Compute by Hand the Continuous Fourier Coefficients

Frequency-domain representation of discrete-time signals

Edmund Lai PhD, BEng , in Practical Digital Signal Processing, 2003

4.1 Discrete Fourier series for discrete-time periodic signals

Any periodic discrete-time signal, with a period of N, can be expressed as a linear combination of N complex exponential functions.

(1) $x\left(n\right)=\sum _{k=0}^{N-1}{c}_{\text{k}}{e}^{\text{j2}\pi \text{kn/N}}$

where

and

Equation 1 is called the discrete-time Fourier series (DTFS).

Given the signal x(n ), the Fourier coefficients can be calculated by

(2) $C_{\text{k}}=\frac{1}{N}\sum _{n=0}^{N-1}x(n)e^{-\text{j}2\pi \text{k}\text{n}/\text{N}}$

Note that these Fourier coefficients are generally complex-valued. They provide a description of the signal in the frequency domain. The coefficient c _k has a magnitude and phase associated with a (normalized) frequency given by

These Fourier coefficients form the (discrete) frequency spectrum of the signal. This normalized frequency can be de-normalized if we know the sampling frequency (F_s ), or

the time lapse (T) between two samples which are related by Since

The denormalized frequency ω takes on values in the range

It is easy to verify that the sequence of coefficients given by equation (2) is periodic with a period of N. This means that the frequency spectrum of a periodic signal is also periodic.

Example: 4.1

Determine the discrete spectrum of a periodic sequence x(n) with a period N=4 given by

$x(n)=\left\{0,1,1,0\right\}\mathrm{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}n}=0,1,2,3$

Solution:

From equation (2), we have

$\begin{array}{l}{c}_k=\frac{1}{4}\sum _{n=0}^{3}x(n)e^{-\text{j}2\text{πkn}/4}\mathrm{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}k}=0,\dots ,3\\ =\frac{1}{4}\left[x(1)e^{-\text{jπk}/2}+x(2)e^{-\text{jπk}}\right]\end{array}$

and

$\begin{array}{l}{c}_0=\frac{1}{4}\left[1+1\right]=\frac{1}{2}\\ c_1=\frac{1}{4}\left[e^{-\text{jπ}/2}+e^{-\text{jπ}}\right]=\frac{1}{4}\left(-1-j\right)\\ c_2=\frac{1}{4}\left[e^{-\text{jπ}}+e^{-\text{j}2\pi }\right]=0\\ c_3=\frac{1}{4}\left[e^{-\text{j}3\pi /2}+e^{-\text{j}3\pi }\right]=\frac{1}{4}\left(-1+j\right)\end{array}$

The magnitude of this discrete spectrum is shown in Figure 4.1.

Figure 4.1. The discrete magnitude squared spectrum of the signal in Example 4.1.

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Fourier Transform

John Semmlow , in Signals and Systems for Bioengineers (Second Edition), 2012

To express the Fourier coefficients in terms of magnitude ( C _m ) and phase (θ _m ) use Equations 3.9 and 3.10. Care must be taken in computing the phase angle to insure that it represents the proper quadrant: ²

$\begin{array}{l}{A}_m=\sqrt{a_m^2+b_m^2}=.187\mathrm{,}\mathrm{.080}\mathrm{,}\mathrm{.054}\mathrm{,}\text{.040}\\ {\theta }_m={\tan }^{-1}\left(\frac{b_m}{a_m}\right)=\text{58}{\mathrm{deg}\mathrm{(}2}^{\text{nd}}\mathrm{)}\mathrm{,}90{\mathrm{deg}\mathrm{(}3}^{\text{rd}}\mathrm{)}\mathrm{,}77{\mathrm{deg}\mathrm{(}2}^{\text{nd}}{\mathrm{)}\mathrm{,}90\mathrm{(}3}^{\text{rd}}\mathrm{)}\mathrm{deg}\\ {\theta }_m=(180-58)=122\mathrm{deg}\text{,}\mathrm{(}-90)=-90\mathrm{deg}\text{,}(180-77)=103\mathrm{deg}\text{,}\mathrm{(}-90)=-90\mathrm{deg}\end{array}$

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Frequency Analysis: The Fourier Series

Luis F. Chaparro , Aydin Akan , in Signals and Systems Using MATLAB (Third Edition), 2019

