Frequency Response Techniques Bode Plots Nyquist Diagram Nichols Charts

The Concept of frequency Response In steady state sinusoidal inputs to a linear system generate sinusoidal responses of the same frequency. even though these responses are of same frequency as the input, they differ in amplitude and phase angle from the input. These differences are the functions of frequency. representation of sinusoids thru complex numbers as phasors.

Mo(w) (w) = Mi(w) M(w) [i(w) + (w)]

M(w) = M0(w) / M(w)(w) = o(w) i(w) We call M(w) as magnitude frequency response & (w) as phase frequency response. the combination of both of them is called as frequency response

Analysis Through Bode Plots

Since G (j*w) is a complex number, we can plot both its magnitude and phase (the Bode plot) or its position in the complex plane (the Nyquist plot).

Magnitude of G(jw) vs w is called as magnitude frequency response. Phase of G(jw) vs w is called as phase frequency response

Plotting the frequency Response lets call system transfer function as G(s) and the frequency response of a system is found as G( jw ) = G(s) | sjw This function G(jw) can be plotted in different ways. 1) as a function of frequency with separate magnitude and phase plots 2) as a polar plot where the phasor length is the magnitude and phasor angle is the phase.

Plotting as separate magnitude & phase plots(1)Called as Bode Diagram

Plotting on the polar plot(2)Called as Nyquist Diagram

Example of plotting the frequency Response(2)Find the analytical expression for the magnitude frequency response and the phase frequency response of the system G(s) = 1 / (s+2)(s+4) . Also plot both the separate magnitude and phase diagrams and the polar plots.Answer: M(w) = 1 / ( 8 w2) 2 + ( 6w) 2 for w 8(w) = - arc tan ( 6w / 8 w2) & for w > 8(w) = - [ + arc tan ( 6w / 8 w2)]

Plotting as separate magnitude & phase plots(2)Bode Diagram

Asymptotic Approximations A line whose distance to a given curve tends to zero. An asymptote may or may not intersect its associated curve.[Ultimately from Greek asumpttos, not intersecting]

Bode plot sketching can be simplified using asymptotes. Straight line approximations can be used to simplify the evaluation of the magnitude and frequency responseX and Y axis are the asymptotes of hyperbola

Bode Plots for First Order Systems G(s) = ( s+a) G(s) = s + a & G(jw) = ( jw + a) = a ( 1 + jw /a) for low frequencies as w a then G(jw) a ( jw/a) = w 90 or just for magnitude G(jw) = 20 log w High frequency approximation is equal to low frequency approximation at the break frequency where w =a the line has a slope of 20dB/dec or 6 dB / octave. the rise in the slope of line begins at w=a. low frequency approximation is called as low frequency asymptote. high frequency approximation is called as high frequency asymptote.

now come to phase response as G(jw) = (a+jw) = a( 1 + jw/a) G(jw) = a tan -1 ( w/a) at w < < a then phase is zero. at w =a the n phase is 45 at w > > a phase is 90 one decade below the break frequency the phase is 0 and one decade above the break frequency the phase is 90. for the table in the coming slides we can enter the accurate values for magnitude and phase as well as the approximations.

Normalization & Scaling we normalize a frequency by factoring out the quantity a and define a new frequency variable s1 i.e to normalize ( s+a) we have a ( s/a + 1) = a ( s1 + 1 ) then the magnitude is divided by the quantity a to yield 0 dB at the break frequency This process is called normalization and scaling. thus the normalized and scaled function is ( s1 + 1) to obtain the original frequency response the magnitude and frequency are multiplied by the quantity a.

More about Asymptotes actual magnitude curve is never greater than 3.01 dB from the asymptotes. Maximum difference in magnitude occurs at break frequency. Max. Phase difference occurs is 5.71 this difference occurs at one decade above and one decade low to the break frequency.

Bode Plots for G(s) = ( s+a)

Asymptotic and actual normalized and scaled Magnitude response Bode Plots for G(s) = ( s+a)

Bode Plots for G(s) = 1 / ( s+a) G(s) = 1 / s + a = 1 / a ( s/a + 1) at low frequencies the asymptote is G(jw) = 20 log 1/a by letting w to 0. at high frequency we need to have w to thus G(jw) = 1 / a ( jw/a ) = 1/w - 90 thus G(jw) = - 20 log w the high frequency approximation equals the low frequency approximation when w =a and decreases for w > a. result is same as the previous one just with the difference that here the slope is negative. Bode log magnitude curve reduces at 20 dB/dec after the break frequency. phase plot is also negative of previous one, starting from zero, and reaching -90 at the rate of -45 deg/dec.

Asymptotic Bode Plots for G(s) = 1 / ( s+a)

Asymptotic Bode Plots for G(s) = s it has only a high frequency asymptote. for s=jw the magnitude is 20 log w Bode magnitude plot is a straight line with +20 dB/dec slope passing thru 0 dB when w =1 phase plot is a constant at + 90 deg.

Asymptotic Bode Plots for G(s) = 1/s it has only a low frequency asymptote. for s=jw the magnitude is -20 log w Bode magnitude plot is a straight line with-20 dB/dec slope passing thru 0 dB when w =1 phase plot is a constant at - 90 deg.

Summary of Asymptotic Bode Plots for all four major functionsFunction NameMagnitude PlotPhase Plots + a At BF gain is zero and after that it increases by the rate of 20 dB/decPhase is 45 at BF and one decade below BF it is zero and increases at the rate of 45 deg/dec and becomes 90 one decade above the BF and continues with this value.1 / s + aAt BF gain is zero and after that it decreases at the rate of - 20 dB/decPhase is -45 at BF and one decade below BF it is zero and decreases at the rate of -45 deg/dec and becomes -90 one decade above the BF and continues with this value.

Summary of Asymptotic Bode Plots for all four major functionsFunction NameMagnitude PlotPhase Plots 1 / s At BF gain is zero and below BF and onwards it increases by the rate of 20 dB/decPhase is fixed at 90 deg.At BF gain is zero and below BF and onwards it decreases by the rate of --20 dB/decPhase is fixed at -90 deg.

Bode Plots for ratio of first order functionsDraw the Bode Plots for the system shown in the figure where G(s) = K ( s + 3) / [ s ( s + 1) ( s + 2) ] Frequency (rad/s)DescriptionPole at 0Pole at -1Pole at -2Zero at -3Total slope (dB/dec)0.1 1 2 3Start pole at 0 start pole at -1 start pole at -2 start pole at -3 -20 -20 -20 -20 0 -20 -20 -20 0 0 -20 -20 0 0 0 20 -20 -40 -60 -40Composite Magnitude Table

Bode Plots for ratio of first order functionsMagnitude Plots

Bode Plots for ratio of first order functionsComposite Phase TableFrequency (rad/s)DescriptionPole at -1Pole at -2Zero at -3

Total slope (deg/dec0.1 0.2 0.3 10 20 30Start pole at -1 start pole at -2 start pole at -3 end pole at -2 end at -2 end at -3 -45 -45 -45 0 -45 -45 -45 0 45 45 0

-45 -90 -45 0 45 0

Bode Plots for ratio of first order functionsPhase Plots

Example Draw the Bode log magnitude and phase plots for the system where G(s) = ( s+20) / ( s+1) ( s+7) ( s+50)Magnitude Plot

Phase Plot

Draw the Bode Plots for the following functions:

G(s) = 1 / s(s+2)(s+4)G(s) = (s+5) / (s+2)(s+4)G(s) = (s+3)(s+5) / (s (s+2)(s+4)