Question

2. In the circuit below, R1 = 10k, R2 : magnitude is applied to the input. Find V(t) at the output and sketch it. Your sketch should be quantitatively correct, that is, it should have labeled V and t axes. Strongly suggested: solve by first finding the Thévenin equivalent of the source voltage and resistor network. 20k and C = 0.1µF. At t = 0, a step function of 10 V Input (problem 3) R, 10V- IN R2 C, OUT 3. Use the technique of complex arithmetic to calculate the frequency response |Vout/Vin|(f) of the circuit in Problem 2. (Note: in this problem, the input is not a step, but a sine-wave source of variable frequency.) Sketch a "Bode plot" of the response, with axes labeled in decibels and log f. (See Horowitz and Hill, p. 50-51, or Hayes, pp. 63-66 for a discussion of Bode plots.)

Answer 1

(2) Given that Ria loke; B-2002; (201 MF Тал i lov oku B 01F ''hevinin's equivalent voltage is the voltage across aokr Resistance. When capacitor is open Granited . .. Applysne vo Hege division Rule - zok r = 20 lu bbbv Vm= 10x OK took = lox equivalent Thevinin's Equivalent Resistance is the Resistance across open Circuited terminal Whan vo Haso is replaced with shart Granit (Internal Resistance of Ideal voltage source=o] loks I zone tree reg= loks in parallel with oke Resistance Такъв е kraf.66 kr RM 30xs ReplaGug The vinin's equivalent Circuit inplace of the given Grauit we get 6.66ks mhmouto O.THF 6-66v

Applying Nodal Analysis at vout weget Vout -bibb - + 6.66k 6 dvout o.lx10. 00 d Vout - vout 666 6.66ktal x103 book to 1x10 ? dvout + 1500 Vout alo! As we know for linear Diff. Eg of the stars ay Py = Q , solution is given by 1. Sode , sødx ye - Jace tc sisoodt u sisoodt vouto e :Jio e ac 150ot Vout.e 4 isoot lo.e tc 1500 alsoot vout a 20o the To find c. we have to apply Initial condition As we know Before tzo, supply given to the Civanitza o vout cot) .0. a out (o-)

2. Vout= vont: 20 e soot vout mi almų 2 To tm 3.33msec t - s time I time constant se til sec 1500 Constants je to 5XL ser 1500 It = 0.6 msec] [52= 3.33 mble 52 -3.33mole