UNIT-VFREQUENCY RESPONSE ANALYSIS

In practice, the performance of a control system is measured more realistically by its time-domain characteristics. The reason is that the performance of most control systems is judged based on the time response due to certain test signals.

In design problems, there are no unified methods of arriving at a designated system that meets time- domain performance specifications. On the other hand, in frequency domain, a wealth of graphical and other techniques are available that are useful for system analysis and design, irrespective of the order of the system.

It is important to realize that there are correlating relations between the frequency- and time-domain performances in linear system so that time-domain properties of the system can be predicted based on the frequencydomain characteristics. With these in mind, we shall study the frequency response analysis of control systems.

Frequency Response of a System

It is well known from linear system theory that, when the input to a linear time invariant system is sinusoidal with amplitude R and frequency o, i.e.,

r(t) R sin ot

the steady-state output of the system, y(t), will be a sinusoid with the same frequency o, but possibly with different amplitude and phase; i.e.

y(t) Y sin (ot )

where Y is the amplitude of the output sine wave and is the phase shift.

Let the transfer function of a SISO system be M(s); the output Y(s) and the input R(s) are related through

Y(s) M(s)R(s)

For sinusoidal steady-state analysis, we replace s by j, and equation (6.3) becomes

Y(j) j) Rj)

By writing Y(j) and M(j) as (similar expression for Rj) also):

Y ( j) Y ( j) Y ( j)

M ( j) M ( j) M ( j)

Y ( j) M ( j) R( j)

and the phase relation:

Y ( j) M ( j) R( j)

Thus, for the input and output signals described by equations (6.1) and (6.2),

Y M ( jo ) R

M ( jo )

Thus, by knowing the transfer function M(s), the frequency response of the system can be obtained.

The frequency response of the loop transfer function G(s)H(s) [G(s) if H(s) is unity] can be plotted in several ways. The two commonly used representations are:a. Bode diagram, or Logarithmic plot.b. Polar plot, or Nyquist plot.

Frequency Response Bode Diagram

A Bode diagram consists of two graphs. One is a plot of the logarithm of the magnitude of a sinusoidal transfer function; the other is a plot of the phase angle; both are plotted against the frequency on a logarithmic scale.

The standard representation of the logarithmic magnitude of G(j) is 20 log |G(j)|, where the base of the logarithm is 10. The unit used in this representation is the decibel (dB). The curves are drawn on a semilog paper, using the log scale for frequency and linear scale for either magnitude (in dB) or phase angle (degrees).

The main advantage of Bode diagrams is that the multiplication of magnitudes can be converted into addition. Furthermore, a simple asymptotic method is available for sketching the approximate curve. Should the exact curve be desired, corrections could be made easily to these basic asymptotic plots.

In Bode diagrams, the frequency ratios are expressed in terms of octaves or decades. An octave is a frequency band from 1 to 21, where 1 is any frequency. A decade is a frequency band from 1 to 101, where 1 is any frequency.

Basic Factors of G(j)H(j):The basic factors that very frequently occur in an arbitrary open-loop transfer function G(j)H(j) are: a. Real Constant: G(s)H(s) = K

GjHj = KMagnitude: |GjHj| (dB) = 20 log10 |K| (dB). Phase angle: GjHj = 0.

The Bode plot for any value of K is shown in Fig. 6.1.

Figure 6.1: Bode plot for gain K.

b. Poles and zeros at the origin: G(s)H(s) = sn

For sn:GjHj=jnMagnitude: |GjHj| (dB) = 20n log10 |j| (dB) = 20n log10 (dB)(6.11) Phase angle: GjHj = 90n (a constant).

For s-n:GjHj=j-nMagnitude: |GjHj| dB = n log10 |j| dB = n log10 (dB)(6.12) Phase angle: GjHj = n (a constant).

The Bode magnitude plots are a straight line in semi log coordinate. The slope of the line is 20n dB/decade i.e. the magnitude change by 20n dB for the frequency change of 10 times. The straight line passes through 0 dB at = 1. The phase angle () of jis constant and equal to 900.The Bode plots are shown in Fig. 6.2.

