1) Consider the circuit shown in figure below. Compute the output waveform for an input of 5kHz
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1 ) Consider the circuit shown in figure below. Compute the output waveform for an input of 5kHz Solution: a) Input is of constant amplitude of 2 V from 0 to 0. ms and 2 V from 0. ms to 0.2 ms. The output for each of these half periods will be ramps i.e a triangular wave. The peak-peak value of the output for the first half cycle is t Vo = Vin dt, voltage output for the op amp integrator RCf 0 0. m Vo = 0 k x 20 n (2 x) dt = 5 x [0. m 0] = V 0 This represents the total change in the output voltage over the first half cycle from 0 to 0. ms. Similarly, integration over the next half cycle produces a positive change of V Ans: Figure (a) 2) To obtain an optimum differentiation choose the values of Rf and C for a 7-kHz input signal a. Rf =.3 kω, C = 0. µf b. Rf = 7 kω, C = 47 µf c. Rf = 8 kω, C = 66 µf d. Rf =.6 kω, C = 33 µf Solution: a) Optimum differentiation condition is T Rf C
2 Given: f = 7 khz, T = /f =.428 x 0-4 Hz ----() Rf C = 0. µ x.3 k =.3 x (2) () (2), satisfies the conditions Ans: Rf =.3 kω, C = 0. µf 3) How passive filters are problematic a) Inductors become large at low frequencies b) Design of inductors are expensive at low frequency applications c) Series resistance of inductors degrade its performance d) All of the above At audio frequencies inductors are problematic due to the its size and bulkiness and hence these are heavy and expensive. Moreover, low frequency applications require more number of turns of wire which in turn adds to the series resistance degrading inductor s performance. Ans: All of the above 4) What is the 0-dB frequency for the circuit given? a) Hz b) 25 MHz c) 0 khz d) 0.26 Hz 0 db frequency, f = 2π RfC f = 2π RfC = = 0.26 Hz 2π x 33 kω x 0.47 µf Ans: 0.26 Hz The circuit shown is a practical differentiator fc = 2π RfC = = 0.26 Hz 2π x 33 kω x 0.47 µf The 0-db frequency is fc = 0.26 Hz In this case RfC >> Rc
3 5) Active filters using op-amps are designed with a) All passive elements (Resistors, Capacitors and Inductors) b) Only Resistors and Inductors c) Only Capacitors and Resistors d) Only Resistors Solution: C) The active filters use capacitor in the feedback loop and thus the use of inductors can be avoided. Hence, an inductor less RC filters can be achieved with op-amps. Ans: Only Capacitors and Resistors 6) Compute the output voltage for the given circuit. a. 2 sin ωt b. 0 V c. - 2 sin ωt d. 8 sin ωt Solution: c) The given circuit is a summing amplifier. Vo = - (V+V2), output voltage of a summing amplifier Vo = - (5sinωt 3 sinωt) = - 2 sinωt Ans: - 2 sinωt
4 7) Consider a differentiator circuit with RF =0 kω, C=26 nf and a cosine wave of 4 Vpp at 500 Hz is applied to it. Then find the appropriate output waveform. Solution: c) (for inverting differentiator) and d) (Simple differentiation) Vin = V cos (2π 500 t) Vo = Rf x C d Vin, voltage output of a opamp differentiator dt Vo = 0 k x 26 x0 9 x d (4 cos 2π 500t) = 3.26 sin 2 π 500t dt The output is a sine wave with no phase shift. Ans: 8) Assume the op-amp given in the circuit is ideal and a triangular wave is given as input to it. Then which among the following will be the output waveform. a) Triangular wave b) Square wave c) Sine wave d) Cosine wave
5 Solution: b) Vo = Rf x C d Vin, voltage output of a opamp differentiator dt Assume amplitude is peak to peak and for a triangular wave it is ± At So dv/dt = ± A, therefore it gives a square wave output Ans: Square wave 9) Consider the circuit shown in figure below, compute the output voltage of U2 op-amp at t = 0.5 s Note: Consider the op-amp is supplied with ± 5 V a) 0 V b) 20 V c) 5 V d) - 5 V Solution: a) The output at U is V0 (let s say) On substitution, So, at t= 0.5, V0 = - 4*0.5 = - 2 V The output at U2 is V02 (let s say) At t = 0.5, V02 = V02 = /0. = 0 V V0 = V t V dt = RC RC V0 = 0.t = 4 t V0 dt = R2C2 R2C2 4 t dt = 4 t 2 2 R2C2
