Second Order Op Amp circuit: Find Vo for t > 0

In summary, the circuit shown in the given image has V_o(t) for t > 0, with V_{IN}\,=\,u(t)\,V, R_!\,=\,R_2\,=\,10\,k\Omega, C_1\,=\,C_2\,=\,100\,\muF. The KCL equations at V_1 and V_2 and their substitutions lead to the solution V_o\,=\,-R_2C_2\left(\frac{V_{IN}\,-\,V_1}{R_1C_2}\,-\,\frac{C_1}{C_2}\,\frac{dv_1}{
  • #1
VinnyCee
489
0

Homework Statement



In the circuit below, determine [itex]v_o(t)[/itex] for t > 0. Let [itex]V_{IN}\,=\,u(t)\,V[/itex], [itex]R_!\,=\,R_2\,=\,10\,k\Omega[/itex], [itex]C_1\,=\,C_2\,=\,100\,\muF[/itex].

http://img249.imageshack.us/img249/3840/problem867cg5.jpg



Homework Equations



[tex]i_c\,=\,C\,\frac{dv_c}{dt}[/tex]



The Attempt at a Solution




Add some node voltages:

http://img413.imageshack.us/img413/2935/problem867part2vn2.jpg

KCL @ [itex]V_1[/itex]:

[tex]\frac{V_{IN}\,-\,V_1}{R_1}\,=\,C_2\,\frac{dv_2}{dt}\,+\,C_1\,\frac{dv_1}{dt}[/tex]

KCL @ [itex]V_2[/itex]:

[tex]C_2\,\frac{dv_2}{dt}\,=\,\frac{V_2\,-\,V_o}{R_2}[/tex]

Substituting for [itex]\frac{dv_2}{dt}[/itex] in the KCL @ [itex]V_1[/itex] equation:

[tex]V_o\,=\,-R_2C_2\left(\frac{V_{IN}\,-\,V_1}{R_1C_2}\,-\,\frac{C_1}{C_2}\,\frac{dv_1}{dt}\right)[/tex]

Now what do I do to solve?
 
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  • #2
Set V2 and dV2/dt = 0. Does that help?
 
  • #3


First, it is important to note that this is a second order op amp circuit, meaning that it contains two energy-storing elements (the capacitors) and will require a second order differential equation to solve. The first step in solving this problem would be to write out the two KCL equations for V1 and V2, as you have done.

Next, you can use the fact that V2 = V1 + Vo (since they are connected in series) to eliminate V2 from the equations and solve for Vo. This will give you a second order differential equation in terms of V1 and Vo.

To solve this equation, you can use standard techniques such as Laplace transforms or circuit analysis methods like the node-voltage method. Once you have solved for V1 and Vo, you can use the relationship Vo = V2 - V1 to find the value of Vo for t > 0.

It is also important to note that since the input voltage is a step function u(t), the output voltage will also be a step function with a time delay due to the capacitors in the circuit. This means that the output voltage will have a transient response before reaching its steady state value.
 

What is a Second Order Op Amp circuit?

A Second Order Op Amp circuit is a type of electronic circuit that uses operational amplifiers (op amps) to perform mathematical operations on input signals. It is called "second order" because it uses two reactive components (such as capacitors or inductors) in addition to resistors, making it a second order system in terms of its transfer function.

How do you find Vo for t > 0 in a Second Order Op Amp circuit?

To find Vo for t > 0 in a Second Order Op Amp circuit, you can use the transfer function of the circuit, which is the output voltage (Vo) divided by the input voltage (Vi). The transfer function can be derived from the circuit's components and their values. Once you have the transfer function, you can substitute the value of t and solve for Vo.

What is the significance of t > 0 in a Second Order Op Amp circuit?

The value of t > 0 in a Second Order Op Amp circuit represents the time after the input signal is applied. This is important because the circuit's behavior and output voltage may change over time as the capacitors and inductors in the circuit charge and discharge. By specifying t > 0, we are focusing on the behavior of the circuit after it has reached a steady state.

What are the applications of Second Order Op Amp circuits?

Second Order Op Amp circuits have a wide range of applications in electronics, including active filters, oscillators, and waveform generators. They are also commonly used in audio equipment, such as amplifiers and equalizers, and in control systems for industrial and scientific instruments.

What are the advantages of using Second Order Op Amp circuits?

Some advantages of using Second Order Op Amp circuits include their ability to provide high gain and precise filtering, their simple and compact design, and their low cost. They also have a wide range of applications and can be easily customized for specific needs by adjusting the values of their components.

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