# Category Dynamics of Flight

## NUMERICAL EXAMPLE

The rudder and aileron angles in a steady truly-banked turn are calculated by way of example for the same general aviation airplane as was used above for the sideslip. The altitude is sea level, the speed is 125 fps (38.1 m/s) and the stability and control derivatives are as in Table 7.2. The solution of (7.8,13) for climb angles between — 10° and +10° shows that the sideslip angle /3 remains less than 1.5° and the rudder and aileron angles are as shown in Figs. 7.24 and 7.25. The value of CL varies over the range of bank angles used from 0.8 to 1.6, so several of the stability derivatives are significantly affected. It is seen that the aileron angle is always positive for a right turn—that is, the right aileron is down (stick to the left), and that the rudder is usually negative (right rudder) although its sign may reverse in a steep climb. The strong ef­fect of the climb angle derives from the fact that the roll rate p is proportional to в and changes sign with it. Thus the terms Clpp and СПрр in the moment equations change sign between climbing and descending and affect the control angles required to produce zero moment.

 Figure 7.24 Rudder angle in turn. • General aviation airplane • Speed 125 fps • Sea level

 • General aviation airplane • Speed 125 fps • Sea level

Finally, it may be remarked that the control angles obtained would have been substantially different had it been stipulated that /3, not Cy, should be zero in the turn. It would not then be possible, however, to satisfy the requirement that the ball be cen­tered in the tum-and-back indicator.

The basic flight condition is steady symmetric flight, in which all the lateral variables /3, p, г, ф are identically zero. Unlike the elevator and the throttle, the lateral controls, the aileron and rudder, are not used individually to produce changes in the steady state. This is because the steady state values of /3, p, г, ф that result from a constant 8a or 8r are not generally of interest as a useful flight condition. There are two lateral steady states that are of interest, however, each of which requires the joint application of aileron and rudder. These are the steady sideslip, in which the flight path is recti­linear, and the steady turn, in which the angular velocity vector is vertical. We look into these below before proceeding to the study of dynamic response to the lateral controls.

The steady sideslip is a condition of nonsymmetric rectilinear translation. It is some­times used, particularly with light airplanes, to correct for cross-wind on landing ap­proaches. Glider pilots also use this maneuver to steepen the glide path, since the L/D ratio decreases due to increased drag at large /3. In this flight condition all the time derivatives in the equations of motion (except xE) and the three rotation rates p, q, r are zero. It is simplest in this case to go back to (4.9,2)-(4.9,6), from which we derive the following:

AY + mgф cos в0 = 0

AL = 0 (7.8,1)

AN = 0

We use (4.9,17) for aerodynamic forces, and for the control forces use the following as a reasonable representation:

(7.8,2)

We now add the assumption that в0 is a small angle and get the resulting equation

(7.8,3)

In this form, v is treated as an arbitrary input, and (5r, 8a, ф) as outputs. (See Exercise 7.6.) Clearly, there is an infinity of possible sideslips, since v can be chosen arbitrar­ily. Note that the other three variables are all proportional to v. We illustrate the steady sideslip with a small general aviation airplane3 of 30-ft (9.14 m) span and a gross weight of 2400 lb (10,675 N). The altitude is sea level and CL = 1.0, corre-

Table 7.2

 c, cn p -0.14 -0.0689 – 0.0917Q 0.01326 + 0.017C2 p -0.039 -0.441 -0.00109 – 0.0966Q r 0.165 -0.0144 + 0.27 CL -0.048 – 0.0238C2 0 -0.0531 0.005 0.117 0.0105 -0.0509

sponding to a speed of 112.3 fps (34.23 m/s), and the wing area is 160 ft2 (14.9 m2). The nondimensional derivatives are given in Table 7.2, from which the numerical system equation is found from (7.8,3) to be (see Exercise 7.5)

 " 280.7 0 2400 “ 4“ “ 2.991 " 755.7 -3821.9 0 = 102.93 -3663.5 359 0 _ -19.394

It is convenient to express the sideslip as an angle instead of a velocity. To do so we recall that /3 = v/u0, with u0 given above as 112.3 fps. The solution of (7.8,4) is found to be

8r//3 = .303 8Jl3 = -2.96 фф = .104

We see that a positive sideslip (to the right) of say 10° would entail left rudder of 3° and right aileron of 29.6°. Clearly the main control action is the aileron displacement, without which the airplane would, as a result of the sideslip to the right, roll to the left. The bank angle is seen to be only 1 ° to the right so the sideslip is almost flat.

