Lossy Inverting Integrator

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In summary, the problem is that you need to differentiate both sides to get a DiffEq that you can solve.
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Hey guys,

I've got this set of problems on RC/RL circuits and differentiators and integrators. I was able to solve all of them, but this one stumped me for some reason.

Here is the problem:

http://img31.imageshack.us/img31/4286/34370298.png [Broken]

Relevant methods:

1) multiplying both sides by e^(-t/RC) and differentiating?

2) integrating, multiplying by e^(-t/RC), and isolating Vout?

Thats the methods I used to solve the other problems, but I'm not even sure how to start on this one, so any hints would be appreciated.

Oh, also, the solution should look something like this (i think?):

Vout(t)=c1+(c2-c1)e^(-(t-t0)/RC) and c1,c2 are constant
 
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  • #2
kushnee said:
Hey guys,

I've got this set of problems on RC/RL circuits and differentiators and integrators. I was able to solve all of them, but this one stumped me for some reason.

Here is the problem:

http://img31.imageshack.us/img31/4286/34370298.png [Broken]

Relevant methods:

1) multiplying both sides by e^(-t/RC) and differentiating?

2) integrating, multiplying by e^(-t/RC), and isolating Vout?

Thats the methods I used to solve the other problems, but I'm not even sure how to start on this one, so any hints would be appreciated.

Oh, also, the solution should look something like this (i think?):

Vout(t)=c1+(c2-c1)e^(-(t-t0)/RC) and c1,c2 are constant

It looks like you need to differentiate both sides to get a DiffEq that you can solve...
 
Last edited by a moderator:
  • #3
Thanks for the tip. So I've tried integrating it and this is what I got:

d/dt[Vout-Vout(t0)]=-[(A-Vout(infinity)+Vout(t0))/RC]

Is this even correct? I'm not sure what method to use to solve this DE and I'm doubting my differentiation skills at this point. I replaced V1 with A, since it's a constant i suppose, and I moved the derivative of Vout(t0) to the left side.

Thanks,
Adam.
 

1. What is a lossy inverting integrator?

A lossy inverting integrator is an electronic circuit that takes an input signal and integrates it over time, producing an output signal that is the integral of the input signal. The term "lossy" refers to the fact that some of the input signal is lost during the integration process, resulting in a less precise output signal.

2. How does a lossy inverting integrator work?

A lossy inverting integrator works by using an operational amplifier (op-amp) and a feedback capacitor. The input signal is applied to the inverting input of the op-amp, and the output of the op-amp is connected to the feedback capacitor. As the input signal changes, the op-amp adjusts its output to keep the inverting input at the same voltage as the non-inverting input. This causes the capacitor to charge or discharge, resulting in an output signal that is the integral of the input signal.

3. What are the applications of a lossy inverting integrator?

Lossy inverting integrators are commonly used in electronic filters, where they can be used to remove unwanted high-frequency signals from a circuit. They are also used in audio processing and data transmission, as well as in control systems where precise integration of a signal is necessary.

4. What are the advantages of using a lossy inverting integrator?

The main advantage of a lossy inverting integrator is its simplicity and low cost. It can also be easily adjusted to integrate over different time periods by changing the value of the feedback capacitor. Additionally, it can be cascaded with other circuits to create more complex systems.

5. Are there any limitations of a lossy inverting integrator?

One limitation of a lossy inverting integrator is that it can introduce noise into the output signal, especially when dealing with high-frequency signals. It is also not suitable for applications where precise signal integration is required, as it can only approximate the integral of the input signal. Additionally, the output signal may be affected by changes in temperature or other environmental factors.

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