Chapter 10 Frequency Response Techniques Frequency Response Techniques

Sinusoidal frequency response: a. system; b. transfer function; c. input and output waveforms The steady-state output sinusoid is The system function is given by

Analytical Expression for Frequency response

Frequency response plots for G(s) = 1/(s + 2): separate magnitude and phase Problem: Find the analytical expression for the magnitude and phase frequency response for a system G(s)=1/(s+2). Plot magnitude, phase and polar diagrams. Solution: substituting s=jw, we get The magnitude frequency response is The phase frequency response is The magnitude diagram is And the phase diagram is

Frequency response plots for G(s)= 1/(s + 2): polar plot The polar plot is a plot of For different values of w

Asymptotic Approximations: Bode plots Consider the transfer function The magnitude frequency response is Converting the magnitude response into dB, we obtain We could make an approximation of each term that would consists only of straight lines, then combine these approximations to yield the total response in dB

a. magnitude plot; b. phase plot. Asymptotic Approximations: Bode plots The straight-line approximations are called asymptotes. We have low frequency asymptote and high frequency asymptote a is called the break frequency Bode plots for G(s) = (s + a): At low frequency when approaches zero The magnitude response in dB is 20log M=20log a where and is constant and plotted in Figure (a) from =0.01a to a. At high frequencies where a The magnitude response in dB is

Table 10.1 Asymptotic and actual normalized and scaled frequency response data for (s + a)

Figure 10.7 Asymptotic and actual normalized and scaled magnitude response of (s + a)

Figure 10.8 Asymptotic and actual normalized and scaled phase response of (s + a)

Normalized and scaled Bode plots for a. G(s) = s; b. G(s) = 1/s;

Normalized and scaled Bode plots for c. G(s) = (s+a); d. G(s) = 1/(s+a);

Closed-loop unity feedback system Example Problem : Draw the Bode plot for the system shown in figure where Solution: The Bode plot is the sum of the Bode plot for each first order term. To make it easeier we rewrite G(s) as

Bode log-magnitude plot for Example: a. components; b. composite

Table 10.2 Bode magnitude plot: slop contribution from each pole and zero in Example

Bode phase plot for Example: a. components; b. composite

Table 10.3 Bode phase plot: slop contribution from each pole and zero in Example 10.2

Bode asymptotes for normalized and scaled G(s) = At low frequencies G(s) becomes The magnitude M, in dB is therefore, At high frequencies And The log-magnitude is The log-magnitude equation is a straight line with slope 40dB/decade The low freq. asymptote and the high freq. asymptote are equal when Thus is called the break frequency for second order polynomial.

Bode asymptotes for normalized and scaled G(s) = a. magnitude; b. phase

Table 10.4 Data for normalized and scaled log-magnitude and phase plots for (s 2 + 2 n s + n 2 ). Mag = 20 log(M/ n 2 ) (table continues) Freq. n Mag (dB) = 0.1 Phase (deg) = 0.1 Mag (dB) = 0.2 Phase (deg) = 0.2 Mag (dB) = 0.3 Phase (deg) = 0.3

Freq. n Mag (dB) = 0.5 Phase (deg) = 0.5 Mag (dB) = 0.7 Phase (deg) = 0.7 Mag (dB) = 1 Phase (deg) = 1 Table 10.4 (continued) Data for normalized and scaled log-magnitude and phase plots for (s 2 + 2 n s + n 2 ). Mag = 20 log(M/ n 2 )

Normalized and scaled log-magnitude response for

Scaled phase response for

Bode Plots for G(s) = 1/ Bode plots for can be derived similarly to those for We find the magnitude curve breaks at the natural frequency and decreases at a rate of -40dB/decade. The phase plot is 0 at low frequencies. At 0.1 begins a decrease of -90 o /decade and continues until, where it levels off at -180 o.

Table 10.5 Data for normalized and scaled log-magnitude and phase plots for 1/(s 2 + 2 n s + n 2 ). Mag = 20 log(M/ n 2 ) (table continues) Freq. n Mag (dB) = 0.1 Phase (deg) = 0.1 Mag (dB) = 0.2 Phase (deg) = 0.2 Mag (dB) = 0.3 Phase (deg) = 0.3

Freq. n Mag (dB) = 0.5 Phase (deg) = 0.5 Mag (dB) = 0.7 Phase (deg) = 0.7 Mag (dB) = 1 Phase (deg) = 1 Table 10.5 (continued) Data for normalized and scaled log-magnitude and phase plots for 1/(s 2 + 2 n s + n 2 ). Mag = 20 log(M/ n 2 )

Normalized and scaled log magnitude response for

Scaled phase response for

Closed-loop unity feedback system Example Problem : Draw the Bode plot for the system shown in figure where Solution: The Bode plot is the sum of the Bode plot for each first order term, and the 2 nd order term. Convert G(s) to show the normalized components that have unity low-frequency gain. The 2 nd order term is normalized by factoring out forming Thus,

Bode magnitude plot for G(s) =(s + 3)/[(s + 2)(s 2 + 2s + 25)]: a. components; b. composite The low frequency value for G(s), found by letting s = 0, is 3/50 or -24.44dB The Bode magnitude plot starts out at this value and continues until the first break freq at 2 rad/s.