We define a “truly banked” turn to be one in which (1) the vehicle angular velocity vector to is constant and vertical (see Fig. 7.22) and (2) the resultant of gravity and centrifugal force at the mass center lies in the plane of symmetry (see Fig. 7.23). This corresponds to flying the turn on the tum-and-bank indicator.[19] [20] It is quite common for turns to be made at bank angles that are too large for linearization of sin ф and cos ф to be acceptable, although all the state variables other than ф and V are small. Thus we turn to the basic nonlinear equations in Sec. 4.7 for this analysis. The large bank angle has the consequence that coupling of the lateral and longitudinal equations oc­curs, since more lift is needed to balance gravity than in level flight. Thus not only the aileron and rudder but the elevator as well must be used for turning at large ф.

 Figure 7.23 Gravity and acceleration in turn.

The body-axis angular rates are given by

(7.8,6)

We now apply the second condition for a truly-banked turn, that is, that the ball be centered in the tum-and-bank indicator. This means that the vector mg — mat, where ac is the acceleration vector of the CG, shall have no у component. But mac is the resultant external force f, so that from (4.5,6)

mg — mac = mg — f = —A

where A is the resultant aerodynamic force vector. Thus we conclude that the aerody­namic force must lie in the xz plane, and hence that Y = 0. We consider the case when there is no wind, so that

(,UE, VE, WE) = (U, V, W)

and choose the body axes so that ax = w = 0. We now use (4.7,1) with all the vari­ables constant and only и and ф not small to get:

Y = —me sin ф + mru = 0 (a)

(7.8,7)

Z = —mg cos ф — mqu (b)

When v is small, a reasonable assumption for a truly banked turn, we also have that и = V, the flight speed. It follows from (7.8,7a) that

rV

sin ф = — g

and with the value of r obtained from (7.8,6)

ojV

tan ф = —- (7.8,8)

g

The load factor nz is obtained from (7.8,7b):

Z qV

n = — —– = cos ф – I—–

mg g

With q from (7.8,6) this becomes

Vco sin ф

n = cos ф ————

g

By using (7.8,8) to eliminate Vco we get

(7.8,9)

We note from (5.1,1) that Z = —L in this case, so that n = LAV = n.. The incremen­tal lift coefficient, as compared with straight flight at the same speed and height, is

We can now write down the equations governing the control angles. From (4.7,2), to first order, L = M = N — 0, so we have the five aerodynamic conditions

Q Cm C„ Су — 0 and Д CL = (n — 1 )CW

On expanding these with the usual aerodynamic derivatives, we get

Сф + Clpp + Clrr + ClsSr + Cls8a = 0 Cm, Aa + Cmqq + CmsA8e = 0 Cnffi + Cnpp + Cnr + Cns8r + Cns8a = 0 Cyfi + Cypp + Cyr + Cys8r = 0

Я„Ла + С/С/ + CLsA8e = (n — 1 )CW

In these relations p, q, r are known from (7.8,6), that is,

(7.8,12)

When (7.8,9) is used to eliminate ф from (7.8,14), and after some routine algebra, the solution for Д<5е is found to be (see Exercise 7.6)

Except for far forward CG positions and low speeds, the angles given by (7.8,15) are moderate. The similarity of this expression to that for elevator angle per g in a pull – up (3.1,6a) should be noted. They are in fact the same in the limit n —> The eleva­

tor angle per g in a turn is therefore not very different from that in a vertical pull-up.

Finally, the lateral control angles are obtained from the solution of (7.8,13).

## RESPONSE TO THE THROTTLE

For the same jet airplane and flight condition as in previous numerical examples, we calculate the response to a step input in the throttle of A8p = |, which corresponds to a thrust increment of 0.05W. The matrix В is given by (7.6,5) and Де is

Дс = [0 l/6f (7.7,13)

The numerical results are shown in Fig. 7.21. Because the model has not included any engine dynamics, the results are not valid for the first few seconds. However, this region is not of much interest in this case. The motion is clearly seen to be dominated by the lightly damped phugoid. The speed begins to increase immediately, before the other variables have time to change. It then undergoes a slow damped oscillation, ul­timately returning to its initial value. The angle of attack varies only slightly, and у makes an oscillatory approach to its final positive value yss. The new steady state is a climb with Ди = Да = 0. When the thrust line does not pass through the CG, the re­sponse is different in several details. Principally, the moment of the thrust causes a

rapid change in a, followed by an oscillatory decay to a new Aass Ф 0, and the speed converges to a new value Дм„ Ф 0.

