Advanced Control Engineering - Chapter 6 doc

Equation 6.14 is a complex quantity of the form a ‡ jb whereImaginary part b ˆ1 ‡ !K!T2T2 6:16Hence equation 6.14 can be plotted in the complex space Argand Diagram toproduce a harmonic

Classical design in the frequency domain 6.1 Frequency domain analysis

Control system design in the frequency domain can be undertaken using a purelytheoretical approach, or alternatively, using measurements taken from the compon-ents in the control loop The technique allows transfer functions of both the systemelements and the complete system to be estimated, and a suitable controller/compen-sator to be designed

Frequency domain analysis is concerned with the calculation or measurement ofthe steady-state system output when responding to a constant amplitude, variablefrequency sinusoidal input Steady-state errors, in terms of amplitude and phaserelate directly to the dynamic characteristics, i.e the transfer function, of the system.Consider a harmonic input

A2

or

Substituting equation (6.6) into (6.3)

For a given value of !, equation (6.9) represents a point in complex space P(!) When

! is varied from zero to infinity, a locus will be generated in the complex space This

locus, shown in Figure 6.2, is in effect a polar plot, and is sometimes called a

harmonic response diagram An important feature of such a diagram is that its shape

is uniquely related to the dynamic characteristics of the system

6.2 The complex frequency approachRelationship between s and j! From equation (6.2)

i(t) ˆ A1ej!td

dt ˆ j!(A1ej!t) ˆ j!i(t)Taking Laplace transforms

or

Hence, for a sinusoidal input, the steady-state system response may be calculated by

substituting s ˆ j! into the transfer function and using the laws of complex algebra

to calculate the modulus and phase angle

6.2.1 Frequency response characteristics of first-order systems

conjugate of (6.13), i.e

G( j!) ˆ K(1 j!T)

(1 ‡ j!T)(1 j!T)

Equation (6.14) is a complex quantity of the form a ‡ jb where

Imaginary part b ˆ1 ‡ !K!T2T2 (6:16)Hence equation (6.14) can be plotted in the complex space (Argand Diagram) toproduce a harmonic response diagram as shown in Figure 6.3

In Figure 6.3 it is convenient to use polar co-ordinates, as they are the modulus andphase angle as depicted in Figure 6.2 From Figure 6.3, the polar co-ordinates are

1 ω2 2 T

+

1 ω2 2 T

∠

Fig 6.3 A point in complex space for a first-order system.

The argument, or phase angle is

–45°

0.707 K

(a) Polar Plot

(c) (b) Rectangular Plot (Frequency Response)

Fig 6.4 Graphical display of frequency domain data for a first-order system.

Table 6.1 Modulus and phase for a first-order system

! (rad/s) jG( j!)j €G( j!) (degrees)

1/T K/p2 45

Using equations (6.18) and (6.21), values for the modulus and phase angle may becalculated as shown in Table 6.1 The results in Table 6.1 may be represented as aPolar Plot, Figure 6.4(a) or as a rectangular plot, Figures 6.4(b) and (c) Since therectangular plots show the system response as a function of frequency, they areusually referred to as frequency response diagrams.

6.2.2 Frequency response characteristics of second-order

From equations (6.26) and (6.27) the modulus and phase may be calculated as shown

in Table 6.2 The results in Table 6.2 are a function of  and may be represented as a

Polar Plot, Figure 6.5, or by the frequency response diagrams given in Figure 6.6

6.3 The Bode diagramThe Bode diagram is a logarithmic version of the frequency response diagrams

illustrated in Figures 6.4(b) and (c), and also Figure 6.6, and consists of

(i) a log modus±log frequency plot

(ii) a linear phase±log frequency plot

The technique uses asymptotes to quickly construct frequency response diagrams by

hand The construction of diagrams for high-order systems is achieved by simple

graphical addition of the individual diagrams of the separate elements in the system

The modulus is plotted on a linear y-axis scale in deciBels, where

jG( j!)j dB ˆ 20 log10jG( j!)j (6:28)The frequency is plotted on a logarithmic x-axis scale

2 K K

Fig 6.5 Polar plot of a second-order system.

6.3.1 Summation of system elements on a Bode diagram

Consider two elements in cascade as shown in Figure 6.7

ω(rad/s)

ω(rad/s) (a) Modulus

R( j!) ˆ G1( j!)G2( j!)

ˆ jG1( j!)kG2( j!)jej( 1 ‡ 2 ) (6:31)Hence

C

R( j!)

ˆ jG1( j!)kG2( j!)jor

C

R( j!)

... ˆs(s2‡ 2s ‡ 4)4 (6: 66) Equation (6. 66) represents a pure integrator and a second-order system of the form

Figure 6. 23(a), curve (i), shows the Bode gain diagram... 6. 13

| (j ) G ω| dB