(a) In the circuit shown in the diagram, initially key K₁ is closed and capacitor has no charge (at time t = 0). Now at time t =…

(a) Capacitor voltage with K₁ switched

Circuit. Let N and M be the top and bottom of the 5 Ω ∥ 1 A pair (1 A driven from M to N). K₁ and the 0.1 A source (driving current from ground into M) are in parallel between M and ground A. The loop closes through N – 95 Ω – 5 V (+ towards B) – C – A.

K₁ closed (0–10 s, and after 18.68 s).

• K₁ shorts M to ground, so the 0.1 A source is short-circuited and plays no part.
• The 1 A source in parallel with 5 Ω is a Norton source. Its Thevenin form is 1 × 5 = 5 V in series with 5 Ω.
• Adding the 95 Ω and the aiding 5 V source gives V_Th = 5 + 5 = 10 V and R_Th = 5 + 95 = 100 Ω.
• The time constant is τ = R_Th C = 100 × 0.1 = 10 s.

K₁ open (10–18.68 s). The only path from M back to ground is through the 0.1 A source, which is in series with the capacitor. The capacitor current is therefore fixed at 0.1 A, since R_Th → ∞. The capacitor charges linearly at dv/dt = I/C = 0.1/0.1 = 1 V/s.

Stage 1 (0–10 s), with v(0) = 0: v(t) = 10(1 − e^(−t/10)) V

v(10) = 10(1 − e^(−1)) = 6.32 V

Stage 2 (10–18.68 s): v(t) = 6.32 + 1 × (t − 10) V

v(18.68) = 6.32 + 8.68 = 15.00 V

Stage 3 (t ≥ 18.68 s): now v exceeds V_Th, so the capacitor discharges exponentially towards 10 V:

v(t) = 10 + (15.00 − 10) e^(−(t − 18.68)/10) V

v(28.68) = 10 + 5.00 e^(−1) = 11.84 V

Each stage starts from the previous end value (capacitor voltage cannot jump); finally v → 10 V.

(b) Op-amp square-wave generator

(i) Waveform and frequency. This is an astable (free-running) relaxation oscillator. It is an inverting Schmitt trigger whose own RC network provides the input, so no external signal is needed.

• Output level. The op-amp output saturates near ±12 V. It feeds V₃ through R₂ = 1 kΩ, and the back-to-back Zeners clamp V₃ at ±(V_Z + V_F) = ±(7.4 + 0.6) = ±8 V. R₂ absorbs the difference, passing about (12 − 8)/1 k ≈ 4 mA.
• Thresholds. Positive feedback through R₃–R₄ gives β = R₄/(R₃ + R₄) = 1/40. So V₂ = ±8/40 = ±0.2 V.
• Capacitor. C₁ charges through R₁ towards ±8 V. When V₁ crosses V₂, the output switches.

For one half-cycle, C₁ goes from −0.2 V to +0.2 V while heading for +8 V:

t₁ = R₁C₁ ln[(8 + 0.2)/(8 − 0.2)] = R₁C₁ ln[(1 + β)/(1 − β)]

With R₁C₁ = 100 kΩ × 0.1 µF = 10 ms, t₁ = 10 ms × ln(1.05128) = 10 ms × 0.05001 = 0.500 ms.

T = 2t₁ = 1.000 ms, so f ≈ 1 kHz (999.8 Hz).

V₃ is a symmetric ±8 V square wave (50 % duty); V₁ is a near-triangular exponential between ±0.2 V.

(ii) Duty cycle D = 0.25 at the same frequency. Replace R₁ by two anti-parallel steering branches between V₃ and V₁:

• D_a in series with R_a, with the anode towards V₃. This branch conducts when V₃ = +8 V and charges C₁.
• D_b in series with R_b, with the anode towards V₁. This branch conducts when V₃ = −8 V and discharges C₁.

With ideal diodes each interval has the same form as before, t = R C₁ ln(1.05128):

• T_high = 0.25 × 1 ms = 0.25 ms, so R_a = 0.25 ms/(0.1 µF × 0.05001) = 50 kΩ.
• T_low = 0.75 ms, so R_b = 150 kΩ.

Because R_a + R_b = 200 kΩ = 2R₁, T stays 1 ms and f stays 1 kHz.

If a 0.6 V diode drop is included, the charging level becomes 7.4 V and ln(7.6/7.2) = 0.05407. The values then become R_a ≈ 46.2 kΩ and R_b ≈ 138.7 kΩ.

(c) Causal x[n] from the pole–zero pattern

Reading the plot. The zeros are at z = −1/2 and z = −1/4. The poles are at z = 1/2 and at z = (1/√2)e^(±jπ/4) = 1/2 ± j1/2, as the labels r = 1/√2 and angle π/4 give. The sketch is not to scale, so I have used the labelled r and angle.

X(z) = G(z + 1/2)(z + 1/4) / [(z − 1/2)(z² − z + 1/2)], with G = 1/4

Since x[n] is causal, the ROC is the region outside the circle |z| = 1/√2.

Partial fractions of X(z)/z:

X(z)/z = −(1/8)/z + (3/2)/(z − 1/2) + A/(z − p) + A*/(z − p*)

Here p = 1/2 + j1/2 and A = −11/16 − j3/16. For example, the coefficient at z = 1/2 is (1/4)(1)(3/4)/[(1/2)(1/4)] = 3/2.

Inverse transform (term by term, causal pairs):

• 1 → δ[n]
• z/(z − a) → aⁿu[n]
• A pⁿ + A*p*ⁿ = 2Re{A pⁿ}

x[n] = −(1/8)δ[n] + (3/2)(1/2)ⁿu[n] + (1/√2)ⁿ[−(11/8)cos(nπ/4) + (3/8)sin(nπ/4)]u[n]

Equivalently, the last term is 1.425(1/√2)ⁿ cos(nπ/4 − 164.7°)u[n].

Check. The formula gives x[0] = 0, x[1] = 0.25, x[2] = 0.5625, x[3] = 0.625, matching long division of X(z); x[0] = 0 because the denominator degree exceeds the numerator's by one.