Phase Lead Compensator Design: Step-by-Step Guide. Input data: Bode Plot Frequency Response and Desired Percentage of Overshoot ($\%OS$). Target audience: Someone asking in Indonesian, likely an engineering student or professional. Goal: Provide a comprehensive, step-by-step explanation. Constraint Checklist & Confidence Score: 1. Explain step-by step? Yes. 2. Subject: Phase Lead Compensator? Yes. 3. Input 1: Bode Plot Frequency Response? Yes. 4. Input 2: Desired %OS? Yes. Structuring the answer: 1. **Introduction/Goal:** Briefly state what we are trying to find ($K_c$). 2. **Step 1: Analyze the Required Bandwidth and Gain (from %OS).** Relate $\%OS$ to desired damping and bandwidth. 3. **Step 2: Find the Cross-over Frequency ($\omega_{gc}$)We need answer in Indonesian. Need explain step by step designing phase lead compensator given Bode plot frequency response and desired %OS. Need infer likely control systems. Need discuss extracting plant transfer function from Bode? Or design using Bode plot data and %OS. Need clarify phase lead compensator design. Desired %OS relates to damping ratio zeta via overshoot formula for second-order: %OS = exp(-ζπ/sqrt(1-ζ²))*100. Then derive phase margin PM = arctan(2ζ/sqrt(sqrt(1+4ζ^4)-2ζ²))? Or approx PM = 100ζ? Need compute ζ and PM. Then use Bode plot to find current phase margin and crossover frequency. Design Gc(s)=Kc * (s+z)/(s+p), p>z to provide phase lead. Need place zero and pole around new gain crossover frequency to maximize phase lead and meet PM. Steps include selecting new crossover where magnitude adjusted by K? Since Bode provided only? Need explain procedure for data provided: from Bode plot determine magnitude/phase at frequencies. If transfer function unknown, can use frequency response data to design? Need handle cases: If Bode plot is open-loop of uncompensated system, with desired %OS. Need design phase lead such that PM sufficient. Need choose Kc to get desired crossover frequency and phase margin. Standard procedure: 1. Convert %OS to ζ. For second-order: ζ = |ln(OS/100)| / sqrt(π² + ln²(OS/100)). Actually overshoot fractional Mp. ζ = -ln(Mp)/sqrt(pi^2 + ln^2(Mp)) with Mp = %OS/100. (positive). 2. Determine desired phase margin PM_d = atan(2ζ/sqrt(sqrt(1+4ζ^4)-2ζ^2)) in degrees? This formula gives PM in rad? Let's derive. For underdamped second order, PM = arctan(2ζ / sqrt(sqrt(1+4ζ^4)-2ζ^2)). If argument? Let's check ζ=0.5: sqrt(sqrt(1+0.25)-? wait sqrt(1+4ζ^4)=sqrt(1+0.25)=1.118; minus 2ζ²= -0.5? Actually formula denominator sqrt(sqrt(1+4ζ^4)-2ζ^2) = sqrt(1.118 -0.5)= sqrt(0.618)=0.786; arctan(1/0.786)=51.8°. Known PM ~ ζ*100=50°, good. Another formula PM = tan^{-1}(2ζ/ sqrt(sqrt(1+4ζ^4)-2ζ²)) gives 51.8. Add safety margin 5-10°. Some use PM = atan(2ζ/sqrt(sqrt(1+4ζ^4)-2ζ²)). Need note %OS only valid for 2nd order no zeros, approximate for higher order; verify via simulation. Alternatively use PM ≈ 100 ζ, but accurate formula. 3. From Bode plot find uncompensated open-loop magnitude and phase. Determine current gain crossover frequency ω_c, PM = 180° + phase at ω_c (if stable), and gain margin. 4. Set desired PM with safety margin: PM_des = PM_required + margin (5-12°). Determine required phase lead φ_lead = PM_des - PM_old (if current at same crossover) maybe plus if changing crossover. But changing Kc changes crossover and phase; need choose new ω_c. Standard design: - If need to keep/alter bandwidth, choose new gain crossover frequency ω_c' where uncompensated phase already maybe lower. Required lead φ_m = PM_des - (180 + ∠G(jω_c'))? Need compute at candidate. - Phase lead compensator: Gc(s)=Kc (s+z)/(s+p) = Kc * (1 + s/z)/(1 + s/p), p>z. It contributes phase φ(ω)=atan(ω/z)-atan(ω/p). Max phase at ω_m=√(z p). φ_m = asin((p/z -1)/(p/z +1))? Or sin φ_m = (1-α)/(1+α) where α=z/p? Need be consistent. Common form: Gc(s)=Kc (s+z)/(s+p) with p>z. Let α = z/p <1. Or Gc=(1+Ts)/(1+βTs), β>1? Need define. If Gc(s)= (s+z)/(s+p) = (1 + s/z)/(1 + s/p). At s=0 gain = z/p = α <1. Need Kc to adjust. Alternatively Gc=Kc (s+z)/(s+p). At high frequency gain = Kc. At low = Kc z/p. Some texts use Gc=Kc (s+z)/(s+p) and choose Kc to satisfy magnitude at ω_m? Need include Kc. The maximum phase: Let a = p/z >1. Gc(s) = Kc (s+z)/(s+p) = Kc/a * (1 + s/z)/(1 + s/(a z)). Phase = atan(ω/z)-atan(ω/(a z)). At ω_m = z√a. φ_m = atan(√a) - atan(1/√a)? Actually ω_m/z=√a; ω_m/(az)=? √a/a=1/√a. So φ_m = arctan(√a)-arctan(1/√a). This equals arcsin((a-1)/(a+1))? Let's compute a=4: arctan2 - arctan0.5=63.435-26.565=36.87°. asin((4-1)/(5)) = asin(0.6)=36.87. yes. Alternatively sinφ_m=(a-1)/(a+1). Or with α=z/p=1/a, sinφ_m=(1-α)/(1+α). Good. Given φ_m, solve a = (1+sinφ_m)/(1-sinφ_m). Then choose ω_m around new crossover ω_c'. Set z = ω_m/√a, p = a z = ω_m√a. Or z = ω_c'/√a, p=√a ω_c'? Let's verify: ω_m = sqrt(z p). If choose ω_m=ω_c', then z=ω_c'/√a, p=ω_c'√a. Yes. - Determine Kc. The compensator magnitude at ω_c' is |Gc(jω_c')| = Kc * sqrt(ω_c'^2+z²)/sqrt(ω_c'^2+p²). Need make |Gc(jω_c')| |G(jω_c')| =1 (new crossover), so Kc = 1/( |G(jω_c')| * factor without Kc ). But since ω_c'=√(zp), factor =? Compute: sqrt(ω_m²+z²)/sqrt(ω_m²+p²). With p=a z, ω_m=z√a. numerator = sqrt(a z²+z²)=z√(a+1). denominator = sqrt(a z²+a² z²?) Wait p²=a² z². plus ω_m²=a z² => a²z²+a z² = a z²(a+1); sqrt= z√a√(a+1). Ratio=1/√a = sqrt(α). So |(s+z)/(s+p)| at ω_m = sqrt(z/p)=1/√a. Then Kc required = √a / |G(jω_c')|. Wait because total = Kc/√a * |G| =1 => Kc=√a/|G|. If using Kc as high-frequency gain? Yes. If use form Gc=Kc * (1+s/z)/(1+s/p), at ω_m magnitude = Kc/√a? Let's carefully: Gc=Kc(s+z)/(s+p). At ω_m, ratio=1/√a, so |Gc|=Kc/√a. Set |Gc||G|=1 => Kc=√a/|G|. Kc often >1 to boost low frequency? But phase lead typically increases gain at high frequencies? Here at ω_m factor <1 (if p>z), so to get crossover, Kc must be >1. The low frequency DC gain = Kc z/p = Kc/a = √a/a /|G| =1/(√a |G|). The high freq gain = Kc=√a/|G|. Okay. But many standard phase lead design set compensator as Gc(s)=Kc * (s+z)/(s+p), with p>z. Kc chosen to satisfy new crossover. Good. Alternative if Bode plot is closed-loop? Problem says "Bode Plot Frequency Response" likely open-loop. Need mention if only experimental Bode no transfer function, can still do frequency shaping graphically: - Use data table to interpolate magnitude/phase. - Compute desired PM, choose crossover frequency where required phase lead feasible and bandwidth desired. Need consider "phase lead compensator" has max phase at ω_m, and to achieve PM at crossover, place zero and pole around crossover (typically zero a decade below, pole a decade above? Actually for maximum phase at new gain crossover, zero at ω_c/√a, pole at ω_c√a). Need also adjust gain so magnitude 0 dB at ω_c. Need iterative because adding compensator changes phase and magnitude; but if we set ω_m=ω_c' and Kc as above, new crossover is ω_c' by construction? Wait the total magnitude at ω_c' equals 1 by choosing Kc. But total magnitude elsewhere may cross 1 at another frequency? Need check monotonic and stability; usually with lead and single crossover. Need verify. - The phase margin at new crossover is PM_new = 180 + ∠G(jω_c') + φ_lead(ω_c'). If ω_m=ω_c', φ_lead=φ_m. So set φ_m = PM_d - (180+∠G(jω_c')) = PM_d - PM_old_at_ω_c'? Careful PM_old_at candidate = 180+∠G(jω_c'). Then required lead φ_m = PM_d - PM_old_at(ω_c'). If negative? Then candidate has enough PM; choose less lead or just gain? Need positive. If current PM insufficient at current crossover, choose candidate where needed lead feasible and not too large (max maybe 60°; for more use multiple stages). Also gain Kc changes crossover, so you must evaluate at chosen ω_c'. Need choose ω_c' such that needed lead less than max (e.g. ≤60°). Could pick ω_c' higher to reduce needed phase? Wait as frequency increases, phase becomes more negative, needed lead increases. Lower frequency phase closer to -90? Need system type? Need consider. For typical lead compensator design, to increase PM, often increase crossover frequency to improve speed; but higher frequency phase more lag, requiring more lead. Or choose lower ω_c' where magnitude <1? Need adjust gain to raise/lower crossover. If current PM too low due to crossing near resonance? Need shape. Let's outline algorithm robustly: A. Data extraction: - Build table of |G(jω)| in dB and ∠G(jω) from plot. Identify: - low-frequency slope/type (N) maybe for steady-state but %OS only transient. - gain crossover frequency ω_co (where magnitude 0 dB). - phase at ω_co; PM = 180 + phase. If phase is less than -180 etc. - gain margin not directly related to %OS but check stability. B. Desired performance: - Convert %OS to ζ (assuming 2nd-order). - Determine desired PM: exact formula or approximate PM_d ≈ 100ζ; add safety margin 5-10° (or 8-12°). PM_req = PM_from_ζ + margin. If desired settling time/bandwidth not specified, choose crossover frequency to achieve speed if needed. If only %OS, PM is primary; crossover frequency affects speed/robustness. Often choose ω_c' near current or slightly higher to meet speed. Need state that %OS alone not enough to uniquely determine bandwidth; many solutions. If only %OS, choose ω_c' based on bandwidth requirement or keep original gain crossover as much as possible. The final answer should mention non-uniqueness. C. Required lead: - At chosen new gain crossover ω_c' (candidate), compute uncompensated phase ∠G(jω_c') and magnitude |G(jω_c')| (from Bode/interpolation). Compute PM_uncomp_at ω_c' = 180 + ∠G(jω_c') (assuming positive; if phase wrapped, handle). Required compensator phase at ω_c': φ_need = PM_des - PM_uncomp_at ω_c' (plus perhaps margin already). Ensure 0<φ_need<~60°. If φ_need>60°, use double lead or choose another ω_c'. If φ_need≤0, lead not needed for phase, just gain or choose different bandwidth; a lead may still be used for bandwidth but not necessary. D. Compute lead network parameters: - Choose max phase φ_m = φ_need (maybe add a few degrees due to approximations and to ensure PM; if placing ω_m at ω_c'). - Solve a = (1+sin φ_m)/(1-sin φ_m) = p/z. Or α=z/p=(1-sinφ)/(1+sinφ). - Set ω_m = ω_c' (for maximum phase at crossover). Then zero z = ω_c'/√a, pole p = a z = ω_c'√a. Alternatively z=α ω_c'? Let's derive: a=p/z. ω_m=√(zp)=ω_c'. z=ω_m/√a; p=ω_m√a. Also α=z/p=1/a. Some forms Gc=Kc(s+z)/(s+p). Good. - Compensator transfer function without gain: G_lead(s)=(s+z)/(s+p). E. Determine gain Kc: - At ω_c', the lead network's magnitude relative to Kc is |(jω_c'+z)/(jω_c'+p)| = sqrt(ω_c'²+z²)/sqrt(ω_c'²+p²). With ω_c'=ω_m=√zp, this equals sqrt(z/p)=1/√a. - Need total magnitude 0 dB: Kc * (1/√a)*|G(jω_c')| =1. So Kc = √a/|G(jω_c')| (linear magnitude, not dB). If using dB: 20 log Kc = -20 log|G(jω_c')| + 20 log√a? Let's derive dB total: 20log|G| +20log|Gc|=0. 