Table 10.6 Magnitude diagram slopes for Example

Bode phase plot for G(s) = (s + 3)/[(s +2)(s 2 + 2s + 25)]: a. components; b. composite

Table 10.7 Phase diagram slopes for Example

Introduction to the Nyquist criterion The Nyquist criterion relates the stability of a closed-loop system to the open-loop frequency response and open-loop pole location. We conclude that The poles of 1+G(s)H(s) are the same as the poles of G(s)H(s), the open-loop system The zeroes of 1+G(s)H(s) are the same as the poles of T(s), the closed-loop system

Mapping Contours concept If we take a complex number on the s-plane and substitute it into a function, F(s), another complex number results. This process is called Mapping. Consider the collection of points, called a contour (shown in Figure as Contour A). Contour A can be mapped through F(s) Into Contour B by substituting each point in contour A into Function F(s) and plotting the resulting complex numbers.

Examples of Contour Mapping If contour A maps clockwise then contour B maps clockwise if: F(s) has just Zeros or Poles that are not encircled by the contour The contour B maps counterclockwise if: F(s) has just Poles that are encircled by contour A

Examples of Contour Mapping If the Pole or Zero of F(s) is enclosed by contour A, then the mapping encircles the origin

Vector Representation of Mapping

Nyquist Contour is a contour which contains the imaginary axis and encloses the right half- place. The Nyquist contour is clockwise. A Clockwise Curve Starts at the origin. Travels along imaginary axis till r = . At r = , loops around clockwise. Returns to the origin along imaginary axis. The Nyquist Contour

a. Turbine and generator; b. block diagram of speed control system for Example 10.4 Sketching a Nyquist diagram

Vector evaluation of the Nyquist diagram for Example 10.4 a. vectors on contour at low frequency; b. vectors on contour around infinity; c. Nyquist diagram The Nyquist Diagram is plotted by substituting the points of the contour shown in Figure (a) into G(s) = 500/[(s+1)(s+3)(s+10)] Its performing complex arithmetic using vectors of G(s) drawn to points of the contour as shown in (a) and (b) The resultant vector, R, found at any point is the product of zero vectors divided by product of pole vectors as in (c)

Vector evaluation of the Nyquist diagram for Example 10.4 a. vectors on contour at low frequency; b. vectors on contour around infinity; c. Nyquist diagram Evaluating G(s) at specified points we get So from A to C the magnitude is finite goes from 50/3 to 0 and the angle goes from 0 to -270 o From C to D the magnitude is 0 while there is an angle change of 540 o so goes from -270 o to 270 o

Vector evaluation of the Nyquist diagram for Example 10.4 The mapping can be explained analytically by Substituting s= j and evaluating G(j) at specified points Multiplying numerator and denominator by the complex conjugate of denominator, we obtain At 0 freq G(j)=500/30= 50/3. Thus Nyquist diagram starts at 50/3 at an angle 0. As increases, the real part remains positive while imaginary part remains negative. At = 30/14, the real part becomes negative. At =43 the Nyquist diagram crosses the negative real axis, the value at the crossing is -0.874. Continuing toward = , the real part is negative and imaginary is positive. At = , G(j)500j/ 3 or approximately 0 at 90 o

Detouring around open-loop poles: a. poles on contour; b. detour right; c. detour left

a. Contour for Example 10.5; b. Nyquist diagram for Example 10.5 Nyquist Diagram for open-loop function with poles on contour Problem: Sketch the Nyquist diagram of unity feedback system where G(s)= (s+2)/ S 2 Solution: The system two poles at origin are on the contour and must be by-passed as in (a). Mapping DE is mirror of AB

Applying the Nyquist Criterion to determine stability a. contour does not enclose closed-loop poles; b. contour does enclose closed-loop poles If a contour A, that encircles the entire right half-plane is mapped through G(s)H(s), then the number of closed loop poles, Z, in the right half- plane equals the number of open-loop poles, P, that are in the right half-plane minus the number of counterclockwise revolutions, N, around -1 of the mapping; that is Z=P-N.

Demonstrating Nyquist stability: a. system; b. contour; c. Nyquist diagram We could set K=1 and position the critical point at -1/K rather than -1. Thus the critical point appears to move closer to the origin as K increases. For this example P=2, and the critical point must be encircled by the Nyquist diagram to yield N=2 and a stable system. The system is marginally stable if the Nyquist diagram intersect the real axis at -1 ( frequency at this point is the frequency at which the root locus crosses the jw axis.)

a. Contour for Example 10.6; b. Nyquist diagram Range of gain for stability via Nyquist criterion Problem: for the unity feedback system with G(s) = K/[s(s+3)(s+5)], find the range for stability, instability and the value of gain for marginal stability. Solution: first set K=1 and sketch Nyquist diagram for the system. Using the contour shown in (a), for all points on the imaginary axis, Next find the point where the diagram intersects the negative real axis by setting the imaginary part equal to zero. We find. and real part = -0.0083. Finally From the contour, P=0; for stability N must equal zero. So the system is stable if the critical point lies outside the contour. K can be increased by 1/0.0083 =120.5 before the diagram encircles -1. Hence for stability K

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Chapter 10 Frequency Response Techniques Frequency Response Techniques