Sine function

In equations (3) and (4) the noise was simulated by sine function 6 = bs, (2nat), where a is the ratio of noise frequency to the carrier frequency of the reference signal. 

The derivation of the D matrix for a given contour is based on first deriving the adiabatic-to-diabatic transformation matrix, A, as a function of s and then obtaining its value at the end of the arbitrary closed contours (when s becomes io). Since A is a real unitary matrix it can be expressed in terms of cosine and sine functions of given angles. First, we shall consider briefly the two special cases with M = 2 and 3. 

Note that we have taken a cosine rather than a sine function for our solution. Substitution of either Eq. (4-2) or the equivalent sine function into Eq. (4-1) gives a true statement (with certain restrictions on co) therefore, both are solutions. Moreover, the sum or difference... 

The difference in exponentials which occurs in Eq. (2.21) is directly related to the hyperbolic sine function... 

Table 2.1 Some Useful Relationships Involving the Hyperbolic Sine and Inverse Hyperbolic Sine Function...

Introduction of the inverse hyperbolic sine function encourages us to take Eq. (2.24) a bit further and derive an expression for 17 itself. Before continuing, let us remember the following ... 

Because the second term in the brackets contains 3v in the sine function, radiation at a frequency which is three times that of the incident radiation is generated. This is referred to as third harmonic generation. The first term in brackets indicates that some radiation of unchanged frequency also results. 

The multiple minima nature of the bending energy and the low barriers for interconversion resemble the torsional energy for organic molecules. An expansion of bend in tenns of cosine or sine functions of the angle is therefore more natural than a... 

Any periodic function can be represented as a series of sine functions having frequencies of co, 2co, 3co, etc. ... 

Mathematical theory shows that any periodic function of time, /(f), can be represented as a series of sine functions having frequencies a>, 2a>, 3ft), 4ft), etc. Function /(f) is represented by the following equation, which is referred to as a Fourier series ... 

Each of these sine functions represents a discrete component of the vibration signature discussed previously. The amplitudes of each discrete component and their phase angles can be determined by integral calculus when the function /(f) is known. Because the subject of integral calculus is beyond the scope of this chapter, the math required to determine these integrals are not presented. A vibration analyzer and its associated software perform this determination using FFT. 

Obtain the momentum equation for an element of boundary layer. If the velocity profile in the laminar region may be represented approximately by a sine function, calculate the boundary-layer thickness in terms of distance from the leading edge of the surface. 

Show that the velocity profile in the neighbourhood of the surface may be expressed as a sine function which satisfies the boundary conditions at the surface and at the outer edge of the boundary layer. 

Obtain the boundary layer thickness and its displacement thickness as a function of the distance from the leading edge of Ihe surface, when the velocity profile is expressed as a sine function. 

Sine-beU An apodization function employed for enhancing resolution in 2D spectra displayed in the absolute-value mode. It has the shape of the first halfcycle of a sine function. 

Each coordination shell contributes a sine function multiplied by an amplitude, which contains the number of neighbors in a coordination shell, Nj, as the most desirable information ... 

Figure 4.11. Left Simulated EXAFS spectrum of a dimer such as Cu2, showing that the EXAFS signal is the product of a sine function and a backscattering amplitude F(k) divided by k, as expressed by Eq. (6). Note that F k)/k remains visible as the envelope around the EXAFS signal xW- Right The Cu EXAFS spectrum of a cluster such as CU2O is the sum of a Cu-Cu and a Cu-O contribution. Fourier analysis is the mathematical tool used to...

Visco-elastic fluids like pectin gels, behave like elastic solids and viscous liquids, and can only be clearly characterized by means of an oscillation test. In these tests the substance of interest is subjected to a harmonically oscillating shear deformation. This deformation y is given by a sine function, [ y = Yo sin ( t) ] by yo the deformation amplitude, and the angular velocity. The response of the system is an oscillating shear stress x with the same angular velocity . 

Before discussing the Fourier transform, we will first look in some more detail at the time and frequency domain. As we will see later on, a FT consists of the decomposition of a signal in a series of sines and cosines. We consider first a signal which varies with time according to a sum of two sine functions (Fig. 40.3). Each sine function is characterized by its amplitude A and its period T, which corresponds to the time required to run through one cycle (2ti radials) of the sine function. In this example the frequencies are 1 and 3 Hz. The frequency of a sine function can be expressed in two ways the radial frequency to (radians per second), which is... 

Fig. 40.3. A composite sine function (solid line) which is the sum of two sine functions of 1 Hz (dotted line) and 3 Hz (dashed line).

The radial frequency co of a periodic function is positive or negative, depending on the direction of the rotation of the unit vector (see Fig. 40.5). co is positive in the counter-clockwise direction and negative in the clockwise direction. From Fig. 40.5a one can see that the amplitudes (A jp) of a sine at a negative frequency, -co, with an amplitude. A, are opposite to the values of a sine function at a positive frequency, co, i.e. = Asin(-cor) = -Asin(co/) = This is a property of an antisymmetric function. A cosine function is a symmetric function because A -Acos(-co/) = Acos(cor) = A. (Fig. 40.5b). Thus, positive as well as negative... 

The approximation is still imperfect and can be improved by adding a third sine function, now with a period 5/(2T ) and an amplitude 1/5, i.e. 

The process of adding sine functions can be continued giving the following expression ... 

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