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## The Bode Diagram - Tutorial by Thomas Bullock

Many users of motion control have heard vendors try to explain certain servo features by using Bode diagrams. A basic understanding of them is necessary in dealing with motion. To begin with, the proper pronunciation is "Bo-dee". just think of Bo Derek (which isn't hard to do). You'll find other similarities between the two (who am I kidding?) like the number 10, as will be seen.

On page 39 of the September/October 1990 issue of Motion Control, Servo Basics were discussed. The block diagram of a basic servo and the closed loop response formula were given as: A Bode diagram, like the one shown below, helps to quantify how well the output, or feedback (F), follows the command (C) by showing the relationship between A and the frequency of excitation. The frequency is normally expressed in radians/sec. ( ) rather than cycles per second (f). Since , it is easy to convert from one to the other. Also, , in seconds, is the time constant of a low pass filter with a bandwidth of f. Once we know the bandwidth ( ) of a servo, we know the time constant of the equivalent filter ( ) and can thus predict its response to a step input. Typically, the Bode diagram is shown on log-log graph paper so that a wide range of frequencies and amplitude levels can be diagrammed. The amplitude is usually expressed in decibels where each 20 decibel increment represents a factor of 10 in amplitude ratio.

The open loop gain A is shown for the usual case of an amplifier/motor combination. You will notice that for every factor of 10 increase in frequency, the amplitude decreases by a factor of 10. A motor is an integrator. If one puts a constant voltage on the input, the motor will run continuously, thereby integrating the position to infinity. If one puts a sine wave alternating signal on the motor input, it will cycle back and forth to the same velocity levels, but the position covered during the excursions wiil vary dramatically with frequency. The higher the frequency, the less time for the excursions to the same velocity levels, and the less distance covered.

Without writing equations, this helps to intuitively explain why A is graphed as shown. One other observation that needs to be made for later use is that the position output lags in phase by 90° from the signal input. Intuitively, this is obvious when one realizes that the motor will continue in the same positional direction until the input changes polarity to start it on its positional excursion in the other direction. The net effect that needs to be understood is that A is really A -90° (it has a gain factor of A and a phase lag of 90°). This Bode plot of A happens to have a gain of 1 (0 DB) at 10 radians/second (approx. 1.6 cycles/second). The gain adjustment would simply shift the A plot up or down.

The closed loop response [F/C = A/(1+A)] can also be plotted on the Bode diagram. When A is very large, and when A is very small, , which makes it easy to plot F/C except in the vicinity of A = 1. At the A = 1 point, the closed loop gain becomes A -90°/(1 + A -90°) (Remember that it is 90° phase shifted). In order to add two quantities that are phase shifted by 90°, it is necessary to view them as a right triangle. In this case, with legs of 1 and a magnitude of  Thus, A/(1+A) becomes at the point where A = 1.

Since 0.707 is -3 DB, the closed loop bandwidth is defined as the point where the magnitude is 3 DB down from where it would be at very low frequencies.

Making these same calculations for other values of A near the A = 1 point (A = 1 /4, 1 /2, 2, 4) allows completion of the closed loop frequency response.

The closed loop response is similar to a simple R-C filter with a time constant of (0.1 seconds in this case). This allows one to predict the servo's response to a step input in position by referring to texts on simple R-C filters.

One can determine the closed loop response to a sine wave input by referring directly to the Bode diagram or by redoing the A/(1 +A) calculation at the frequency of interest. Suppose we want to know how the servo would respond to a 0.1 radians/second sine wave input. You would use A = 1 00 from the Bode plot and calculate: The output sine wave would be slightly smaller than the input with a small phase shift.

The real beauty of the Bode diagram is its use in analyzing compensation techniques and the effects of loads and machine resonances. Loads and resonances limit the gain one can use without incurring instability. Compensation techniques allow one to increase A at low frequencies while maintaining the frequency bandwidth (the frequency where A = 1). These compensation techniques are the things that the vendors normally tout.

The purpose of this month's column has been to provide a basic understanding of Bode diagrams so you can better understand vendor offerings. If you comprehend what has been written here, you will know more about Bode diagrams than I do about women after 31 years of marriage (which means we both still have a lot to learn). Certainly this new found knowledge will not make you a "10" as a servo expert, but there aren't many Bo Dereks around, either.