4.3.4 Reflection and Even and Odd Periodic Signals

If the Fourier series of $x(t)$ , periodic with fundamental frequency ${\mathrm{\Omega }}_0$ , is

$x(t)=\sum _k{X}_k{e}^{jk{\mathrm{\Omega }}_{0}t},$

then the one for its reflected version $x(-t)$ is

(4.26) $x(-t)=\sum _m{X}_m{e}^{-jm{\mathrm{\Omega }}_{0}t}=\sum _k{X}_{-k}{e}^{jk{\mathrm{\Omega }}_{0}t},$

so that the Fourier coefficients of $x(-t)$ are $X_{-k}$ (remember that m and k are just dummy variables). This can be used to simplify the computation of Fourier series of even and odd signals.

For an even signal $x(t)$ , we have $x(t)=x(-t)$ and as such $X_k=X_{-k}$ and therefore $x(t)$ is naturally represented in terms of cosines, and a dc term. Moreover, since in general $X_k=X_{-k}^{⁎}$ then $X_k=X_{-k}^{⁎}=X_{-k}$ , as such these coefficients must be real-valued. Indeed, the Fourier series of $x(t)$ is

(4.27) $\begin{array}{ccc}\hfill x(t)& \hfill =\hfill & X_0+\sum _{k=-\infty }^{-1}{X}_k{e}^{jk{\mathrm{\Omega }}_{o}t}+\sum _{k=1}^{\infty }{X}_k{e}^{jk{\mathrm{\Omega }}_{o}t}=X_0+\sum _{k=1}^{\infty }{X}_k[e^{jk{\mathrm{\Omega }}_{o}t}+e^{-jk{\mathrm{\Omega }}_{o}t}]\hfill \\ \hfill & \hfill =\hfill & X_0+2\sum _{k=1}^{\infty }{X}_k\cos (k{\mathrm{\Omega }}_{0}t).\hfill \end{array}$

That the Fourier series coefficients need to be real-valued when $x(t)$ is even can be shown directly:

$\begin{array}{ccc}\hfill X_k& \hfill =\hfill & \frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}x(t)e^{-jk{\mathrm{\Omega }}_{0}t}dt\hfill \\ \hfill & \hfill =\hfill & \frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}x(t)\cos (k{\mathrm{\Omega }}_{0}t)dt-j\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}x(t)\sin (k{\mathrm{\Omega }}_0)dt\hfill \\ \hfill & \hfill =\hfill & \frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}x(t)\cos (k{\mathrm{\Omega }}_{0}t)dt=X_{-k}\hfill \end{array}$

because $x(t)\sin (k{\mathrm{\Omega }}_{0}t)$ is odd and the corresponding integral is zero, and because the $\cos (k{\mathrm{\Omega }}_{0}t)=\cos (-k{\mathrm{\Omega }}_{0}t)$ .

It will be similar for odd signals for which $x(t)=-x(-t)$ , or $X_k=-X_{-k}$ , in which case the Fourier series has a zero dc value and sine harmonics. The $X_k$ are purely imaginary. Indeed, for $x(t)$ odd

$\begin{array}{ccc}\hfill X_k& \hfill =\hfill & \frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}x(t)e^{-jk{\mathrm{\Omega }}_{0}t}dt\hfill \\ \hfill & \hfill =\hfill & \frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}x(t)[\cos (k{\mathrm{\Omega }}_{0}t)-j\sin (k{\mathrm{\Omega }}_0)]dt\hfill \\ \hfill & \hfill =\hfill & \frac{-j}{T_0}{\int }_{-T_0/2}^{T_0/2}x(t)\sin (k{\mathrm{\Omega }}_{0}t)dt\hfill \end{array}$

since $x(t)\cos (k{\mathrm{\Omega }}_{0}t)$ is odd and their integral is zero. The Fourier series of an odd function can thus be written as

(4.28) $x(t)=2\sum _{k=1}^{\infty }(j{X}_k)\sin (k{\mathrm{\Omega }}_{0}t).$

According to the even and odd decomposition, any periodic signal $x(t)$ can be expressed as

$x(t)=x_e(t)+x_o(t)$

where $x_e(t)$ is the even and $x_o(t)$ is the odd component of $x(t)$ . Finding the Fourier coefficients of $x_e(t)$ , which will be real, and those of $x_o(t)$ , which will be purely imaginary, we would then have $X_k=X_{ek}+X_{ok}$ since

