Presentation on theme: "Bode Magnitude Plots Constructed Bode Actual Bode"— Presentation transcript:
1Bode Magnitude Plots Constructed Bode Actual Bode Professor Walter W. OlsonDepartment of Mechanical, Industrial and Manufacturing EngineeringUniversity of ToledoBode Magnitude Plots
2Outline of Today’s Lecture ReviewPoles and ZerosPlotting Functions with Complex NumbersRoot LocusPlotting the Transfer FunctionEffects of Pole PlacementRoot Locus Factor ResponsesFrequency ResponseReading the Bode PlotComputing Logarithms of |G(s)|Bode Magnitude Plot Construction
3Root LocusThe root locus plot for a system is based on solving the system characteristic equationThe transfer function of a linear, time invariant, system can be factored as a fraction of two polynomialsWhen the system is placed in a negative feedback loop the transfer function of the closed loop system is of the formThe characteristic equation isThe root locus is a plot of this solution for positive real values of KBecause the solutions are the system modes, this is a powerful design tool
4Root LocusThe root locus plot for a system is based on solving the system characteristic equationThe transfer function of a linear, time invariant, system can be factored as a fraction of two polynomialsWhen the system is placed in a negative feedback loop the transfer function of the closed loop system is of the formThe characteristic equation isThe root locus is a plot of this solution for positive real values of KBecause the solutions are the system modes, this is a powerful design tool
5The effect of placement on the root locus Imaginary axisReal Axisjwsjwdwns = -zwnsin-1(z)The magnitude of the vector topole location is the natural frequencyof the response, wnThe vertical component (the imaginarypart) is the damped frequency, wdThe angle away from the vertical is theinverse sine of the damping ratio, z
6The effect of placement on the root locus Imaginary axisReal Axisjwsjwdwns = -zwnsin-1(z)The magnitude of the vector topole location is the natural frequencyof the response, wnThe vertical component (the imaginarypart) is the damped frequency, wdThe angle away from the vertical is theinverse sine of the damping ratio, z
7Frequency ResponseGeneral form of linear time invariant (LTI) system was previously expressed asWe now want to examine the case where the input is sinusoidal. Theresponse of the system is termed its frequency response.
8Frequency ResponseThe response to the input is amplified by M and time shifted by the phase angle.To represent this response we need two curves:one for the magnitude at any frequency andone for phase shiftThese curves when plotted are called the Bode Plot of the system
9Reading the Bode Plot Amplifies Attenuates Input Response Note: The scale for w is logarithmicThe magnitude is in decibelsAmplifiesAttenuatesdecadealso, cycleInputResponse
10What is a decibel?The decibel (dB) is a logarithmic unit that indicates the ratio of a physical quantity relative to a specified or implied reference level. A ratio in decibels is ten times the logarithm to base 10 of the ratio of two power quantities.(IEEE Standard 100 Dictionary of IEEE Standards Terms, Seventh Edition, The Institute of Electrical and Electronics Engineering, New York, 2000; ISBN ; page 288)Because decibels is traditionally used measure of power, the decibel value of a magnitude, M, is expressed as 20 Log10(M)20 Log10(1)=0 … implies there is neither amplification or attenuation20 Log10(100)= 40 decibels … implies that the signal has been amplified 100 times from its original value20 Log10(0.01)= -40 decibels … implies that the signal has been attenuated to 1/100 of its original value
11NoteThe book does not plot the Magnitude of the Bode Plot in decibels.Therefore, you will get different results than the book where decibels are required.Matlab uses decibels.
