Reason(R): The variation in the gain of the system has no effect on the phase margin of the system. Which are these points? View Answer, 8. b) 0° Frequency range of bode magnitude and phases are decided by : a) Determine the transfer function, H(s) = Vout(s)/Vin(s) b) Sketch the Bode plots of the phase and the magnitude. Make both the lowest order term in the numerator and denominator unity. a) Both A and R are true but R is correct explanation of A Like Reply. 2. Make both the lowest order term in the numerator and denominator unity. The numerator is an order 0 polynomial, the denominator is order 1. Bode Magnitude Plot b) -40 dB/decade (25 points) Solve each problem below. Bode Plots (Bode Magnitude and Phase Plots) - Topic wise Questions in Control Systems ( from 1987) 2003 1. c) Natural frequency and damping ratio This data is useful while drawing the Bode plots. Many common system behaviors produce simple shapes (e.g. The second frequency domain analysis method uses Fourier’s Theorem to compute the process’ Bode plot indirectly. This Bode plot is called the asymptotic Bode plot. The phase is negative for all ω. From $\omega = \frac{1}{\tau}$ rad/sec, it is having a slope of 20 dB/dec. For a conditionally stable type of system as in Fig. d) 90° The Zero degrees line itself is the phase plot for all the positive values of K. Consider the open loop transfer function $G(s)H(s) = s$. The approximate phase of the system response at 20 Hz is : hwmadeeasy Uncategorized 1 Minute. 2. Show that the Nyquist Plot of G(s) = 1 s+a is a semicircle of radius 1 2a and centre (1 2a;0). Similarly, you can draw the Bode plots for other terms of the open loop transfer function which are given in the table. c) 90° This Bode plot is called the asymptotic Bode plot. Bode plots for ratio of ﬁrst/second order factors Problem: Draw the Bode plots for G(s) = s + 3 (s + 2)(s2 + 2s + 25) Solution: We ﬁrst convert G(s) showing each term normalized to a low-frequency gain of unity. Nyquist plot of the transfer function s/(s-1)^3 Bode plot of s/(1-s) sampling period .02 Generate a root locus plot: root locus plot for transfer function (s+2)/(s^3+3s^2+5s+1) At $\omega = 1$ rad/sec, the magnitude is 0 dB. The problem lies with the stimulus frequency, its amplitude (to avoid saturation) and the switching period. Consider the starting frequency of the Bode plot as 1/10 th of the minimum corner frequency or 0.1 rad/sec whichever is smaller value and draw the Bode plot upto 10 times maximum corner frequency. The format is a log frequency scale on … A straight line segment that is tangent to the phase plot … c) Resonant frequencies of the second factors Simply divide each amplitudein the output’s Bode plot by the corresponding amplitude in the input’s Bode plot. Nichol’s chart is useful for the detailed study and analysis of: i. The Bode plot starts at −24.44dB and con-tinue until the ﬁrst break frequency at 2rad/s, yielding -20dB/decade slope downwards un-til the next break frequency at 3rad/s, which causes +20dB/decade slope upwards, which when added to the previous -20dB, gives a net For the negative values of K, the horizontal line will shift $20\: \log K$ dB below the 0 dB line. Tag: Bode plot solved problems 10.87 The differential gain of a MOS amplifier is 100 V/Vwith a dominant pole at 10 MHz. We pick a point, IG(j. Jun 29, 2015 #9 WBahn said: In general, no. d) -180° (s 10)(s 200) 10(s 2) (s) + + + H = 2. s(s 10)2 500 H(s) + = 3. s(s 10s 1000) View Answer, 4. a) Closed loop frequency response b) Origin and +1 The approximate Bode magnitude plot of a minimum phase system is shown in figure. The constant N loci represented by the equation x^2+x+y^2=0 is for the value of phase angle equal to: c) Close loop and open loop frequency responses In the most general terms, a Bode plot is a graph of system frequency response. c) A is true but R is false The magnitude curve breaks at the natural frequency and de- creases at a rate of 40dB/dec. d) open loop and Close loop frequency responses Consider the open loop transfer function $G(s)H(s) = K$. This set of Control Systems Multiple Choice Questions & Answers (MCQs) focuses on “Bode Plots”. If $K < 1$, then magnitude will be negative. We know the form of the magnitude plot, but need to "lock' it down in the vertical direction. Several examples of the construction of Bode plots are included here; click on the transfer function in the table below to jump to that example. What is a Bode Plot. a constant of 6, a zero at s=-10, and complex conjugate poles at the roots of s 2 +3s+50. 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