$\begin{array}{ccc}\hfill & \hfill \hfill & x_e(t)=0.5[x(t)+x(-t)]\Rightarrow X_{ek}=0.5[X_k+X_{-k}],\hfill \\ \hfill & \hfill \hfill & x_o(t)=0.5[x(t)-x(-t)]\Rightarrow X_{ok}=0.5[X_k-X_{-k}].\hfill \end{array}$

Reflection: If the Fourier coefficients of a periodic signal $x(t)$ are $\{X_k\}$ then those of $x(-t)$ , the time-reversed signal with the same period as $x(t)$ , are $\{X_{-k}\}$ .

$\underset{\_}{\text{Even periodic signal}x(t):}$ Its Fourier coefficients $X_k$ are real, and its trigonometric Fourier series is

(4.29) $x(t)=X_0+2\sum _{k=1}^{\infty }{X}_k\cos (k{\mathrm{\Omega }}_{0}t).$

$\underset{\_}{\text{Odd periodic signal}x(t):}$ Its Fourier coefficients $X_k$ are imaginary, and its trigonometric Fourier series is

(4.30) $x(t)=2\sum _{k=1}^{\infty }j{X}_k\sin (k{\mathrm{\Omega }}_{0}t).$

Any periodic signal $x(t)$ can be written $x(t)=x_e(t)+x_o(t)$ , where $x_e(t)$ and $x_o(t)$ are the even and the odd components of $x(t)$ then

(4.31) $X_k=X_{ek}+X_{ok}$

where $\{X_{ek}\}$ are the Fourier coefficients of $x_e(t)$ and $\{X_{ok}\}$ are the Fourier coefficients of $x_o(t)$ or

(4.32) $\begin{array}{ccc}\hfill & \hfill \hfill & X_{ek}=0.5[X_k+X_{-k}],\hfill \\ \hfill & \hfill \hfill & X_{ok}=0.5[X_k-X_{-k}].\hfill \end{array}$

Example 4.5

Consider the periodic pulse train $x(t)$ , of fundamental period $T_0=1$ , shown in Fig. 4.4. Find its Fourier series.

Figure 4.4. Train of rectangular pulses.

Solution: Before finding the Fourier coefficients, we see that this signal has a dc component of 1, and that $x(t)-1$ (zero-average signal) is well represented by cosines, given its even symmetry, and as such the Fourier coefficients $X_k$ should be real. Doing this analysis before the computations is important so we know what to expect.

The Fourier coefficients are obtained either using their integral formula or from the Laplace transform of a period. Since $T_0=1$ , the fundamental frequency of $x(t)$ is ${\mathrm{\Omega }}_0=2\pi$ (rad/s). Using the integral expression for the Fourier coefficients we have

$\begin{array}{ccc}\hfill X_k& \hfill =\hfill & \frac{1}{T_0}{\int }_{-T_0/4}^{3T_0/4}x(t)e^{-j{\mathrm{\Omega }}_{0}kt}dt={\int }_{-1/4}^{1/4}2e^{-j2\pi kt}dt\hfill \\ \hfill & \hfill =\hfill & \frac{2}{\pi k}\left[\frac{e^{j\pi k/2}-e^{-j\pi k/2}}{2j}\right]=\frac{\sin (\pi k/2)}{(\pi k/2)},k\ne 0\hfill \\ \hfill X_0& \hfill =\hfill & \frac{1}{T_0}{\int }_{-T_0/4}^{3T_0/4}x(t)dt={\int }_{-1/4}^{1/4}2dt=1,\hfill \end{array}$

which are real as we predicted. The Fourier series is then

$x(t)=\sum _{k=-\infty }^{\infty }\frac{\sin (\pi k/2)}{(\pi k/2)}{e}^{jk2\pi t}.$

To find the Fourier coefficients with the Laplace transform, let the period be $x_1(t)=x(t)$ for $-0.5\le t\le 0.5$ . Delaying it by 0.25 we get $x_1(t-0.25)=2[u(t)-u(t-0.5)]$ with a Laplace transform