20log|Gc|=20logKc +20log(1/√a). Set = -20log|G|. So 20logKc = -20log|G| +20log√a = -20log|G| +10log a. Wait 20log(1/√a)= -10 log a. Set 20logG +20logKc -10log a=0 =>20logKc = -20logG +10log a. Yes. Kc=√a/|G|. In dB: Kc_dB = -G_dB +10 log10 a? Since |G|dB=20log|G|, Kc_dB=20logKc = -|G|dB +10log a. Need be careful if G_dB at ω_c'. So if G_dB is -10 dB and a=4, Kc_dB=10+6.02? Wait formula -(-10)+6.02=16.02 dB. Kc=√4/|G|=2/0.316=6.325=16.02 dB yes. - Then final compensator Gc(s)=Kc (s+z)/(s+p). Or equivalent Kc' (1+Ts)/(1+βTs) with β etc. Need show conversion if needed: Gc(s)=Kc (s+z)/(s+p). Multiply by? To express as K * (1+s/z)/(1+s/p), set K = Kc z/p? Actually Gc=Kc (s+z)/(s+p). =Kc z(1+s/z)/[p(1+s/p)] = Kc(z/p)*[(1+s/z)/(1+s/p)]. If define Kc2=Kc z/p = (1/√a|G|)? Hmm. Some design procedures choose Gc=Kc (s+z)/(s+p) with Kc as gain; okay. Or choose Gc=K * (s+z)/(s+p) and calculate K = |G|^{-1}√a. Need be clear. F. Verify: - Construct compensated open-loop L(s)=Gc(s)G(s). If G(s) unknown, combine Bode plot graphically: add lead phase curve and magnitude curve to original Bode. Determine new 0 dB crossing and PM. Need iterate because Kc shifts magnitude; but our chosen Kc ensures at ω_c' total 0 dB. Need check other crossings. If PM below desired, increase φ_m or adjust ω_c' or increase Kc? Actually Kc affects crossover, not PM at chosen because total at chosen zero. Changing Kc would change chosen crossover; but design chosen Kc exactly. Increase φ_m increases phase lead and changes a; with ω_m fixed, zero/pole closer/further changes magnitude factor sqrt(z/p)=1/√a, so Kc also changes; total at ω_c' remains zero. But total curve changes. Check. - Check gain margin, closed-loop step response simulation. Adjust safety margin. - If no plant model, use Nichols chart or Bode graphical trial-and-error. Need also maybe include step "from desired %OS to damping ratio and then to PM" and "if Bode plot given for closed-loop?": If Bode plot frequency response of plant (open-loop) likely. If closed-loop frequency response given? The design cannot directly from closed-loop Bode? Need mention: ensure Bode plot is open-loop G(jω); if closed-loop only, first derive open-loop or use frequency-domain closed-loop specs (Mr, bandwidth) to infer PM? But question says "Bode Plot Frequency Response" maybe they have plant Bode plot (frequency response). Let's answer assuming open-loop. Need maybe include formulas in Indonesian. Use "p > z" or "zero lebih dekat ke origin daripada pole" to create lead. The lead compensator increases phase (positive phase shift) in range between zero and pole. It also increases gain at high frequencies. Need specify if using "phase lead compensator" form: G_c(s)=K_c \frac{s+z}{s+p},\quad p>z>0 or G_c(s)=K_c \frac{1+\alpha T s}{1+T s}, \alpha>1? Let's avoid ambiguity. Many Indonesian controls use Gc(s)=Kc * (1 + aTs)/(1+Ts), with a>1? Need recall. Phase lead: Gc(s)=Kc * (s+z)/(s+p), p>z. Or Gc(s)=Kc * (α(1+Ts))/(1+αTs)? Let's choose one and explicitly state. Let's verify phase lead sign for Gc=(s+z)/(s+p) with p>z. At s=jω, phase = atan(ω/z) - atan(ω/p). Since p>z => ω/p<ω/z => atan(ω/z)>atan(ω/p), positive. Good. Compute max phase: Gc(jω) = (jω+z)/(jω+p). φ = atan(ω/z)-atan(ω/p). Let a=p/z. dφ/dω zero at ω_m=√(zp)=z√a. φ_m = arcsin((p-z)/(p+z))? Let's derive sign: earlier we had φ_m=atan(√a)-atan(1/√a). arcsin((a-1)/(a+1)) = arcsin((p/z -1)/(p/z+1)) = arcsin((p-z)/(p+z)). Good. Some formula: sin φ_m = (p/z -1)/(p/z+1). Need solve a? Let r=p/z=a. sin φ_m=(r-1)/(r+1). Solve r=(1+sin φ_m)/(1-sin φ_m). Yes. Good. Magnitude at ω_m: |Gc(jω_m)| =? Gc without Kc = (jω+z)/(jω+p). At ω_m=√(z p)=√(a z²)=z√a. Numerator magnitude sqrt(ω_m²+z²)=z√(a+1). Denominator sqrt(ω_m²+p²)= sqrt(a z² + a²z²)=z√(a(1+a))? Wait p=az => p²=a²z². Sum = a z² + a²z² = a z²(1+a). sqrt = z√a√(1+a). Ratio =1/√a. Good. Since a>1 factor <1. So to have total zero dB at ω_c'=ω_m, need Kc/√a * |G(jω_c')|=1 => Kc=√a/|G|. But if the original Bode plot magnitude at ω_c' is e.g. -20dB, Kc positive. The lead's low freq gain is Kc/a = 1/(√a|G|) maybe less than 1? It may reduce DC gain; but lead typically used with additional gain to maintain steady-state. If need preserve low-frequency gain (steady-state error), use K_lead =? Standard design often set Kc =1 and choose compensator DC gain? Let's examine. In many standard phase lead design using Gc(s)=Kc * (1+αTs)/(1+Ts), α>1? Wait if α>1 with zero at -1/(αT) and pole at -1/T? Let's map. Standard "lead compensator": Gc(s)=Kc * (s+z)/(s+p), z>p? Hmm no. Let's recall from Ogata: Lead compensator: Gc(s)=Kc * (s+z)/(s+p), p>z. Transfer function has zero closer to origin than pole. Yes. It increases gain at high frequency, so gain at s=0 is Kc z/p = α? If Kc chosen for desired error, high frequency gain larger by factor p/z. Good. There is also form Gc(s)=Kc * (1+Ts)/(1+αTs), α<1? If zero at -1/T and pole at -1/(αT), zero closer to origin (since T<αT? Actually pole farther left? zero -1/T, pole -1/(αT) with α<1 => pole - larger magnitude? e.g. zero -1, pole -5; p>z). DC gain Kc. High freq gain Kc/α. This form sets DC gain Kc and increases high frequency. Equivalent to our form? Our form Kc(s+z)/(s+p). DC gain Kc z/p. If set DC gain Kc_form = Kc z/p, then Kc_form*(1+Ts)/(1+αTs)? Let's map: Gc=Kc*(s+z)/(s+p) = Kc*z(1+s/z)/[p(1+s/p)] = Kc*(z/p)*(1+s/z)/(1+s/p). Define K_dc=Kc z/p. Then Gc=K_dc*(1+s/z)/(1+s/p). This is K*(1+Ts)/(1+αTs) with T=1/z, α = p/z? Wait denominator 1+s/p = 1+αTs? If T=1/z, αT =? Need α T =1/p? But T=1/z. α=z/p <1. Yes α=z/p. G=K_dc (1+Ts)/(1+αTs), α<1. DC gain K_dc. Good. Some textbooks use α>1 by putting zero at -1/(αT), pole at -1/T? Let's not confuse. We can stick to explicit form. Need discuss if Bode plot only and no transfer function: How to compute "current phase margin" from Bode plot? At gain crossover frequency ω_c where L(jω) = 0 dB. PM = 180° + ∠L(jω_c). If the plot is a Bode plot of uncompensated open-loop. Desired PM from %OS. Required lead at candidate. Need choose new crossover frequency. But if we don't have mathematical G(s), we can still compute Kc using magnitude from Bode at ω_c'. But we need "phase of G" at candidate from Bode. Use interpolation. Potential issue: If we choose ω_c' where original magnitude is -x dB, adding lead with Kc chosen to make total 0dB at that frequency. But the lead's phase at ω_c' is φ_m. However, adding lead also changes magnitude and phase at all frequencies. The actual new gain crossover might not exactly be ω_c' if original magnitude curve not flat? Wait by construction total magnitude at ω_c' is 1. But there could be another crossing; typically one. Need verify. But is it correct to set Kc = √a/|G(jω_c')|? This ensures |GcG| =1 at ω_c'. However, the phase at ω_c' is ∠G + φ_m. If choose φ_m = PM_d - (180+∠G), then PM = PM_d. Good. So design meets PM exactly at chosen frequency. But is that the actual gain crossover? If total magnitude =1 at ω_c' and no other crossings, yes. If another crossing, could be different. Need verify. But standard Bode lead design sometimes choose a new crossover frequency ω_c' where the compensated magnitude is 0 dB, but you don't know Kc until choose ω_m and gain. Actually if using Gc=Kc(s+z)/(s+p) and set max phase at ω_m=ω_c', the magnitude factor at ω_c' is 1/√a. Thus Kc determined by original magnitude at ω_c'. So yes. Alternative design: If want to maintain the same low-frequency gain/steady-state (set Kc=1 or pre-specified), then you cannot choose arbitrary ω_c' because Kc fixed. In that case choose lead network so that |G(jω_c')|*|Gc(jω_c')|=1 with Kc=1, which determines a and ω_c'. But with %OS and Bode plot, typical procedure: 1. From desired PM, set φ_m = PM_d - PM_old? At original crossover? If adding compensator without changing Kc? Let's examine standard design in Ogata for Gc=Kc(s+z)/(s+p) with Kc chosen to satisfy steady-state error and p>z. Steps: - Determine Kc to satisfy steady-state error. - Plot Bode of KcG(s). - Determine required phase lead φ_m = γ_des - γ_current + (5-12°). γ_current at new gain crossover? Since gain crossover changes due to lead, often first find frequency where uncompensated magnitude is -20 log(1/√a)? Actually iterative. - Determine attenuation factor 1/α? - Set ω_m = new gain crossover frequency, where |KcG(jω_m)| = √α? Hmm because lead magnitude at max =1/√a if DC gain? Let's recall Ogata: Gc(s)=Kc α (1+Ts)/(1+αTs), α<1? Wait. There is a known procedure: For Gc(s)=Kc * (s+z)/(s+p), p>z. If choose max phase at new crossover ω_c', the magnitude of lead at ω_c' is √(z/p) = √α (where α=z/p<1). If Kc is fixed, the total magnitude at ω_c' is |KcG| * √α. To be 0 dB, we need |KcG(jω_c')| = 1/√α = √(p/z). That means original (after Kc) magnitude at ω_c' should be positive dB equal to 10 log(p/z). This is used to locate ω_c' from Bode: find frequency where magnitude of KcG is -? Let's derive dB: total =20log|KcG| +20log√α =20log|KcG| +10log α. Since α<1, term negative. Set 0 =>20log|KcG| = -10log α = 10log(1/α)=10log(p/z). So original magnitude at ω_c' is +10log(p/z) dB. But p/z=a. So original magnitude =10log a dB. Wait if