## Step-Function Response

The response of the airplane to a sudden movement of the elevator is shown by the step response. This requires a solution in the time domain as distinct from the pre­ceding solution, which was in the frequency domain. Time domain solutions are com­monly obtained simply by integrating (7.1,4) by a Runge-Kutta, Euler, or other inte­gration scheme, the choice being dependent on the order of the system, accuracy required, computer available, and so on. The software used[18] for the example to follow does not integrate the equations, but instead uses an alternative method. It inverts the transfer function using the Heavyside expansion theorem (A.2,10). For the same jet airplane and flight condition as in the preceding example, the control vector for ele­vator input is с = [Д5е 0]T and A and В are as before. (Note that only the first col­umn of В is needed.) Time traces of speed, angle of attack, and flight path angle are shown in Figs. 7.19 and 7.20 for two time ranges when the elevator displacement is one degree positive, that is, down.

It is seen from Fig. 7.19, which shows the response during the first 10 sec, that only the angle of attack responds quickly to the elevator motion, and that its variation is dominated by the rapid, well-damped short-period mode. By contrast, the trajec­tory variables, speed, and flight path angle, respond much more slowly. Figure 7.20, which displays a 10 min time span, shows that the dynamic response persists for a very long time, and that after the first few seconds it is primarily the phugoid mode that is evident.

The steady state that is approached so slowly has a slightly higher speed and a slightly smaller angle of attack than the original flight condition—both changes that would be expected from a down movement of the elevator. The flight path angle is seen to be almost unchanged—it increased by about one-tenth of a degree. The reason for an increase instead of the decrease that would be expected in normal cruising flight is that at this flight condition the airplane is flying below its minimum-drag speed.

If the reason for moving the elevator is to establish a new steady-state flight con­dition, then this control action can hardly be viewed as successful. The long lightly damped oscillation has seriously interfered with it. Clearly, longitudinal control, whether by a human or an automatic pilot, demands a more sophisticated control ac­tivity than simply moving it to its new position. We return to this topic in Chap. 8.

Phugoid Approximation

We can get an approximation to the transfer functions by using the phugoid ap­proximation of Sec. 6.3. The differential equation is (6.3,6) with control terms added, that is,

л8е m Z* m MSe 0

frequency responses calculated with the above approximate transfer func­tions are shown on Figs. 7.15 to 7.18. It is seen that the phugoid approximation is ex­act at very low frequencies and the short-period approximation is exact in the high – frequency limit. For frequencies between those of the phugoid and short-period modes, one approximation or the other can give reasonable results.

## Responses to Elevator and Throttle

RESPONSE TO ELEVATOR

When the only input is the elevator angle ASe the system reduces to

where bfj are the elements of B. Solving for the ratios GuSe(s) = Au/A8e and so on yields the four transfer functions.1 Each is of the form (7.2,8). For this case the char-

‘In the subscripts for the transfer function symbols, the symbol Д is omitted in the interest of sim­plicity.

acteristic polynomial is the left side of (6.2,6) (with s replacing A), and the numerator polynomials are

NuSe = -0.000188s3 – 0.2491s2 + 24.68s +11.16

NwSe = -17.85s3 – 904.0s2 – 6.208s – 3.445

NqSe = – 1.158s3 – 0.3545s2 – 0.003873s (?’7’2)

NeSe = -1.158s2 – 0.3545s – 0.003873

There are two other response quantities of interest, the flight path angle у and the load factor nz. Since в0 = 0, Ав = в, and Ay = Ав — Да, it readily follows that

GySe = Gm – GaSe (7.7,3)

nz = -Z/W (7.7,4)

It is equal to unity in horizontal steady flight, and its incremental value during the re­sponse to elevator input is

Anz = —AZJW = —(ZuAu + Zww + Zqq + Z^w + ZSeASe)/W

After taking the Laplace transform of the preceding equation and dividing by A8e, we get the transfer function for load factor to be

Gnzse = ~(ZUGU S’ + ZwGwSe + ZqGqSe + sZivG„K + Z Se)/W (7.7,5)

The total gain and phase of the frequency responses calculated by (7.5,6) from five of the above six transfer functions (for Ли, w, q, Ay, and Anz) are shown in Figs. 7.14 to 7.18.

