MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL)
4th Edition
ISBN: 9781266368622
Author: NEAMEN
Publisher: MCG
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Chapter 15, Problem 15.67P
To determine
The range of the frequency of the oscillation and the range of the duty cycle.
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Chapter 15 Solutions
MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL)
Ch. 15 - Design a twopole lowpass Butterworth filter with a...Ch. 15 - Consider the switchedcapacitor circuit in Figure...Ch. 15 - Prob. 15.3EPCh. 15 - (a) Design a threepole highpass Butterworth active...Ch. 15 - Prob. 15.2TYUCh. 15 - Prob. 15.3TYUCh. 15 - Simulate a 25M resistance using the circuit in...Ch. 15 - Design the phaseshift oscillator shown in Figure...Ch. 15 - Design the Wienbridge circuit in Figure 15.17 to...Ch. 15 - Prob. 15.5TYU
Ch. 15 - Prob. 15.6TYUCh. 15 - Prob. 15.6EPCh. 15 - Redesign the street light control circuit shown in...Ch. 15 - A noninverting Schmitt trigger is shown m Figure...Ch. 15 - For the Schmitt trigger in Figure 15.30(a), the...Ch. 15 - Prob. 15.9TYUCh. 15 - Prob. 15.8EPCh. 15 - Prob. 15.9EPCh. 15 - Consider the 555 IC monostablemultivibrator. (a)...Ch. 15 - The 555 IC is connected as an...Ch. 15 - Prob. 15.10TYUCh. 15 - Prob. 15.11TYUCh. 15 - Prob. 15.12TYUCh. 15 - Prob. 15.12EPCh. 15 - Prob. 15.13EPCh. 15 - (a) Consider the bridge amplifier in Figure 15.46...Ch. 15 - Prob. 15.14EPCh. 15 - Prob. 15.15EPCh. 15 - Prob. 15.16EPCh. 15 - Prob. 1RQCh. 15 - Prob. 2RQCh. 15 - Consider a lowpass filter. What is the slope of...Ch. 15 - Prob. 4RQCh. 15 - Describe how a capacitor in conjunction with two...Ch. 15 - Sketch a onepole lowpass switchedcapacitor filter...Ch. 15 - Explain the two basic principles that must be...Ch. 15 - Prob. 8RQCh. 15 - Prob. 9RQCh. 15 - Prob. 10RQCh. 15 - Prob. 11RQCh. 15 - What is the primary advantage of a Schmitt trigger...Ch. 15 - Sketch the circuit and explain the operation of a...Ch. 15 - Prob. 14RQCh. 15 - Prob. 15RQCh. 15 - Prob. 16RQCh. 15 - Prob. 17RQCh. 15 - Prob. 18RQCh. 15 - Prob. D15.1PCh. 15 - Prob. 15.2PCh. 15 - The specification in a highpass Butterworth filter...Ch. 15 - (a) Design a twopole highpass Butterworth active...Ch. 15 - (a) Design a threepole lowpass Butterworth active...Ch. 15 - Prob. 15.6PCh. 15 - Prob. 15.7PCh. 15 - Prob. 15.8PCh. 15 - A lowpass filter is to be designed to pass...Ch. 15 - Prob. 15.10PCh. 15 - Prob. 15.11PCh. 15 - Prob. D15.12PCh. 15 - Prob. D15.13PCh. 15 - Prob. D15.14PCh. 15 - Prob. 15.15PCh. 15 - Prob. 15.16PCh. 15 - Prob. 15.17PCh. 15 - Prob. 15.18PCh. 15 - A simple bandpass filter can be designed by...Ch. 15 - Prob. 15.20PCh. 15 - Prob. 15.21PCh. 15 - Prob. D15.22PCh. 15 - Prob. 15.23PCh. 15 - Consider the phase shift oscillator in Figure...Ch. 15 - In the phaseshift oscillator in Figure 15.15, the...Ch. 15 - Consider the phase shift oscillator in Figure...Ch. 15 - Prob. 15.27PCh. 15 - Prob. 15.28PCh. 15 - Prob. 15.29PCh. 15 - Prob. 15.30PCh. 15 - Prob. 15.31PCh. 15 - A Wienbridge oscillator is shown in Figure P15.32....Ch. 15 - Prob. 15.33PCh. 15 - Prob. D15.34PCh. 15 - Prob. D15.35PCh. 15 - Prob. 15.36PCh. 15 - Prob. 15.37PCh. 15 - Prob. D15.38PCh. 15 - Prob. 15.39PCh. 15 - Prob. 15.40PCh. 15 - Prob. 15.41PCh. 15 - For the comparator in the circuit in Figure...Ch. 15 - Prob. 15.43PCh. 15 - Prob. 15.44PCh. 15 - Prob. 15.45PCh. 15 - Consider the Schmitt trigger in Figure P15.46....Ch. 15 - The saturated output voltages are VP for the...Ch. 15 - Consider the Schmitt trigger in Figure 15.30(a)....Ch. 15 - Prob. 15.50PCh. 15 - Prob. 15.52PCh. 15 - Prob. 15.53PCh. 15 - Prob. 15.54PCh. 15 - Prob. 15.55PCh. 15 - Prob. 15.56PCh. 15 - Prob. 15.57PCh. 15 - Prob. D15.58PCh. 15 - Prob. 15.59PCh. 15 - The saturated output voltages of the comparator in...Ch. 15 - (a) The monostablemultivibrator in Figure 15.37 is...Ch. 15 - A monostablemultivibrator is shown in Figure...Ch. 15 - Prob. D15.63PCh. 15 - Design a 555 monostablemultivibrator to provide a...Ch. 15 - Prob. 15.65PCh. 15 - Prob. 15.66PCh. 15 - Prob. 15.67PCh. 15 - Prob. 15.68PCh. 15 - An LM380 must deliver ac power to a 10 load. The...Ch. 15 - Prob. 15.70PCh. 15 - Prob. D15.71PCh. 15 - Prob. 15.72PCh. 15 - (a) Design the circuit shown in Figure P15.72 such...Ch. 15 - Prob. 15.74PCh. 15 - Prob. 15.75PCh. 15 - Prob. 15.76PCh. 15 - Prob. D15.77PCh. 15 - Prob. 15.78P
