Second Order Op-amp Circuit

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Homework Statement:

the rate of change of the output of both op-amps when the initial energy stored in the circuit is zero and the signal jumps instantaneously from 0 to Vg.

Relevant Equations:

SEE ATTACHED IMAGE
242722


I was reading about this 2nd order op-amp circuit which is essentially a cascaded integrator and got confused with the explanation of the book regarding the rate of change of the outputs. The book said that when the initial energy stored in the circuit is zero then

242715


this rate of change is zero when Vo1(0) = 0(initial energy is zero). Now going to the rate of change of the first of amp,
242716

would this also be equal to zero? The reason I asked this is when I derived for this rate of change I ended up with this

242717

the input Vg is a non zero quantity therefore this rate of change can not be zero. But the initial energy stored is zero so this should not be the case. This got me confused. Please enlighten me. TIA!
 
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Answers and Replies

  • #2
collinsmark
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Problem Statement: the rate of change of the output of both op-amps when the initial energy stored in the circuit is zero and the signal jumps instantaneously from 0 to Vg.
Relevant Equations: SEE ATTACHED IMAGE

View attachment 242722

I was reading about this 2nd order op-amp circuit which is essentially a cascaded integrator and got confused with the explanation of the book regarding the rate of change of the outputs. The book said that when the initial energy stored in the circuit is zero then

View attachment 242715

this rate of change is zero when Vo1(0) = 0(initial energy is zero). Now going to the rate of change of the first of amp,
View attachment 242716
would this also be equal to zero? The reason I asked this is when I derived for this rate of change I ended up with this

View attachment 242717
the input Vg is a non zero quantity therefore this rate of change can not be zero. But the initial energy stored is zero so this should not be the case. This got me confused. Please enlighten me. TIA!
Your equations look correct to me. :smile:

You are correct that the rates of change of voltages, [itex] \frac{dv_{o1}}{dt} [/itex] and [itex] \frac{dv_o}{dt} [/itex] are zero when the input is 0 V, and both capacitors are uncharged (zero initial energy).

But it doesn't stay that way. As soon as the input is changed from 0 to [itex] v_g [/itex], then [itex] \frac{d v_{o1}}{dt} [/itex] is no longer 0.

You can make a similar statement about the next stage too.
 

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