Question

I want the answer for Part B I have the answer for Part A-Q1 I uploaded it

2 H I F ua 212 0.5% on Y; 4Ω Figure 1 PART A: MATHEMATICAL MODEL AND TIME DOMAIN ANALYSIS (5%) 1) For the circuit of Figure 1, determine the transfer function relating the output voltage (s) to the input voltage V(s). Assuming zero initial condition, obtain the output response vo(t) to an input step with 6V amplitude, that is vi(t)6 u(t) V. Then plot the output response against time. 2) 3) Now, consider only the second-order poles (and with numerator of 50) of the transfer function obtained in Q1, determine the output response vo(t) to an input v(t)6 u(t)Vfor the system in a. open-loop configuration, b. closed-loop configuration with a unity feedback. Plot the output responses. 4) Compare and discuss the characteristics of the responses obtained in Q3 and Q5. PART B: FREQUENCY RESPONSE ANALYSIS (5%) 1) Using the transfer function obtained in Part A 01, for an input voltage vit)6 sin wt V obtains the amplitude and phase of the frequency response. Use frequency from 0.001 100 rad/s. Represent the data in a table. 2) Plot the exact magnitude (in decibels) and phase (in degrees) of the response against the frequency via Bode diagram. 3) Now, sketch the asymptotic Bode diagram for the same transfer function in Part A 01. 4) Next, consider only the second-order poles (and with numerator of 50) of the transfer function obtained in Part A Q1, sketch the asymptotic Bode diagram. 5) Analyse the closed-loop characteristics of the system from the Bode diagrams plotted in Q3 and Q4. Q1: At loop 1: -V(s) + Ls, (s) + R14(s) + R, (h(s)-4(s)) = 0 (2s 6)1s) (4)12(s) V(s) At loop 2: -0.5(R,(s) 2(s) +Rala(s)+R22(s) R21(s)0 (-5),(s) + ( 14) 4(s) = 0 1ド 2 H 0.5 (2s +6) (-4) MI 6 28s66 -5) 5+14) 4Ω !(2-56) vios, 5V(s) skis (28s +66) 50V (s) (s) = (25s+466) る(s)-1012 (s) => Vo(s) 50 50s (28s +0+66)(28s 665+6)

Answer 1

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