(a) Given a second-order linear time-invariant system G(s) with a relative degree of 2. G(s) admits a zero steady-state error for…

(a) For a unity-feedback system to have zero steady-state error to a step, the open-loop transfer function must be type 1. Since G(s) is second order and has relative degree 2, write G(s)=K/[s(s+a)], a>0. The closed-loop transfer function is T(s)=G/(1+G)=K/(s²+a s+K). The characteristic equation is s²+a s+K=0, so ωₙ=√K and ζ=a/(2√K). Hence ζωₙ=a/2. For an underdamped second-order step response, the 2% settling time is tₛ≈4/(ζωₙ). Given tₛ=4 s, 4=4/(a/2)=8/a, so a=2 s⁻¹. The unit-ramp steady-state error is eₛₛ=1/Kv, where Kv=lim s→0 sG(s)=K/a. Given eₛₛ=0.1, K/a=10, so K=10a=20 s⁻². Therefore G(s)=20/[s(s+2)].

With a delay, the loop transfer function is L(s)=exp(-sT)G(s). Sustained oscillation requires L(jω)=-1. The magnitude condition is |G(jω)|=1. Now |G(jω)|=20/[ω√(ω²+4)], so 20/[ω√(ω²+4)]=1. Let x=ω². Then x(x+4)=400, i.e. x²+4x-400=0. The positive root is x=-2+2√101, so ω=√(2√101-2) rad/s≈4.2544 rad/s. The phase of G(jω) is ∠G(jω)=-π/2-arctan(ω/2). The phase condition is ∠G(jω)-ωT=-π (mod 2π). The smallest positive delay is obtained from ωT=π/2-arctan(ω/2)=arctan(2/ω). Thus T=arctan(2/√(2√101-2))/√(2√101-2) s. Numerically, arctan(2/ω)=arctan(0.4701)≈0.4394 rad, so T=0.4394/4.2544≈0.1033 s. At this delay, the loop transfer has magnitude 1 and phase -π, so the closed-loop characteristic has a pair of roots on the imaginary axis. This is the first positive delay for which oscillation occurs. T≈0.1033 s.

(b) A frequency counter measures N cycles in a gate time τ, so f=N/τ. A ±1 LSD error is a ±1 count error, giving ±1/τ Hz. For the usual 1 s gate, this is ±1 Hz. The time-base accuracy ±1×10⁻⁶ is a fractional error, so it contributes ±f×10⁻⁶ Hz. Taking worst-case addition, the absolute error is ±(1+f×10⁻⁶) Hz and the percentage error is (1/f+10⁻⁶)×100%.

• f=100 Hz: absolute error=±(1+0.0001) Hz=±1.0001 Hz; percentage error=±1.0001%.
• f=1 MHz=10⁶ Hz: absolute error=±(1+1) Hz=±2 Hz; percentage error=±0.0002%.
• f=100 MHz=10⁸ Hz: absolute error=±(1+100) Hz=±101 Hz; percentage error=±0.000101%.

If the gate time is τ s, replace 1 Hz by 1/τ Hz in the count term. These are worst-case percentage errors, not rms statistical errors. The time-base error is proportional to the measured frequency, so its absolute value increases with frequency while its percentage remains ±0.0001%. The count error is fixed for a fixed gate, so its percentage decreases as frequency increases. A better time base reduces the fractional part; a longer gate reduces the count-error part.

(c) The 11 kV alternator rating is line-to-line. The phase rms voltage is 11000/√3 V. The peak phase voltage is Vₘ=√2×11000/√3=11000√(2/3) V≈8981.5 V.

(i) For the ideal undamped LC recovery transient, the maximum voltage across the circuit-breaker contacts is taken as 2Vₘ. Thus Vmax=2×11000√(2/3)=22000√(2/3) V≈17.96 kV.

(ii) L=10 mH=0.01 H and C=0.01 µF=1×10⁻⁸ F. The transient angular frequency is ωₙ=1/√(LC)=1/√(0.01×1×10⁻⁸)=1/√(1×10⁻¹⁰)=1×10⁵ rad/s. The frequency is fₙ=ωₙ/(2π)=10⁵/(2π) Hz≈15.915 kHz. The 50 Hz fundamental does not set this transient frequency; the LC network does.

(iii) For the standard undamped recovery voltage v(t)=Vₘ(1-cos ωₙt), the maximum is 2Vₘ at t=π/ωₙ. The average RRRV is the rise 2Vₘ divided by this time, (2Vₘ)/(π/ωₙ)=(2/π)Vₘωₙ. Substituting values, average RRRV=(2/π)×11000√(2/3)×10⁵ V/s≈5.717×10⁸ V/s=0.5717 kV/µs.

(iv) The maximum RRRV is the maximum slope of v(t)=Vₘ(1-cos ωₙt), i.e. Vₘωₙ sin ωₙt, so its maximum is Vₘωₙ. Thus maximum RRRV=11000√(2/3)×10⁵ V/s≈8.981×10⁸ V/s=0.898 kV/µs. These RRRV values are undamped estimates; circuit resistance would lower them.

(d) Use 50 MVA as the common base. The alternator reactance is Xa=0.15 pu. The reactor is rated 35 MVA, 10%, so on the 50 MVA base its reactance is Xr=0.10×50/35=1/7 pu≈0.142857 pu. Consider a fault at the feeder connected to the junction of one alternator and its reactor. The local alternator contributes through Xa. Each of the three remote alternators contributes through its own Xa+Xr to the common bus, and then through the local reactor Xr to the fault. One remote branch is Xa+Xr=3/20+1/7=41/140 pu. Three such branches in parallel give 41/420 pu. Adding the local reactor gives 41/420+1/7=101/420 pu. With all alternator emfs shorted for the Thevenin impedance, the Thevenin reactance at the fault is the parallel combination of the local alternator and this remote path: Xth=(3/20)∥(101/420). Since 3/20=63/420, Xth=(63×101)/(420×164)=303/3280 pu≈0.09238 pu. The symmetrical fault MVA is Sfault=50/Xth=50×3280/303=164000/303 MVA≈541.25 MVA. Each feeder circuit breaker rating≈541.3 MVA at the bus voltage. If a standard breaker rating is to be selected, choose the next standard value, for example 600 MVA. If the bus voltage is V kV, the corresponding current rating is 541.25/(√3 V) kA.

(e) Let Pdot be the probability of a dot. The probability of a dash is one-third of that, so Pdash=Pdot/3. Since Pdot+Pdash=1, Pdot+Pdot/3=1, giving Pdot=3/4 and Pdash=1/4.

(i) The information in a dot is Idot=-log₂(3/4)=log₂(4/3)≈0.4150 bit. The information in a dash is Idash=-log₂(1/4)=2 bit.

(ii) The entropy is H=(3/4)log₂(4/3)+(1/4)×2. Since log₂(4/3)=2-log₂3, H=(3/4)(2-log₂3)+1/2=2-(3/4)log₂3≈0.8113 bit/symbol.

(iii) A dot lasts 10 ms and a dash lasts 3×10=30 ms. The same 10 ms is allowed between symbols, so a dot occupies 10+10=20 ms and a dash occupies 30+10=40 ms. The average time per symbol is (3/4)×20+(1/4)×40=25 ms=0.025 s. The average rate of information is H/0.025=40[2-(3/4)log₂3]=80-30log₂3≈32.45 bit/s.