6 0) Choose the correct option for the following circuit a) A Low-pass filter b) An All-pass filter c) A Band-pass filter d) A High-pass filter Solution: b) The transfer function of the above circuit is Vo/ Vi = (- j2πrfc)/(+ j2πrfc) From equation, it is obvious that the amplitude of Vo /Vi is unity, that is Vo = Vi throughout the useful frequency range and the phase shift between the input and output voltages is a function of frequency. Hence it is an all-pass filter. Ans: An All-pass filter ) Determine the roll-off for the filter circuit shown below a) 40 db/decade b) 60 db/decade c) 20 db/decade d) 80 db/decade Solution: c) The circuit shown is a first order band-pass filter, and thereby roll-off equals 20 db/decade Ans: 20 db/decade 2) Consider the circuit shown in the figure below. If input V =3 Vp-p sine wave, compute the output voltage of the op-amp U2
7 a) 3 Vp-p Sine wave b).5 Vp-p sine wave c) 3 V DC d).5 V DC The circuit shown is a peak detector which captures the peak value of the input voltage (V). When V is positive D is reverse biased, D2 is forward biased and no current flows in the feedback resistor R. Since, the output of U2 is feedback to U, thus the output voltage of U2 follows the input voltage (V) because the outer feedback loop drives the inputs of U to a virtual short (V+ = V ). Additionally, the output voltage U2 = the voltage on capacitor C because U2 is configured as a voltage follower. Hence, C is charged to this voltage by the output current of U through the D2. The use of R2 prevents the U from exceeding its short circuit output current and isolates U from the capacitance of C which prevents ringing or even oscillations. This state holds as long as the input voltage is positive and increasing. Whereas, when the input voltage decreases. D2 is reverse biased when the input voltage decreases because the output of U (anode of D2) drops below the cathode voltage of D2 which is equal to the previous peak voltage stored on C. The outer feedback loop is broken in this state and the output of U attempts to snap to the negative rail voltage. D is forward biased in this state and provides local feedback to U which clamps the anode of D2 at one diode drop below the input voltage. The hold state is maintained until the input voltage over comes the capacitor voltage which is equal to the output voltage. The D clamp reduces the transition time from the hold state back to the tracking state By this analysis, it is clear that the circuit behaves a peak detector. Hence the output voltage is the peak of input signal. i.e.5 V DC 3) For the circuit shown in the question 2, select the correct operation of the circuit a) Simple low pass filter b) Half-wave rectifier with filter c) Full-wave rectifier with filter d) Peak detector
8 Please refer the previous solution. The circuit is a peak detector. 4) Match the frequency response curve with respective filter type. High pass 2. Band pass 3. Low pass 4. Band-reject a) -d, 2-c, 3-a, 4-b b) -c, 2-d, 3-a, 4-d c) -b, 2-a, 3-c, 4-d d) -a, 2-b, 3-c, 4-d Ans: c) -b, 2-a, 3-c, 4-d 5) When a second order high pass filter and second order low pass sections are cascaded, the resultant filter is a a) 80 db/decade high-pass filter b) 60 db/decade low-pass filter c) 20 db/decade band reject filter d) 40 db/decade band-pass filter The resultant will be band pass configurations, and being second order filter, it has a 40 db/decade roll-off. Ans: 40dB/decade band-pass filter
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