The exact solutions show that the responses in the “trajectory” variables и and у are dominated entirely by the large peak at the low-frequency Phugoid mode. Be­cause of the light damping in this mode, the resonant gains are very large. The peak |G„J of nearly З X 104 means that a speed amplitude of 100 fps would result from an elevator angle amplitude of about 100/(3 X 104) rad, or about 0.2°. Similarly, at reso­nance an amplitude of 10° in у would be produced by an elevator amplitude of about 5°. For both of these variables the response diminishes rapidly with increasing fre­quency, becoming negligibly small above the short-period frequency. The phase an­gle for u, Fig. 7.14b, is zero at low frequency, decreases rapidly to near -180° at the phugoid frequency (very much like the lightly damped second-order systems of Fig. 7.13) and subsequently at the short-period frequency undergoes a second drop char­acteristic of a heavily damped second-order system. The “chain” concept of high-or­der systems (Sec. 7.2) is well exemplified by this graph.

By contrast, the attitude variables w and q show important effects at both low and high frequencies. The complicated behavior of w near the phugoid frequency indi­cates the sort of thing that can happen with high-order systems. It is associated with a pole/zero pair of the transfer function being close to one another. Again, above the short-period frequency, the amplitudes of both w and q fall off rapidly.

The amplitude of the load factor Anz has a very large resonant peak at the phugoid frequency, almost 100/rad. It would not take a very large elevator amplitude at this frequency to cause structural failure of the wing!

## Longitudinal Response

To treat this case we need the matrices A and В of (7.1,4). A is given in (4.9,18), but we have not yet given В explicitly. On the right side of (4.9,18) we have the product ВДс given by

A Z,

(m – ZJ

AM, M* AZ,

Iy Iy (m – Zw)

0

We now have to specify the control vector c and the corresponding aerodynamic forces and moment. For longitudinal control, we assume here that the available con­trols are well enough represented by

c = [8e 8p]T (7.6,2)

and that the incremental aerodynamic forces and moment that result from their actua­tion are given by a set of control derivatives XSe and so on, in the form

 AX, *5 Xs ‘ °e Op as; AZ, = Zs Zs Oe Op AS AM, MSe Mgp P

Additional elements can be added to c and to (7.6,3) if the situation requires it.

The use of constant derivatives, as in (7.6,3), to describe the force output of the propulsion system in response to throttle input does not allow for any time lag in the buildup of engine thrust since it implies that the thrust is instantaneously proportional to the throttle position. This is not unreasonable for propeller airplanes, but it is not a good model for jets in situations when the short-term response is important, as for ex­ample in a balked landing. To allow for this effect when the system is modeled in the Laplace domain, one can use control transfer functions instead of control derivatives. That is, one can replace, for example, XSp by Gxs (5). If the system model is in the time domain, the same result can be obtained by adding an additional differential equation and an additional variable. This latter method is illustrated in the example of Sec. 8.5.

By substituting (7.6,3) into (7.6,1) we derive the matrix В to be

Ms, M„ZSe MSp M^ZSp

Iy ly(m – ZJ Iy Iy(m – ZJ
0 0

With A and В known, we can compute the desired transfer functions and responses. (In this example the elevator angle is in radians, and English units are used for all other quantities). We calculate responses for the same jet transport as was used previ­ously in Sec. 6.2, with A given by (6.2,1). The nondimensional elevator derivatives are:

CXSc = -3.818 X КГ6 CJ6 = -0.3648 Cmit=~ 1.444

from which the dimensional derivatives are calculated as

= CxspulS = -3.717 Z5<, = CzspulS = -3.551 X 105 MSr = CmSr hpu20Sc = -3.839 X 107

For the throttle, we arbitrarily choose a value of XsJm = 0.3 g when Sp = 1, and ZSp and MSp = 0. With these values we get for the matrix B:

(7.6,5)

## FREQUENCY RESPONSE OF A SECOND-ORDER SYSTEM

The transfer function of a second-order system is given in (7.3,9). The frequency-re­sponse vector is therefore

 Me’*

From the modulus and argument of (7.5,11), we find that

1

{[1 – (o>/&>„)2]2 + 4£2(cu/wn)2}l/2

 Figure 7.18 Frequency-response functions, elevator angle input. Jet transport cruising at high altitude, (a) Load factor amplitude. (b) Load factor phase.

