(a) For the circuit shown in the figure, obtain the value of voltage across 0·5 Ω and 2·5 Ω resistors using nodal current…

(a) Nodal analysis: voltages across the 0.5 Ω and 2.5 Ω resistors

Take the centre node (earth symbol) as reference, 0 V. The 12 V source has its positive end at the reference, so V₂ = −12 V. Read the polarities from the figure: V_x = V₃ − V₂ = V₃ + 12 (across 0.5 Ω) and V_y = V₁ − V₂ = V₁ + 12 (across 2.5 Ω). The 0.2V_y element is a dependent voltage source with its positive end at V₄, and the 0.5V_x element is a dependent current source flowing from the reference node towards V₄. The 2 A source feeds V₃ from the reference node.

Unknowns: V₁, V₃, V₄.

(i) Constraint of the dependent voltage source: V₄ − V₁ = 0.2V_y = 0.2(V₁ + 12), so V₄ = 1.2V₁ + 2.4.

(ii) KCL at node V₃: 2 = (V₃ + 12)/0.5 + (V₃ − V₄)/2, which gives 2.5V₃ − 0.5V₄ = −22.

(iii) KCL at the supernode (V₁, V₄): currents leaving are V₁/1 + (V₁ + 12)/2.5 + (V₄ − V₃)/2 and the source 0.5V_x = 0.5(V₃ + 12) enters V₄. So V₁ + 0.4(V₁ + 12) + 0.5(V₄ − V₃) − 0.5(V₃ + 12) = 0, that is 1.4V₁ − V₃ + 0.5V₄ = 1.2.

Substituting V₄ = 1.2V₁ + 2.4 into (ii) and (iii): 2.5V₃ − 0.6V₁ − 1.2 = −22 and 1.4V₁ − V₃ + 0.6V₁ + 1.2 = 1.2, so V₃ = 2V₁. Then 5V₁ − 0.6V₁ = −20.8, giving V₁ = −4.727 V, V₃ = −9.455 V, V₄ = −3.273 V.

Hence the voltage across the 0.5 Ω resistor is V_x = V₃ + 12 = 2.545 V (28/11 V), and across the 2.5 Ω resistor is V_y = V₁ + 12 = 7.273 V (80/11 V).

(b) Discrete-time LTI system

(i) Transfer function. Taking the z-transform: H(z) = 1/(1 − (3/4)z⁻¹ + (1/8)z⁻²) = z²/[(z − 1/2)(z − 1/4)], with ROC |z| > 1/2 because the system is causal.

(ii) Impulse response. H(z)/z = z/[(z − 1/2)(z − 1/4)] = 2/(z − 1/2) − 1/(z − 1/4). So H(z) = 2/(1 − 0.5z⁻¹) − 1/(1 − 0.25z⁻¹) and h[n] = [2(1/2)ⁿ − (1/4)ⁿ]u[n]. Check: h[0] = 1, h[1] = 3/4, as the difference equation gives.

(iii) Step response. S(z) = H(z)·z/(z − 1), so S(z)/z = z²/[(z − 1)(z − 1/2)(z − 1/4)] = (8/3)/(z − 1) − 2/(z − 1/2) + (1/3)/(z − 1/4). Therefore s[n] = [8/3 − 2(1/2)ⁿ + (1/3)(1/4)ⁿ]u[n]. Check: s[0] = 1 and s[1] = 7/4 = h[0] + h[1]. As n grows, s[n] tends to 8/3 = H(1), the dc gain.

(c) Differential pair: changes in V_X, V_Y and V_X − V_Y

Both bases are at V_CM, so the tail current splits equally, I_C1 = I_C2 = I_EE/2, independent of the collector resistors (early effect neglected). Thus V_X = V_CC − (I_EE/2)R_C1 and V_Y = V_CC − (I_EE/2)R_C2.

(i) V_CC rises by ΔV, R_C1 = R_C2. ΔV_X = ΔV_Y = ΔV, and Δ(V_X − V_Y) = 0.

(ii) I_EE changes by ΔI, R_C1 = R_C2 = R_C. ΔV_X = ΔV_Y = −(ΔI/2)R_C, and Δ(V_X − V_Y) = 0.

(iii) R_C1 = R_C2 + ΔR. ΔV_X = −(I_EE/2)ΔR, ΔV_Y = 0, and V_X − V_Y = −(I_EE/2)ΔR.

The pair rejects common-mode changes in (i) and (ii), but a resistor mismatch converts the common-mode current into a differential output in (iii).