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- Imaginary Axis (seconds) 1 6. Root locus for a closed-loop system with L(s) = is shown below. s(s+4)(s+6) 15 10- 0.89 0.95 0.988 0.988 -10 0.95 -15 -25 0.89 20 Root Locus 0.81 0.7 0.56 0.38 0.2 5 10 15 System: sys Gain: 239 Pole: -0.00417 +4.89 Damping: 0.000854 Overshoot (%): 99.7 Frequency (rad/s): 4.89 System: sys Gain: 16.9 Pole: -1.57 Damping: 1 Overshoot (%): 0 Frequency (rad/s): 1.57 0.81 0.7 0.56 0.38 0.2 -20 -15 -10 -5 5 10 Real Axis (seconds) From the values shown in the figure, compute the following. a) Range of K for which the closed-loop system is stable. b) Range of K for which the closed-loop step response will not have any overshoot. Note that when all poles are real, the step response has no overshoot. c) Smallest possible peak time of the system. Note that peak time is the smallest when wa is the largest for the dominant pole. d) Smallest possible settling time of the system. Note that peak time is the smallest when σ is the largest for the dominant pole.arrow_forwardFor a band-rejection filter, the response drops below this half power point at two locations as visualised in Figure 7, we need to find these frequencies. Let's call the lower frequency-3dB point as fr and the higher frequency -3dB point fH. We can then find out the bandwidth as f=fHfL, as illustrated in Figure 7. 0dB Af -3 dB Figure 7. Band reject filter response diagram Considering your AC simulation frequency response and referring to Figure 7, measure the following from your AC simulation. 1% accuracy: (a) Upper-3db Frequency (fH) = Hz (b) Lower-3db Frequency (fL) = Hz (c) Bandwidth (Aƒ) = Hz (d) Quality Factor (Q) =arrow_forwardP 4.4-21 Determine the values of the node voltages V1, V2, and v3 for the circuit shown in Figure P 4.4-21. 29 ww 12 V +51 Aia ww 22. +21 ΖΩ www ΖΩ w +371 ①1 1 Aarrow_forward
- 1. What is the theoretical attenuation of the output voltage at the resonant frequency? Answer to within 1%, or enter 0, or infinity (as “inf”) Attenuation =arrow_forwardWhat is the settling time for your output signal (BRF_OUT)? For this question, We define the settling time as the period of time it has taken for the output to settle into a steady state - ie when your oscillation first decays (aka reduces) to less than approximately 1/20 (5%) of the initial value. (a) Settling time = 22 μs Your last answer was interpreted as follows: Incorrect answer. Check 22 222 What is the peak to peak output voltage (BRF_OUT pp) at the steady state condition? You may need to use the zoom function to perform this calculation. Select a time point that is two times the settling time you answered in the question above. Answer to within 10% accuracy. (a) BRF_OUT pp= mVpp As you may have noticed, the output voltage amplitude is a tiny fraction of the input voltage, i.e. it has been significantly attenuated. Calculate the attenuation (decibels = dB) in the output signal as compared to the input based on the formula given below. Answer to within 1% accuracy.…arrow_forwardmy previous answers for a,b,d were wrong a = 1050 b = 950 d=9.99 c was the only correct value i got previously c = 100hz is correctarrow_forward