A representative vector plot of Me“p, for damping ratio £ = 0.4, is shown in Fig. 7.11, and families of M and cp are shown in Figs. 7.12 and 7.13. Whereas a single pair of curves serves to define the frequency response of all first-order systems (Fig. 7.10), it takes two families of curves, with the damping ratio as parameter, to display the char­acteristics of all second-order systems. The importance of the damping as a parame­ter should be noted. It is especially powerful in controlling the magnitude of the reso­nance peak which occurs near unity frequency ratio. At this frequency the phase lag is by contrast independent of £, as all the curves pass through <p = —90° there. For all
values of £, Л/ —> 1 and <p —» 0 as ш/шп 0. This shows that, whenever a system is driven by an oscillatory input whose frequency is low compared to the undamped natural frequency, the response will be quasistatic. That is, at each instant, the output will be the same as though the instantaneous value of the input were applied stati­cally.

The behavior of the output when £ is near 0.7 is interesting. For this value of £, it is seen that (p is very nearly linear with шІшп up to 1.0. Now the phase lag can be in­terpreted as a time lag, т = (<р/2тг)Т = <p/a> where T is the period. The output wave form will have its peaks retarded by т sec relative to the input. For the value of £ un­der consideration, ср/(шІшп) Ф it/2 or <p/w = ттІ2шп = Tn, where Tn = 2тг/ш, п the un­damped natural period. Hence we find that, for £ = 0.7, there is a nearly constant time lag t = Tn, independent of the input frequency, for frequencies below reso­nance.

The “chain” concept of higher-order systems is especially helpful in relation to frequency response. It is evident that the phase changes through the individual ele­ments are simply additive, so that higher-order systems tend to be characterized by greater phase lags than low-order ones. Also the individual amplitude ratios of the el­ements are multiplied to form the overall ratio. More explicitly, let

G(s) = G,(i) • G2(s) G„(s)

be the overall transfer function of n elements. Then

G(ioS) = G,(7ft>) • G2(ia>) Gn(i(o)

= (A, M, • K2M2 ■■■

= KMei4>

n

so that KM — PI KrMr

r= 1

n

r= 1

On logarithmic plots (Bode diagrams) we note that

n

log KM = ^ log KrMr (7.5,14)

r= 1

Thus the log of the overall gain is obtained as a sum of the logs of the component gains, and this fact, together with the companion result for phase angle (7.5,13) greatly facilitates graphical methods of analysis and system design.

RELATION BETWEEN IMPULSE RESPONSE AND FREQUENCY RESPONSE

We saw earlier (7.3,4), that h{t) is the inverse Fourier transform of G(iw), which we can now identify as the frequency response vector. The reciprocal Fourier transform relation then gives

G(iw)= h(t)e~iuj’ dt

j — oo

228 Chapter 7. Response to Actuation of the Controls-Open Loop

that is, the frequency response and impulsive admittance are a Fourier transform pair.

## FREQUENCY RESPONSE OF FIRST-ORDER SYSTEM

The first-order transfer function, written in terms of the time constant T is

1

7ТТ/Г

whence

К = lim G(s) = T

s^O

ш/шп

(a)

Figure 7.13 Frequency-response curves—second-order system.

 10 2 10“1 10° 101 to (rad/sl to)

№)

Figure 7.14 Frequency-response functions, elevator angle input. Jet transport cruising at high altitudes, (a) Speed amplitude, (h) Speed phase.

The frequency response is determined by the vector G(ico)

T

G(ico) = KMe’v = ———

1 + itoT

whence

1 — icoT

From (7.5,9), M and <p are found to be

1

(1 + co2T2)m

ip = — tan-1 шТ

A vector plot of Mei<p is shown in Fig. 7.9. This kind of diagram is sometimes called the transfer-function locus. Plots of M and <p are given in Figs. 7.10a and b. The ab­scissa is fT or log wT where / = ш/ітг, the input frequency. This is the only parame-

 Figure 7.16 Frequency-response functions, elevator angle input. Jet transport cruising at high altitude, (a) Pitch-rate amplitude. (b) Pitch-rate phase.

ter of the equations, and so the curves are applicable to all first-order systems. It should be noted that at w = 0, M = 1 and <p = 0. This is always true because of the definitions of К and G(s)—it can be seen from (7.4,5) that G(0) = K.