$e^{-0.25s}{X}_1(s)=\frac{2}{s}(1-e^{-0.5s})$

so that $X_1(s)=2(e^{0.25s}-e^{-0.25s})/s$ and therefore

$X_k=\frac{1}{T_0}\mathcal{L}[x_1(t)]{|}_{s=jk{\mathrm{\Omega }}_0}=\frac{2}{jk{\mathrm{\Omega }}_0{T}_0}2j\sin (k{\mathrm{\Omega }}_0/4)$

and for ${\mathrm{\Omega }}_0=2\pi$ , $T_0=1$ , we get

$X_k=\frac{\sin (\pi k/2)}{\pi k/2}k\ne 0.$

To find $X_0$ (the above equation gives zero over zero when $k=0$ ) we can use L'Hôpital's rule or use the integral formula as before. As expected, the Fourier coefficients coincide with the ones found before.

The following script is used to find the Fourier coefficients with our function fourierseries and to plot the magnitude and phase line spectra.

Notice that in this case:

1.

The $X_k$ Fourier coefficients of the train of pulses are given in terms of the $\sin (x)/x$ or the sinc function. This function was presented in Chapter 1. Recall the sinc

•

is even, i.e., $\sin (x)/x=\sin (-x)/(-x)$ ,

•

has a value at $x=0$ that is found by means of L'Hôpital's rule because the numerator and the denominator of sinc are zero for $x=0$ , so

$\underset{x\to 0}{\lim }\frac{\sin (x)}{x}=\underset{x\to 0}{\lim }\frac{d\sin (x)/dx}{dx/dx}=1,$

•

is bounded, indeed

$\frac{-1}{x}\le \frac{\sin (x)}{x}\le \frac{1}{x}.$

2.

Since the dc component of $x(t)$ is 1, once it is subtracted it is clear that the rest of the series can be represented as a sum of cosines:

$x(t)=1+\sum _{k=-\infty ,k\ne 0}^{\infty }\frac{\sin (\pi k/2)}{(\pi k/2)}{e}^{jk2\pi t}=1+2\sum _{k=1}^{\infty }\frac{\sin (\pi k/2)}{(\pi k/2)}\cos (2\pi kt).$

3.

In general, the Fourier coefficients are complex and as such need to be represented by their magnitudes and phases. In this case, the $X_k$ coefficients are real-valued, and in particular zero when $k\pi /2=\pm m\pi$ , m an integer (or when $k=\pm 2,\pm 4,\cdots$ ). Since the $X_k$ values are real, the corresponding phase would be zero when $X_k\ge 0$ , and ±π when $X_k<0$ . In Fig. 4.5 we show a period of the signal, and the magnitude and phase line spectra displayed only for positive values of frequency (with the understanding that the magnitude and the phase are even and odd functions of frequency).

Figure 4.5. Top: period of x(t) and real X _k vs. kΩ₀; bottom: magnitude and phase line spectra.

4.

The $X_k$ coefficients and its squares (see Fig. 4.5), which are related to the power line spectrum, are:

$\begin{array}{ccc}\hfill k\hfill & X_k=X_{-k}\hfill & X_k^2\hfill \\ \hfill 0\hfill & 1\hfill & 1\hfill \\ \hfill 1\hfill & 0.64\hfill & 0.41\hfill \\ \hfill 2\hfill & 0\hfill & 0\hfill \\ \hfill 3\hfill & -0.21\hfill & 0.041\hfill \\ \hfill 4\hfill & 0\hfill & 0\hfill \\ \hfill 5\hfill & 0.13\hfill & 0.016\hfill \\ \hfill 6\hfill & 0\hfill & 0\hfill \\ \hfill 7\hfill & -0.09\hfill & 0.008.\hfill \end{array}$

We notice that the dc value and 5 harmonics, or 11 coefficients (including the zero values), provide a very good approximation of the pulse train, and would occupy a bandwidth of approximately $5{\mathrm{\Omega }}_0=10\pi$ (rad/s). The power contribution, as indicated by $X_k^2$ after $k=\pm 6$ is about 3.3% of the signal power. Indeed, the power of the signal is

$P_x=\frac{1}{T_0}{\int }_{-T_0/4}^{3T_0/4}{x}^2(t)dt={\int }_{-0.25}^{0.25}4dt=2$

and the power of the approximation with 11 coefficients is

$X_0^2+2\sum _{k=1}^5|X_k{|}^2=1.9340,$

corresponding to $96.7\text{\%}{P}_x$ . A very good approximation! □

Example 4.6

Consider the periodic signals $x(t)$ and $y(t)$ shown in Fig. 4.6. Determine their Fourier coefficients by using the symmetry conditions and the even-and-odd decomposition. Verify your results for $y(t)$ by computing its Fourier coefficients using their integral or Laplace formulas.