- V₁(t) ww ZRI ZLI ZL2 ZTH Zci VTH Zc21 Figure 8. Circuit diagram showing calculation approach for VTH and Z TH we want to create a blackbox for the red region, we want to use the same input signal conditions as previously the design of your interference ector circuit: Sine wave with a 1 Vpp, with a frequency of 100 kHz (interference) Square wave with 2.4Vpp, with a frequency of 10 kHz (signal) member an AC Thevenin equivalent is only valid at one frequency. We have chosen to calculate the Thevenin equivalent circuit (and therefore the ackbox) at the interference frequency (i.e. 100 kHz), and the signal frequency (i.e. 10 kHz) as these are the key frequencies to analyse. Your boss is assured you that the waveform converter module has been pre-optimised to the DAB Receiver if you use the recommended circuit topology.arrow_forwardVs(t) + v(t) + vi(t) ZR ZL Figure 1: Second order RLC circuit Zc + ve(t) You are requested to design the circuit shown in Figure 1. The circuit is assumed to be operating at its resonant frequency when it is fed by a sinusoidal voltage source Vs (t) = 2sin(le6t). To help design your circuit you have been given the value of inductive reactance ZL = j1000. Assume that the amplitude of the current at resonance is Is (t) = 2 mA. Based on this information, answer the following to help design your circuit. Use cartesian notation for your answers, where required.arrow_forwardWhat is the attenuation at the resonant frequency? You should use the LTSpice cursors for your measurement. Answer to within 1% accuracy, or enter 0, or infinity (as "inf") (a) Attenuation (dB) = dB Check You may have noticed that it was significantly easier to use frequency-domain "AC" simulation to measure the attenuation, compared to the steps we performed in the last few questions. (i.e. via a time-domain "transient" simulation). AC analysis allows us to observe and quantify large scale positive or negative changes in a signal of interest across a wide range of different frequencies. From the response you will notice that only frequencies that are relatively close to 100 kHz have been attenuated. This is the result of the Band-reject filter you have designed, and shows the 'rejection' (aka attenuation) of any frequencies that lie in a given band. The obvious follow-up question is how do we define this band? We use a quantity known as the bandwidth. A commonly used measurement for…arrow_forward
- V₁(t) ww ZRI ZLI ZL2 ZTH Zci VTH Zc21 Figure 8. Circuit diagram showing calculation approach for VTH and Z TH we want to create a blackbox for the red region, we want to use the same input signal conditions as previously the design of your interference ector circuit: Sine wave with a 1 Vpp, with a frequency of 100 kHz (interference) Square wave with 2.4Vpp, with a frequency of 10 kHz (signal) member an AC Thevenin equivalent is only valid at one frequency. We have chosen to calculate the Thevenin equivalent circuit (and therefore the ackbox) at the interference frequency (i.e. 100 kHz), and the signal frequency (i.e. 10 kHz) as these are the key frequencies to analyse. Your boss is assured you that the waveform converter module has been pre-optimised to the DAB Receiver if you use the recommended circuit topology.arrow_forwardVs(t) + v(t) + vi(t) ZR ZL Figure 1: Second order RLC circuit Zc + ve(t) You are requested to design the circuit shown in Figure 1. The circuit is assumed to be operating at its resonant frequency when it is fed by a sinusoidal voltage source Vs (t) = 2sin(le6t). To help design your circuit you have been given the value of inductive reactance ZL = j1000. Assume that the amplitude of the current at resonance is Is (t) = 2 mA. Based on this information, answer the following to help design your circuit. Use cartesian notation for your answers, where required.arrow_forwardFor a band-rejection filter, the response drops below this half power point at two locations as visualised in Figure 7, we need to find these frequencies. Let's call the lower frequency-3dB point as fr and the higher frequency -3dB point fH. We can then find out the bandwidth as f=fHfL, as illustrated in Figure 7. 0dB Af -3 dB Figure 7. Band reject filter response diagram Considering your AC simulation frequency response and referring to Figure 7, measure the following from your AC simulation. 1% accuracy: (a) Upper-3db Frequency (fH) = Hz (b) Lower-3db Frequency (fL) = Hz (c) Bandwidth (Aƒ) = Hz (d) Quality Factor (Q) =arrow_forward
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