\(\dfrac{d^{2}y}{dt^{2}}+y=0\) can be solved using an Operational Amplifier. Which one is preferred ?
OPAMP as an Integrator
The analogue-computer rule is always the same: solve differential equations with integrators, never with differentiators.
Step 1 — rearrange for the highest derivative.
\(\dfrac{d^{2}y}{dt^{2}}=-y\)
Step 2 — integrate down instead of differentiating up. Feed \(\ddot{y}\) into one integrator to obtain \(\dot{y}\), then into a second to obtain \(y\), and feed that back (inverted) to the input of the first. The loop then satisfies the equation by construction and oscillates as \(y=A\sin t+B\cos t\), the true solution. Initial conditions enter simply as the initial charge on each integrating capacitor.
Why the differentiator is rejected — the real reason is noise. The two circuits have opposite frequency behaviour:
| Integrator (C in feedback) | Differentiator (C at input) | |
|---|---|---|
| Transfer function | \(-\dfrac{1}{sRC}\) | \(-sRC\) |
| Gain vs frequency | Falls at 20 dB/decade | Rises at 20 dB/decade |
| Effect on noise | Smooths and averages it out | Amplifies every spike |
| Stability | Inherently stable | Prone to ringing and oscillation |
Because a differentiator's gain grows without limit with frequency, the small high-frequency noise always present on a real signal is amplified far more than the signal itself, and a chain of two differentiators would be hopeless. The integrator does the opposite: it attenuates high frequencies, so noise is progressively suppressed as the solution is built up.
A practical footnote. A real op-amp integrator needs a large resistor across the feedback capacitor to stop the input offset current charging it into saturation, and a real differentiator needs a small series input resistor to limit its high-frequency gain. Both fixes exist precisely because of the behaviour described above.
Hence, the preferred configuration is the OPAMP as an integrator.
Similar Questions
Assertion (A) : An Op-Amp is a direct coupled high gain amplifier.
Reason (R) : It consists of one or more differential amplifiers and usually followed by a level translator and push pull stage.
The input to a differentiator is –5 V. Its output will be
Match the following :
List – I List – II a. h-parameters i. O/P voltage varies as the slope of i/p voltage b. differentiator ii. Noise division c. half-wave rectifier iii. Function of a Q point d. integrator iv. series diode clipper Codes :
Assertion (A) : Op-Amp is used for sensor circuit.
Reason (R) : A small signal amplifier amplify weak measured signals.
For an inverting comparator circuit acting as a Schmitt Trigger, as shown in figure below, the expression of Hysteresis Voltage (Vny) is given by :
Consider the following statements :
(A) The output voltage of a summing amplifier (inverting configuration) with three inputs VA, VB and VC and input resistors RA, RB and RC is \(V_{o}=\left(1+\dfrac{R_{F}}{R_{A}R_{B}R_{C}}\right)\left[\dfrac{V_{A}}{R_{A}}+\dfrac{V_{B}}{R_{B}}+\dfrac{V_{C}}{R_{C}}\right]\)
(B) In a subtractor circuit, the output voltage is equal to voltage applied to non-inverting terminal minus voltage applied to inverting terminal
(C) The narrow band pass filter is called a Notch filter
(D) VCO is also called as frequency to voltage
(E) The all pass filter provides unity-gain with predictable phase shifts for different input frequenciesChoose the most appropriate answer from the options given below :
Consider the following circuit, the switch S1 allows the output to switch between two ranges of amplitudes from 0-0.1 V and 0-1 V. Arrange these values of R1, R2 and R3 in increasing order.
(A) Value of R1
(B) Value of R2
(C) Value of R3Choose the most appropriate answer from the options given below :
Match List I with List II
LIST I LIST II A. Butterworth filter of order '2' I. Impedance matching B. Buffer II. CMRR = ∞ (infinity) C. Schmitt Trigger III. Positive feedback D. Ideal OPAMP IV. 40 dB/decade roll off Choose the correct answer from the options given below:
The given operational amplifier circuit corresponds to which electronic circuit application ?
Statements in connection to Op-Amp applications are :
A. If we use a square wave generator followed by integrator circuit we get a triangular wave at the output
B. The logarithmic amplifier called a log-amplifier or a logger, is basically a current to voltage converter.
C.
is a first order high pass filter with voltage followerD. If we use a square wave generator followed by a clipping circuit then we get a saw-tooth wave generator.
Choose the correct answer from the options given below:
Important Questions from Op-Amp and Its Applications
What is the typical value of open-loop voltage gain, AVOL, for a 741 op-amp?
An ideal Op-Amp is an ideal
- In an ideal OP-AMP, which one is correct?
Which of the following statements about the Op-Amp differential amplifiers is INCORRECT?
The total output offset voltage of an operational amplifier is a function of these effects.
