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The Shifting Theorem

one of the elementary building blocks is recognized, except for an additive constant. Thus, suppose that s always appears added to a constant factor a, that is F s + a). Then it is easy to show that the original time function was multiplied by the exponential exp(-af) and [Pg.360]

To prove this, write the transform integral explicitly [Pg.360]


Figure 2 Flow diagram of the DHT with N=8, P=3. Broken lines represent transfer factors -1 while full lines represent unity transfer factor. The crossover boxes perform the sign reversal called for by the shift theorem which also requires the sine and cosine factors Sn, Cn. Figure 2 Flow diagram of the DHT with N=8, P=3. Broken lines represent transfer factors -1 while full lines represent unity transfer factor. The crossover boxes perform the sign reversal called for by the shift theorem which also requires the sine and cosine factors Sn, Cn.
To derive the decomposition formula we require two theorems, the shift theorem and the similarity theorem. The shift theorem states that if/(r) has DHT H(v) then/(r+ a) has DHT... [Pg.183]

A proof of this relation may be found in Bracewell (1978). Note that the spectral variable used in this and the next chapter is the same as that defined in Eqs. (7) and (8). Now consider a spatial distribution /(x) and its Fourier spectrum F(w) that come close to satisfying the equality in Eq. (4). We may take Ax and Aw as measures of the width, and hence the resolution, of the respective functions. To see how this relates to more realistic data, such as infrared spectral lines, consider shifting the peak function /(x) by various amounts and then superimposing all these shifted functions. This will give a reasonable approximation to a set of infrared lines. To discuss quantitatively what is occurring in the frequency domain, note that the Fourier spectrum of each shifted function by the shift theorem is given simply by the spectrum of the unshifted function multiplied by a constant phase factor. The superimposed spectrum would then be... [Pg.267]

The complex susceptibility x( ) yielded by Eq. (9), combined with Eq. (22) when the small oscillation approximation is abandoned, may be calculated using the shift theorem for Fourier transforms combined with the matrix continued fraction solution for the fixed center of oscillation cosine potential model treated in detail in Ref. 25. Thus we shall merely outline that solution as far as it is needed here and refer the reader to Ref. 25 for the various matrix manipulations, and so on. On considering the orientational autocorrelation function of the surroundings ps(t) and expanding the double exponential, we have... [Pg.142]

Now in view of the shift theorem for Fourier transforms as applied to one-sided Fourier transforms [38], namely. [Pg.142]

The complex susceptibility of a ferrofluid in a weak applied field may be written directly from Eqs. (109) and (110) and the Langevin equations (98) and (99) [taking note of Eq. (102)] using the shift theorem for one-sided Eourier transforms, Eq. (30). Thus... [Pg.165]

Hence using the shift theorem for Fourier transforms... [Pg.398]

Using the shift theorem [to -> (to — ajls)] in the latter equation, we have... [Pg.138]

There are several theorems that are useful in obtaining Laplace transforms of various functions. The first is the shifting theorem-. [Pg.183]

The Fourier transform of (7.50) is obtained with the help of the shift theorem (see Appendix B), which shows that the Fourier transform of 8(—z — t) is equal to exp(z ) times the Fourier transform of < (—z). We thus obtain... [Pg.252]

The shift theorem states that if the function f(t) has the Fourier transform F(w), then the function f(t-a) has the transform F(uj) exp(-iu)a). Its derivation is also quite simple. We go through it here for illustration. [Pg.48]

Due to the assumption that the injection concentration profile is a Dirac delta function and the output profile is simply shifted by the retention time, we find, using the shift theorem from Table 16.1 and the transform of the Dirac delta function in... [Pg.441]

The implication of the shift theorem is that all that happens to the transform of a time-shifted signal is a phase shift in the frequency domain, linear in m with slope equal to the amoimt of shift, represented by the Nothing... [Pg.217]

To imderstand the shift theorem in the frequency domain, we can invoke the duality of the time and frequency domains as discussed in Section A.5.1. Instead of saying that we multiplied the sine wave by the window, we could say that we multiplied the window by the sine wave. Doing this shifted the transform of the window from zero frequency up to the frequency of the sinusoid. This is sometimes called heterodyning, or shifting a spectrum by multiplication by a sinusoid in the time domain. [Pg.217]

The shifting theorem could be used to good effect except for the term 1 /s. The transforms in Appendix D shows that (1/5)e equals 7 (2v ), which is directly applicable to the present problem if we can replace s with (5 -l- 1) in the multiplier. To do this, we inspect the process of integration with respect to ... [Pg.447]

The last term can be simplified using the shifting theorem (equation (3.30)) to obtain... [Pg.102]

Here we have in fact derived a rule known as the Shift theorem ... [Pg.397]

It is these additional V-dependent terms which can clearly cause a break-down of the shift theorem which says... [Pg.38]

The following relation is called the shift theorem for convolution with the delta function. [Pg.353]

By applying the shift theorem given in Equation (D25) to each of the delta functions with different n values in Equation (D28), B J v) is finally given as... [Pg.356]

In the descriptions of Fourier self-deconvolution and Figure 6.6, the convolution theorem in Equation (D6b) and the shift theorem for convolution with the delta function in Equation (D25) are repeatedly used. In Equation (6.2), the Lorentz profile centering at V = 0 is shifted to the same profile centering at v = vq by the shift property of the delta function. [Pg.357]

Example 11.8. Use the shifting theorem to obtain the Laplace transform of the function... [Pg.150]

There are several theorems pertaining to Laplace transforms, and we present a few of them without proofs. The shifting theorem implies ... [Pg.150]

In order to model spin dependent reactivity, where reaction is only possible through the singlet channel, the authors make use of the exponential model of the form P(t) = p(l — to calculate the probability of reaction. Here, the P(t) relaxes exponentially towards the asymptotic value of p (the probability of being in a singlet state), p in this expression is the inverse of the spin relaxation time. Using the shift theorem of Laplace transform, the authors derive an expression for v(s) to be... [Pg.58]


See other pages where The Shifting Theorem is mentioned: [Pg.632]    [Pg.710]    [Pg.142]    [Pg.302]    [Pg.48]    [Pg.53]    [Pg.85]    [Pg.96]    [Pg.632]    [Pg.360]    [Pg.366]    [Pg.394]    [Pg.448]    [Pg.476]    [Pg.7]    [Pg.168]   


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