Figure 4.6. Non-symmetric periodic signals.

Solution:

The given signal $x(t)$ is neither even nor odd, but the advanced signal $x(t+0.5)$ is even with a fundamental period of $T_0=2$ , ${\mathrm{\Omega }}_0=\pi$ . Let $z(t)=x(t+0.5)$ and call its period between −1 and 1

$z_1(t)=2[u(t+0.5)-u(t-0.5)],$

so that its Laplace transform is

$Z_1(s)=\frac{2}{s}[e^{0.5s}-e^{-0.5s}],$

which gives the Fourier coefficients

$Z_k=\frac{1}{2}\frac{2}{jk\pi }[e^{jk\pi /2}-e^{-jk\pi /2}]=\frac{\sin (0.5\pi k)}{0.5\pi k}$

after replacing s by $jk{\mathrm{\Omega }}_o=jk\pi$ and dividing by the fundamental period $T_0=2$ . These coefficients are real-valued as corresponding to an even function. The dc coefficient is $Z_0=1$ . Then

$x(t)=z(t-0.5)=\sum _k{Z}_k{e}^{jk{\mathrm{\Omega }}_0(t-0.5)}=\sum _k\underset{X_k}{\underbrace{\left[Z_k{e}^{-jk\pi /2}\right]}}{e}^{jk\pi t}.$

The $X_k$ coefficients are complex since $x(t)$ is neither even nor odd.

The signal $y(t)$ is neither even nor odd, and cannot be made even or odd by shifting. A way to find its Fourier series is to decompose it into even and odd signals. The even and odd components of a period of $y(t)$ are shown in Fig. 4.7. The even and odd components of a period $y_1(t)$ between −1 and 1 are

Figure 4.7. Even and odd components of the period of y(t), −1 ≤t ≤ 1.

$\begin{array}{ccc}\hfill y_{1e}(t)& \hfill =\hfill & \underset{\text{rectangular pulse}}{\underbrace{[u(t+1)-u(t-1)]}}+\underset{\text{triangular pulse}}{\underbrace{[r(t+1)-2r(t)+r(t-1)]}},\hfill \\ \hfill y_{1o}(t)& \hfill =\hfill & \underset{\text{triangular pulse}}{\underbrace{[r(t+1)-r(t-1)-2u(t-1)]}}-\underset{\text{rectangular pulse}}{\underbrace{[u(t+1)-u(t-1)]}}\hfill \\ \hfill & \hfill =\hfill & r(t+1)-r(t-1)-u(t+1)-u(t-1).\hfill \end{array}$

Thus the mean value of $y_e(t)$ is the area under $y_{1e}(t)$ divided by 2 or 1.5. For $k\ne 0$ , $T_0=2$ and ${\mathrm{\Omega }}_0=\pi$ ,

$\begin{array}{ccc}\hfill Y_{ek}& \hfill =\hfill & \frac{1}{T_0}{Y}_{1e}(s){|}_{s=jk{\mathrm{\Omega }}_0}=\frac{1}{2}{[\frac{1}{s}(e^s-e^{-s})+\frac{1}{s^2}(e^s-2+e^{-s})]}_{s=jk\pi }\hfill \\ \hfill & \hfill =\hfill & \frac{\sin (k\pi )}{\pi k}+\frac{1-\cos (k\pi )}{{(k\pi )}^2}=0+\frac{1-\cos (k\pi )}{{(k\pi )}^2}=\frac{1-{(-1)}^k}{{(k\pi )}^2}k\ne 0,\hfill \end{array}$

which are real as expected. The mean value of $y_o(t)$ is zero, and for $k\ne 0$ , $T_0=2$ and ${\mathrm{\Omega }}_0=\pi$ ,

$\begin{array}{ccc}\hfill Y_{ok}& \hfill =\hfill & \frac{1}{T_0}{Y}_{1o}(s){|}_{s=jk{\mathrm{\Omega }}_0}=\frac{1}{2}{[\frac{e^s-e^{-s}}{s^2}-\frac{e^s+e^{-s}}{s}]}_{s=jk\pi }\hfill \\ \hfill & \hfill =\hfill & -j\frac{\sin (k\pi )}{{(k\pi )}^2}+j\frac{\cos (k\pi )}{k\pi }=0+j\frac{\cos (k\pi )}{k\pi }=j\frac{{(-1)}^k}{k\pi }k\ne 0,\hfill \end{array}$

which are purely imaginary as expected.

