For the network shown in Figure 4(a)(i) and its excitation function shown in Figure 4(a)(ii), find the response v(t) using convolution by:

(i) 's' domain approach.

(ii) time domain analytical approach.

(iii) graphical convolution approach.

(b)(i)

Question

For the network shown in Figure 4(a)(i) and its excitation function shown in Figure 4(a)(ii), find the response v(t) using convolution by:

(i) 's' domain approach.

(ii) time domain analytical approach.

(iii) graphical convolution approach.

(b)(i) A 240 V, 50 Hz single phase supply is connected to a full controlled converter to control the speed of a 10 kW, 220 V separately excited dc motor. The rated current of motor at full load is 25 A, armature resistance is 0·4 ohm and machine constant is 0·3 V/rpm. Calculate the speed of motor when converter is operating at an angle α = 50°, assuming continuous armature current.

(b)(ii) Describe the performance requirements of a chopper circuit that can perform the chopping functions in any modulation technique.

(c) The amplifier shown in Figure 4(c) is biased to operate at I_D = 1 mA and g_m = 1 mA/V. Neglecting r_o,

(i) determine the midband gain.

(ii) determine the value of C_S that places f_L at 10 Hz.

UPSC Answer Check · Accepted Answer

(a)(i) By the s-domain convolution theorem, with zero initial capacitor voltage, v(t)=L⁻¹{H(s)E(s)}. For R=1 Ω and C=1 F, RC=1 s, so H(s)=1/(1+sRC)=1/(1+s). The pulse of height V volts is e(t)=V[u(t)-u(t-1)], with u(t) the unit step, so E(s)=V(1-exp(-s))/s. Hence V_o(s)=V(1-exp(-s))/[s(1+s)]. Since L⁻¹{1/[s(1+s)]}=1-exp(-t), v(t)=V[(1-exp(-t))u(t)-(1-exp(-(t-1)))u(t-1)] volts. Thus v(t)=0 for t<0 s; V(1-exp(-t)) for 0≤t≤1 s; V(exp(-(t-1))-exp(-t))=V(e-1)exp(-t) for t≥1 s. Final: v(t)=V(1-exp(-t))u(t)-V(1-exp(-(t-1)))u(t-1), in volts.

(a)(ii) Time-domain convolution: h(t)=exp(-t)u(t) s⁻¹, with t in seconds. v(t)=∫ h(τ)e(t-τ)dτ. Since e(t-τ)=V for t-1≤τ≤t, the effective limits are max(0,t-1) to t. For 0≤t≤1 s: v(t)=V∫ exp(-τ)dτ from 0 to t = V(1-exp(-t)). For t≥1 s: v(t)=V∫ exp(-τ)dτ from t-1 to t = V(exp(-(t-1))-exp(-t)). For t<0 s, v(t)=0. Final: same piecewise response as (a)(i).

(a)(iii) Graphical convolution: keep h(τ)=exp(-τ)u(τ) s⁻¹ fixed and slide the flipped rectangular pulse. The response is the area under h(τ)e(t-τ). For 0<t<1 s the overlap is 0≤τ≤t, giving V(1-exp(-t)). For t>1 s the overlap is t-1≤τ≤t, giving V(exp(-(t-1))-exp(-t)). At t=1 s, v=V(1-exp(-1)) volts. Final: same piecewise response.

(b)(i) Assume the standard single-phase fully controlled bridge converter, ideal devices, continuous armature current, and motor current equal to rated full-load 25 A. V_m=240√2 V=339.4 V. The average dc output voltage is V_dc=(2V_m/π)cosα=(480√2/π)cos50° V=138.9 V. The back emf is E_b=V_dc-I_aR_a=138.9-25(0.4)=128.9 V. With K=0.3 V/rpm, N=E_b/K=128.9/0.3=429.6 rpm. Final: N≈430 rpm.

(b)(ii) A chopper that can implement any modulation technique must:
switch over the full duty range 0<D<1 with accurate on/off timing over a wide frequency range;
use fast devices with low propagation delay, low on-state drop, and adequate voltage/current ratings;
give high efficiency, low output ripple, low EMI, and good line/load regulation;
provide freewheeling paths, snubbers, sensing, and protection against overcurrent, overvoltage, short-circuit, and thermal overload;
have a reliable gate-drive/control interface for PWM, PFM, current-mode, or hysteresis control.
Final: these are the required performance characteristics.

(c)(i) From Figure 4(c), RD=10 kΩ and RS=6 kΩ. At midband CS is a short, so the source is at AC ground. Neglecting ro, A_v=V_o/V_i=-gmRD=-(1 mA/V)(10 kΩ)=-10 V/V. Final: A_v=-10 V/V.

(c)(ii) Using the resistance-seen-by-capacitor method, set Vi=0. The source node sees RS in parallel with 1/gm. Since 1/gm=1 kΩ, R_eq=6 kΩ||1 kΩ=6/7 kΩ=857.1 Ω. Source-node KCL gives v_s(1/R_S+sC_S)=gm(V_i-v_s), and V_o=-gmR_D(V_i-v_s), so A(s)=A_mid(s+1/(C_S·R_S))/(s+1/(C_S·R_eq)). Thus f_z=1/(2πC_S·R_S) and f_p=1/(2πC_S·R_eq), with f_z/f_p=1/7. The exact 3-dB condition |A/A_mid|=1/√2 gives f_L=f_p√(1-2/49)=f_p√(47/49). Setting f_L=10 Hz gives C_S=√47/(120000π) F=18.2 μF. The common pole approximation f_L≈f_p gives C_S=7/(120000π) F=18.6 μF. Final: C_S≈18.2 μF exact; 18.6 μF by standard approximation.

For the network shown in Figure 4(a)(i) and its excitation function shown in Figure 4(a)(ii), find the response v(t) using…

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