40dB r 9+20dB -20dB 40dB 1og【] Figure 11.3 Bode magnitude functions for(1)K=5, (2)1/s, and (3)s In the last form of the expression, t,=1/@, and tp=1/@ tp is a time constant of the system and s=-0, is the corresponding natural frequency. Because it is understood that Bode diagrams are limited to sinusoidal steady state frequency response analysis, one can work directly from the transfer function H(s) rather than resorting to the formalism of making the substitution s= jo Bode frequency response curves(gain and phase) for H(s) are generated from the individual contributions of the four terms K, st,+ 1, 1/s, and 1/(st, 1).As described in the following paragraph, the frequency response effects of these individual terms are easily drawn. To obtain the overall frequency response curves for the transfer function, the curves for the individual terms are added together. The terms used as the basis for drawing Bode diagrams are found from factoring N(s)and D(s), the and denominator polynomials of the transfer function. The factorization results in four standard forms.These are(1)a constant K, (2)a simple s term corresponding to either a zero( if in the numerator)or a pole(if in the denominator)at the origin;(3)a term such as(st 1)corresponding to a real valued (nonzero) pole or zero; and(4)a quadratic term with a possible standard form of [(s/o)2+(25/@ )s 1] corresponding to a pair of complex conjugate poles or zeros. The Bode magnitude and phase curves for these possibilities are isplayed in Figs. 11.3-11.5. Note that both decibel magnitude and phase are plotted semilogarithmically. The frequency axis is logarithmically scaled so that every tenfold, or decade, change in frequency occurs over an equal distance. The magnitude axis is given in decibels. Customarily, this axis is marked in 20-dB increments Positive decibel magnitudes correspond to amplifications between input and output that are greater than one (output amplitude larger than input). Negative decibel gains correspond to attenuation between input and utput Figure 11.3 shows three separate magnitude functions. Curve I is trivial; the Bode magnitude of a constant is simply the decibel-scaled constant 20 logo K, shown for an arbitrary value of K= 5(20 logo 5= 13.98) Phase is not shown. However, a constant of k>0 has a phase contribution of 0 for all frequencies For K< 0, the contribution would be t180(Recall that-cos 0=cos(0+ 180) Curve 2 shows the magnitude frequency response curve for a pole at the origin(1/s). It is a straight line with a slope of -20 dB/decade. The line passes hrough 0 dB at @=0 rad/s The phase contribution of a simple pole at the origin is a constant-90o, independent of frequency. The effect of a zero at the origin(s) is shown in Curve 3. It is again a straight line that passes through 0 dB at @=0 rad/s; however, the slope is +20 dB/decade. The phase contribution of a simple zero at ent of frequen e 2000 by CRC Press LLC© 2000 by CRC Press LLC In the last form of the expression, tz =1/wz and tp =1/wp. tp is a time constant of the system and s = –wp is the corresponding natural frequency. Because it is understood that Bode diagrams are limited to sinusoidal steady￾state frequency response analysis, one can work directly from the transfer function H(s) rather than resorting to the formalism of making the substitution s = jw. Bode frequency response curves (gain and phase) for H(s) are generated from the individual contributions of the four terms K¢, stz + 1, 1/s, and 1/(stp + 1).As described in the following paragraph, the frequency response effects of these individual terms are easily drawn. To obtain the overall frequency response curves for the transfer function, the curves for the individual terms are added together. The terms used as the basis for drawing Bode diagrams are found from factoring N(s) and D(s), the numerator and denominator polynomials of the transfer function. The factorization results in four standard forms. These are (1) a constant K; (2) a simple s term corresponding to either a zero (if in the numerator) or a pole (if in the denominator) at the origin; (3) a term such as (st + 1) corresponding to a real valued (nonzero) pole or zero; and (4) a quadratic term with a possible standard form of [(s/wn)2 + (2z/wn)s + 1] corresponding to a pair of complex conjugate poles or zeros. The Bode magnitude and phase curves for these possibilities are displayed in Figs. 11.3–11.5. Note that both decibel magnitude and phase are plotted semilogarithmically. The frequency axis is logarithmically scaled so that every tenfold, or decade, change in frequency occurs over an equal distance. The magnitude axis is given in decibels. Customarily, this axis is marked in 20-dB increments. Positive decibel magnitudes correspond to amplifications between input and output that are greater than one (output amplitude larger than input). Negative decibel gains correspond to attenuation between input and output. Figure 11.3 shows three separate magnitude functions. Curve 1 is trivial; the Bode magnitude of a constant K is simply the decibel-scaled constant 20 log10 K, shown for an arbitrary value of K = 5 (20 log10 5 = 13.98). Phase is not shown. However, a constant of K > 0 has a phase contribution of 0° for all frequencies. For K < 0, the contribution would be ±180° (Recall that –cos q = cos (q ± 180°). Curve 2 shows the magnitude frequency response curve for a pole at the origin (1/s). It is a straight line with a slope of –20 dB/decade. The line passes through 0 dB at w = 0 rad/s. The phase contribution of a simple pole at the origin is a constant –90°, independent of frequency. The effect of a zero at the origin (s) is shown in Curve 3. It is again a straight line that passes through 0 dB at w = 0 rad/s; however, the slope is +20 dB/decade. The phase contribution of a simple zero at s = 0 is +90°, independent of frequency. Figure 11.3 Bode magnitude functions for (1) K = 5, (2) 1/s, and (3) s