Finally, the Fourier series coefficients of $y(t)$ are

$Y_k=\{\begin{array}{cc}{Y}_{e0}+Y_{o0}=1.5+0=1.5\hfill & k=0,\hfill \\ Y_{ek}+Y_{ok}=(1-{(-1)}^k)/{(k\pi )}^2+j{(-1)}^k/(k\pi )\hfill & k\ne 0.\hfill \end{array}\text{\hspace{1em}□}$

Example 4.7

Find the Fourier series of the full-wave rectified signal $x(t)=|\cos (\pi t)|$ shown in Fig. 4.8. This signal is used in the design of dc sources; the rectification of an ac signal is the first step in this design.

Figure 4.8. Full-wave rectified signal x(t) and one of its periods x ₁(t).

Solution: Given that $T_0=1$ , ${\mathrm{\Omega }}_0=2\pi$ , the Fourier coefficients are given by

$X_k={\int }_{-0.5}^{0.5}\cos (\pi t)e^{-j2\pi kt}dt,$

which can be computed by using Euler's identity. We would like to show that it can easily be done using the Laplace transform.

A period $x_1(t)$ of $x(t)$ can be expressed as

$x_1(t-0.5)=\sin (\pi t)u(t)+\sin (\pi (t-1))u(t-1)$

(show it graphically!) and using the Laplace transform we have

$X_1(s)e^{-0.5s}=\frac{\pi }{s^2+{\pi }^2}[1+e^{-s}],$

so that

$X_1(s)=\frac{\pi }{s^2+{\pi }^2}[e^{0.5s}+e^{-0.5s}].$

For $T_0=1$ and ${\mathrm{\Omega }}_0=2\pi$ , the Fourier coefficients are then

(4.33) $X_k=\frac{1}{T_0}{X}_1(s){|}_{s=j{\mathrm{\Omega }}_{0}k}=\frac{\pi }{{(j2\pi k)}^2+{\pi }^2}2\cos (\pi k)=\frac{2{(-1)}^k}{\pi (1-4k^2)},$

since $\cos (\pi k)={(-1)}^k$ . From Equation (4.33), the dc value of the full-wave rectified signal is $X_0=2/\pi$ . Notice that the Fourier coefficients are real given the signal is even. The phase of $X_k$ is zero when $X_k>0$ , and π (or −π) when $X_k<0$ .

A similar MATLAB script to the one in Example 4.5 can be written. In this case we only plot the spectra for positive frequencies—the rest are obtained by the symmetry of the spectra. The results are shown in Fig. 4.9. □

Figure 4.9. Period of full-wave rectified signal x(t) and its magnitude and phase line spectra.

Example 4.8

Computing the derivative of a periodic signal enhances the higher harmonics of its Fourier series. To illustrate this consider the train of triangular pulses $y(t)$ (see left figure in Fig. 4.10) with fundamental period $T_0=2$ . Let $x(t)=dy(t)/dt$ (see right figure in Fig. 4.10), find its Fourier series and compare $|X_k|$ with $|Y_k|$ to determine which of these signals is smoother, i.e., which one has lower-frequency components.

Figure 4.10. Train of triangular pulses y(t) and its derivative x(t). Notice that y(t) is a continuous function while x(t) is discontinuous.

