Circuits II EECS 3220 Lecture notes on making Bode plots Definitions Network Transfer Function: The function H s Xout s X in s where X out represents the voltage or current response of the network to X in, which is a voltage or current stimulus to the network. The frequency variable s is complex frequency (in radians/s) if Laplace analysis is used, or it is frequency j (in radians/s) if phasor analysis is used. This transfer function is expressed as the quotient of numerator and denominator polynomials in s. Frequency Response Plot: Graphs of 20 log 10 H j and H j versus frequency. These are typically placed on linear scales when the range of frequency is narrow, or semi-log scales when the range of frequency is large. The frequency response plot is based on a steady-state sinusoid and may be measured or calculated data. Bode Plot: An approximation to the exact frequency response plot made using only semi-log graph paper and a straight edge. The magnitude portion of a Bode plot consists only of straight lines having slopes of k 20 db/decade, where k is an integer which may be negative, zero, or positive. The angle portion of the Bode plot consists only of straight lines having slopes of k 45 degrees/decade, where again k is an integer. Decade: A ten-to-one range of frequency. On a log scale, each 10X change in frequency requires the same distance on the scale. Octave: A two-to-one range of frequency. (In music, eight notes are spaced out over an octave. If the half-steps are also included, there are twelve steps in an octave, evenly spaced on a logarithmic scale in frequency.) Within a small round-off error, the slope 20 db/decade is equal to 6 db/octave. Zero: A root of the numerator s-polynomial of the transfer function. It is typical to find zeros with either a positive or a negative real part. Pole: A root of the denominator s-polynomial of the transfer function. It is typical to find poles with a negative real part. A pole having a positive real part indicates an unstable system, which is unlikely to be practical. Singularities: A collective term for the poles and zeros of a transfer function. 1 of 6
Network Analysis A linear circuit having L, R, C, coupled inductors and linearly-dependent sources may be analyzed for its transfer function by replacing each element with its frequency-domain equivalent impedance (R, sl, 1/sC, etc.). The outcome invariably can be expressed in the form: H s Xout s X in s K hf s m as m 1 z s n bs n 1 p wherem n (1) By factoring (1) to determine its poles and zeros, (1) can be rewritten in the following more-insightful form: H s Xout s X in s K hf (s z 1 )(s z 2 ) (s z m ) (s p 1 )(s p 2 ) (s p n ) (2) When this form is obtained, a great deal can be known about the network behavior and its frequency response function by merely stating the list of its singularities (poles and zeros). Equation (2) assumes that all of the singularities are purely real. Many circuits will produce a pair of poles or zeros which are complex conjugates. In these cases, it is more helpful to just keep the conjugate pair together as the second-order term: s 2 bs c, where b 2 < 4c. This discussion will consider only real-valued poles and zeros. A quick check on your network function (1) or (2) can be made by looking back at the circuit itself. The constant "K" in these expressions is the high-frequency limit of the transfer function, which can often be verified by directly examining the circuit itself. (At high frequency, inductors become open circuits, and capacitors short circuits.) Be aware that these functions can be reorganized to display their low-frequency limits instead, such as the following: H s Xout s 1 z s X in s K 1 1 z s 2 (1 s lf zm ) 1 s p 1 1 s p 2 1 s pn (3) Another quick check on validity of your transfer function is to count the number of independent energy-storing elements (capacitors and inductors) in your circuit. This number is equal to n, the number of poles. Note that two capacitors in parallel are not independent, and only add one pole. There are also obscure ways for non-independance; for example, three capacitors in a loop will add two to the number of circuit poles, not three. 2 of 6
Bode Plots To make a quick and reasonably accurate frequency response plot from a transfer function, gather the following information: 1) a list of the poles, 2) a list of the zeros, and 3) the magnitude and phase values at one or more convenient frequencies. For example, if your transfer function is in the form of (2), its high-frequency (hf) magnitude is the magnitude of K hf, and its hf phase is either 0 or 180 degrees, depending on the sign of K hf. If your transfer function happens to be in the form of (3), then the lf magnitude and phase are readily apparent. In the case where you don't have either (2) or (3), you can estimate the magnitude and phase at some frequency in the middle range. This can best be done for a frequency which is at least one decade away from any other singularity frequency. If all else fails, an exact calculation could be done for any one convenient