(a) Determine the time signal that corresponds to the following bilateral Laplace transform and the ROCs given below by using the…

(a) Inverse bilateral Laplace transform

Partial fractions:

X(s) = A/(s + 2) + (Bs + C)/(s² + 2s + 5), so 4s² + 8s + 10 = A(s² + 2s + 5) + (Bs + C)(s + 2).

• s = −2: 16 − 16 + 10 = 5A, so A = 2
• s² terms: 4 = A + B, so B = 2
• s terms: 8 = 2A + 2B + C = 8 + C, so C = 0 (constant term check: 5A + 2C = 10 ✓)

Writing s² + 2s + 5 = (s + 1)² + 2² and 2s = 2(s + 1) − 2:

X(s) = 2/(s + 2) + 2(s + 1)/[(s + 1)² + 4] − 2/[(s + 1)² + 4]

Poles: s = −2 and s = −1 ± j2. Transform pairs used (a pole to the left of the ROC gives a right-sided term; a pole to the right of the ROC gives a left-sided term with a minus sign):

• 1/(s + 2) ↔ e^(−2t)u(t) for Re(s) > −2, or −e^(−2t)u(−t) for Re(s) < −2
• (s + 1)/[(s + 1)² + 4] ↔ e^(−t)cos2t u(t) for Re(s) > −1, or −e^(−t)cos2t u(−t) for Re(s) < −1
• 2/[(s + 1)² + 4] ↔ e^(−t)sin2t u(t) for Re(s) > −1, or −e^(−t)sin2t u(−t) for Re(s) < −1

(i) ROC Re(s) < −2 (left of all poles, every term left-sided):

x(t) = −[2e^(−2t) + 2e^(−t)cos2t − e^(−t)sin2t] u(−t)

(ii) ROC Re(s) > −1 (right of all poles, every term right-sided):

x(t) = [2e^(−2t) + 2e^(−t)cos2t − e^(−t)sin2t] u(t)

(iii) ROC −2 < Re(s) < −1 (right of −2, left of −1 ± j2):

x(t) = 2e^(−2t)u(t) − [2e^(−t)cos2t − e^(−t)sin2t] u(−t)

Only case (ii) is causal, and since its ROC contains the jω-axis it is also stable; cases (i) and (iii) are non-causal and, as their ROCs exclude the jω-axis, unstable.

(b) Op-amp triangular/square-wave generator

Reading the circuit:

• I₁ has its inverting input grounded and its non-inverting input fed from R₁ (from output A) and R₂ (from node D). It is a non-inverting Schmitt trigger: its output is at +V_sat or −V_sat depending on the sign of V₊ = (R₂V_A + R₁V_D)/(R₁ + R₂).
• R₃ and the back-to-back Zeners Z₁, Z₂ clamp node D to ±V_D, where V_D = V_Z + V_F ≈ 3.3 + 0.7 = 4.0 V (one Zener in breakdown, the other forward-biased). R₃ limits the Zener current to about (12 − 4)/2 kΩ ≈ 4 mA.
• I₂ with R₄ and C₁ is an inverting integrator: dV_A/dt = −V_D/(R₄C₁).
• I₃ takes the square wave at D on its non-inverting input and, with R₅ and R₆ as the gain-setting resistors (as the question intends), amplifies it by (1 + R₆/R₅) = 2. So B is a square wave, not a copy of A.

Working: when D = +V_D, A ramps down. V₊ stays positive until R₂V_A + R₁V_D = 0, i.e. V_A = −(R₁/R₂)V_D. At that instant I₁ switches to −V_sat, D becomes −V_D, and A ramps up until V_A = +(R₁/R₂)V_D, when I₁ switches back. The loop therefore oscillates on its own.

• Peak of the triangle at A: V_Ap = (R₁/R₂)V_D = 1 × 4 = ±4 V
• Square wave at B: ±(1 + R₆/R₅)V_D = ±8 V (B is high while A is falling)
• Half period: the ramp covers 2V_Ap at a slope V_D/(R₄C₁), so T/2 = 2(R₁/R₂)R₄C₁ and T = 4(R₁/R₂)R₄C₁ = 4 × 1 × 2 kΩ × 1 µF = 8 ms, f = 125 Hz. The frequency does not depend on the Zener voltage.

Changes for ±6 V at A and B and f = 500 Hz (keeping R₂ = R₅ = 2 kΩ and C₁ = 1 µF):

• Output A: (R₁/R₂) × 4 = 6, so R₁/R₂ = 1.5 and R₁ = 3 kΩ.
• Output B: (1 + R₆/R₅) × 4 = 6, so R₆/R₅ = 0.5 and R₆ = 1 kΩ.
• Frequency: f = R₂/(4R₁R₄C₁) = 500 Hz gives R₄ = 1/(4 × 1.5 × 10⁻⁶ × 500) = 333 Ω (use 330 Ω).

If the Zener forward drop is ignored (V_D = 3.3 V), the same method gives R₁/R₂ = 6/3.3 = 1.82 (R₁ ≈ 3.64 kΩ), R₆/R₅ = 0.82 (R₆ ≈ 1.64 kΩ) and R₄ = 1/(4 × 1.82 × 10⁻⁶ × 500) = 275 Ω. Note also that, if the I₃ stage is read literally as drawn (the R₅–R₆ junction tied to B), it is a voltage follower and B = V_D; ±6 V at B would then need Zeners of about 5.3 V instead of a resistor change.

(c) Logic equations and truth table

The first NAND gives N = (D₀D₁)′, which is also brought out as Ȳ₃. The top-right gate has three inputs: N, Ȳ₁ and Ȳ₂.

• Ȳ₃ = (D₀D₁)′
• Ȳ₁ = (D₀·N)′ = [D₀(D₀′ + D₁′)]′ = (D₀D₁′)′ = D₀′ + D₁
• Ȳ₂ = (D₁·N)′ = [D₁(D₀′ + D₁′)]′ = (D₀′D₁)′ = D₀ + D₁′
• Ȳ₀ = (N·Ȳ₁·Ȳ₂)′. Since (D₀′ + D₁′)(D₀′ + D₁) = D₀′ and D₀′(D₀ + D₁′) = D₀′D₁′, Ȳ₀ = (D₀′D₁′)′ = D₀ + D₁

Concise form (with D₁ as the MSB, mᵢ the minterms): Ȳ₀ = m₀′, Ȳ₁ = m₁′, Ȳ₂ = m₂′, Ȳ₃ = m₃′.

Truth table (D₁ D₀ → Ȳ₀ Ȳ₁ Ȳ₂ Ȳ₃):

• 0 0 → 0 1 1 1
• 0 1 → 1 0 1 1
• 1 0 → 1 1 0 1
• 1 1 → 1 1 1 0

Exactly one output is LOW for each input combination, so the circuit is a 2-to-4 line decoder with active-low outputs.