The Bode magnitude plot of the transfer functionK(1+ as)(1+0.25s)G(s) =s(1+)(1+ bs)(1+ )%3Dis shown in figure below. Notethat – 6dB/octave = -20DB/decade.OdB/0ctave-6dB/Octave6dB/0ctavedB-odB/OctaveOdB/Octave|-12dB/0ctave0.01 224 36+ o (rad/s)Figure 1: Bode magnitude plot3) What is the value of K rounded to 2 decimal points?8.

Using bode plot we will write expression for magnitude , with this slope the magnitude is zero at 8…

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

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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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
Find the frequency gain
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
= b) A three-stage amplifier system operated at T 25°C and 1 MHz bandwidth which is determined by an LC-tuned circuit. The first stage has a power gain of 10dB and an NF of 5dB, the second has a power gain of 20dB and NF=10dB, and the third stage has a power gain of 30dB and NF=10dB. The output load is 2000. The input noise is generated by a 20k resistor. Calculate: i) The noise voltage and power at the input and the output of this system assuming ideal noiseless amplifiers. ii) The overall noise figure for the system. iii) The actual output noise voltage and power.
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
Q. Lag compensator..
Answer all
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
Q-1: Determine the 3 dB high frequencies and midband gain of a cascode circuit shown in Figure, the parameters are: V+ = 10 V, V- = -10 V, Rs = 0.1ΚΩ, R1 = 42.5kΩ, R₂ = 20.5k2, R3 = 28.3 k2, RĘ = 6 kN, Rc = 10 kN, R₁ = 15 k2, and C₁ = 0. The transistor parameters are:ß= 150, 9m1 = 9m2 = 9m 39.2 mA/V, C1 = = C2 C36 pF, and C₁ = 6 pF. = V+ -0% CB= Rs ww Cci R₂ R₂ Ccz 2₂ 2₁ ww RE www R₁ =CL CE
Q5- Sketch the complete Bode plot for the transfer function margins and the crossover frequencies on your plot. 37(7s+2) s² (2s+7) (7s² +8s+8) mark the KS

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