(a) Design a PI controller so that a unity feedback system having open loop transfer function G(s)H(s) = 10 / ((s + 2) (s + 3))…

(a) PI controller for ζ = 0.6 and ω_n = 4 rad/s (20 marks)

Set-up. Take the PI controller as G_c(s) = K_p + K_i/s = (K_p s + K_i)/s, with unity feedback. The open-loop transfer function is

G_c(s)G(s)H(s) = 10(K_p s + K_i)/[s(s + 2)(s + 3)].

Characteristic equation. Setting 1 + G_cGH = 0 gives

s(s + 2)(s + 3) + 10(K_p s + K_i) = 0, that is, s³ + 5s² + (6 + 10K_p)s + 10K_i = 0.

Desired dominant pair. For ζ = 0.6 and ω_n = 4 rad/s, the pair of poles satisfies s² + 2ζω_n s + ω_n² = s² + 4.8s + 16 = 0, so s = −2.4 ± j3.2. The system is third order, so the third pole is placed at s = −p and the characteristic polynomial must equal

(s² + 4.8s + 16)(s + p) = s³ + (4.8 + p)s² + (16 + 4.8p)s + 16p.

Matching coefficients.

• s²: 4.8 + p = 5, so p = 0.2.
• s: 6 + 10K_p = 16 + 4.8 × 0.2 = 16.96, so K_p = 1.096.
• s⁰: 10K_i = 16 × 0.2 = 3.2, so K_i = 0.32.

The controller is G_c(s) = 1.096 + 0.32/s = 1.096(s + 0.292)/s. Its zero at s = −0.292 lies close to the third closed-loop pole at s = −0.2, so the two nearly cancel and the complex pair s = −2.4 ± j3.2 governs the response, as required. The integral action also makes the loop type 1 with zero steady-state error for a step input.

Characteristic equation of the compensated system.

s³ + 5s² + 16.96s + 3.2 = 0, which factorises as (s² + 4.8s + 16)(s + 0.2) = 0.

(b)(i) 8253/8254 counter 0 in Mode 3 and Mode 2 (10 marks)

Count. The clock is 1 MHz and the output must be 1 kHz, so N = 1 MHz/1 kHz = 1000 = 03E8H. The low byte is E8H and the high byte is 03H.

Control word. The format is D7 D6 = SC1 SC0 (counter select), D5 D4 = RW1 RW0 (read/write mode), D3 D2 D1 = M2 M1 M0 (mode) and D0 = BCD. For counter 0 (SC = 00), low byte then high byte (RW = 11) and binary counting (BCD = 0):

• Mode 3: 0 0 1 1 0 1 1 0 = 36H
• Mode 2: 0 0 1 1 0 1 0 0 = 34H

Program for Mode 3 (square wave).

• MVI A, 36H
• OUT 2DH (write the control word)
• MVI A, E8H
• OUT 2BH (low byte of the count into counter 0)
• MVI A, 03H
• OUT 2BH (high byte of the count)
• HLT

Program for Mode 2 (rate generator). It is the same except that the control word is 34H:

• MVI A, 34H
• OUT 2DH
• MVI A, E8H
• OUT 2BH
• MVI A, 03H
• OUT 2BH
• HLT

Output. In Mode 3, the count is even, so the output is a symmetrical square wave of period 1000 μs, high for 500 clock periods and low for 500. In Mode 2, the output stays high and goes low for one clock period (1 μs) once every 1000 clocks, giving a 1 kHz pulse train of narrow negative pulses, and the counter reloads automatically each time.

(b)(ii) Interfacing an 8255 with the 8085, mode-0 control word and program (10 marks)

Hardware scheme. The 8255 is I/O-mapped. The address/data lines AD0–AD7 of the 8085 are connected to D0–D7 of the 8255, and the 74LS373 latch, enabled by ALE, separates the low-order address A0–A7. Address lines A0 and A1 go directly to the A0 and A1 pins of the 8255 to select the register, and A2–A7 are decoded with IO/M to produce the chip select. RD, WR and RESET OUT of the 8085 go to the corresponding pins of the 8255.

With A7–A2 = 000000 and IO/M = 1 the chip select is active (low), and A1A0 chooses the register: 00 = Port A (00H), 01 = Port B (01H), 10 = Port C (02H) and 11 = control word register (03H), as the question states.

Control word. For the mode-set word (D7 = 1), Group A in Mode 0 (D6 D5 = 00), Port A input (D4 = 1), Port C upper input (D3 = 1), Group B in Mode 0 (D2 = 0), Port B output (D1 = 0) and Port C lower input (D0 = 1):

1 0 0 1 1 0 0 1 = 99H

Program.

• MVI A, 99H
• OUT 03H (configure the ports)
• IN 00H (first number from Port A)
• MOV B, A (keep it in register B)
• IN 02H (second number from Port C)
• ADD B (A = sum of the two numbers)
• OUT 01H (display the sum on Port B)
• HLT

The sum is an 8-bit result; a carry beyond bit 7 appears in the carry flag and is not shown on Port B.

(c) Thermistor constants and resistance range (10 marks)

Constants. Temperatures must be in kelvin, so T₀ = 273 K (ice point) and T₁ = 60 + 273 = 333 K. Only the product pR₀ can be found from the data, so R₀ is taken as the resistance at the ice point, R₀ = 4290 Ω, which makes the relation R_T = pR₀e^(b/T) fit the ice-point reading. At 273 K, 4290 = p × 4290 × e^(b/273), so p = e^(−b/273). At 333 K,

980 = p × 4290 × e^(b/333), so 980/4290 = e^(b(1/333 − 1/273)).

Taking logarithms, ln(0.2284) = −1.4767 = b × (−6.60 × 10⁻⁴), so

b = 2237 K, and p = e^(−2237/273) = e^(−8.195) = 2.76 × 10⁻⁴.

Resistance at 70 °C and 120 °C. R_T = 4290 × e^(2237(1/T − 1/273)).

• At 70 °C (T = 343 K): 1/343 − 1/273 = −7.475 × 10⁻⁴, the exponent is −1.672, so R = 4290 × 0.1878 = 806 Ω.
• At 120 °C (T = 393 K): 1/393 − 1/273 = −1.1185 × 10⁻³, the exponent is −2.502, so R = 4290 × 0.0819 = 351 Ω.

Range. The resistance to be measured lies between about 351 Ω and 806 Ω, a change of about 455 Ω. The resistance falls as the temperature rises, since the thermistor is of negative temperature coefficient. The sensitivity (1/R)(dR/dT) = −b/T² is about −1.9% per K at 70 °C and −1.45% per K at 120 °C, so the response is strongly non-linear.