Unit circle function

The literature on ergodic theory contains an interesting theorem concerning the spectrum of the Frobenius-Perron operator P. In order to state this result, we have to reformulate P as an operator on the Hilbert space L P) of all square integrable functions on the phase space P. Since and, therefore, / are volume preserving, this operator P L P) —+ L r) is unitary (cf. [20], Thm. 1.25). As a consequence, its spectrum lies on the unit circle. 

To reduce an angle to the first quadrant of the unit circle, that is, to a degree measure between 0° and 90°, see Table 1-1. For function values at major angle values, see Tables 1-2 and 1-3. Relations between functions and the sum/ difference of two functions are given in Table 1-4. Generally, there will be two angles between 0° and 360° that correspond to the value of a function. 

The same argument implies that the analytic continuation of q x) in the unit circle is meromorphic. The argument is based on the functional equation (4.16) which is equivalent with the continued fraction (8 ). The continuation of q x) is derived from the continuous fraction by a minor variation of the reasoning used above. We note the following shortcut. 

When co = ir/0. ZG(jco) = -k. On the polar plot, the dead time function is a unit circle. 

A sampled-data system is stable if all the roots of its characteristic equation (the poles of its transfer function) lie inside the unit circle in the z plane. 

The stability of this openloop system will depend on the values of the poles of the openloop transfer function. If all the p, lie inside the unit circle, the system is openloop stable. 

We want to find out if the function F( ) = 1 + has any zeros or roots outside the unit circle in the z plane. If it does, the system is closedloop unstable. 

Note that this function has no poles outside the unit circle, so P = 0. 

Exercise 4.34 (SU (2) and the unit quaternions) Recall the functions f, fj and f)i,from Exercise 2.8. Show that the restrictions off, f and fk to the unit circle T are group homomorphisms whose range lies in the unit quaternions. Call their images T, Tj and Tk, respectively. Write the images ofTj, Tj azjJ Tk under the homomorphism explicitly a5 3 x 3 matrices. 

Exercise 4.41 Thought experiment draw the graph ofy = sinx/or x in the interval [—tt, tt]. Now wrap the paper on which the graph is drawn around a cylinder so that the x — axis forms a circle, with the point (rr, 0) meeting the point —tt, 0). What shape does the graph of sin form (Hint consider the restrictions to the unit circle of the functions f and f introduced in Section 4.4.)... 

The left-hand side of this equation is the slope of eie, while the right-hand side represents the same function rotated by 90 degrees, so the tangent line turns out to be perpendicular to the radius vector, therefore forming a circle in the complex plane. Furthermore, since the modulus of e e is always unity, the corresponding circle is the unit circle. Euler s formula then asserts that ... 

For the sake of simplicity, we will consider an analytic function / = f(z) of a single complex variable z. It is well known that except for the integer functions and the meromorphic functions defined in the entire complex plane, such an analytic function usually has a natural domain restricted by a boundary of singularities, over which it cannot be analytically continued. This means that even in the case of complex scaling v = v(rj)—for certain values of 11—the transformation may take the variable rjz outside the domain of analyticity, and the operation is then meaningless. As an example, we may consider the analytic functions / = f(z) which are defined only within the unit circle z < 1, with this circle as natural boundary, and it is then evident that the complex scaling is meaningful only for 7 < 1. 

The normal Fokker-Planck equation for the time evolution of the probability density function on the unit circle in configuration space is then... 

Fig. 4 General synthetic strategy to obtain block copolymers with azobenzene-containing side-groups via polymeranalogous reaction (open circles) protected functional groups, (filled circles) functional groups, (rectangles) azobenzene chromophore units...

If any of the poles of G (z) lie outside the unit circle, the process is unstable. Similarly, for the combination hold-process we examine the location of poles of the combined pulse transfer function... 

From the analysis above we conclude that the function fk(nT) is bounded when the root pk lies inside or on the unit circle in the complex plane (see Figure 29.11), where its magnitude is smaller or equal to unity. Extending these results to all the terms within the braces of eq. (29.27), we conclude that ... 

In this chapter we have learned some of the mathematics and notation of i transforms and developed transfer function descriptions of sampled-data systems. The usefulness of MATLAB in obtaining transfer funetions and step responses has been illustrated. The stability region is the interior of the unit circle in the z plane. 

The numerator of the transfer function tells us that the output is delayed by a factor which is a function of the length of the tube. All these zeros lie at 0, and can be ignored for purposes of determining the frequency response of the transfer function. The poles are evenly spaced on the unit circle at intervals of n/5. The first is at 7i/10 = 0.314, which when converted to a real frequency value with Equation 10.29, gives 500Hz. Subsequent resonances occur every lOOOHz after that, i.e. 1500Hz, 2500Hz and so on. 

If the tube is completely lossless, all the poles will lie on the unit circle, the formants will not be damped and hence have infinite amplitude. The spectrum of this will therefore be extremely spiky and it is this property that gives rise to the name, in that the spectrum appears as a pattern of lines rather than a smooth function. 

Figure 10.20 Plots of a first-order HR filter, with a = 0.8, 0.7, 0.6 and 0.4 (a) time-domain impulse response, (b) magnitude spectrum of frequency response, (c) pole-zero plot and (d) magnitude z-domain transfer function. As the length of decay increases, the frequency response becomes sharper. Because only a single coefficient is used, there will be one pole, which will always lie on the real axis. As i - 1, the impulse response will have no decay and the pole will lie at 1.0. Because the pole Hes on the unit circle, it will he in the frequency response, and hence there will be an infinite value for the frequency at this point in the spectrum.

Figure 4.90 summarizes the separation of the modulated sample temperature into two components, one is in-phase with T, the other, 90° out-of-phase, i.e., the in-phase component is described by a sine function, the out-of-phase curve by a cosine function. The figure also shows the description of the time-dependent T as the sum of the two components. The sketch of the unit circle links the maximum amplitudes of the two components. Furthermore, the standard addition theorem of trigonometric functions in the box at the bottom expresses that T is a phase-shifted sine curve. 

Proposition 7 Consider the Kalman filter mechanism (10.4)-(l 0.5) in steady state, and suppose that the matrix F — FKH has all its eigenvalues inside the unit circle. Then, the functions Cn satisfy the dynamic-programming scheme ... 

Fig. 17 Comparison of the experimental and simulated persistence lengths normalized to the length of one repeat unit, as function of the number-average side chain degree of polymerization Both the simulated and the experimental persistence lengths were derived from the simulated or measured cross-sectional radii of gyration, /fgc, according to Eq. (5) simulation (solid curve), measurements in toluene (red circles and triangles), measurements in cyclohexane black squares)...

The z-domain transfer function is shown to be the ratio of two Mth-order polynomials in z, namely, JV(z) and D z). The values of z for which N z) = 0 are termed the zeros of the filter, whereas those for which D z) = 0 are the poles. The poles of such an FIR filter are at the origin, that is, z = 0, in the z plane. The positions of the zeros are determined by the weighting coefficients, that is, foji, fc = 0,1,..., M. The poles and zeros in the z plane for the simple moving average filter are shown in Fig. 8.96. The zeros, marked with a circle, are coincident with the unit circle, that is, the contour in the z plane for which... 

The frequency response of the HR filter is the value of the system function evaluated on the unit circle on the complex plane, that is, with z = where / varies from 0 to 1, or from -1/2 to 1/2. The variable / represents the digital frequency. For simplicity, we write H f) for H(z) z=expU2nf)- Therefore... 

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