lead has low gain at DC? This seems if original magnitude is positive at chosen ω_c' (above 0 dB), adding lead attenuates at ω_m? But lead factor at max is <1? Hmm if Gc=Kc(s+z)/(s+p), at ω_m factor without Kc =√(z/p)=1/√a <1. So adding lead attenuates at ω_m? That seems odd: phase lead network should have magnitude >1 at high frequencies relative to low? Let's check Bode of (s+z)/(s+p) with p>z. At low frequency magnitude = z/p <1 (if no Kc), at high frequency =1. It increases from z/p to 1. At ω_m, magnitude =√(z/p) =1/√a, which is less than 1 but greater than DC. If Kc set to DC gain? Let's map to standard form with DC gain Kc: Gc=Kc (1+Ts)/(1+αTs) with α<1. At high frequency gain Kc/α > Kc. At ω_m, magnitude =Kc/√α? Actually T=1/z, αT=1/p? α=z/p. G=Kc (1+Ts)/(1+αTs). At ω_m =1/√(α T²)? =√(zp)? Compute magnitude =Kc*√(1+(ωT)²)/√(1+(αTω)²). At ω_m, ωT=1/√α. numerator=√(1+1/α)=√((α+1)/α); denom=√(1+α)=? αTω = α*(1/√α)=√α; square=α, denom=√(1+α). Ratio= [√((α+1)/α)]/√(1+α)=1/√α. So magnitude=Kc/√α >Kc. Wait earlier for Kc(s+z)/(s+p), DC gain=Kc α. The equivalent DC gain K_dc=Kc α. So standard form G=K_dc (1+Ts)/(1+αTs). At ω_m, magnitude=K_dc/√α = (Kc α)/√α = Kc√α. So without Kc in our form factor is√α, but Kc multiplies. The total compensator magnitude at ω_m is Kc√α. If Kc=DC gain? No, in standard form DC gain Kc_standard=Kc_our α. At ω_m magnitude=Kc_standard/√α. If Kc_standard fixed to steady-state, then total at ω_m is Kc_std/√α. To make total =1, need original G magnitude = √α/Kc_std? Let's derive correctly. Let's settle definitions: Our explicit Gc = Kc (s+z)/(s+p). At s=0: Gc(0)=Kc z/p = Kc/ a (a=p/z). At high freq: Kc. At ω_m=√(zp): magnitude =Kc *√(z/p)=Kc/√a. If Kc is chosen as overall gain (high-frequency asymptote), then compensator has high-freq gain Kc and DC gain Kc/a. Adding lead raises high-frequency gain relative to low. Many texts use Gc(s)=Kc α (1+Ts)/(1+αTs), with α<1? Let's map: Our Gc = Kc (s+z)/(s+p). Let z=1/T? p=1/(αT)? p>z => 1/(αT)>1/T => α<1. Gc = Kc * [ (s+1/T)/(s+1/(αT)) ] = Kc * [ (1+Ts)/(1+αTs) ]? Multiply numerator/denominator by T? (s+1/T)= (1/T)(1+Ts); (s+1/(αT))=(1/(αT))(1+αTs). Ratio = α (1+Ts)/(1+αTs). So Gc=Kc * α (1+Ts)/(1+αTs). If define Kc' = Kc α as DC gain, Gc=Kc' (1+Ts)/(1+αTs). At ω_m, magnitude=Kc'/√α. Since Kc'=Kc α, this equals Kc α/√α=Kc√α. Good. In standard design with DC gain fixed Kc', the max magnitude is Kc'/√α > DC. That makes sense: lead increases magnitude at high frequency. So if design using Kc' as DC gain (often to meet steady-state), choose ω_c' where uncompensated magnitude |G(jω_c')| = ? total = Kc' * factor * |G| =1. factor at ω_m=1/√α. So |G(jω_c')| = √α/Kc'. In dB: |G|_dB = 10 log α -20log Kc'? This is negative? α<1 so 10logα negative. Hmm. But the user didn't specify steady-state error; only Bode plot and %OS. So we can use design where Kc adjusts to get crossover, no need preserve DC. But a control system design should maybe maintain steady-state? Desired %OS only transient. Need mention if there is steady-state error requirement, use Kc to satisfy it first; then design zero/pole with new crossover location based on magnitude condition. Let's formulate both? Maybe answer with general step-by-step, highlighting that %OS gives PM, then use Bode to place lead. Need not overly complicate. Let's parse Indonesian: "Jelaskan secara step by step cara membuat phase lead compensator jika data yang disediakan adalag Bode Plot Frequency Response dan %OS yang diinginkan" Translation: Explain step by step how to make a phase lead compensator if the data provided are Bode Plot Frequency Response and desired %OS. They likely are asking for design procedure from Bode plot and percent overshoot. We should provide educational explanation. Need likely from control systems lab/assignment. Need detail: determine ζ from %OS; determine desired phase margin; from Bode plot find phase margin and gain crossover; calculate φ_m; calculate alpha; calculate T; determine Kc; form Gc; verify. Need maybe include example? Could be useful. But no actual plot/numeric values. We can give template and formulas. We can mention if Bode plot has actual numbers, substitute. Let's structure final: 1. Siapkan data Bode plot: - Plot magnitude dB vs log ω and phase vs log ω. - Tabel/intersep: ω_c0 = frequency where magnitude uncompensated crosses 0 dB; PM0=180°+∠G(jω_c0). - Identify low-frequency slope/type and gain, check stability. Also note desired bandwidth? Since %OS only gives damping, not speed. There are infinite solutions; choose ω_c' (new crossover) based on desired speed/constraints. If no bandwidth spec, usually choose ω_c' around ω_c0 or slightly larger, but ensure enough phase lead ≤60°. Alternatively choose frequency where required lead reasonable. Need state. 2. Convert %OS to ζ: Mp = %OS/100. ζ = -ln(Mp)/√(π² + ln²(Mp)). If OS ≤0? handle. Then desired PM: PM_ζ = atan(2ζ/√(√(1+4ζ⁴)-2ζ²)) in degrees. Or approx PM≈100ζ. Add safety margin 5°-12°: PM_d=PM_ζ+margin. Note formula for 2nd-order. For higher order, use as starting design and verify. 3. Determine needed lead angle at chosen ω_c': PM_uncomp(ω_c')=180°+∠G(jω_c') (using phase in degrees; unwrap continuous). φ_need = PM_d - PM_uncomp(ω_c'). If φ_need ≤0, lead network not needed for phase; choose other ω or just gain. If >60°, use double-stage lead or choose lower ω_c' where phase less lag (but magnitude conditions). 4. Calculate lead parameters: Let φ_m = φ_need (or slightly larger, e.g. +5° due to rounding). a = p/z = (1+sin φ_m)/(1-sin φ_m). Note a>1. α=z/p=1/a<1. Place maximum phase at new gain crossover: ω_m=ω_c' (usually; alternatively ω_m=ω_c' to maximize PM at crossover). Then z = ω_c'/√a = α ω_c'? Wait if a=p/z, √a =? z=ω_c'/√a = ω_c'√(z/p)? Let's express in terms of α=z/p: a=1/α. √a=1/√α. z=ω_c'/√a = ω_c'√α. p=√a ω_c' = ω_c'/√α. Check with earlier: a=p/z, ω_m=√(zp). z=ω_m/√a. Since √a=√(p/z). α=z/p=1/a. √α=√(z/p)=1/√a. z=ω_m√α. Yes. p=ω_m/√α. Good. Need avoid error. Let's present either: a=p/z = (1+sinφ)/(1-sinφ). z=ω_c'/√a, p=ω_c'√a. Or using α=z/p=(1-sinφ)/(1+sinφ): z=α? Wait if α=z/p, and ω_m²=zp. Solve p? z=ω_m√α? Let's verify with α=z/p<1. p=z/α. ω_m²=z*(z/α)=z²/α => z=ω_m√α. p=ω_m/√α. Yes. So both. Gc_without_gain(s)=(s+z)/(s+p). 5. Determine gain Kc: Need total magnitude at ω_c' be 0 dB: |Gc(jω_c')| |G(jω_c')| =1. With ω_c'=ω_m, |(jω_c'+z)/(jω_c'+p)| =√(z/p)=√α=1/√a. Thus Kc = √a / |G(jω_c')| = 1/(√α |G(jω_c')|). (If G magnitude from Bode in dB, convert: |G|=10^(G_dB/20).) In dB: 20 log Kc = -G_dB(ω_c') + 10 log a. Or equivalently = -G_dB -10log α? Let's derive with α: Kc=1/(√α |G|). Kc_dB=-20log|G| -10log α. Since α<1, -10logα positive. a=1/α, 10log a=-10logα, so Kc_dB=-G_dB+10log a. Good. Final Gc(s)=Kc (s+z)/(s+p). If want standard DC-gain form, Gc(s)=Kdc (1+s/z)/(1+s/p) with Kdc=Kc z/p = Kc/ a = 1/(√a |G|). Check Kdc: At DC gain = Kc z/p = (√a/|G|)*(1/a)=1/(√a |G|). At ω_m magnitude of Gc = Kdc/√α? α=1/a. Kdc/√α = [1/(√a|G|)] *√a =1/|G|, which multiplied by |G|=1. Good. So standard DC gain Kdc = 1/(√a |G|). But note this may not preserve original DC gain; it's whatever to meet crossover. If steady-state matters, set Kdc to required and use different crossover condition. Need mention. 6. Verify graphically/analytically: - Add lead magnitude and phase to original Bode: 20log|GcG| = G_dB + Kc_dB + 20log√((ω²+z²)/(ω²+p²)); phase = ∠G + atan(ω/z)-atan(ω/p). - Find new actual 0 dB crossing. Check PM =180°+phase at crossing ≥PM_d; gain margin. If not met, iterate (adjust ω_c' or φ_m). - Simulate step response, measure %OS; adjust margin/parameters. If %OS too high, increase PM (move ω_c' to where less lag or increase lead angle; use double lead if >60°); if response too slow/fast, adjust ω_c'/bandwidth. Note %OS alone doesn't determine speed. 7. Special note if Bode plot only closed-loop: need open-loop G(jω). If only closed-loop Bode, cannot directly design; use closed-loop specs (Mr, bandwidth) or reconstruct plant. If Bode is plant open-loop, above applies. Also if compensator inserted in forward path and unity feedback; if not, open-loop transfer L(s) = Gc(s)G(s) at crossover. Need maybe include "step by step" in numbered list. Provide formulas clearly. Maybe include a small illustrative example? The user didn't provide numeric plot. We can create a generic example? Could be useful but may be overkill. Desired oververbosity 5. We can include a short "contoh alur" with hypothetical numbers to illustrate substitution. But final should be concise enough. Maybe include "jika misalnya %OS=10% ..." to clarify. Let's consider adding a mini example with arbitrary values: %OS=16.3% -> ζ=0.5 -> PM≈51.8° + margin=56.8°. Suppose at chosen ω_c'=5 rad/s, Bode gives ∠G=-140° and |G|=-6dB. PM_uncomp=40°. φ_need=16.8°. a=(1+sin16.8)/(1-sin16.8)=? sin16.8=0.289; a=1.289/0.711=1.813. z=5/√1.813=5/1.347=3.71; p=5*1.347=6.735. Kc=√a/|G|; |G|=10^-6/20=0.501; Kc=1.347/0.501=2.688. Gc=2.688(s+3.71)/(s+6.735). Check. But we need avoid confusion. A generic example helps. But maybe the problem expects "how to design phase lead from Bode and %OS" not actual numbers. We can include "Template perhitungan". Need be accurate with PM formula. Let's verify PM formula from ζ. For second-order closed-loop with open-loop G(s)=ω_n²/[s(s+2ζω_n)], phase margin PM = arctan(2ζ/√(√(1+4ζ^4)-2ζ²)). Derive: At gain crossover ω_c=ω_n√(√(1+4ζ^4)-2ζ²)? Actually for G(s)=ω_n²/[s(s+2ζω_n)], magnitude =1: ω²( (ω²)+(2ζω_n)²? Wait |G|=ω_n²/[ω√(ω²+4ζ²ω_n²)]? Set 1 => ω^4+4ζ²ω_n²ω²-ω_n^4=0 => (ω²/ω_n²)^2+4ζ²(ω²/ω_n²)-1=0. Let x=ω²/ω_n² = -2ζ²+√(4ζ^4+1). PM=180+∠G(jω_c). ∠G=-90°-atan(ω/(2ζω_n))? PM=90° - atan(ω_c/(2ζω_n)) = atan(2ζω_n/ω_c). Let x=ω_c²/ω_n² = √(1+4ζ^4)-2ζ². Then tan PM=2ζ/√x. PM=atan(2ζ/√(√(1+4ζ^4)-2ζ²)). Yes. Good. Need handle "desired %OS" formula: Mp = exp(-ζπ/√(1-ζ²)) for 0<ζ<1. Solve ζ = |ln(Mp)| / √(π²+ln²(Mp)). Actually ln(Mp) negative. Use -ln(Mp). Yes. Need note "PM_desired = PM_formula + margin". The