## EFFECT OF POLES AND ZEROS ON FREQUENCY RESPONSE

We have seen (7.2,9) that the transfer function of a linear/invariant system is a ratio of two polynomials in 5, the denominator being the characteristic polynomial. The roots of the characteristic equation are the poles of the transfer function, and the roots of the numerator polynomial are its zeros. Whenever a pair of complex poles or zeros lies close to the imaginary axis, a characteristic peak or valley occurs in the ampli­tude of the frequency-response curve together with a rapid change of phase angle at the corresponding value of u>. Several examples of this phenomenon are to be seen in the frequency response curves in Figs. 7.14 to 7.18. The reason for this behavior is readily appreciated by putting (7.2,9) in the following form:

G ~ Zi) • (s – z2) ••• (s – zj G – A,) • (s – A2) ••• (x – AJ

where the A, are the characteristic roots (poles) and the г, are the zeros of G(s). Let

(s – zk) = pke’ak (s – k) = гкефк

where p, r, a, /3 are the distances and angles shown in Fig. 7.8b for a point s = іш on the imaginary axis. Then

 (a)

 Figure 7.10 Frequency-response curves—first-order system.

When the singularity is close to the axis, with imaginary coordinate со’ as illustrated for point S on Fig. 1.8b, we see that as со increases through со’, a sharp minimum oc­curs in p or r, as the case may be, and the angle a or /3 increases rapidly through ap­proximately 180°. Thus we have the following cases:

1. For a pole, in the left half-plane, there results a peak in |g| and a reduction in (p of about 180°.

2. For a zero in the left half-plane, there is a valley in g and an increase in <p of about 180°.

3. For a zero in the right half-plane, there is a valley in |g| and a decrease in <p of about 180°.

## Frequency Response

When a stable linear/invariant system has a sinusoidal input, then after some time the transients associated with the starting conditions die out, and there remains only a steady-state sinusoidal response at the same frequency as that of the input. Its ampli­tude and phase are generally different from those of the input, however, and the ex­pression of these differences is embodied in the frequency-response function.

Consider a single input/response pair, and let the input be the sinusoid a, cos cot. We find it convenient to replace this by the complex expression c = A, e,eut, of which ax cos cot is the real part. A, is known as the complex amplitude of the wave. The re­sponse sinusoid can be represented by a similar expression, x = Аге1Ш, the real part of which is the physical response. As usual, x and c are interpreted as rotating vectors whose projections on the real axis give the relevant physical variables (see Fig. 7.8a).

From Table A. 1, item 8, the transform of c is

A,

c = ——–

s — ico

 Figure 7.8 (a) Complex input and response. (b) Effect of singularity close to axis.

The function G(s) is given by (7.2,9) so that

m

X 1 (i – ico)f(s)

The roots of the denominator of the r. h.s. are

A, • • • A„, іш

so that the application of the expansion theorem (A.2,10) yields the complex output

«+1 г (^ – K)N(s) I

*(0 =AX 7———– ■ ,v4

r=l L (S – l0>)f(S) J,= A,

Since we have stipulated that the system is stable, all the roots A, ••• A„ of the charac­teristic equation have negative real parts. Therefore eK, t —» 0 as t —> °° for r = 1 ••• n, and the steady-state periodic solution is

N(iw)

x(t) = At——ela“, t-* oo f(iw)

or

x(t) = AtG(iw)eia“
= А2еш

Thus

A2 = AtG(ito) (7.5,3)

is the complex amplitude of the output, or

Giito) = — (7.5,4)

At

the frequency response function, is the ratio of the complex amplitudes. In general, G{ito) is a complex number, varying with the circular frequency w. Let it be given in polar form by

G{iw) = KMeiv (7.5,6)

where К is the static gain (7.4,5). Then

An.

— = KMe‘v (7.5,7)

Ai

From (7.5,7) we see that the amplitude ratio of the steady-state output to the input is! A2M| = KM: that is, that the output amplitude is a2 = КМаъ and that the phase re-

 Locus of Me* (semicircle)

lation is as shown on Fig. 7.8a. The output leads the input by the angle <p. The quan­tity M, which is the modulus of G(ioj) divided by K, we call the magnification factor, or dynamic gain, and the product KM we call the total gain. It is important to note that M and q> are frequency-dependent.

Graphical representations of the frequency response commonly take the form of either vector plots of Me,<p (Nyquist diagram) or plots of M and cp as functions of fre­quency (Bode diagram). Examples of these are shown in Figs. 7.9 to 7.13.