1 Poles and zeros Bode plots by example Andrea Pacelli Department of Electrical and Computer Engineering SUNY at Stony Brook pacelli@ece.sunysb.edu First edition, February 21 Copyright c 21 Andrea Pacelli All Rights Reserved A transfer function can be usually written in the form! "!# where the quantities $ are called zeros and the!#$ are the poles of the transfer function. % is called the low-frequency or dc gain. For simplicity, we will consider only real poles and zeros. Poles of stable systems are always negative. Only unstable systems (e.g., oscillators) have positive poles (or more precisely, complex poles with a positive real part). Zeros can be positive or negative. Zeros in passive structures (e.g., connections of resistors and capacitors) are always negative, however active circuits (e.g., with transistors) can have positive zeros. The Bode plot can be quickly found if one remembers the following golden rules: 1. A negative pole contributes a slope of '& ( db/decade to the magnitude, and a phase of %)*( +. 2. A negative )*( zero contributes a slope of +, of,. 3. A positive zero contributes a slope of, of %)*( +. & ( &( db/decade to the magnitude, and a phase db/decade to the magnitude, and a phase We will illustrate the above rules by a number of examples. In all cases we express frequencies in radians per second, to get rid of the & - and thus save a lot of printer ink. 1
2 Examples 2.1 A single negative pole./ 4(5 The pole at 3 6( 5 1-1 -2-3 -4-5 -1-2 -3-4 -5-6 -7-6 -8-7 1 1 1 2 1 3 1 4 1 5 1 6-9 causes the magnitude plot to point downward at '& ( db/decade. The phase starts out at zero, then shifts to %)*( + at the pole frequency. Note how the transition from ( + to )( + takes about three decades of frequency. 2.2 A single negative zero./ 6 4( 5 7( 5 7 6 5 4 3 2 1 1-1 1 1 1 2 1 3 1 4 1 5 1 6 The zero at 3 causes the magnitude plot to climb upward at, db/decade. The )*(*+ phase starts out at zero, then shifts to, at the zero frequency. The transfer function tends to infinity at high frequency, which is very unusual. In practice, there will always be some pole somewhere up there in the high frequency range to catch the magnitude and bring it down to reasonable values (as in example 2.5). 9 8 7 6 5 4 3 2 &( 2
2.3 One negative pole and one negative zero./ 4(8 4(5 The pole at 3 zero at 3 6( 5 9( 8 1-1 -2-3 -4-5 -1-2 -3-4 -5-6 -7-6 -8-7 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8-9 causes the magnitude plot to point downward at '& ( db/decade. The catches the curve and restores the magnitude to zero slope. Between db. The the pole and the zero the curve has suffered a total attenuation of ( 8 4( 5 6:( phase starts out at (+, shifts to %)*( + at the pole frequency, then goes back to (*+ at the zero frequency. 2.4 Two negative poles./ 6( 8 4( 5 ; 14 12 1 8 6 4 2-2 -4-6 -8-1 -12-14 -16-2 1 2 1 3 1 4 1 5 1 6 1 7 1 8-18 The transfer function starts out at dc with a gain of (<8= >&( db. Each subsequent pole bends the slope of the magnitude down by &( db/decade, and shifts the phase by %)*( +. Note that the final slope of the curve is %?*( db/decade, and the final phase is @A*( +. 3
2.5 One negative zero and one negative pole./ 4( 5 4( 8 7 6 5 4 3 2 1 1-1 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 9 8 7 6 5 4 3 2 Like in example 2.3, but with the pole and the zero interchanged. The positive slope imparted by the zero is canceled by the later pole. Same for the phase. Between the zero and the pole, the transfer function climbs by a factor of (<8B( 5 :*( db. 2.6 Two negative poles, one negative zero./ 6( 8 4(*C 4( 5 ; 14 12 1 8 6 4 2-2 -4-6 -8-1 -12-14 As in example 2.4, but at a frequency 3-2 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9-16 D( C a zero corrects the effect of the second pole, so that the final slope of the curve is &( db/decade and the final phase is %)*(E+. Note, however, that with respect to the case of a single pole, there is an additional attenuation of ( C 4(8% 6&( db between the second pole and the zero (note the zig-zag of the curve between ( 8 and (*C, with a short section at %?*( db/decade). Also note that, due to the close @A*( + proximity of the second pole and the zero, the phase does not even get close to, in fact it does not exceed @F(+. 4
, 2.7 Two negative poles, one positive zero./ 6( 8 4( C 4(5" ; 14 12 1 8 6 4 2 2 15 1 5-5 -1-15 -2 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9-2 As in example 2.6, but the positive zero shifts the phase even further, all the way to &*G( +H )*( +. 2.8 One negative pole, one positive zero./ 4( 8 4(5 1-1 -2-3 -4-5 -2-4 -6-8 -1-12 -14-6 -16-7 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8-18 Another example of positive zero. Positive zeros are typically associated with changes in the sign of the transfer function. Note how the gain starts out as real and positive, and ends up as real and negative ( @A*( + phase). Real examples are typically of the opposite kind, with the transfer function starting out as negative (as in an inverting amplifier) and ending up as positive, due to an input-output capacitive coupling. 5
3, 2.9 One zero in the origin, two negative poles 4( 5 8 7 6 5 4 1 3-2 2-4 -6 1-8 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8-1 The zero in the origin makes the transfer function vanish at low frequencies. One can imagine the zero being located&( somewhere off the left edge of the plot. The magnitude starts out by climbing at, db/decade, then flattens out at the first pole frequency Ï( 5 J(*8 &(. At the second pole at 3, the magnitude dives again and decreases at db/decade. This is a typical bandpass function. The important feature of the transfer function is the plateau between the two poles, where the phase is close to zero and the input is simply transmitted to the output. The value of the transfer function at the plateau can be found as follows. On the plateau, the frequency is higher than the first pole, but 4( 5LK 4(8NM lower than the second pole: i.e., 3, 3. Therefore,./,O 4( 5,O1*3 B( 8 P 1*3 8 6 4 2 1*3 B( 5 ; D( 5 Q:*( The above can be more directly seen if one writes the transfer function as./ 6( 5 ( 5 Note that even though the two poles are spaced by three decades of frequency, the phase is not very flat on the plateau. This could cause serious problems if phase accuracy is desired, e.g., in a digital communication system. dbr 6
2.1 One pole in the origin, one negative zero, one negative pole./ S( 8 4( T 1-1 8 6 4-2 -3-4 -5-6 -7 2-8 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8-9 Like the zero in the previous example, the pole in the origin can be seen as sitting somewhere off the left edge of the plot. The magnitude starts with a slope of &( db/decade and the phase starts at %)*(+. At the zero frequency of ( T, there is a compensation, the slope becomes flat and the phase gets closer to (*+. However, a second pole restores both the slope and the phase to their initial values, leaving only a small plateau of the magnitude in the frequency range (T to (8. The magnitude at the plateau is (E84( THÜ?E( db. 7