margin accounts for approximations, especially if system order >2. Usually 5°-12°. Need discuss choosing new crossover frequency ω_c'. This is tricky. Since %OS only defines PM, not crossover. Need choose based on desired bandwidth/settling time. If not specified, common choices: - Use original gain crossover frequency ω_c0 (if possible) to avoid changing speed too much. But adding lead changes magnitude at ω_c0; to keep ω_c0 as crossover, Kc chosen accordingly. However the lead phase at ω_c0 might not be maximum if we set ω_m=ω_c0? If we set ω_m=ω_c0 and Kc to make magnitude 0 at ω_c0, yes PM at ω_c0 equals PM_old + φ_m. But adding lead also changes magnitude at other frequencies; original ω_c0 remains 0dB by construction. Could be okay. But if original PM insufficient and φ_need at ω_c0 is e.g. 40°, lead can provide. This increases high-frequency gain and may shift crossover? Wait at ω_c0 total magnitude is 0dB exactly. Since original magnitude at ω_c0 was 0dB; lead factor at ω_c0=1/√a<1; to compensate, Kc>1, so total magnitude at low frequencies? At ω_c0 =1. But adding Kc>1 and lead factor <1, total at other frequencies might be >0dB at lower frequencies? Let's check original G decreases with frequency. If Kc>1, total magnitude at low freq may be boosted, potentially another crossover? But original magnitude at ω_c0=0; for ω<ω_c0, original >0. Kc>1 and lead factor near α (at low) maybe? total = Kc*α? Wait Gc(0)=Kc α=1/√a? Actually Kc=√a/|G(ω_c0)|, with |G|=1 at original crossover, so Kc=√a. DC gain Gc=Kc α=√a*(1/a)=1/√a<1. At low freq total = original (high) * Gc(low). Original at low may be large; product could be >1. So another crossover at lower frequency may exist. Need verify. Standard design often shifts crossover to higher frequency to avoid low-frequency gain increase? Let's analyze: If original G has high DC gain and crosses 0dB at ω_c0. Add lead with max at ω_c0 and Kc=√a (if |G|=1) has DC gain 1/√a (<1). At very low frequency, original G >>1 (if type 1 etc), product huge. It will cross 0dB at lower frequency where original G drops to √a. So actual gain crossover may be lower, not at ω_c0. Oops! The condition |GcG|=1 at ω_c0 doesn't guarantee it's the first/only crossover. If product >1 below, it crosses earlier. The actual PM may be at lower frequency where phase is less lag (better PM) or worse? For typical low-pass, lower frequency phase less lag, PM higher. But crossover frequency shifts lower (slower). Need verify. Standard lead design chooses new gain crossover frequency where compensated magnitude is 0dB and often this is at ω_m. To ensure, they set ω_m to new crossover, but then use the condition |G(jω_m)| = √α? Wait with Kc chosen? Let's derive if Kc is set to maintain steady-state or fixed. But if Kc adjustable, we can choose Kc to make total 0dB at ω_c'. However low frequency product may cross earlier. To avoid, maybe choose ω_c' as the actual first crossing, which means the uncompensated magnitude at ω_c' must satisfy some condition based on Kc. But Kc also sets low-frequency gain. This is complex. Let's recall standard phase-lead design from Bode for desired PM, with steady-state error satisfied first (Kc fixed): - Set Kc to meet error. - Plot Bode of uncompensated KcG. - Find required phase lead φ_m = PM_des - PM_old + margin at new crossover? Actually PM_old at original crossover? No, because adding lead changes crossover. Procedure: 1. Determine φ_m = PM_des - PM_old + margin (where PM_old is at original gain crossover? But adding lead changes magnitude and crossover; not exact.) 2. Calculate α = (1-sin φ_m)/(1+sin φ_m)? Or attenuation factor? Note: For lead Gc=Kc (1+Ts)/(1+αTs) with α<1? Wait. 3. Find frequency ω_m where magnitude of uncompensated KcG is -10 log(1/α)? Let's derive. If compensator is Gc=Kc_lead? Need use standard form where DC gain is Kc. Suppose Gc(s)=Kc (1+Ts)/(1+αTs), α<1 (zero at -1/T, pole at -1/(αT), p>z; DC gain Kc). At ω_m=1/(T√α)? Let's find max phase: Tω_m =1/√α. Magnitude at ω_m = Kc * [√(1+(ω_m T)^2)/√(1+(αω_m T)^2)] = Kc * [√(1+1/α)/√(1+α)] = Kc/√α. Wait compute: numerator sqrt((α+1)/α); denominator sqrt(1+α); ratio=1/√α. Yes. So compensator gain at ω_m is Kc/√α > Kc. To make total 0dB at ω_m, need |KcG(jω_m)| = √α. In dB: |KcG|_dB = 20log√α =10logα (negative since α<1). So find frequency on uncompensated magnitude plot where magnitude is -? 10logα dB. For example φ_m=40°, α=(1-sin40)/(1+sin40)=? (0.357)/(1.643)=0.217; 10logα=-6.63 dB. So find frequency where uncompensated (after Kc) magnitude is -6.63 dB. Set ω_m there. Then T=1/(ω_m√α). Zero z=1/T=ω_m√α; pole p=1/(αT)=ω_m/√α. This matches our earlier with α=z/p? Let's map: In Gc=Kc(1+Ts)/(1+αTs), zero z=1/T, pole p=1/(αT), p>z. α=z/p. At ω_m=1/(T√α)=√(zp). Magnitude at ω_m = Kc/√α. To set total =1, need |G(jω_m)| = √α/Kc. If Kc=1 (or pre-set), then original magnitude =√α (negative dB). This is standard. But if Kc is not fixed? In the design, Kc is usually set to meet steady-state, then α and T chosen to meet PM. The new crossover ω_c' becomes ω_m. The final compensator includes Kc as DC gain (not the high-frequency asymptote). If the user only has Bode plot of G (not including Kc), and %OS, they might set Kc=1? But then low-frequency gain/steady-state may not be considered. Or they can choose Kc later. Need decide. Let's read wording: "cara membuat phase lead compensator jika data yang disediakan adalah Bode Plot Frequency Response dan %OS yang diinginkan". In labs, they may have a Bode plot of a plant (uncompensated open-loop) and desired %OS. They want to design a phase lead compensator. Usually steps in control textbooks for lead compensator with Bode plot: 1. Determine Kc from steady-state error. But no steady-state error specified. Maybe they just design for %OS, so choose Kc to meet magnitude crossover? Or Kc maybe not needed? Phase lead compensator formula often Gc(s)= (s+z)/(s+p) times gain? They may set gain to 1? Hmm. Let's consider typical lab question: Given Bode plot frequency response and desired overshoot, design phase lead compensator. They might expect: - Calculate damping ratio from %OS. - Calculate desired phase margin. - From Bode plot, determine current phase margin and gain crossover frequency. - Calculate required phase lead φ_m = desired PM - current PM + margin. - Determine α from sinφ_m = (1-α)/(1+α) or similar. - Find new gain crossover frequency where uncompensated magnitude = -20log(1/√α)? Need formula depending on compensator form. - Calculate T = 1/(ω_c√α)? - Construct Gc(s)= (1+αTs)/(1+Ts) or etc. This is classic. Let's reconstruct textbook procedure for "lead compensator design using Bode plot": Compensator: Gc(s) = Kc * α * (1+Ts)/(1+αTs)? Or Gc(s)=Kc * (1+Ts)/(1+αTs) with α<1? Let's recall from Ogata "Modern Control Engineering" phase lead compensator: Gc(s) = Kc * α * (1+Ts)/(1+αTs), α<1? Wait I recall a formula: Gc(s)=Kc α (1+Ts)/(1+αTs) where α<1? Let's check DC gain = Kc α? Actually at s=0, Gc(0)=Kc α. High frequency =Kc. The zero at -1/T, pole at -1/(αT), p>z. This matches our explicit Gc=Kc(s+z)/(s+p) with z=1/T, p=1/(αT), because Gc=Kc(s+1/T)/(s+1/(αT)) = Kc * α(1+Ts)/(1+αTs). Yes. So Ogata uses Gc=Kc α(1+Ts)/(1+αTs) maybe with Kc chosen to satisfy error? Let's recall: "Gc(s)=Kc α (1+Ts)/(1+αTs)"? At s=0 gain=Kc α. If Kc chosen for error, DC gain is Kc α? Hmm maybe they set Kc=1? Let's verify memory: In Ogata, lead compensator: Gc(s)=Kc (s+z)/(s+p) = Kc [ (s+1/T)/(s+1/(αT)) ] = Kc [ (1+Ts)/(1+αTs) ]? Wait they might define Gc(s)=Kc * (s+1/T)/(s+1/(αT)) = Kc * α(1+Ts)/(1+αTs). If they want DC gain = Kc, they set Gc(s)=Kc * (1+Ts)/(1+αTs)? But at DC that's Kc, not Kc α. Let's derive: Gc(s)=Kc * (1+Ts)/(1+αTs) has zero -1/T, pole -1/(αT). At s=0 gain=Kc. At high frequency gain=Kc/α. This is a lead compensator with DC gain Kc. Its zero is at -1/T and pole at -1/(αT) (p>z) because α<1. This is simpler. It is equivalent to Kc(s+1/T)/(s+1/(αT))? Let's check: Kc*(1+Ts)/(1+αTs) = Kc * [T(s+1/T)]/[αT(s+1/(αT))] = (Kc/α) (s+1/T)/(s+1/(αT)). So high-frequency gain Kc/α. If define K_high=Kc/α. Our explicit Gc=K_high(s+z)/(s+p) with z=1/T, p=1/(αT). Then DC gain=K_high z/p= (Kc/α)*α=Kc. Good. So standard form Gc=Kdc(1+Ts)/(1+αTs), α<1, has DC gain Kdc. At ω_m=1/(T√α), magnitude = Kdc/√α? Let's compute: Gc(jω_m) magnitude = Kdc * sqrt(1+(ω_m T)^2)/sqrt(1+(αω_m T)^2). ω_m T=1/√α. numerator sqrt(1+1/α)=√((α+1)/α); denom sqrt(1+α); ratio =1/√α. So yes magnitude=Kdc/√α. Phase max φ_m = arcsin((1-α)/(1+α))? Let's derive: Gc=Kdc(1+Ts)/(1+αTs). Phase = atan(ωT)-atan(αωT). At ω_m=1/(T√α): atan(1/√α)-atan(√α). Since α<1. sin φ_m =? Formula: tan φ_m =? At max, sin φ_m = (1-α)/(1+α). Let's check α=0.217: (1-0.217)/(1.217)=0.643; arcsin=40°. Yes. So α = (1 - sinφ_m)/(1 + sinφ_m). Good. This is likely the standard formula. The explicit Gc=Kc(s+z)/(s+p) with p>z corresponds to Kc=K_high and α=z/p. But standard form with DC gain Kdc=Kc_high α? Let's be consistent. Maybe present standard form Gc(s)=K_c \frac{1+Ts}{1+\alpha Ts}, 0<α<1. This is widely recognized as lead compensator with DC gain K_c. Then: - Max phase φ_m = arcsin((1-α)/(1+α)). - Solve α=(1 - sin φ_m)/(1 + sin φ_m). - Frequency of max phase ω_m=1/(T√α). - Magnitude at ω_m = K_c/√α? Wait K_c is DC gain. Yes. - To make ω_m the new gain crossover: |Gc(jω_m)G(jω_m)|=1 => |G(jω_m)| = √α / K_c? If K_c not specified? If K_c chosen to meet steady-state or set to 1? Hmm. If only %OS and Bode plot G, we can choose K_c (DC gain) to achieve magnitude condition. But then there are two unknowns K_c and T/α. Actually α from φ_m. T from ω_m. K_c from magnitude condition: K_c = √α / |G(jω_m)|. Wait total = K_c/√α * |G| =1 => K_c = √α/|G|. Note K_c is DC gain of compensator. If original Bode G at ω_m has negative dB, K_c may >? If |G| less than √α, K_c>1. This design sets DC gain to achieve crossover at ω_m. But does it preserve steady-state? No. But no steady-state requirement given. This is a valid design for %OS and bandwidth (ω_m chosen). However, note if G has high DC gain, adding K_c might change low-frequency. But we choose K_c possibly less than 1 if |G| large at ω_m. Wait if ω_m chosen where |G| maybe large (negative? if above 0dB then |G|>1), K_c =√α/|G| <1. That may reduce DC gain and could eliminate low-frequency crossover. Need choose ω_m such that final actual crossover. But standard procedure with K_c fixed? Let's think. If we use standard form Gc=Kdc(1+Ts)/(1+αTs), and choose ω_m as new gain crossover, we need total magnitude at ω_m=1. This gives Kdc=√α/|G(ω_m)|. But as noted, total magnitude at lower frequencies may also cross. If original G is large at low frequency (type 1 integrator), and Kdc chosen maybe reduces DC enough to make only one crossing? Let's analyze type 1: G(jω) ~ K/(jω). At low freq |G| huge. At ω_m chosen, |G| maybe <1? For a typical uncompensated system, gain crossover ω_c0 where |G|=1. If choose new crossover ω_m > ω_c0 (faster), |G(ω_m)|<1. Then Kdc=√α/|G| >√α >? >1? Actually |G|<1, so Kdc>√α. At low freq, total |GcG| ~ Kdc * |G| ~ Kdc K/ω. It will cross 0dB at frequency ω_low = Kdc K (if type1) vs original ω_c0 = K. Since Kdc>1, new low-freq crossover is higher than original? Wait type1 |G|=K/ω. Original ω_c0=K. Compensated low-freq: |GcG| ~ Kdc * K/ω. Cross at ω=Kdc K > original. But this low-frequency approximation valid below zero. If ω_m > original, the product may cross at Kdc K which is > original and maybe near ω_m. Let's compute actual: At high freq, lead gain tends Kdc/α? Wait Gc DC Kdc, high freq Kdc/α >Kdc. At low freq Gc≈Kdc. Total low freq ≈Kdc*K/ω. At high freq Gc≈Kdc/α, G≈K/ω, total ≈(Kdc/α)K/ω. The lead increases gain at high freq, so crossover may be at higher frequency than original. It might still be at ω_m if chosen appropriately. Actually adding lead increases gain at high frequency; with Kdc>1, gain increases across all? At low freq Gc=Kdc, at high Kdc/α>Kdc. So total magnitude is boosted at all frequencies relative to original if Kdc>1. Thus 0dB crossing shifts higher. At ω_m, total set to 1. For frequencies below ω_m, total might be >1 if original slope -20dB/dec and crossover lower? Let's calculate type1 with Kdc chosen so total at ω_m=1. At ω_m, Gc magnitude =Kdc/√α. Need Kdc=√α/|G(ω_m)|. If |G(ω_m)|=K/ω_m. Kdc=√α ω_m/K. Total at arbitrary ω: Gc magnitude varies from Kdc at low to Kdc/α at high. For ω<ω_m, Gc maybe less than at ω_m if ω_m is above? For lead, magnitude increases with frequency. At lower ω, Gc < Kdc/√α. But G is larger. Need product. Let's use asymptotic for type1: |GcG| ≈ Kdc * K/ω for ωp, ≈(Kdc/α)K/ω; between, increases? At ω_m, set 1. If ω<ω_m, Gc smaller than at ω_m but G larger. Product? For pure integrator and lead, product = Kdc K * |1+jωT|/(ω|1+jαωT|). At ω_m total=1. Could there be lower crossing? Let's evaluate ω→0 product ~Kdc K/ω →∞. So yes product starts ∞, decreases, crosses 0dB at some lower frequency, then due to lead flattening maybe could cross again? Wait if product at low ∞ and at ω_m=1, if product monotonically decreasing, then it would be >1 for all ω<ω_m and =1 at ω_m. No lower crossing. The actual first crossing from ∞ to below 1 occurs at ω_m. Because product might decrease monotonically until lead phase? Let's check product asymptote: For ω1 indeed. So no lower crossing. I earlier confused original Gc low freq with original G: total at low = Kdc K/ω. If Kdc>1, the crossover where this asymptotic equals 1 is ω=Kdc K, which is > original K. At ω_m we set total=1. If ω_m > original, Kdc=√α ω_m/K. Then Kdc K =√α ω_m. This is less than ω_m (since √α<1). But this is the frequency where low-frequency asymptote crosses 1. At ω_m total=1. The product at ω=KdcK might be affected by lead and not exactly 1. Let's numerically: α=0.217, √α=0.466. choose ω_m=10, K=1. Kdc=4.66. At ω=KdcK=4.66, low-freq total=1. But actual Gc at ω=4.66: T=1/(ω_m√α)=1/(10*0.466)=0.214. Gc magnitude=Kdc sqrt(1+(ωT)^2)/sqrt(1+(αωT)^2). ωT=4.66*0.214=1; numerator sqrt2; αωT=0.217*4.66*0.214? Wait αωT =? ω_mT=1/√α=2.146; at ω=4.66, ωT=1. Denom sqrt(1+0.217²)=sqrt1.047=1.023; ratio sqrt2/1.023=1.383. Gc mag=4.66*1.383=6.445. G mag=K/ω=0.2146. Total=1.383. So actual crosses later? At ω_m=10: ωT=2.146; Gc=4.66*sqrt(1+4.604)/sqrt(1+0.217*4.604?) wait αωT=0.217*2.146=0.466; square=0.217; denom sqrt1.217=1.103; numerator sqrt5.604=2.367; ratio=2.145? Actually 2.367/1.103=2.146 =1/√α. Gc=4.66*2.146=10.0. G=0.1. total=1. Good. At ω=4.66 total=1.383 >1. So product remains >1 below ω_m. It crosses at ω_m? It might cross from above to below at ω_m? But at high ω, Gc→Kdc/α=4.66/0.217=21.47; G=1/ω; total→21.47/ω, so for ω>ω_m, total <1? At ω_m=10, total=1; for ω>10, total<1. So indeed product decreases? But we found at ω=4.66 total>1; at ω_m=10 total=1. So product decreases from ∞ to 1 at ω_m then below. So ω_m is first crossing from above to below. Good. So no lower crossing. The low-freq asymptote crossing at 4.66 was overestimate because lead adds extra gain? Wait product at 4.66 >1. It crosses at 10. Good. So choosing Kdc=√α/|G(ω_m)| works to make ω_m the actual crossing if original G monotonic and lead phase shape? Likely. Thus standard design with Gc=Kdc(1+Ts)/(1+αTs) and Kdc chosen to satisfy |GcG|=1 at ω_m works. But typical textbook with Kc fixed to steady-state finds ω_m from |KcG| = √α? Let's compare: If Kc fixed (DC gain), then to make ω_m crossover, we need |KcG(ω_m)| = √α? Because Gc magnitude at ω_m = Kc/√α. Set total =1 => |KcG| =√α. Yes. So find frequency on uncompensated after Kc where magnitude =√α (in linear), i.e., 20log|KcG| =10logα (negative dB). If Kc not fixed but we choose to meet %OS and some ω_m, we instead choose ω_m first and calculate Kc = √α/|G(ω_m)|. But then Kc is DC gain. Note if we want maintain steady-state, Kc is set first and ω_m determined by magnitude condition. Which approach is more common in exam? Since only Bode plot and %OS, they might set Kc=1? Or they may not care about steady-state and set Kc to satisfy magnitude at chosen ω_c. Hmm. Let's see typical design steps for lead compensator from Bode plot and desired PM, assuming compensator Gc(s)=Kc α (1+Ts)/(1+αTs)? There is a known algorithm: 1. From desired PM, calculate φ_m = PM_des - PM_old + ε. (PM_old at original crossover? But new crossover unknown.) 2. Calculate α = (1 - sin φ_m)/(1 + sin φ_m). (α<1) 3. Determine new gain crossover frequency ω_c from uncompensated Bode plot where |G(jω_c)| = -20log(1/√α)? Actually because at max phase, magnitude of lead is 1/√α if DC gain 1? Wait if compensator Gc(s)= (1+Ts)/(1+αTs), DC gain=1. At ω_m, magnitude=1/√α. To make total 0dB, need |G(jω_m)| = √α. In dB: -20log(1/√α)? Let's derive: total = |G|*(1/√α)=1 => |G|=√α. In dB |G| = 20log√α =10logα. Since α<1, negative. So find frequency where uncompensated magnitude is 10logα dB (negative). Example α=0.1, find where |G|=-10dB. This is common: "find frequency where magnitude of uncompensated system is -20 log(1/√α)"? Since 1/√α>1, -20log(1/√α)= -10log(1/α)=10logα. Yes. 4. Set ω_m=ω_c. Then T=1/(ω_c√α). Gc(s)= (1+Ts)/(1+αTs). This assumes DC gain Kc=1. But if steady-state error needs Kc, include Kc as separate gain: Gc(s)=Kc(1+Ts)/(1+αTs). If Kc chosen first to meet error, then the magnitude condition uses KcG: find ω_c where |KcG|=√α. If Kc=1, above. Then final. This design uses Kc as a pre-specified gain (often set to satisfy error). If no error spec, Kc=1 is arbitrary but yields a lead compensator with unity DC gain. However adding lead with unity DC gain increases high-frequency gain; it may alter crossover. The new crossover ω_c is chosen as frequency where original magnitude is √α (negative dB) so that after lead gain 1/√α it becomes 0dB. This is elegant and doesn't require choosing ω_m arbitrarily; it's determined by α. But if desired %OS only, we need choose φ_m. But φ_m calculation uses PM_old at what frequency? Standard procedure uses "PM_old" at original gain crossover? Let's derive. In this design, the new gain crossover is not original. The required lead angle φ_m is determined from desired PM minus the uncompensated phase at the new gain crossover? But that's unknown until α chosen. Standard approximation: first compute required φ_m = PM_des - PM_old + margin, where PM_old is phase margin at original gain crossover? Actually adding lead changes crossover to lower frequency (where original magnitude is negative dB), so uncompensated phase at new crossover is less negative (more positive) than at original? Wait original crossover is where |G|=0dB. New crossover where |G|=√α<1 (negative dB), which for low-pass is at higher frequency? Magnitude decreases with frequency; |G| <1 at frequencies above original crossover. So new crossover is higher frequency (since original |G|=1; to get |G|<1, go to higher ω). At higher frequency, phase is more negative, so PM_old at new crossover is worse than at original. Adding lead compensator (with DC gain 1) increases magnitude at high frequency; to keep 0dB at new crossover, the original magnitude must be below 0dB. So crossover shifts higher. Thus the phase lag at new crossover is greater, so required lead φ_m should be PM_des - PM at new crossover + margin. But standard approximation uses PM_old at original? Let's recall: Since lead compensator increases phase and magnitude, the new crossover is at higher frequency. The phase of uncompensated at new crossover is not known until α. Many textbook design procedures use iterative or use Nichols chart. However an approximate method: - Determine φ_m = PM_des - PM_old + ε where PM_old is at original gain crossover. This underestimates required lead because new crossover has more lag. But if the gain crossover does not shift much? But it does shift. Hmm. Actually standard lead design from Bode: "Determine the necessary phase lead angle φ_m to be added to the system, including a safety margin. The phase margin required is PM_des. Since the gain crossover frequency will be shifted, we first evaluate the phase margin at the new gain crossover after choosing α? This is circular." Ogata procedure: 1. Assume Gc(s)=Kc α (1+Ts)/(1+αTs)? Determine Kc from steady-state. 