In the op amp circuit of Fig. 8.107 , determine vo(t) for t>0 . Let vin =u(t) V, R1=R2=10 k Ω, C

step 1 Hello everyone, let us do the following question. In the following circuit, we have to determine

In the op amp circuit of Fig. 8.107 , determine vo(t) for...

Key Concepts

Operational Amplifier (Op Amp) Behavior

This concept covers the essential characteristics of op amps including high gain, high input impedance, and low output impedance. Understanding how an ideal op amp functions—particularly its behavior in feedback configurations—is crucial when analyzing circuits that employ op amps to amplify or process signals.

Transient Analysis in Electrical Circuits

Transient analysis deals with the response of circuits to changes from one steady state to another, particularly when reactive elements like capacitors or inductors are involved. This analysis is important in understanding how the circuit output evolves over time immediately after a change, such as the application of a step input, before reaching a new equilibrium.

RC Circuit Dynamics and Time Constants

This concept involves the behavior of resistor-capacitor (RC) networks, where the time constant (? = RC) determines the rate at which voltages in the circuit charge or discharge. The time constant is a key parameter in predicting how quickly a voltage will rise or fall, which is critical when analyzing circuits with capacitive components undergoing transient changes.

Step Input Response

The step input, typically represented as a unit step function u(t), is a fundamental excitation used in circuit analysis to study dynamic behavior. Analyzing how a circuit responds to a sudden input change provides insights into its stability and transient characteristics, which is essential in control systems and signal processing.

Laplace Transform Techniques in Circuit Analysis

Laplace transforms are a powerful mathematical tool used for converting time-domain differential equations into algebraic equations in the s-domain. This technique simplifies the analysis of circuits with reactive components, allowing for the computation of transient responses and system behaviors, especially when initial conditions are present.

Transcript

00:01 Hello everyone, let us do the following question.

00:03 In the following circuit, we have to determine the value for v node.

00:07 This is node 1 and this is node 2.

00:11 At first, we will solve this equation at node 1.

00:15 We will solve this equation at node 1.

00:18 We will get the equation as b input minus b1 divided by r1 is equal to c1, d by dt.

00:27 This can be written as b1 minus v1.

00:30 B node plus c2 d by d t this can be written as v1 minus 0 this is equation number 1 now we will solve this equation at node now we will solve this equation and node 2 we will get c2 d by d2 v1 minus 0 is equal to d0 minus v node by r2 or from this we will get value of d1 by d t is equal to minus v0 by c2 into r2 this is equation number two now we will solve this for from now we will say that from equation number one and equation two from equation one and equation two we will get v i0 minus v1 is equal to r1 c1 c2 c2 d b node by d t minus r1 c1 d b not by d t minus r1 v not by r2 next it will solve this equation for v1 this can be written as b i 0 plus r1 c1 c2 r2 d b b b not by d t plus r1 c1 d b b b b b b b b b b b b b b b 0 by d t plus r1 by r2 into v0 now we will say this will be equation number 3 this will be equation number 3 now from second and third equation now from second and third we will get minus v0 c2 by c2 into r2 is equal to dv1 by d t is equal to dv 1 by d t is equal to to d b input by d t plus r1 c1 c2 r2 d b b not by d t plus r1 c1 double derivative v0 by d t square plus r1 r1 r2 d b not by d t after that if we will solve this equation we will get d no d b squared by d t squared here plus 1 by r2 in the bracket we will get 1 by c1 plus 1 by c2 d b nod by d t plus v nod c1 c2 r1 is equal to minus 1 by r1 c1 d b input by d t but we know that c1 c2 r1 r2.

03:33 This can be written as 10 raised to power minus 4, 10 raised to power 4, 10 raised to power 4.

03:42 This is equation number 1.

03:45 And if we'll solve this further, we will get 1 by r2.

03:50 In the bracket, we will write 1 by c1 plus 1 by c2.

03:54 Value of this can be written as 2, r2 into c1.

03:59 So we will get 2.

04:01 R1 is 10 raised to 5.

04:02 4, c1 is 10 raise to par minus 4, value of this can be written as 2.

04:08 Now if we will solve this equation, we will get this differential equation as d squared double derivative of v0 plus 2, db0 by d t plus v nod is equal to minus d b input by d t.

04:28 This equation can be written as s square plus 2s plus 1 is equal to 0.

04:35 From this we will get 2 roots...

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