ladder gives $V_2/V_1 = \dfrac{0.8}{s^3+2s^2+2s+1}$, the required Butterworth response. **Answer:** $R_1 = 1\ \Omega$, $L_1 = 6.387$ H, $C_2 = 0.3608$ F, $L_3 = 2.170$ H, $R_2 = 4\ \Omega$ (normalized to $\omega_c = 1$ rad/s). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does |rho| = 1 indicate? A. Perfect matching B. Maximum power transfer C. Total reflection, as in the stopband or at a transmission zero D. A purely resistive load Answer: C. |rho|^2 is the fraction of available power reflected; 1 means none reaches the load. (Key point: Meaning of rho) 2. For R1 = 1 and R2 = 4, what is the largest possible |t(0)|^2? A. 1 B. 0.8 C. 0.36 D. 0.64 Answer: D. |t(0)|^2 = 4R1R2/(R1+R2)^2 = 16/25 = 0.64. (Key point: DC transmission limit) 3. How is rho(s) chosen for the unequal-termination Butterworth design? A. Zeros from the right half plane, denominator any polynomial B. Zeros at the origin C. Zeros on the circle s^6 = 0.36 in the left half plane, with Butterworth denominator B(s) D. Zeros at infinity Answer: C. The zeros of rho(s)rho(-s) lie on a circle of radius 0.8434; left-half-plane ones are taken with the Hurwitz denominator B(s). (Key point: Choosing rho(s)) 4. The continued fraction of Z1 ends with 4. What does this confirm? A. The load is R2 = 4 ohm, as required B. The source is 4 ohm C. The filter is a high pass D. The sign of rho is wrong Answer: A. The final remainder is the load; 4 ohm matches R2, confirming this sign choice of rho. 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