Report

Professor Walter W. Olson Department of Mechanical, Industrial and Manufacturing Engineering University of Toledo Bode Phase Plots Outline of Today’s Lecture Review Frequency Response Reading the Bode Plot Computing Logarithms of |G(s)| Bode Magnitude Plot Construction Phase Phase Computations Full Bode Plot System Identification Using Bode Plots for System Identification Frequency Response General form of linear time invariant (LTI) system was previously expressed as (n) s n 1 n 2 m m 1 m 2 y a1 y a2 y ... an 2 y an 1 y an y b0 u b1 u b2 u ... bn 2u bn 1u bnu n a1s n 1 a2 s n 2 ... an 2 s 2 an 1 2 an y0e st b0 s m b1s m1 b2 s m2 ... bn 2 s 2 bn 1s bn e st b s bs s a s m y (t ) y (0)e st 0 m 1 1 n 1 n 1 b2 s m2 ... bn 2 s 2 bn 1s bn a2 s n 2 ... an 2 s an 1 2 an 2 e st The transfer function form is then b0 s m b1s m1 b2 s m2 ... bn 2 s 2 bn 1s bn b( s ) y ( t ) G ( s )u ( t ) G ( s ) n s a1s n 1 a2 s n 2 ... an 2 s 2 an 1 2 an a(s) We now want to examine the case where the input is sinusoidal. The response of the system is termed its frequency response. s i u(t ) eit cos i i sin i y (t ) G (i )eit Me i t where M G (i ) is the magnitude or gain of G i and Im G i tan is the phase angle or argument of G i Re G i 1 Reading the Bode Plot Amplifies Attenuates Input Response The magnitude is in decibels decade also, cycle Note: The scale for is logarithmic What 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 0-7381-2601-2; 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 attenuation 20 Log10(100)= 40 decibels … implies that the signal has been amplified 100 times from its original value 20 Log10(0.01)= -40 decibels … implies that the signal has been attenuated to 1/100 of its original value Computing Logarithms of G(s) 20Log10 K constant 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 down 20 pLog10 i 20 pLog10 When this is plotted on a semilog graph ( the abscissa) this is 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) Computing Logarithms of G(s) 20 Log10 i a 20 Log10 2 a 2 a 20 Log10 2 a 2 20 Log10a constant a 20 Log10 2 a 2 20 Log10 a is called the break frequency for this factor For 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) Computing Logarithms of G(s) 20 Log10 i 2n i n 20 Log10 2 n n 2 n 2 2 2 4 2n 2 2 20 Log10 n 2 2 4 2n 2 2 20Log10n 2 40Log10n constant 20 Log10 2 n 2 2 2 4 2n 2 2 20 Log10 2 40 Log10 wn is called the break frequency for this factor For 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) Bode 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. Example Plot the Bode magnitude plot of ( s 0.2)( s 2 2s 25) G s 10 s( s 3)( s 2 4s 16) Break Frequency Factor Effect Cum value Cum Slope dB/dec 0.01 K=10 20Log10(10)=20 20 … 0.01 s Line -20db/dec Thru (1,0) 20-slope for two decades (40) =60 -20 0.2 s+0.2 +20Log10(.2)= -13.98 60+6.02=46.02 0 3 s+3 -20Log10(3)= -9.54 46.02-9.54= 36.48 -20 4 s2+4s+16 -40Log10(4)= - 24.08 36.48- 24.08= 12.4 -60 5 s2+2s+25 +40Log10(5)= 27.96 32.4+27.96= 40.36 -20 Example Constructed Bode Actual Bode Phase s i u(t ) eit cos i i sin i y (t ) G (i )eit Me i t where M G (i ) is the magnitude or gain of G i and Im G i tan is the phase angle or argument of G i Re G i 1 For a sinusoidal input, phase represents the lag of the system or, alternatively, the processing time of the system to produce an output from the input Phase is measured as an angle A cycle of the input is consider to take 2p radians or 360 degrees Phase is the angular distance it takes for the output to represent the input Thus it is normal that as the frequency increases that the phase also increase In the case where the phase exceeds 180 degrees, the output appears to “lead” the input. This is particularly evident in the range of 270 to 360 degrees. Phase Im G i G ( s) tan Re G i 1 As with magnitude there are 4 factors to consider which can be added together for the total phase angle. We will consider, in turn, K s n ( s a ) ( s 2 2n s n 2 ) The sign will be positive if the factor is in the numerator and negative if the factor is in the denominator Phase Computations K is the angle of a real number; therefore it is always 0 (or 180 if the number were negative) sn (in ) n If n is an even number, then i n will either be +1 or -1, both real numbers n an odd number, i n will be -1: sn 180 the angle of complete turns (n 4 ) 2 n for an even number, i n will be +1: sn 0 the angle of complete turns (n 4 ) 2 If n is an odd number, then i n will either be i or - i, both pure imaginary numbers for n 1 an odd number, i n will be i: s n 90 the angle of complete turns (n 4 ) 2 n 1 for an even number, i n will be - i: s n 270 the