12Sketching Semilog Paper First, determine how many cycles you will need:Ideally, you will need at least one cycle below the smallest zero or pole andAt least one cycle above the largest pole or zeroExampleKnowing how many cycles needed, divide the sketch area into these regions
13Sketching Semilog Paper 0.11.010100Frequency w rpsFor this example assume
14Sketching Semilog Paper For each cycle,Estimate the 1/3 point and label this 2 x the starting point for the cycleEstimate the midway point and label this 3 * the starting point for the cycleAt the 2/3 point, label this 5 x * the starting point for the cycleHalf way between 3 and 5 place a tic representing 4Halfway between 5 and 10 place a tic and label this 7Halfway between 5 and 7, place a tic for 6Divide the space between 7 and 10 into thirds and label these 8 and 9These are not exact but useful for sketching purposes0.11.010100Frequency w rps0.33302518.104.22.16805070
15Computing Logarithms of |G(s)| Therefore, in plotting the magnitude portion of the Bode plot, we cancompute each term separately and then add them up for the result
17Computing Logarithms of G(s) Since this does not vary with the frequency it a constant that will be added to all of the other factors when combined and has the effect of moving the complete plot up or downWhen this is plotted on a semilog graph (w the abscissa) thisis a straight line with a slope of 20p (p is negative if the sp term is in the denominator of G(s)) … without out any other terms it would pass through the point (w,MdB) = (1,0)p is often called the “type” of the system
18Static Error Constants If the system is of type 0 at low frequencies will be level.A type 0 system, (that is, a system without a pole at the origin,) will have a static position error, Kp, equal toIf the system is of type 1 (a single pole at the origin) it will have a slope of -20 dB/dec at low frequenciesA type 1 system will have a static velocity error, Kv, equal to the value of the -20 dB/dec line where it crosses 1 radian per secondIf the system is of type 2 ( a double pole at the origin) it will have a slope of -40 dB/dec at low frequenciesA type 2 system has a static acceleration error,Ka, equal to the value of the -40 dB/dec line where it crosses 1 radian per second
19Computing Logarithms of G(s) a is called the break frequency for this factorFor frequencies of less than a rad/sec, this is plotted as a horizontal line at the level of 20Log10 a,For frequencies greater than a rad/sec, this is plotted as a line with a slope of ± 20 dB/decade, the sign determined by position in G(s)
20+ Example Assume we have the transfer function To compute the magnitude part of the Bode plot+
21Example Bode plot of Constructed Plot Actual Bode Magnitude Plot Our plotting technique produces an “asymptotic Bode Plot”
22Computing Logarithms of G(s) wn is called the break frequency for this factorFor frequencies of less than wn rad/sec, this is plotted as a horizontal line at the level of 40Log10 wn,For frequencies greater than wn rad/sec, this is plotted as a line with a slope of ± 40 dB/decade, the sign determined by position in G(s)
23Example + Construct a Bode magnitude plot of Note: there are two lines here!
25CorrectionsYou seen on the asymptotic Bode magnitude plots, there were deviations at the break frequencies.Note that these could be either positive of negative corrections dependingon whether or not the term is in the numerator or denominatorFor (s+a) type termsFor quadratic termspertains to a phase correction which willbe discussed next class
26Bode Plot Construction When constructing Bode plots, there is no need to draw the curves for each factor: this was done to show you where the parts came from.The best way to construct a Bode plot is to first make a table of the critical frequencies and record that action to be taken at that frequency.You want to start at least one decade below the smallest break frequency and end at least one decade above the last break frequency. This will determine how semilog cycles you will need for the graph paper.This will be shown by the following example.
2720-slope for two decades (40) =60 ExamplePlot the Bode magnitude plot ofBreakFrequencyFactorEffectGain CumvalueCumSlope dB/dec0.01K=1020Log10(10)=2020…sLine -20db/decThru (1,0)20-slope for two decades (40) =60-200.2s+0.2+20Log10(.2)=-13.98=46.023s+3-20Log10(3)=-9.54=36.484s2+4s+16-40Log10(4)==12.4-605s2+2s+25+40Log10(5)=27.96=40.36
29Note: This form has certain advantages: 1) the time constants of the 1st order terms can be directly read2) When constructing Bode plots the part of the curves up to thebreak frequencies are 0 (20Log101=0). The level parts have beentaken up in the constant gain, Kt
30Summary Frequency Response Reading the Bode Plot Computing Logarithms of |G(s)|Bode Magnitude Plot ConstructionNext: Bode Phase Plots