Solution: A period of $y(t)$ , $-1\le t\le 1$ , is given by $y_1(t)=r(t+1)-2r(t)+r(t-1)$ with a Laplace transform

$Y_1(s)=\frac{1}{s^2}[e^s-2+e^{-s}],$

so that the Fourier coefficients are given by ( $T_0=2$ , ${\mathrm{\Omega }}_0=\pi$ ):

$Y_k=\frac{1}{T_0}{Y}_1(s){|}_{s=j{\mathrm{\Omega }}_{o}k}=\frac{1}{2{(j\pi k)}^2}[2\cos (\pi k)-2]=\frac{1-\cos (\pi k)}{{\pi }^2{k}^2}k\ne 0,$

which is equal to

(4.34) $Y_k=0.5{\left[\frac{\sin (\pi k/2)}{(\pi k/2)}\right]}^2$

using the identity $1-\cos (\pi k)=2{\sin }^2(\pi k/2)$ . By observing $y(t)$ we deduce its dc value is $Y_0=0.5$ (verify it!).

Let us then consider the periodic signal $x(t)=dy(t)/dt$ (shown on the right of Fig. 4.10) with a dc value $X_0=0$ . For $-1\le t\le 1$ , its period is $x_1(t)=u(t+1)-2u(t)+u(t-1)$ and

$X_1(s)=\frac{1}{s}[e^s-2+e^{-s}],$

which gives the Fourier series coefficients

(4.35) $X_k=j\frac{{\sin }^2(k\pi /2)}{k\pi /2},$

since $X_k=\frac{1}{2}{X}_1(s){|}_{s=j\pi k}$ .

For $k\ne 0$ we have $|Y_k|=|X_k|/(\pi k)$ , so that as k increases the frequency components of $y(t)$ decrease in magnitude faster than the corresponding ones of $x(t)$ : $y(t)$ is smoother than $x(t)$ . The magnitude line spectrum $|Y_k|$ goes faster to zero than the magnitude line spectrum $|X_k|$ as $k\to \infty$ (see Fig. 4.11).

Figure 4.11. Magnitude and phase spectra of triangular signal y(t) and its derivative x(t). Ignoring the dc values, the magnitudes {|Y _k |} decay faster to zero than the magnitudes {|X _k |}, thus y(t) is smoother than x(t).

Notice that in this case $y(t)$ is even and its Fourier coefficients $Y_k$ are real, while $x(t)$ is odd and its Fourier coefficients $X_k$ are purely imaginary. If we subtract the average of $y(t)$ , the signal $y(t)$ can be clearly approximated as a series of cosines, thus the need for real coefficients in its complex exponential Fourier series. The signal $x(t)$ is zero-average and as such it can be clearly approximated by a series of sines requiring its Fourier coefficients $X_k$ to be imaginary. The following script does the computations and plotting. □

Example 4.9

Integration of a periodic signal, provided it has zero mean, gives a smoother signal. To see this, find and compare the magnitude line spectra of a sawtooth signal $x(t)$ , of fundamental period $T_0=2$ , and its integral

$y(t)={\int }_{-\infty }^{t}x(t)dt$

shown in Fig. 4.12.

Figure 4.12. Sawtooth signal x(t) and its integral y(t). Notice that x(t) is a discontinuous function while y(t) is continuous.

Solution: Before doing any calculations it is important to realize that the integral would not exist if the dc is not zero as it would accumulate as t grows. Indeed, if $x(t)$ had a Fourier series

$\begin{array}{ccc}\hfill & \hfill \hfill & x(t)=X_0+\sum _k{X}_k{e}^{jk{\mathrm{\Omega }}_{0}t}\text{the integral would be}\hfill \\ \hfill & \hfill \hfill & {\int }_{-\infty }^{t}x(\tau )d\tau ={\int }_{-\infty }^t{X}_{0}d\tau +\sum _k{X}_k{\int }_{-\infty }^t{e}^{jk{\mathrm{\Omega }}_0\tau }d\tau \hfill \end{array}$

and the first integral would continuously increase.

Using the following script we can compute the Fourier series coefficients of $x(t)$ and $y(t)$ . A period of $x(t)$ is

$x_1(t)=t\text{w}(t)+(t-2)\text{w}(t-1)0\le t\le 2$

where $\text{w}(t)=u(t)-u(t-1)$ is a rectangular window.

The following script gives the basic code to find the Fourier series of the two signals.

Ignoring the dc components, the magnitudes $\{|Y_k|\}$ of $y(t)$ decay a lot faster to zero than the magnitudes $\{|X_k|\}$ of $x(t)$ as shown in Fig. 4.13. Thus the signal $y(t)$ is smoother than $x(t)$ as $x(t)$ has higher frequency components than $y(t)$ . The discontinuities in $x(t)$ cause its higher frequencies.