angle of complete turns (n 4 ) 2 1 Examples: s 90 s 1 90 s 2 -180 s s 3 = -270 s 5 90 360 450 for Phase Computations ( s a ) (i a ) arctan a This now involves the variable, , which is plotted on the semilog scale Where a, arctan 0 a Where a, arctan 900 a Where a 10a, arctan a line from 0 to 90 10 a 2 ( s 2 2n s n 2 ) (n 2 2 i 2n ) arctan 2 n 2 n This now also involves the variable, Where Where Where n , arctan 0 a n , arctan 1800 a a 10a, arctan a line from 0 to 180 with a correction for 10 a Example Plot the Bode Phase Plot for G( s ) 14 s2 s2 s 9 s 2 180 determines the starting phase angle s2 s 9 , n 9 3 2n 1 0.1667 The slope will be -90 per decade as the term is in the denominator for a total transition of -180 or starting at -180 and ending at -360 3 Starting from , 180 , draw a line to 30, 360 10 Then, because the damping ratio is small, sketch in a correction using the chart as a guideline Example Plot the Bode Phase Plot for Again a table is useful: n ( s 0.2)( s 2 2s 25) G s 10 s( s 3)( s 2 4s 16) Frequency Factor Effect Cum Value Cum Slope 0 s-1 constant -90 0.02 s+0.2 break up at 45/dec 45/dec 0.3 s+3 Break down at -45/dec 0 0.4 ( s 2 4s 16) 4 0.5 Break down at -90/dec 0.5 ( s 2 2s 25) 5 0.2 -90 - -90/dec Break up at 90/dec 0 2 s+0.2 take out 45/dec -45/dec 30 s+3 take out -45/dec 0 40 ( s2 4s 16) take out -90/dec 90/dec take out 90/dec 0 50 ( s 2 2s 25) total exponent 3-4=-1 >> -90 -90 Example Asymptotic Bode Phase Actual Bode Phase Full Bode Plot Constructed Bode Actual Bode Asymptotic Bode Phase Actual Bode Phase Matlab Command bode(sys) System Identification It is not unusual for a field engineer to be shown a piece of equipment and then asked if he can put a control system on it or replace the control system for which there are no parts. The task of determining how an unknown structure responds is called “System Identification”. To identify a system, there are many tools are your disposal First and foremost, what should the system structure look like? Motors are often first order transfer functions (G( s) K ) which you then sa attempt to identify the constants Perform step tests and see what the response looks like Perform tests with sinusoidal outputs and use the Bode plot to identify the system Apply statistical/time series methods such as ARMAX and RELS Using Bode Plots for System Identification The overall order of the system will be the high frequency phase divided by 90 degrees The exponent of the “s” term will be the slope on the magnitude plot at the lowest frequency divided by 20 Alternatively, the exponent of “s” is the lowest frequency phase divided by 90 degrees. The system gain constant (Kt) in dB will be the height value at the extension of the “s” term line on the magnitude plot to where it crosses1 rps Starting from the left (the lowest frequency) on the magnitude plot, determine the structural components using the change in slopes in increments of 20 degrees either up or down Then by using the intersection of the lines at those places match to the test curve, determine the break frequencies Write the transfer function in the form s2 s s s 2 z1s s 2 2 s 1 1 ... 1 1 2 1 ... b b b 2 2 nz1 nz 2 m nz1 nz 2 G( s) Kt 1 2 2 2 s 2 s s 2 s s s s s p 1 1 ... 1 p1 1 p 2 1 ... 2 2 np1 np 2 np1 np 2 a1 a2 aq Example -40 dB/dec Kt=35 dB -60 dB/dec 20 dB/dec 40 dB/dec 60 dB/dec 80 dB/dec -40 dB/dec -80 dB/dec -40 dB/dec 0.52 2.2 4 20 exp of s = -2 overall order =3 Example -40 dB/dec -40 dB/dec 35 Log10 K t 20 K t 10 G ( s )test 35 20 Kt=35 dB -60 dB/dec 20 Log10 Kt 35 -80 dB/dec 56.23 2 2 z s s s 1 1 2 20 2.2 20 56.23 2 2 s s s s2 1 2 p 1 4 0.52 4 -40 dB/dec 0.52 Total order of G( s) test (1 2) (2 1 2) 2 2.2 4 20 Doesn't match phase! There is a mistake.! exp of s = -2 overall order =-3 Example -40 dB/dec Kt=35 dB -60 dB/dec 20 Log10 K t 35 Log10 K t K t 10 35 20 -40 dB/dec 35 20 -80 dB/dec -60 dB/dec -40 dB/dec 56.23 s s 1 1 2.2 20 G ( s )test 56.23 2 2 s s s 2 s 1 2 p 1 4 0.52 4 0.52 Total order of G ( s )test (1 1) (2 1 2) 3 is fairly large 4 20 0.6 exp of s = s -2 2.2 s 20 s 2 s 0.52 s 2 4.8s 16 G ( s )test 10.63 2.2 overall order =-3 Example s 2.2 s 20 s s 0.52 s 2 4.8s 16 s 3 s 9 G( s )actual 25 2 s s 0.4 s 2 4s 16 G( s )test 10.63 2 Summary Phase For a sinusoidal input, phase represents the lag of the system or, alternatively, the processing time of the system to produce an output from the input Phase Computations K Full Bode Plot s n ( s a ) System Identification ( s 2 2n s n 2 ) The task of determining how an unknown structure responds is called “System Identification”. Using Bode Plots for System Identification Next Class: Laplace Transforms