frequency value. This initial process places one starting point on each of your magnitude and phase graphs. The magnitude graph should be done first. In many cases, this is all that is needed. To get started, examine your list of singularities. Example 1: List of Singularities (in rad/s) number 1 2 3 4 Pole (rad/s) -1-10 -10-1,000 Zero (rad/s) 0-100 1,000 Note that there are 4 poles, all with negative real parts because the system is stable, and 3 zeros, some negative and some positive. The list of zeros has been padded by adding a fourth zero at "infinity." This is the same as multiplying the term 1 s into the numerator of the transfer function, which does not change the transfer function at all. However, the concept of a "zero at infinity" will be useful in making the Bode plot. Example 1 was taken from the following transfer function, which is not presented in an orderly form: H s 100 s s 1 1 1 s 10 2 1 s 100 1 1 s 1000 s 1000 (4) Don't jump to the conclusion that 100 is either the lf or the hf value of H (it isn't). However, you can verify the pole/zero list in the table. The magnitude plot will be 20 log 10 H s s j, which implies adding terms of the form 20 log 10 zero j and subtracting terms of the form 20 log 10 pole j. Therefore, as the frequency on the Bode plot passes through the value corresponding to the magnitude of a zero, there is an upturn in slope of +20 db/dec. Likewise, as the frequency passes through the value corresponding to the magnitude of a pole, there is a downturn in slope of -20 db/dec. 3 of 6
If a single valid point on the magnitude graph can be located, and the slope of the plot as it passes through that point is known, the entire plot can be filled in by starting from this point and drawing straight lines, adjusting their slopes up and down as (the frequency magnitudes of) the zeros and poles are passed. Note that zeros of either sign cause an upturn, and poles cause a downturn in slope. In the case of (4), the low-frequency magnitude can be found by taking the limit for small s (low frequency). lim H s 100 s so H j s 0 100 (5) Therefore, the magnitude plot of Example 1 may be started with the point ( = 0.1, +20 db). Equation (5) also indicates that the magnitude is zero at zero frequency, but this location cannot be found on a Bode plot. The best way to think of a single zero at 0 frequency is as an upturn in slope of 20 db/dec far off the left side of the graph. The far left side of the plot then appears as a rising line with this slope, passing through the point (0.1, +20 db), and changing slope at the frequency corresponding to each singularity on the list as it is encountered. Note that the zero corresponding to 1000 rad/s cancels the effect of the pole corresponding to that same frequency. The slope at the far right side of the graph should be -20 db/dec. This could be thought of as the effect of the single zero at "infinity," a location off the far right side of the graph. The initial slope on the far left side of plot will be +p 20 db/dec, where p is the number zeros at 0. Likewise, the ultimate slope at the far right side of the plot will -q 20 db/dec, where q is the number of zeros at "infinity" needed to make the list of zeros equal in length to the list of poles. 4 of 6
The phase graph is plotted in a similar manner. In this case, what is being plotted is the summation of phase terms from the numerator of H, minus phase terms from the denominator of H. The form is the following: H j arctan( z 1 ) arctan( z 2 ) arctan( p 1 ) arctan( p 2 ) (6) Note that the negative real zeros will contribute between 0 and 90 degrees to the result, positive real zeros will contribute 0 to -90 degrees, and negative real poles will contribute 0 to -90 degrees. The asymptotic behavior for a negative zero is 0 degrees up to z 10, then an upward slope of +45 degrees/decade to z 10, then the final contribution of +90 degrees. The contribution of a negative zero is +45 degrees at the frequency corresponding to the zero. The asymptotic behavior for a positive zero is 0 degrees up to, then an downward slope of -45 degrees/decade to z 10, then the final contribution of -90 degrees. The contribution of a positive zero is -45 degrees at the frequency corresponding to the zero. The asymptotic behavior for a negative pole is 0 degrees up to p 10, then a downward slope of -45 degrees/decade to p 10, then the final contribution of -90 degrees. The contribution of a negative pole is -45 degrees at the frequency corresponding to the pole. The composite phase curve can be obtained by finding one correct starting point, and then extending the graph in both directions using the slope changes given above. In this case, slopes change at one-tenth and ten-times the frequency corresponding to the singularity, so in many practical systems several singularities may be contributing to the net result at a given frequency. The Bode plot of magnitude and phase of the transfer function data given in Example 1 is shown on page 6. z 10 5 of 6
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