The Bode magnitude plot of the transfer functionK(1+ as)(1+0.25s)G(s) :s(1+)(1+ bs)(1 + )is shown in figure below.Notethat – 6dB/octave = -20dB/decade.OdB/Octave-6dB/0ctave6dB/Octave-6dB/0ctaveOdB/0ctave|-12dB/Octave0.01224 36→이(ad/s)Figure 1: Bode maguitude plot1) What is the value of a rounded to 2 decimalpoints?

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Answered: The Bode magnitude plot of the transfer…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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QUESTIONS: 1. What is the Transfer function? 2. What do the BODE diagram and poles-zeros graph tell us? What is their main purpose? 3. Write the transfer function of figure 1.3 for the given values, add the BODE diagram and poles-zeros graph. 4. Take the K1=1 N/m, K2=2 N/m, fv=1 N-s/m, M1=M2=2 kg, and find the transfer function of figure 2.1. 5. Implement the transfer function of figure 2.1 to Matlab and plot the BODE and poles-zeros graph.
Q.17: For the circuit shown if RL varies from 5 to 102. 1. Did the value of "R" effect on resonance, prove it using the locus diagram. 2. Prove that there is no resonance case in the VT Rc 30 RL XL circuit. 3. Find the new value of "Rc" that makes the circuit at resonance. (Ans.: Rc=4.92, & 82) ww
Series and parallel circuits are made from the same components as follows. A coil of resistance 300 ohm and inductance 100mH and a 4000pF capacitor are connected (i) in series and (ii) in parallel. (A) Find for each connection: (i) the resonant frequency, (ii) the Q-factor (iii) the impedance at resonance and (iv) draw the respective circuits and response curves.
9. A transfer function is given by 100 H (f) = 1 + j(f/1000) Sketch the asymptotic magnitude and phase Bode plots to scale. What is the value of the half-power frequency?
In the circuit in the figure, Vcc = 18 V and RL = 2.2 ko, while C₁ = C₂ = CE = 0.1 µF. It is also known that for parasitic and wiring capacitors, Cwi = 15 pF, Cwo = 21 pF, Cbc = 27 pF, Cbe = 817 pF and Cce = 25 pF. Accordingly, what is the bandwidth of the system? NOTE-1: The output impedance of the transistor will be neglected in ro calculations. NOTE-2: The high frequency dependence of hfe,, will be neglected. C1 Vcc .470 02 0.91 Q .30 B = 100 C2 CE www RL va
1) In the amplifier of Figure 1, g--6.667 mS, Ip-0.5 mA, 2- -0.1 V and CL-31.83 pF. Neglect all other capacitances; Obtain the transfer function VeuVin (s). VDD 333M a) VHE M2 Vout Vin Vout VinoEM, CL Figure 1 b) Determine pole and zero frequencies of the transfer function. there タイCし c) Calculate the low frequency gain in dB. Construct the Bode plot of Voul Vin. d) 40 -20 2 34 10 20 10000 100000 100M 100 1000 Frequency (kHz) Magnitude (dB) 20
For the low-pass active filter shown below, the transfer function is expressed as: V. Zc V: ZR + Zc where: 106 Zc ja rad w = angular frequency in ZR = 2000 ZR V. Vị - Zc Vo can be written as: Vi 4-A) Show that the Vo 500 Vị 500 + jw Vo What should be the value of w so that the magnitude of Vi is ? 4-B) 4-C) With w = 1250 rad/s, and V; = 9ej1.8 – j10 volts, evaluate the magnitude and phase of V. Vo (Hint: evaluate 2 in polar form, and Vị in polar form, then V, = 2x V; ) Vị 4-D) When 4 modules of this filter are cascaded, the total transfer function will be (2) . Use de Moivre's Theorem to find the expression of the total transfer function in polar form. Then, find determine the value of w so that the magnitude of the total transfer function is ? V10
Ex. 1400. See Fig1400a. MdB= -5 dB, Phdeg=-142.42 deg, w1=12 rad/sec, and w2=1200 rad/sec. The Bode plots of GH are given. Determine (1) closed- loop gain margin (dB), and (2) the closed-loop phase margin (deg) ans:2
753. For the transfer function (TF) 4s/(s+9) and resulting Bode Plot, calculate the gains (dB) at the corner, one decade below, and one decade above. Also calculate the phase (deg) of the TF at the same three frequencies. Last answer is w_corner (rad/sec) ans:7 Repeat Ex. 753 using asymptotic approximations. ans:6
Ex. 1340. Refer to Figure setH(1340). Given: w1=1 rad/sec and G(s) = 34/s. Determine asymptotic (also actual): G1,G2,Ph1, mc where gains are in dB, frequencies are in rad/sec, and phases are in degrees. mc=Bode Plot type code. ans:4
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The Bode magnitude plot of the transfer function K(1+ as)(1 + 0.25s) G(s) : s(1+ )(1+ bs)(1+ ) 8 36 is shown in figure below. Notethat – 6dB/octave = -20dB/decade. OdB/0ctave -6dB/0ctave 6dB/Octave dB -6dB/0ctave OdB/0ctave |-12dB/0ctave 0.01 2 24 36 →이(ad/s) Figure 1: Bode magnitude plot