2. Consider the RLC circuit in Homework 1, Question 3. The values of R, L, and C are given as: R = 0.9 . L = 1 H and C = 0.3 F. + V₁(t) L V(t) + (a) Consider the input x(t) = V,(t) = sin(3t+ 10) and the output y(t) = V₁(t). (i) Find the input phasor X. X=-10j e (ii) Using the phasor-domian circuit analysis, find the output phasor Y. Y = (iii) Find the output y(t). y(t) = | (b) Consider the input x(t) = V,(t) = cos(14t+5) and the output y(t) = Ic(t). (i) Using the phasor-domian circuit analysis, find the output phasor Y. Y=☐ (ii) Find the output y(t). v(t) = - || Ic (t) R C

Find vC(t) and iL(t) for the circuit below and plot both waveforms for t = -1 s to t = 5 s

SOLVE STEP BY STEP IN DIGITAL FORMAT DONT USE CHATGPT 1. Referring to the circuit in the following figure, find: a. i(0*)&v(0*) b. di(0*)/dt & dv(0*)/dt c. i(∞) & v(∞0) 12 V (+ t = 0 692 2 H ele 492 0.4 F www + p -

Draw SD( state diagram of state equation). See attached example

C(s) R(s) s+5 to get an overshoot of 15% and a settling time at 2% of 0.5 seconds, The value of damping ratio is Select one: O a. 0.707 O b. 0.417 c. 0,517 O d. 0.69

Q2) Sketch the following waveforms a) V:(t) = Su(t+2) + 8u(t) - 6u(t-4) b) V2(t) = 50e 200 c) V:(t) = r(t+4) – r(t-3)

Differential equations Find the steady-state current in an LRC circuit when L = 1/2h , R = 20Ω, C = 0.001f and  E(t) = 100sen60t+200cos40t V. Dont skip steps pls

A second order circuit's impulse and step response The figure below shows a series second order RLC circuit. The system input is a step voltage source vin(t) = u(t) and the output is the voltage across resistor R. + Vin(t) C m L i(t). RVout (t) Series RLC As shown in the section regarding second order LCCDES impulse response, the Kirchhoff's voltage law equation can be written as di ½ i(t)dt + L + Ri = vix(t) = u(t). di Differentiating results: + di² Ldt LC d²i Rdi 1 + i= =(t). The circuit's characteristic equation can be written as: = 0. R + = + s+ L LC A system's impulse response can be found using the residue function. Ex: for the LCCDE b₂ d²v + b₁ dy b₁1/4 + bo= αι +ao, dx di [R, P] = residue ([a1 a0], [b2 b1 b0]) Vectors [a1 a0] and [b2 b1 b0] are the input (right hand side) and output (left hand side) coefficients respectively. Vector R contains the residues and vector P contains the roots (poles) of the characteristic equation. If the impulse response has the form h(t) = Aet +…

please do all parts

a series connected rlc circuit is shown the switch opens at t=0 iL(t=0+)= ic(t=0+)= vc(t=0+)= vL(t=0+)= diL/dt(t=0+)= dvc/dt(t=0+)= vL(t=inf.)= vR(t=inf.)= vc(t=inf.)=