2. Evaluate φ_m = PM_des - PM_old + ε, where PM_old is the phase margin at the original gain crossover? Then calculate α. Then find new gain crossover where |KcG|=1/√α? Wait. Let's search memory. Standard design for lead compensator: Gc(s)=Kc (1+Ts)/(1+αTs), α<1? Or Kc α? The procedure from Ogata: (1) Determine Kc to satisfy steady-state error. (2) Using the Kc determined, draw Bode plot of KcG(s). (3) Determine the phase margin of KcG(s). If PM not enough, determine the required phase lead φ_m = desired PM - PM + margin. (Here PM at gain crossover of KcG(s).) (4) Determine α from sin φ_m = (1-α)/(1+α). (5) Determine the frequency where magnitude of KcG(s) is -20log(1/√α)? Wait. (6) Set ω_m as new gain crossover. Calculate T = 1/(ω_m√α). (7) Gc(s)=Kc α(1+Ts)/(1+αTs)? Need. Actually if Gc(s)=Kc α (1+Ts)/(1+αTs) with DC gain Kc α? Hmm. Let's derive for Gc=Kc(1+Ts)/(1+αTs) with Kc fixed. At ω_m, |Gc|=Kc/√α. To make |GcG|=1, need |G(ω_m)|=√α/Kc. So find frequency where uncompensated |G| = √α/Kc. If using KcG, need |KcG|=√α. In dB: 20log|KcG| =10logα. This matches step. The phase at new crossover is ∠G(ω_m)+φ_m. The design condition for PM: 180+∠G(ω_m)+φ_m = PM_des. Thus φ_m = PM_des -180 -∠G(ω_m). But φ_m also determined by α. The procedure using original PM (at original ω_c) to find φ_m is an approximation because ω_m not yet known. But after calculating α and ω_m, one should verify PM. If not enough, iterate. Some textbooks suggest choosing φ_m = desired PM - current PM + margin, then find ω_m where |G| condition; because adding lead shifts crossover, they add margin to account. This is standard. Given the question likely from a lab or class, they expect a straightforward procedure. We can present a robust step-by-step that includes iteration. We can mention two approaches: - Approach A (recommended iterative, with chosen new crossover ω_c'): choose ω_c' based on speed; compute required lead at that frequency; set zero/pole around it; choose gain to make 0dB at ω_c'; verify. This is conceptually clear and directly uses Bode plot data at chosen frequency. It avoids circularity because you choose ω_c' first. This is valid and easy to explain. It uses Kc as high-frequency gain in explicit form or DC gain in standard form. It may yield a compensator that meets PM at chosen crossover by construction (if no other crossings). This might be best for answer. - Approach B (classic Bode): use φ_m from desired PM and original PM, compute α, find ω_m where |G| condition (with Kc fixed or unity), set T. But since only %OS no bandwidth, need choose Kc? Hmm. We can provide "Langkah ringkas" with chosen new crossover. That is more general and avoids ambiguity. Need mention "karena %OS hanya menentukan redaman, tidak menentukan kecepatan; pilih ω_c' secara engineering atau ikuti spesifikasi bandwidth." Then design. Let's make final answer comprehensive but not overly verbose. Use Indonesian. Include equations. Let's plan final: Title: Desain phase lead compensator dari Bode plot + %OS. Assumptions: - Bode plot is open-loop plant G(jω) in forward path with unity negative feedback. - Controller in series: Gc(s)G(s). - Desired closed-loop dominated by second-order; %OS used as approximation. For higher order verify simulation. - If there is steady-state error spec, Kc must be set first; otherwise Kc adjusted to achieve crossover. Mention. Define compensator: Gc(s)=Kc * (s+z)/(s+p), p>z>0. Equiv Gc(s)=Kdc*(1+Ts)/(1+αTs), α=z/p<1? Wait equivalence: Kc(s+z)/(s+p) = Kc*(s+z)/(s+p). Let z=1/T? p=1/(αT), α=z/p? Let's map to Kdc form: Let α = z/p <1. Then Gc(s)=Kc (s+z)/(s+p). Since p=z/α? Wait p>z => α=z/p<1. Gc=Kc * [ (s+z)/(s+z/α) ] = Kc * [ (1+s/z)/(1+s/(z/α))? ] = Kc * [ (1+s/z)/(1+ (α s)/z) ]? Let's use T=1/z. Then z=1/T. p=1/(αT). Gc=Kc*(s+1/T)/(s+1/(αT)) = Kc * α (1+Ts)/(1+αTs). So DC gain=Kc α. If we instead use standard Gc=Kdc(1+Ts)/(1+αTs) with Kdc=Kc α. At ω_m, magnitude=Kdc/√α. The explicit Kc here is high-frequency gain. To avoid confusion, present standard form Gc(s)=K_c \frac{1+T s}{1+\alpha T s}, 0<α<1. This is a lead compensator because zero at -1/T and pole at -1/(αT) (zero lebih dekat origin? Let's check: zero -1/T, pole -1/(αT). Since α<1, 1/(αT) >1/T, pole more negative/farther left. On negative real axis, zero is to the right of pole? For stability, both negative. Zero at -1/T (closer to origin), pole at -1/(αT) (farther left). So zero closer to origin than pole? Actually on real axis, origin at 0; zero at -1/T, pole at -1/(αT) more negative. So zero is closer to origin. Lead zero-pole pair: zero closer to origin than pole. Yes. In time domain, derivative action. At s=0, gain=Kc (DC gain). At high frequency, gain=Kc/α>Kc. This is a lead compensator. This form is standard and easier. Use Kc as DC gain. But if Kc chosen to achieve crossover, we set Kc=√α/|G(ω_c')|? Wait with this form, at ω_m, |Gc|=Kc/√α. Set total=1 => Kc=√α/|G|. Yes Kc is DC gain. Note Kc may be less than 1 if |G|>√α. That's fine. But if steady-state error, Kc should be set from error; then choose ω_m where |G(ω_m)| = √α/Kc. Need present both. Let's use Gc(s)=K_c \frac{1+T s}{1+\alpha T s}, α<1. The zero z=1/T, pole p=1/(αT). Max phase at ω_m=1/(T√α). Max phase φ_m=arcsin((1-α)/(1+α)). Inverse: α=(1 - sin φ_m)/(1+sin φ_m). Magnitude at ω_m: |Gc(jω_m)|=K_c/√α. Note DC gain K_c. This form is likely expected. But the question says "phase lead compensator" often written Gc(s)=Kc (1+aTs)/(1+Ts)? Let's check sign. If someone uses Gc(s)=Kc (1+αTs)/(1+Ts) with α>1? Let's avoid by defining clearly. We can also give explicit form to avoid α ambiguity: Gc(s)=K_c (1+T s)/(1+\alpha T s), 0<α<1. This is a lead because α<1. If they use opposite α, they can adjust. Good. Now steps with this form: 1. Dari %OS: Mp=%OS/100; ζ = -ln(Mp)/√(π²+ln²(Mp)). PM_d = atan(2ζ/√(√(1+4ζ^4)-2ζ²)) + margin (5°–12°). Or approx PM_d ≈100ζ + margin. Note in degrees. 2. Baca Bode plot G(jω): - Tentukan ω_c0 (0 dB crossing) dan PM0=180°+∠G(jω_c0). - Tabelkan |G|dB dan ∠G vs ω untuk interpolasi. - Tentukan ω_c' (new crossover) based on desired speed/bandwidth. If no spec, choose ω_c' around ω_c0 or where needed lead ≤60°. Need check phase at ω_c': PM_uncomp=180°+∠G(jω_c'). Compute required lead at that frequency: φ_m = PM_d - PM_uncomp(ω_c'). If φ_m<0 choose lower ω_c'? Actually if PM_uncomp already >PM_d, no lead needed. If φ_m>60°, choose another ω_c' or cascade two lead networks. Note if using margin already included in PM_d. - Since adding lead changes magnitude and may shift crossover, this is initial choice; verify later. 3. Calculate α from φ_m: sinφ_m = (1-α)/(1+α) => α=(1-sinφ_m)/(1+sinφ_m), with 0<α<1. (If φ_m negative set zero; if >60 use two stages). For double lead: design each with φ_m/2? Actually for two identical stages, total phase =2φ_single; set φ_single=φ_m/2? Max phases add if same ω_m; total φ=2 asin((1-α)/(1+α)). Solve α per stage from sin(φ_m/2) = (1-α)/(1+α). Or use formula. Mention. 4. Place max phase at desired new gain crossover: ω_m=ω_c'. For Gc form, ω_m=1/(T√α) => T=1/(ω_c'√α). Zero z=1/T=ω_c'√α; pole p=1/(αT)=ω_c'/√α. Check p>z. (Equivalent explicit zero/pole.) 5. Determine K_c (DC gain) so that total magnitude at ω_c' is 0 dB: |Gc(jω_c')|=K_c/√α (because ω_c'=ω_m). Condition: K_c/√α * |G(jω_c')| =1. So K_c = √α / |G(jω_c')|. If Bode gives G_dB, |G|=10^(G_dB/20). In dB: K_c,dB =20logK_c = -G_dB(ω_c') +20log√α = -G_dB +10logα. Note 10logα negative, so Kc dB = -G_dB - positive. Wait earlier for explicit form Kc_high dB = -G_dB +10log a. Here Kc is DC gain of standard form. At ω_m lead gain =Kc/√α; total: G_dB + Kc_dB +20log(1/√α)=0. 20log(1/√α) = -20log√α = -10logα. So Kc_dB = -G_dB +10logα? Let's solve: G_dB + Kc_dB -10log? Wait carefully: Gc(jω_m) magnitude = Kc/√α. In dB: 20log|Gc| =20logKc -20log√α = Kc_dB -10logα. Since α<1, -10logα positive. Set total: G_dB + (Kc_dB -10logα) =0 => Kc_dB = -G_dB +10logα. Yes 10logα negative. So Kc_dB = -G_dB + 10 log10 α. For example α=0.217, G_dB=0? Kc_dB=-6.63? But if G=1 at ω_m and lead gain at ω_m=Kc/√α, to total 0 dB we need Kc/√α=1 => Kc=√α<1, Kc_dB=10logα=-6.63 dB. Correct. If G_dB=-6.63, Kc_dB=-(-6.63)-6.63=0 dB. Good. Formula works. But note Kc can be <1, reducing DC gain. If steady-state error spec requires Kc fixed, do not use this; instead choose ω_m where G_dB = -Kc_dB +10logα? Let's derive fixed Kc: total at ω_m: G_dB + Kc_dB -10logα =0 => G_dB = -Kc_dB +10logα. Or |G(ω_m)| = √α/Kc. We'll mention. 6. Final compensator: Gc(s)=K_c \frac{1+T s}{1+\alpha T s}. Or in zero-pole form: Gc(s)=K_c \frac{s+1/T}{s+1/(\alpha T)}? Wait check: Gc=Kc(1+Ts)/(1+αTs). At s=0 gain Kc. To express as Kc'? If we write Gc(s)=Kc (s+z)/(s+p) with p>z, need multiply factor. Let's not convert unless needed. If using explicit form Gc(s)=K' (s+z)/(s+p), K'=Kc/α? Since Gc=Kc(1+Ts)/(1+αTs)=Kc * [T(s+1/T)]/[αT(s+1/(αT))] = (Kc/α)*(s+z)/(s+p). Here z=1/T, p=1/(αT). DC gain=(Kc/α)*(z/p)= (Kc/α)*α=Kc. Good. So if someone expects K(s+z)/(s+p), K=Kc/α. Note Kc is DC gain. Need mention to avoid confusion. We can present both: Gc(s)=K_c \frac{1+Ts}{1+\alpha T s} = \left(\frac{K_c}{\alpha}\right)\frac{s+z}{s+p} with z=1/T=ω_c'√α, p=1/(αT)=ω_c'/√α. Wait check z: T=1/(ω_c'√α); z=1/T=ω_c'√α. p=1/(αT)=ω_c'/(α√α)? Wait recalc! Let's compute p for standard form Gc=Kc(1+Ts)/(1+αTs). Pole at s=-1/(αT). With T=1/(ω_m√α). Then p=1/(αT)=1/(α * 1/(ω_m√α)) = ω_m√α/α = ω_m/√α. Yes. Zero z=1/T=ω_m√α. Good. p=ω_m/√α. Since α<1, p>z. Good. K_high = Kc/α? In explicit K(s+z)/(s+p), K_high = Kc/α? Let's verify DC gain: K_high*z/p = (Kc/α)*(ω_m√α)/(ω_m/√α)= (Kc/α)*α =Kc. Good. At ω_m, explicit K_high factor ratio √(z/p)=√α. Magnitude =K_high√α = (Kc/α)*√α=Kc/√α. Correct. Good. But earlier explicit form Gc=Kc_high(s+z)/(s+p) with Kc_high=√a/|G| etc. Here K_high=Kc/α. Since Kc=√α/|G|, K_high= (√α/|G|)/α =1/(√α|G|). That's large. Equivalent to previous explicit form with a=1/α? Previously we had Gc=K_high(s+z)/(s+p), p>z, a=p/z=1/α. At ω_m, magnitude=K_high/√a =K_high√α. Set total=1 => K_high=1/(√α|G|). Yes. Good. So if final using explicit form, K_high = 1/(√α |G(ω_c')|). Note no √a? Since a=1/α, √a=1/√α. Previous Kc=√a/|G|=1/(√α|G|). Same. Good. This might confuse with Kc standard. We'll define clearly. 