Figure 6.2: Bode diagrams for (a)G(j) = 1/j(b)G(j) = j.

c. Simple zeros and poles: G(s)H(s) = (1+sT)1GjHj= + jT1

Magnitude:|GjHj| (dB) = 20 log10 |1 + jT| (dB)= 20 log10 + 2T2] (dB)(6.11)

To obtain asymptotic approximation we consider both very large and very small values of . For low frequencies, such that T > 1,

20 log

12T 2

12T 2

20 log1 0 dB

20 logT dB

At =1/T, log magnitude = 0 dB while at =10/T, log magnitude = 20 dB. Thus, the value of

20 logT

increases/decreases with 20 dB/decade. Hence, the magnitude plot can be approximated

by two straight-line asymptotes, one a straight line at 0 dB for the frequency range 0 < < 1/T and the other a straight line with slope 20 dB/decade for the frequency range 1/T < < . The frequency,=1/T, at which the two asymptotes meet is called the corner frequency or break frequency.

Phase angle: GjHj = tan1 T .At corner frequency, GjHj = 45. The phase plot can be approximated by a straight line passing through 0at one decade below corner frequency and 90at one decade above corner frequency. The Bode plots are shown in Fig. 6.3 and Fig.An advantage of the Bode diagram is that for reciprocal factors, for example the factor 1/(1+jT), the log-magnitude and phase angle curves need only be changed in sign, since

20 log

11jT

20 log1jT

and phase angle of 1/(1+jT) = tan1 T

= (phase angle of

(1+jT). Figure 6.3: Bode plot for (1+jT).

Figure 6.4: Log-magnitude curve (with asymptotes) for 1/(1+jT).

For low frequencies such that /n > 1, the log-

magnitude becomes 20 log

2

240 log

dB. The equation for the high frequency asymptote is a

nnstraight line with a slope of 40 dB/decade.

The frequency n is the corner frequency. The two asymptotes just derived are independent of the value of . Fig. 6.5 shows exact curves with the straight-line asymptotes and the exact phase angle curves.

Phase angle of the quadratic factor is:

Figure 6.5: Bode plot for Eqn. (6.12) and (6.13).

Example 6.1: Sketch the Bode plot for the following function:

G(s)

1

s 2

1

1

1 1

Solution:

G(s)

,s 2

G( j)

j2

2 1

j

2

2

Magnitude: 20 log G( j) 20 log1 - 20log

1

2

G( j) tan1 2

2

-5

-10

Phase (deg); Magnitude (dB)-15

-20

-25

0

-20

To: Y(1)-40

-60

-8010-1100101

Frequency (rad/sec)

Figure 6.6: Bode plots for the system in Example 6.1.

General Procedure for Plotting the Bode Diagrams: Rewrite the sinusoidal transfer function as a product of the basic factors discussed above. Identify the corner frequencies associated with these basic factors. Draw the asymptotic log-magnitude curves with proper slopes between the corner frequencies considering all the basic factors together. The exact curve, which lies very close to the asymptotic curve, can be obtained by adding contributions from all the factors and proper corrections. Phase-angle curve can be drawn by adding the phase-angle curves of individual factors.

Polar Plot (Nyquist Plot)

The polar plot of a sinusoidal transfer function G(j) is a plot of the magnitude of G(j) versus the phase angle of G(j) on polar coordinates as is varied from zero to infinity. Note that, in polar plots, a positive (negative) phase angle is measured counterclockwise (clockwise) from the positive real axis. The polar plot is very often called the Nyquist plot in control system engineering. An example of such a plot is shown in Fig. 6.7. Each point on the polar plot of G(j) represents the terminal point of a vector at a particular value of . In the polar plot, it is important to show the frequency graduation of the locus.

Example of a Polar Plot.

If approaches infinity, the magnitude approaches 0 and the phase angle approaches 900. The polar plot of this transfer function is a semicircle as the frequency is varied from 0 to . It is shown in Fig. 6.8. The center is located at 0.5 in the real axis and the radius is equal to 0.5. The lower semicircle corresponds to 0 , and the upper semicircle corresponds to 0 .

Figure 6.8: (a) Polar plot of 1/(1+jT) ; (b)Same plot in X-Y plane.

The polar plot of the transfer function 1+jT is simply the upper half of the straight line passing through the point (1, 0) in the complex plane and parallel to the imaginary axis as shown in Fig. 6.7.

Figure 6.7: Polar plot of 1+jT.

4 nQuadratic factors: G(s)H (s) 12s /

s /

2 1

nnG( j)H ( j) 12j/

j/

2 1

nThe low and high frequency portions of the polar plot of the following transfer function

GjH ( j)

1

n12j/

nj/ 2

are given, respectively, bylim GjH ( j) 1000

and

lim GjH ( j) 01800

Thus, the high frequency portion is tangent to the negative real axis. The polar plots are shown in Fig. 6.8.

*Phase angle of the quadratic factor is the same as Eqn. (6.13):

tan1

2n2

1 n

1

12j/Figure 6.8: Polar plots ofn

nj/2

for > 0.

