A triangular pulse is given at the input of an integrator circuit, the output will be of the form
Parabolic
An integrator raises the degree of its input by one, so a first-order ramp becomes a second-order curve.
The circuit relation :
\(V_{o}=-\dfrac{1}{RC}\int V_{in}\,dt\)
Now integrate a triangular wave. A triangle is made of straight-line segments, so on each segment the input is a linear function of time:
\(V_{in}=kt\qquad\Rightarrow\qquad V_{o}=-\dfrac{1}{RC}\int kt\,dt=-\dfrac{kt^{2}}{2RC}\)
The \(t^{2}\) is the answer: each segment produces a parabolic arc, and the output is a succession of parabolic curves — option 3.
The chain of waveforms is worth memorising, because these questions recur in both directions :
| Input | After integration | After differentiation |
|---|---|---|
| Impulse | Step | Doublet |
| Step | Ramp | Impulse |
| Square | Triangular | Impulses (spikes) |
| Triangular | Parabolic | Square |
| Sine | −Cosine (lags 90°) | Cosine (leads 90°) |
Option 1 is the trap for reading the table backwards : a square wave is what a triangle differentiates to, not what it integrates to. Option 2 would require the integrator to leave the waveform unchanged, which only happens for a sine wave (and even then the phase shifts by 90°).
Why the degree always rises by one. Integration is the inverse of differentiation, and each integration adds a power of t. A triangle is piecewise degree 1; its integral is piecewise degree 2, a parabola. Integrate again and the result is cubic.
Two practical points about the real circuit. First, a large resistor must be placed across the feedback capacitor, or the op-amp's input offset current will charge it steadily until the output saturates — the circuit then integrates only above \(f=\dfrac{1}{2\pi R_{sh}C}\). Second, the gain falls at 20 dB per decade, so an integrator attenuates high-frequency noise — the opposite of a differentiator, and the reason integrators rather than differentiators are used in analogue computing.
Hence, the output is parabolic.
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
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 input resistance of an ideal Op-Amp is
The input impedance of an ideal Op-amp is ______.