7. Verification: - Plot compensated open-loop Bode: |GcG| and ∠GcG. - Find actual 0 dB crossing. Compute PM=180°+phase. Check ≥ desired PM. Check gain margin and stability. - Simulate step response; measure %OS. Adjust margin/ω_c'/α if needed. - If PM too low, increase φ_m/α (smaller α) or choose ω_c' where uncompensated phase less lag; if lead angle too large >60°, use two lead stages. - If response too slow, increase bandwidth (ω_c') and re-design. - If steady-state error not acceptable, set Kc from error first and find ω_m from magnitude condition instead; then redesign T/α. Need include "cara membaca Bode" details: - Use log scale. For interpolation, linear between points in log frequency and dB. Phase interpolate linearly in log ω? Phase vs log ω can be interpolated linearly between points; for better, use local model. For exam, read from graph. - PM at candidate: phase might be shown as -180° to -270°; unwrap. PM = 180 + phase (if phase at 0dB). For compensated PM = 180 + ∠GcG at new crossing. If phase below -180, PM negative. Need maybe discuss "jika Bode plot yang diberikan adalah closed-loop frequency response". Add a note: If the Bode plot is closed-loop, first convert to open-loop or use closed-loop specs; the steps above assume open-loop plant. If only closed-loop Bode, you can estimate bandwidth and resonance peak Mr to infer ζ/PM? For second-order, Mr = 1/(2ζ√(1-ζ²)) for ζ<0.707; bandwidth etc. But %OS relates to ζ. However phase lead design typically uses open-loop Bode. Mention. Need maybe include "why phase lead compensator?" It increases phase margin and bandwidth. Good. Let's decide final answer length. Desired oververbosity 5. We can give numbered steps and formulas. Include a short example? Maybe no need, but a small illustrative "contoh perhitungan PM" might help. The question asks "jelaskan step by step". A list is enough. But adding "contoh singkat" with hypothetical numbers can clarify. However final might get long. Desired oververbosity 5 means moderate. We can include a concise example at end if not too long. But maybe skip to avoid confusion with arbitrary numbers. Instead include "jika angka dari plot dimasukkan..." Maybe mention one example in a parenthetical. Need be careful with "Bode Plot Frequency Response" maybe they mean the Bode plot is given as data, not transfer function. We can say use graph to read magnitude/phase at selected ω. If only graph no equation, design graphically: add curves. But formulas still need |G(jω_c')| and ∠G(jω_c'). Can read from plot. If we set ω_c'=ω_m and Kc, we can compute compensator curves and overlay. Good. Let's draft final in Indonesian: ``` Asumsi: Bode plot adalah open-loop G(jω), feedback unity, sistem dominan orde dua. %OS hanya memberi redaman; pemilihan bandwidth (ωc) bebas/spesifikasi tambahan. Bentuk kompensator lead: Gc(s)=Kc(1+Ts)/(1+αTs), 0<α<1 (zero -1/T, pole -1/(αT), p>z). ``` Then steps: 1. Hitung ζ from %OS. Mp=OS/100. ζ = -ln(Mp)/sqrt(pi^2+ln^2 Mp). 2. Hitung PM desired. PM_zeta = atan(2ζ/sqrt(sqrt(1+4ζ^4)-2ζ²)) (deg). PM_d=PM_zeta+margin (5-12 deg). If %OS=... maybe PM≈100ζ. 3. Baca Bode: - Cari ω_c0 where |G|=0dB; PM0=180+∠G. - Buat tabel/interpolasi. - Pilih ω_c' (new gain crossover) sesuai bandwidth/kecepatan. If not specified, choose near ω_c0 or such that φ_need ≤60°. (Adding lead tends increase bandwidth; if want same speed choose appropriate.) 4. Hitung lead needed at ω_c': ∠Gc(ω_c') = φ_need = PM_d - (180° + ∠G(ω_c')). If φ_need≤0: lead not needed for PM; choose other ω_c' or just gain. If >60: use double lead or choose other ω. 5. Compute α: α=(1 - sin φ_need)/(1+sin φ_need) (0<α<1). For double: use φ per stage = φ_need/2, α per stage similarly; total Gc product. 6. Compute T and zero/pole: Set ω_m=ω_c' (max phase at new crossover). T = 1/(ω_c'√α). zero z=1/T=ω_c'√α; pole p=1/(αT)=ω_c'/√α. Gc without gain = (1+Ts)/(1+αTs). 7. Compute gain Kc: At ω_c', |Gc(jω_c')|=Kc/√α. Need Kc/√α * |G(jω_c')|=1. Kc=√α/|G(jω_c')|. (If using dB: Kc_dB= -G_dB(ω_c') +10log10 α; because 20log(1/√α)=-10logα? Wait let's re-evaluate formula for Kc in standard form. Set total: G_dB + Kc_dB -10log10(1/α)? Wait I might have a sign error in formula presentation. Let's derive carefully one more time with standard Gc=Kc(1+Ts)/(1+αTs). At ω_m=1/(T√α), magnitude of Gc = Kc/√α. In dB: 20log|Gc| =20logKc -20log√α = Kc_dB -10logα. Since α<1, -10logα positive. Set total: G_dB + Kc_dB -10logα =0. Wait " -10logα" because 20log(1/√α) = -20log√α = -10logα. So Gc_dB = Kc_dB +20log(1/√α) = Kc_dB -10logα. Yes total: G_dB + Kc_dB -10logα =0 => Kc_dB = -G_dB +10logα. Since 10logα negative. For example α=0.1: Kc=√0.1/|G|. If G=1, Kc=0.316, Kc_dB=-10. Formula: -0+10log0.1=-10. Correct. So Kc_dB = -G_dB + 10 log10 α. Good. Note that 10logα is negative. Some might expect +? Let's test G=-10dB, α=0.1: Kc=√0.1/0.316=0.316/0.316=1 =>0dB. Formula: -(-10)+10log0.1=10-10=0. Good. So formula correct. We can write Kc=10^{(-G_dB +10logα)/20}? Actually Kc_dB = -G_dB +10logα. Then Kc=10^(Kc_dB/20). But simpler use linear. Alternatively in explicit zero-pole form Gc=K_high(s+z)/(s+p), K_high=1/(√α|G|). Since Kc/α? Wait if we use standard form, Kc is DC gain. If user expects Kc in Gc=Kc(s+z)/(s+p), then Kc_high = Kc/α? Need clarify. We can present final both. Final: Gc(s)=Kc * (1+Ts)/(1+αTs), with Kc as above. Equivalent zero-pole: Gc(s)= (Kc/α) * (s+z)/(s+p), z=ω_c'√α, p=ω_c'/√α. ``` 8. Verifikasi: - Plot/overlay Bode kompensasi: GcG. Cari actual 0dB crossing. PM_actual=180+∠GcG(jω_c,actual). Check PM_actual≥PM_d. Check GM. - Simulasi closed-loop step; measure %OS. Adjust. - If actual crossover different due to other crossings, iterate with new ω_c'. - If steady-state error spec, set Kc from error first; then find ω_m where |G(jω_m)|=√α/Kc (or |KcG|=√α). Then T=1/(ω_m√α). Do not use Kc formula. - If phase lead >60°, cascade two lead networks (or use lag-lead). Need include "Mengapa ω_m=ω_c'?" It maximizes PM at new crossover. Good. Let's ensure sign in phase lead needed: PM_uncomp =180+∠G. If ∠G=-120, PM_uncomp=60. Desired PM=50? φ_need=-10? Actually if desired lower than current, no lead. If desired PM=70, φ_need=10. Good. But if ∠G at candidate -170, PM_uncomp=10; desired 60; φ_need=50. Good. Need mention phase angle wrapping. If ∠G is shown as -180 to -270, unwrap. For example -190 equivalent +170? In Bode, phase usually continues decreasing. PM=180+phase. At 0dB, if phase=-190, PM=-10 unstable. Use unwrapped. If using wrapped, be cautious. Need maybe mention "margin safety": add 5-12 degrees because %OS formula assumes ideal second-order and no zeros. Use PM_d = PM_zeta + (5° to 10°). If using exact formula PM_zeta, margin maybe 5-12. If approximate PM=100ζ, already approximate, add margin. Need maybe mention "gain margin" not directly from %OS but check stability. "Bode plot frequency response" includes magnitude and phase; use PM and GM. If system has integrators, low frequency etc. %OS relates to damping; PM should be positive. If plant unstable or nonminimum phase, lead design tricky; verify. Let's consider if "Bode plot frequency response" might be closed-loop frequency response from which they want to design compensator? If closed-loop Bode given and desired %OS, one can convert %OS to ζ and then to closed-loop frequency domain specs (Mr, bandwidth) and design? But phase lead compensator is designed on open-loop. If only closed-loop Bode of current closed-loop, one can estimate current damping from Mr and bandwidth, then modify? That's unusual. The phrase "Bode Plot Frequency Response" likely means they have a Bode plot of the plant frequency response. We can include a caveat: "Pastikan Bode plot adalah open-loop G(jω). Jika yang diberikan closed-loop, langkah PM tidak langsung; perlu open-loop atau gunakan Mr/bandwidth." Good. Need maybe use "persen OS" relation to damping ratio formula. Good. Let's write final in a clear step-by-step manner. Use markdown. Keep formulas readable. Use "ω" and "°". Use "linear" vs "dB". Potential issue: In step 4, if we choose ω_c' before calculating φ_need, but adding lead changes phase at ω_c' and magnitude. We set ω_m=ω_c' to maximize lead. However the actual gain crossover may shift if total magnitude at ω_c'=0 and no other crossing. We choose Kc to make total at ω_c'=0. So actual crossover should be ω_c' if no other crossings. But as we saw for type1 with Kdc chosen, product monotonically decreasing and crosses at ω_m. Is that always? If original G has resonances, maybe multiple crossings. Verification step covers. So it's valid. Need mention "jika hasil verifikasi tidak memenuhi, ulangi dengan ω_c' baru." Good. Let's also present "Jika menggunakan bentuk Gc(s)=K(s+z)/(s+p) dengan p>z" maybe as a note because many write this. We can map from standard form: From Gc=Kc(1+Ts)/(1+αTs) with α=z/p? Wait in standard form, zero z=1/T, pole p=1/(αT). Here α=z/p. Yes α=z/p. Kc is DC gain. Equivalent Gc(s)=Kc(1+Ts)/(1+αTs) = (Kc/α)(s+z)/(s+p). So if writing as Gc(s)=K(s+z)/(s+p), p>z, then: K = Kc/α = 1/(α? let's derive from Kc=√α/|G| => K=√α/(α|G|)=1/(√α|G|). Wait √α/α=1/√α. So K=1/(√α|G|). Good. z = ω_c'√α. p = ω_c'/√α. Check K? At ω_m, explicit factor |(s+z)/(s+p)|=√(z/p)=√α. Magnitude Gc=K√α = [1/(√α|G|)]*√α =1/|G|. Total with G =1. Good. This explicit form has DC gain K*z/p = [1/(√α|G|)]*α = √α/|G| = Kc. Good. So if user uses explicit form, K = 1/(√α|G(ω_c')|). But wait if Kc standard =√α/|G|, K = Kc/α = (√α/|G|)/α =1/(√α|G|). Yes. This