Question

Problem 1. (40 pts): A student in ME 345 shows you the following circuit with R=1.0 k 2 and C = 1.0 uF Vout Vin c= a. What kind of circuit is this? What is the order of this dynamic system? b. Using KVL/KCL, derive the differential equation and put in standard form. What is the static sensitivity K of this system? c. What is the cutoff frequency (@c) of this circuit? What is the time constant (c)? How are Oc and related to one another? d. For a sudden change in input DC voltage Vin (step input) from 0.0 V to a steady state value of 1.0 V, how long will it take for the output to reach 95% of its final value? e. If Vin is initially 3.0 V DC and instantaneously decreases to 1.0 V at t=0 s, calculate Vout at t = 0.225 s.

Answer 1

The circuit given in the Question is "RC Low Pass Filter" of "First Order" as there is only one load component present which is 'Capacitor' , and then by using KVL property the differential equation is calculated and by observing first order differential equation the 'Time constant' and 'static sensitivity (K) ' is calculated and then by using Charging equation of capacitor the 'elapsed time' is calculated.

there are 6 different pictures in the solution ,so for complete solution download all(6) pictures.

solution. Relkr - Luf Yout (a.) The above circuit is RC Low Pass Filter Circuit The order of this filter is" First Order" This type of filter is tware known generally as a "First order filter" or "one-pole filter", because it has only to one" reactive component, the capacitor, in the circuit. (b) i R = 1 ks = 10²2 Vout Ic=luf = 1x1067 Coop using KUL in the loop shown in circuit :-

(c) The cutoff frequency of RC low pass filter (2.5 is given by :- Time constant, Fo= RC ) . Relation between I and we :- [w= 1 (d) Equation of charging of capacitor (Yout= Vc = Vin [.- & Re] where, Ve is the 'voltage across the capacitor? Vin is the 'supply voltage! t is the 'elapsed time since the application of the supply voltage RC is the time constant of the RC Charging circuit o According to the question time to be when Vout= 95% of Vin & & Vout = 0.95 Vin

- Vin & iR & į Sidt zo differentiating with respect to t:- - during + R dies + į = 0 R dit 4 Differential equation form

The capacitive Reactance is given as :- is calculated as: and the circuit Impedance Z= √R²+ X² RC Potential Divider Vout = Winx te ye - Vink Now, from differential equation form

dvin Ja dt ac dirt i = rolling Here, RC=t= (2=k= time constant of the circuit static Sensitivity /

» 0.95 Vin = Vin {i- é t/Re] * 0.95 = -e-tipe * etike -0.05 taking 'log' both side :- t e = ln (0.05) ata RC [-2.996] t= RC * 2.996 t = 1000 * 1x 106 x 2.996 It = 0.002996 sec It= 2.996 msec