Next, consider the following transfer function:

nGjH ( j) 12j/

nj/ 2

2

2

1

2 j

n

n

The low-frequency portion of the curve is:lim GjH ( j) 1000

and the high-frequency portion of the curve is: lim GjH ( j) 1800The general shape of the polar plot is shown in Fig. 6.9.

nFigure 6.9: Polar plot of 12j/j/2 for > 0.

Figure 6.10: Polar plot of G(s) in Example 6.2.

Example 6.3: Draw polar plot for the system with G(s)

10

s(s 1)(s 2)

Solution:

G( j)

10

j( j1)( j2)

10(2 1) (2 2)G( j)

G( j) 90tan1 tan1 2

The plot is shown in Fig. 6.11.

Figure 6.11: Polar plot for the system in Example 6.3.

NYQUIST STABILITY TEST THE CAUCHY CRITERION

The Cauchy criterion (from complex analysis) states that when taking a closed contour in the complex plane, and mapping it through a complex function G(s), the number of times, N, that the plot of G(s) encircles the origin is equal to the number of zeros, Z, of G(s) enclosed by the frequency contour minus the number of poles, P, of G(s) enclosed by the frequency contour.N = Z PEncirclements of the origin are counted as positive if they are in the same direction as the original closed contour or negative if they are in the opposite direction.

When studying feedback control, we are not as interested in G(s)H(s) as in the closed-loop transfer function G(s)H(s)/[1+G(s)H(s)]

If 1+G(s)H(s) encircles the origin, then G(s)H(s) will enclose the point -1. Since we are interested in the closed-loop stability, we want to know if there are any closed-loop poles (zeros of 1+G(s)H(s)) in the right-half plane.

The Nyquist Stability Criterion usually written as Z = P + N, where>Z is the number of right hand plane poles for the closed loop system (or zeros of 1+G(s)H(s))>P is the number of open-loop poles (in the RH side of the s-plane) of G(s)H(s) (or poles of 1+G(s)H(s)), and>N is the number of clockwise encirclements of (-1,0)

A feedak otol sste is stale if ad ol if the ue of oute-clockwise encirclements of the critical point (-1,0) by the GH polar plot is equal to the number of poles of GH ith positie eal pats. (Nyquist Stability Criterion Definition)

Example:4.5 Consider the unity feedback applied to the following systemG(s)=K/[s(s+3)(s+5)]4.6 The loop transfer function isG(j)H(j)=K/[s(s+3)(s+5)]|K=1,s= j4.7 The number of open-loop poles in the RH side of the s-plane, P =

4.8 For stability Z = 0, therefore N must also be 4.9 From Nyquist diagram it can be seen that K can be increased bybefore the Nyquist diagram encircles -1.4.10 For marginal stability, K =

RELATIVE STABILITY GAIN AND PHASE MARGIN

4.11 K is a variable (constant) gain4.12 G(s) is the plant under consideration

Gain margin is defined as the change in open loop gain required to make the system unstable. Systems with greater gain margins can withstand greater changes in system parameters before becoming unstable in closed loop. Phase margin is defined as the change in open loop phase shift required to make a closed loop system unstable.

Stability Analysis with Bode Plot

Bode plot is a very useful graphical tool for the analysis and design of linear control systems.

Advantages of the Bode Plot over Nyquist plot:

1. Gain crossover, phase crossover, gain margin, and phase margin are more easily determined on the Bode plot.2. For design purposes, the effects of adding controllers and their parameters are more easily visualized on the Bode plot.

Identifying Marginal Values from Bode Plot

The gain margin is the difference between the magnitude curve and 0dB at the point corresponding tothe feue that gies us a phase of the phase oss oe feue, p).

The phase margin is the difference in phase betwee the phase ue ad at the poitcorresponding to the frequency that gives us a gain of 0dB (the gain cross over frequency, g).

Figure 6.12: Determination of GM and PM from the Bode plot.

Example 6.4: Consider the loop transfer function given as

L(s)

2500

s(s 5)(s 50)

From the provided Bode diagram, find the GM and PM and corresponding frequencies.

Solution:

Figure 6.13: Bode diagram for Example 6.4.

Fig. 6.14 illustrates PM and GM of a stable and unstable system in Bode diagrams.

Figure 6.14: Phase and Gain margins for stable and unstable systems.

Stability Analysis with Nyquist Plot

OPEN LOOPCLOSED LOOP

Gain Margin, GMThe change in open-loop gain, expressed in dB, required at -180 to make the closed- loop system unstable. A good range is 2