is high-frequency gain. Need clearly label. Let's avoid using Kc ambiguously. Use K_dc for standard form and K_high for explicit form. But final answer can say "Jika dosen/pustaka memakai bentuk Gc(s)=K(s+z)/(s+p), gunakan ...". Good. Let's decide final notation: Use standard lead: Gc(s)=K_c \frac{1+Ts}{1+\alpha Ts}, \quad 0<\alpha<1. This is clean. Then note equivalent zero-pole form. But many readers may expect K(s+z)/(s+p). To be safe, present both at start: ``` Gc(s)=K_c \frac{1+Ts}{1+\alpha Ts} = \frac{K_c}{\alpha}\frac{s+z}{s+p}, \quad p>z>0, \quad \alpha=\frac{z}{p}<1, \ z=\frac1T,\ p=\frac1{\alpha T}. ``` Wait in standard form, α =? Let's verify: Gc=Kc(1+Ts)/(1+αTs). Zero at -1/T, pole at -1/(αT). z=1/T, p=1/(αT). Then z/p = [1/T]/[1/(αT)] = α. Yes α=z/p. Good. The coefficient in explicit form is Kc/α. Good. At ω_m=√(zp)=1/(T√α). φ_m=arcsin((p-z)/(p+z))? With α=z/p, sinφ_m=(1-α)/(1+α). Yes. So present. Now formula for Kc (DC gain) when designing to meet crossover: At ω_c'=ω_m, |Gc(jω_c')|=Kc/√α. Set total=1: Kc=√α/|G(jω_c')|. Wait earlier with standard form DC gain Kc. At ω_m magnitude =Kc/√α. Set =1/|G| => Kc=√α/|G|. Yes. Note Kc is DC gain. Good. But in explicit form coefficient K=Kc/α =1/(√α|G|). Good. Let's double-check with α=0.217, |G|=-? Suppose |G|=0.5 at ω_m. Kc=0.466/0.5=0.932. At ω_m lead gain Kc/√α=0.932/0.466=2.0. Total=1.0. DC gain=0.932. High freq gain=Kc/α=4.29. Explicit K=Kc/α=4.29. At ω_m explicit factor √α=0.466, magnitude=4.29*0.466=2.0. Good. If using Kc=1/(√α|G|)? That's explicit K. Good. So in final, when I say "Kc" maybe use subscript "K_{dc}". To avoid confusion: Let $G_c(s)=K_{dc}\frac{1+Ts}{1+\alpha Ts}$. Then $K_{dc}$ is DC gain. At $\omega_c'$, $|G_c|=K_{dc}/\sqrt{\alpha}$, so $K_{dc}=\sqrt{\alpha}/|G(j\omega_c')|$. If using zero-pole $G_c(s)=K\frac{s+z}{s+p}$, then $K=K_{dc}/\alpha=1/(\sqrt{\alpha}|G(j\omega_c')|)$. This is clear. But wait: If $K_{dc}=\sqrt{\alpha}/|G|$, and $|G|$ from Bode might be >1 (positive dB) at chosen ω_c'. If ω_c' below original crossover, |G|>1, Kdc<√α<1. This reduces DC gain significantly. But adding lead has high freq gain Kdc/α = [√α/|G|]/α =1/(√α|G|). If |G|>1, high freq gain might still >1 if |G|<1/√α? etc. Fine. Need mention "Jika ada spesifikasi error steady-state, $K_{dc}$ harus ditentukan dari error, bukan dari rumus di atas. Setelah $K_{dc}$ ditetapkan, cari $ω_m$ pada Bode plot $K_{dc}G(jω)$ yang magnitude-nya $\sqrt{\alpha}$ (atau $G$ dB = -Kdc_dB+10logα), lalu $T=1/(ω_m√α)$." Let's derive correctly: With Kdc fixed, total at ω_m = |Kdc G| * (1/√α) =1 => |Kdc G| =√α. In dB: Kdc_dB + G_dB =10logα. So G_dB =10logα - Kdc_dB. Yes. Or find frequency where |KdcG(jω_m)|=√α. This is standard. Then T. This is useful if they set Kc to meet error. Include as note. Need maybe mention "phase lead compensator increases bandwidth and gain at high frequencies; if noise is an issue, be careful." Let's also address "Data Bode plot and %OS" but no transfer function: We can say "Jika G(s) tidak diketahui, cukup baca |G(jω)| dan ∠G(jω) dari grafik. Untuk ω yang tidak tepat di grid, interpolasi log-linear." Good. Let's write final answer now. Use Indonesian. It should be explanatory and step-by-step. Include final "ringkasan rumus". Keep moderate length. Use headings. Let's produce final. Berikut langkah praktis mendesain **phase lead compensator** dari data **Bode plot open-loop** $G(j\omega)$ dan spesifikasi **% overshoot (%OS)**. Saya asumsikan sistem feedback negatif unity dan Bode plot yang diberikan adalah Bode plot plant/open-loop $G(j\omega)$, bukan closed-loop. Jika yang diberikan closed-loop, perlu didapatkan/open-loop-nya dulu. Bentuk kompensator lead yang umum: \[ G_c(s)=K_c\frac{1+Ts}{1+\alpha Ts},\qquad 0<\alpha<1 \] atau ekuivalen dengan bentuk zero-pole: \[ G_c(s)=\frac{K_c}{\alpha}\frac{s+z}{s+p},\quad p>z>0 \] dengan \[ z=\frac1T=\omega_m\sqrt{\alpha},\qquad p=\frac1{\alpha T}=\frac{\omega_m}{\sqrt{\alpha}},\qquad \omega_m=\frac1{T\sqrt{\alpha}}. \] Di sini $\omega_m$ adalah frekuensi saat fase lead maksimum. Kompensator lead memberi fase positif pada rentang $\omega$ di antara zero dan pole, serta menaikkan gain pada frekuensi tinggi. --- ## 1. Ubah %OS menjadi damping ratio $\zeta$ Untuk sistem orde dua underdamped: \[ M_p=\frac{\%OS}{100} \] \[ \zeta=\frac{-\ln(M_p)}{\sqrt{\pi^2+\ln^2(M_p)}} \] dengan $0<\zeta<1$. ## 2. Hitung desired phase margin, $PM_d$ Hubungan $\zeta$ dan phase margin untuk sistem orde dua: \[ PM_{\zeta}=\arctan\left( \frac{2\zeta}{\sqrt{\sqrt{1+4\zeta^4}-2\zeta^2}} \right) \] dalam derajat. Sering juga dipakai aproksimasi: \[ PM_{\zeta}\approx 100^\circ \zeta \] Karena %OS hanya aproksimasi orde dua, tambahkan **safety margin** sekitar $5^\circ$ sampai $12^\circ$: \[ PM_d=PM_{\zeta}+\text{margin} \] Tujuannya agar hasil simulasi tidak terlalu overshoot akibat dinamika orde tinggi/zero-pole tambahan. ## 3. Baca Bode plot plant $G(j\omega)$ Dari Bode plot, tentukan: - Gain crossover awal $\omega_{c0}$, yaitu frekuensi saat $|G(j\omega)|_{dB}=0$. - Phase margin awal: \[ PM_0=180^\circ+\angle G(j\omega_{c0}) \] - Buat tabel atau interpolasi nilai $|G(j\omega)|_{dB}$ dan $\angle G(j\omega)$ pada berbagai frekuensi. Kalau hanya ada grafik, interpolasi log-linear antara titik-titik yang ada. Catatan: karena %OS hanya menentukan redaman/overshoot, tidak menentukan kecepatan respons. Jadi ada banyak solusi. Pilih frekuensi crossover baru $\omega_c'$ berdasarkan kebutuhan bandwidth/kecepatan atau, jika tidak ada spesifikasi, pilih di sekitar $\omega_{c0}$ dan pastikan lead yang dibutuhkan tidak terlalu besar. ## 4. Tentukan besar phase lead yang diperlukan pada $\omega_c'$ Misalkan kita memilih frekuensi gain crossover baru $\omega_c'$. Dari Bode plot, baca: \[ PM_{uncomp}(\omega_c')=180^\circ+\angle G(j\omega_c') \] Maka phase lead yang harus diberikan kompensator pada frekuensi tersebut adalah: \[ \phi_{need}=PM_d-PM_{uncomp}(\omega_c') \] Jika $\phi_{need}\le 0$, berarti phase margin di frekuensi itu sudah cukup; lead tidak diperlukan untuk phase margin. Jika $\phi_{need}>60^\circ$, sebaiknya gunakan **double-stage lead compensator** atau pilih $\omega_c'$ lain karena satu stage lead sulit memberi lead lebih dari sekitar $60^\circ$. ## 5. Hitung parameter $\alpha$ Untuk phase lead maksimum $\phi_m$: \[ \sin\phi_m=\frac{1-\alpha}{1+\alpha} \] maka \[ \alpha=\frac{1-\sin\phi_m}{1+\sin\phi_m} \] dengan $0<\alpha<1$. Ambil $\phi_m\approx \phi_{need}$, atau beri tambahan kecil $2^\circ$–$5^\circ$ untuk keamanan. ## 6. Letakkan fase maksimum di frekuensi crossover baru Agar phase margin maksimum di $\omega_c'$, set: \[ \omega_m=\omega_c' \] Kemudian: \[ T=\frac{1}{\omega_c'\sqrt{\alpha}} \] Zero dan pole kompensator: \[ z=\frac1T=\omega_c'\sqrt{\alpha} \] \[ p=\frac1{\alpha T}=\frac{\omega_c'}{\sqrt{\alpha}} \] Jadi bagian dinamik kompensator: \[ \frac{1+Ts}{1+\alpha Ts} \] atau \[ \frac{s+z}{s+p} \] dengan $p>z$. ## 7. Hitung gain kompensator Pada frekuensi $\omega_m=\omega_c'$, magnitude kompensator lead adalah: \[ |G_c(j\omega_c')|=\frac{K_c}{\sqrt{\alpha}} \] untuk bentuk $G_c=K_c\frac{1+Ts}{1+\alpha Ts}$, dengan $K_c$ adalah gain DC kompensator. Agar total open-loop menjadi $0\text{ dB}$ di $\omega_c'$: \[ |G_c(j\omega_c')|\,|G(j\omega_c')|=1 \] sehingga: \[ \frac{K_c}{\sqrt{\alpha}}|G(j\omega_c')|=1 \] \[ K_c=\frac{\sqrt{\alpha}}{|G(j\omega_c')|} \] Jika Bode plot dalam dB: \[ K_{c,dB} = -G_{dB}(\omega_c') +10\log_{10}\alpha \] dengan $G_{dB}=20\log_{10}|G|$. Jadi final kompensator: \[ \boxed{ G_c(s)=K_c\frac{1+Ts}{1+\alpha Ts} } \] atau jika memakai bentuk zero-pole: \[ G_c(s)=K\frac{s+z}{s+p} \] maka \[ K=\frac{K_c}{\alpha} = \frac{1}{\sqrt{\alpha}\,|G(j\omega_c')|} \] dan \[ z=\omega_c'\sqrt{\alpha},\qquad p=\frac{\omega_c'}{\sqrt{\alpha}}. \] ## 8. Verifikasi dengan Bode plot hasil kompensasi Bentuk open-loop terkompensasi: \[ L(s)=G_c(s)G(s) \] Gambarkan ulang Bode plot $L(j\omega)$ atau overlay kurva kompensator pada Bode plot plant: \[ |G_cG|_{dB}=|G_c|_{dB}+|G|_{dB} \] \[ \angle G_cG=\angle G_c+\angle G \] Kemudian: 1. Cari gain crossover aktual, yaitu frekuensi saat $|G_cG|=0\text{ dB}$. 2. Hitung phase margin aktual: \[ PM_{actual}=180^\circ+\angle G_cG(j\omega_{c,actual}) \] 3. Pastikan: \[ PM_{actual}\ge PM_d \] 4. Cek gain margin dan stabilitas. 5. Simulasikan respons step closed-loop, ukur %OS aktual. Jika overshoot masih terlalu besar, naikkan $PM_d$ atau ubah $\omega_c'$. Jika respons terlalu lambat, naikkan $\omega_c'$ dan ulangi desain. --- ## Jika ada spesifikasi steady-state error Jika $K_c$ sudah ditetapkan lebih dulu untuk memenuhi error steady-state, jangan memakai rumus $K_c=\sqrt{\alpha}/|G|$. Langkahnya menjadi: 1. Tentukan $K_c$ dari spesifikasi error. 2. Hitung $\alpha$ dari $\phi_m$. 3. Cari frekuensi $\omega_m$ pada Bode plot $K_cG(j\omega)$ yang memenuhi: \[ |K_cG(j\omega_m)|=\sqrt{\alpha} \] atau dalam dB: \[ K_{c,dB}+G_{dB}(\omega_m)=10\log_{10}\alpha \] 4. Set: \[ T=\frac{1}{\omega_m\sqrt{\alpha}} \] 5. Bentuk: \[ G_c(s)=K_c\frac{1+Ts}{1+\alpha Ts} \] 6. Verifikasi PM, GM, dan respons step. --- ## Ringkasan alur 1. $\%OS \rightarrow \zeta$ 2. $\zeta \rightarrow PM_d = PM_{\zeta}+\text{margin}$ 3. Baca Bode plant: $|G(j\omega)|$ dan $\angle G(j\omega)$ 4. Pilih $\omega_c'$ (gain crossover baru) 5. Hitung $\phi_{need}=PM_d-[180^\circ+\angle G(j\omega_c')]$ 6. Hitung $\alpha=\frac{1-\sin\phi_{need}}{1+\sin\phi_{need}}$ 7. Set $\omega_m=\omega_c'$, hitung $T=\frac1{\omega_c'\sqrt{\alpha}}$, $z=\omega_c'\sqrt{\alpha}$, $p=\frac{\omega_c'}{\sqrt{\alpha}}$ 8. Hitung $K_c=\frac{\sqrt{\alpha}}{|G(j\omega_c')|}$ 9. Bentuk $G_c(s)=K_c\frac{1+Ts}{1+\alpha Ts}$ 10. Verifikasi Bode plot terkompensasi, phase margin, gain margin, dan simulasi step. Jika $\phi_{need}>60^\circ$, gunakan dua stage lead secara cascade, misalnya masing-masing memberi $\phi_{need}/2$, lalu kalikan kedua transfer function kompensatornya.

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