(a) An underdamped second order system having a transfer function of the form M(s) = (Kωₙ²)/(s² + 2ξωₙ s + ωₙ²) has a frequency…

(a) Gain K and damping factor ξ from the frequency response

Magnitude. With u = ω/ω_n, M(jω) = Kω_n²/[(ω_n² − ω²) + j2ξω_nω], so |M(jω)| = K/√[(1 − u²)² + (2ξu)²]

Gain K. At ω = 0, |M(j0)| = K. The plot starts at 1.0, so K = 1.

Damping factor from the resonant peak. Setting d/du of the denominator to zero gives the resonant frequency and peak: ω_r = ω_n√(1 − 2ξ²), M_r = K/(2ξ√(1 − ξ²)) (valid for ξ < 0.707) With M_r = 2.5 and K = 1: 2ξ√(1 − ξ²) = 0.4 ⇒ 4ξ²(1 − ξ²) = 0.16 ⇒ ξ⁴ − ξ² + 0.04 = 0 ξ² = [1 − √(1 − 0.16)]/2 = 0.0417 ⇒ ξ = 0.204 The other root, ξ² = 0.958 (ξ = 0.979), is rejected because a system with ξ > 0.707 has no resonant peak.

Reading the peak exactly at ω_n. The figure marks the peak at ω_n. At u = 1, |M(jω_n)| = K/(2ξ); setting this to 2.5 gives ξ = 0.2. The two readings agree closely because, for light damping, ω_r = ω_n√(1 − 2 × 0.204²) = 0.958ω_n, almost at ω_n.

Result: K = 1, ξ ≈ 0.2 (0.204 from the resonant-peak formula). The system is lightly underdamped, and its step response would overshoot by about 52% (e^(−πξ/√(1 − ξ²)) = 0.52).

(b) CRT beam speed, deflection sensitivity, deflection factor and input voltage

Data. V_a = 3000 V; plate length l = 0.025 m; plate spacing d = 0.005 m; distance from plate centre to screen L = 0.30 m; amplifier gain A = 100; e/m = 1.759×10¹¹ C/kg.

(i) Beam speed. ½mv² = eV_a v = √(2eV_a/m) = √(2 × 1.759×10¹¹ × 3000) = √(1.055×10¹⁵) = 3.25×10⁷ m/s This is about 0.11c, so the non-relativistic formula is adequate.

(ii) Deflection sensitivity. Between the plates the electron has transverse acceleration a = eV_d/(md) for a time t = l/v. It leaves the plates with slope v_y/v = eV_dl/(mdv²) = V_dl/(2dV_a). Projecting this straight line back, it appears to come from the centre of the plates, so the screen deflection is D = L × V_dl/(2dV_a) = LlV_d/(2dV_a) S = D/V_d = Ll/(2dV_a) = (0.30 × 0.025)/(2 × 0.005 × 3000) = 2.5×10⁻⁴ m/V = 0.25 mm/V

(iii) Deflection factor. G = 1/S = 1/(2.5×10⁻⁴) = 4000 V/m = 4 V/mm

(iv) Input voltage for 5 cm deflection. Voltage on the plates: V_d = D × G = 50 mm × 4 V/mm = 200 V Input to the amplifier: V_in = V_d/A = 200/100 = 2 V The overall sensitivity including the amplifier is 25 mm/V.

Check that the beam clears the plates. The deflection at the plate exit is y₁ = V_dl²/(4dV_a) = 200 × 0.025²/(4 × 0.005 × 3000) = 2.08 mm. This is less than d/2 = 2.5 mm, so a 5 cm deflection is achievable.

(c) 8085 program: add two bytes, store sum and carry

Memory map. Operands at 4200H and 4201H; low byte of the sum at 4202H; carry at 4203H.

LXI H, 4200H ; HL points to the first operand MOV A, M ; A ← [4200H] INX H ; HL = 4201H ADD M ; A ← A + [4201H]; CY = carry out of bit 7 INX H ; HL = 4202H (INX does not affect flags) MOV M, A ; [4202H] ← low byte of the sum MVI A, 00H ; A = 00H (MVI does not affect CY) ACI 00H ; A ← 00H + 00H + CY, so A = carry INX H ; HL = 4203H MOV M, A ; [4203H] ← carry (00H or 01H) HLT ; stop

Example. [4200H] = 9AH and [4201H] = 89H: 9AH + 89H = 123H, so [4202H] = 23H and [4203H] = 01H. With [4200H] = 25H and [4201H] = 14H the sum is 39H, so [4203H] = 00H.

An equivalent version clears a register (MVI C, 00H) before the addition, uses JNC SKIP / INR C after ADD, and stores C at 4203H. XRA A must not be used to clear A after ADD, because it resets CY before the carry has been saved.