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Texas A & M University
Department of Mechanical Engineering
MEEN 651 Control System Design
Dr. Alexander G. Parlos
Fall 2003
Lecture 9: Frequency Response Analysis
Bode Plots of First and Second Order Systems
The objective of this lecture is to provide you with some background on the use of transfer functions to perform the so-called frequency response analysis of dynamic systems, also known as analysis based on Bode plots. This method will be later used in the design of control systems.
Frequency Response
Consider a system with transfer function
G(s) = Y U^ ((ss)) , (1)
where the input is u(t) = Uosin(ωt), (2)
with a Laplace transform of
U (s) = Uoω s^2 + ω^2.^ (3) Considering zero ICs, the output of the system is
Y (s) = G(s) (^) s 2 U (^) +oω ω 2. (4)
Assuming that G(s) has distinct poles we can perform a partial fraction expansion and take the inverse Laplace transform. The result of the partial fraction expansion can be written as Y (s) = (^) s −α^1 a 1
(^) s −α^2 a 2
· · · + (^) s −αn a n
(^) s +α 0 jω + α
∗ 0 s − jω ,^ (5)
Figure 1: Linear System Response to Sinusoidal Input.
where a 1 , a 2 ,.. ., an are the poles of G(s). Taking the inverse Laplace transform of equation (5) results in
y(t) = α 1 e−a^1 t^ + α 2 e−a^2 t^ + · · · + αne−ant^ + 2|α 0 |sin(ωt + φ), (6)
where φ = tan−^1 Im Re((αα^0 ) 0 )^
If the system is stable, the exponential will all die out and the response of the system will be y(t) = 2|α 0 |sin(ωt + φ). (8)
This equation indicates that if a linear system is excited by a sinusoidal input, then its response is also sinudoidal with the same frequency, but with possibly different amplitude and phase. This is shown in Figure 1. From equation (4) we can see that the response of the system after the exponential terms have decayed can also be expressed as
y(t) = UoAsin(ωt + φ), (9)
can be written in polar form as follows
G(jω) = r^1 e
jθ (^1) r 2 ejθ 2 r 3 ejθ^3 r 4 ejθ^4 r 5 ejθ^5 = ( r^1 r^2 r 3 r 4 r 5 )ej(θ^1 +θ^2 −θ^3 −θ^4 −θ^5 ). (12)
So, |G(jω)| = (^) rr^1 r^2 3 r 4 r 5
or log 10 |G(jω)| = log 10 r 1 + log 10 r 2 − log 10 r 3 − log 10 r 4 − log 10 r 5. (14)
Also, < G(jω) = θ 1 + θ 2 − θ 3 − θ 4 − θ 5. (15) Bode plots are usually expressed in terms of log |G| versus log ω and φ versus log ω. A particularly important unit to remember is that of decibels, or dBs, defined as |G|dB = 20 log 10 |G|. In general, we can express a transfer function in terms of its poles and zeros, in the following form
KG(s) = K (s − z 1 )(s − z 2 ) · · · (s − p 1 )(s − p 2 ) · · ·.^ (16) The transfer function of equation (16) can be transformed to the so-called Bode form, as follows: KG(jω) = K 0 ( (jωτjωτ^1 + 1)(jωτ^2 + 1)^ · · · a + 1)(jωτb + 1)^ · · ·^
where K 0 and the τ ’s are all related to the K, the poles and zeros of the transfer function (16). For example, suppose that
KG(jω) = K 0 jωτ^1 + 1 (jω)^2 (jωτa + 1)
Then, log |KG(jω)| = log |K 0 | + log |jωτ 1 + 1| − log |(jω)^2 | − log |jωτa + 1|, (19)
or in dBs,
|KG(jω)|dB = 20 log |K 0 | + 20 log |jωτ 1 + 1| − 20 log |(jω)^2 | − 20 log |jωτa + 1|. (20)
So, it is sufficient to analyze the most commonly encountered transfer function terms. These terms can be classified as follows:
- Class 1: Singularities at the origin, K 0 (jω)n,
Figure 3: Magnitude of (jω)n.
- Class 2: First order terms, (jωτ + 1)±^1 ,
- Class 3: Second order terms, [( (^) ωjωn )^2 + 2ζ( (^) ωjωn ) + 1]±^1
Class 1:
Since log K 0 |(jω)n| = log K 0 + n log |jω|, (21)
the magnitude plot of the Class 1 term is a straight line with slope n. A value of n = 1 denotes 20 dB/dec, so the slope is in multiples of this value. Examples of different Class 1 terms are depicted in Figure 3. The phase of (jω)n^ is n × 90 deg; this is a horizontal line.
Class 2:
The sketch of this term’s magnitude is obtained by looking at its asymptotes at low and
Figure 5: Phase of jω 10 + 1 for τ = 10.
of the phase of the term (jωτ + 1) with respect to the horizontal axis. The overall phase varies from 0 deg to −90 deg.
Class 3:
This term behaves like the Class 2 term, but there are some differences in the details. In particular,
- the break point is now at ω = ωn,
- the magnitude changes with slope of +2 or +40 dB/dec when the term is in the numerator and −2 or − 40 dB/dec when the term is in the denominator,
- the phase changes from 0 deg to +180 deg when the term is in the numerator and 0 deg or −180 deg when the term is in the denominator.
- the values of the magnitude and phase near the break point depend on the value of ζ. A rough sketch of the magnitude when the term appears in the denominator can be obtained by observing that in addition to the above rules,
|G(jω)| = 1 2 ζ , at ω = ωn. (22)
No such easy rule exists for the phase plot, however, the phase also depends on ζ. In the limiting cases, when ζ = 0 the phase plot is a step function from 0 deg to ±180 deg at ω = ωn, whereas when ζ = 1 it can be treated like two first order terms (Class 2 terms). For other values of ζ the phase is in between these two limits. An example of a magnitude and phase plot for Class 3 systems is shown in Figure 6. The magnitude and phase plots of the Class 3 term in the numerator can be obtained from the equivalent plots in the denominator by considering the mirror image of the magnitude and phase plots with respect to the horizontal axis.
The Composite Curve
When a dynamic system is composed of many poles and zeros, plotting the Bode plot requires that we plot the individual component Bode plots and then combine them into a composite plot. The composite curve is the sum of the individual curves and a sketch can be easily obtained by hand. More accurate Bode plots are obtained using Matlab. NOTE: Please, read carefully the Summary of Bode Plot Rules on page 379 of the text. Examples of how to sketch the composite Bode plot of transfer functions will be provided during the problem solving session.
Nonminimum-phase Systems
A system with one or more zeros in the RHP behaves very differently that its counterpart with zeros in the LHP. For example, consider the transfer functions
G 1 (s) = 10 (^) ss + 10+ 1, (23)
G 2 (s) = 10 (^) ss + 10−^1. (24)
The magnitude curves for these two transfer functions is identical. The phase curve, however, is different. Both are shown in Figure 7. The phase for G 2 (s) starts at 180 degrees, instead of 0 and this makes it nonminimum-phase.
Reading Assignment
Read pages 364-386 of the textbook. Read the examples in Handout E.17 posted on the course web page.
Figure 7: Bode plot for minimum and nonminimum-phase systems (a) magnitude; (b) phase.
