Texas A & M University

Department of Mechanical Engineering

MEEN 364 Dynamic Systems and Controls

Dr. Alexander G. Parlos

Lecture 16: 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 (s)

U(s), (1)

where the input is

u(t) = Uosin(ωt), (2)

with a Laplace transform of

U(s) =Uoω

s2 + ω2 . (3)

Considering zero ICs, the output of the system is

Y (s) = G(s)Uoω

s2 + ω2 . (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) =α1

s + a1+

α2

s + a2+ · · ·+ αn

s + an+

α0

s + jω+

α∗0s− jω

, (5)

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Figure 1: Linear System Response to Sinusoidal Input.

where a1, a2, . . ., an are the poles of G(s). Taking the inverse Laplace trans-

form of equation (5) results in

y(t) = α1e−a1t + α2e

−a2t + · · ·+ αne−ant + 2|α0|sin(ωt + φ), (6)

where

φ = tan−1Im(α0)

Re(α0). (7)

If the system is stable, the exponential will all die out and the response of

the system will be

y(t) = 2|α0|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)

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Figure 2: Definitions of Bandwidth and Resonant Peak.

where

A = |G(jω)| = |G(s)|s=jω, (10)

and

φ = tan−1Im[G(jω)]

Re[G(jω)]=< G(jω). (11)

Equations (10) and (11) allow us to compute the magnitude A and phase φ

of the system G(s).

A very important specification in system performance is that of band-

width (BW). This is the frequency above which the system output does not

track the system input, if the latter is a sinusoid. Assume a feedback look

with unity feedback. Then a typical closed-loop transfer function Bode plot

is shown in Figure 2, which also depicts the concept of the BW and resonant

peak, Mr.

Bode Plot Techniques

The idea of Bode plotting is to express the log of the system magnitude and

its phase as a function of the input signal frequency. Following some simple

rules we can plot the frequency response of some fairly complex transfer

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functions.

The principle of the Bode plots resides in the fact that a transfer function

G(s) for s = jω can be written in polar form as follows

G(jω) =r1e

jθ1r2ejθ2

r3ejθ3r4ejθ4r5ejθ5= (

r1r2

r3r4r5)ej(θ1+θ2−θ3−θ4−θ5). (12)

So,

|G(jω)| = r1r2

r3r4r5, (13)

or

log10 |G(jω)| = log10 r1 + log10 r2 − log10 r3 − log10 r4 − log10 r5. (14)

Also,

< G(jω) = θ1 + θ2 − θ3 − θ4 − θ5. (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 log10 |G|.In general, we can express a transfer function in terms of its poles and

zeros, in the following form

KG(s) = K(s− z1)(s− z2) · · ·(s− p1)(s− p2) · · · . (16)

The transfer function of equation (16) can be transformed to the so-called

Bode form, as follows:

KG(jω) = K0(jωτ1 + 1)(jωτ2 + 1) · · ·(jωτa + 1)(jωτb + 1) · · · , (17)

where K0 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) = K0jωτ1 + 1

(jω)2(jωτa + 1). (18)

Then,

log |KG(jω)| = log |K0|+ log |jωτ1 + 1| − log |(jω)2| − log |jωτa + 1|, (19)

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or in dBs,

|KG(jω)|dB = 20 log |K0|+20 log |jωτ1 +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, K0(jω)n,

• Class 2: First order terms, (jωτ + 1)±1,

• Class 3: Second order terms, [( jωωn

)2 + 2ζ( jωωn

) + 1]±1

Class 1:

Since

log K0|(jω)n| = log K0 + 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 asymp-

totes at low and high frequencies. In particular,

• for ωτ ¿ 1, jωτ + 1 ∼= 1,

• for ωτ À 1, jωτ + 1 ∼= jωτ .

If we call the frequency value of ω = 1τ the break point then we observe that

for frequencies below the break point the magnitude is approximately con-

stant, whereas for frequencies above the break point the magnitude behaves

like a Class 1 term K0(jω)n. An example is shown in Figure 4. The slope at

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Figure 3: Magnitude of (jω)n.

high frequencies is 20 dB/dec. The magnitude plot of the term (jωτ + 1)−1

can be easily obtained as the mirror image of the magnitude of term (jωτ +1)

with respect to the horizontal axis.

The phase curve can be obtained by using similar low and high frequency

asymptotes. In particular

• for ωτ ¿ 1, < 1 = 0 deg,

• for ωτ À 1, < jωτ = 90 deg,

• for ωτ ∼= 1, < (jωτ + 1) = 45 deg,

The overall phase varies from 0 deg to +90 deg. An example of a phase plot

is shown in Figure 5. The phase plot of the term (jωτ + 1)−1 can be easily

obtained as the mirror image of the phase of the term (jωτ +1) with respect

to the horizontal axis. The overall phase varies from 0 deg to −90 deg.

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Figure 4: Magnitude of jω 10 + 1 for τ = 10.

Figure 5: Phase of jω 10 + 1 for τ = 10.

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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 nu-

merator 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

2ζ, 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

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Figure 6: Magnitude and phase of second order term.

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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

G1(s) = 10s + 1

s + 10, (23)

G2(s) = 10s− 1

s + 10. (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 G2(s)

starts at 180 degrees, instead of 0 and this makes it nonminimum-phase.

Reading Assignment

See separate file on textbook reading assignments depending on the text

edition you own. Read the examples in Handout E.17 posted on the course

web page.

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Figure 7: Bode plot for minimum and nonminimum-phase systems (a) magnitude; (b) phase.

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