Figure 4.13. Periods of the sawtooth signal x(t) and its integral y(t) and their magnitude and phase line spectra.

As we will see in Section 4.5.3, computing the derivative of a periodic signal is equivalent to multiplying its Fourier series coefficients by $j{\mathrm{\Omega }}_{0}k$ , which emphasizes the higher harmonics—differentiation makes the resulting signal rougher. If the periodic signal is zero mean, so that its integral exists, the Fourier coefficients of the integral can be found by dividing them by $j{\mathrm{\Omega }}_{0}k$ so that now the low harmonics are emphasized—integration makes the resulting signal smoother.  □

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Fourier Series

Enrique A. González-Velasco , in Fourier Analysis and Boundary Value Problems, 1995

§2.8 Integration of Fourier Series

In §2.2 the Fourier coefficients were obtained under the assumption that the equality in (1) held and that term by term integration of the series was possible. Once in possession of the Fourier coefficients these hypotheses were no longer necessary and were discarded. The convergence of the Fourier series was established under a different set of conditions. At this moment we have not yet shown whether or not a Fourier series can be integrated term by term. ¹ In what follows we present a particular case of a theorem discovered in 1906 by Lebesgue, which asserts that the Fourier series of an integrable function can always be integrated term by term, even if the Fourier series itself diverges. The particular version that we state has been selected because it is very easy to prove and sufficient for most applications. For a proof of the more general case refer to Exercise 4.11 of Chapter 6.

Theorem 2.8

Let $f:\left(-\pi ,\pi \right)\to ℝ$ be a piecewise continuous function that has Fourier series

$\frac{1}{2}{a}_0+{\displaystyle \sum _{n=1}^{\infty }\left(a_{n}cosnx+b_{n}sinnx\right)}$

on (–π, π). If a and x are points in [–π, π] then

${\displaystyle {\int }_a^{x}f=\frac{1}{2}{a}_0\left(x-a\right)+{\displaystyle \sum _{n=1}^{\infty }{\displaystyle {\int }_a^x\left(a_{n}cos\mathit{nt}+b_{n}sin\mathit{nt}\right)dt,}}}$

whether or not the Fourier series of f converges.

Proof

It is enough to prove the theorem for a =   0. The general case then follows from the equation

${\displaystyle {\int }_a^{x}f={\displaystyle {\int }_0^{x}f-{\displaystyle {\int }_0^{a}f}}}$

and the fact that two convergent series of functions can be subtracted term by term. Then to establish the desired identity it suffices to prove that the Fourier series on (–π, π) of the function $F:\left[-\pi ,\pi \right]\to ℝ$ defined by

$F\left(x\right)={\displaystyle {\int }_0^{x}f-\frac{1}{2}{a}_{0}x}$

is

${\displaystyle \sum _{n=1}^{\infty }{\displaystyle {\int }_0^x\left(a_{n}cos\mathit{nt}+b_{n}sint\right)dt}}$

and that F equals its Fourier series.

First observe that the extension of F to ℝ as a periodic function of period 2π is continuous on ℝ because F is clearly continuous on [–π, π] and because, recalling the definition of a ₀,

$F\left(\pi \right)={\displaystyle {\int }_0^{\pi }f-\frac{1}{2}{a}_0\pi =a_0\pi -{\displaystyle {\int }_{-\pi }^{0}f-\frac{1}{2}{a}_0\pi ={\displaystyle {\int }_0^{-\pi }f+\frac{1}{2}{a}_0\pi =F\left(-\pi \right).}}}$

Also,

$F^{\prime }\left(x\right)=f\left(x\right)-\frac{1}{2}{a}_0$

at every point x at which f is continuous. That is, F' is piecewise continuous on (–π, π), and, by Lemma 2.2, F satisfies the conditions of Theorem 2.2. Hence, F(x) is equal to the sum of its Fourier series on (–π, π):

$F\left(x\right)=\frac{1}{2}{A}_0+{\displaystyle \sum _{n=1}^{\infty }\left(A_{n}cosnx+B_{n}sinnx\right)}$

for any x in ℝ. For n >   0 the Fourier coefficients in this equation can be computed using integration by parts.

$\begin{array}{l}{A}_n=\frac{1}{\pi }{\displaystyle {\int }_{-\pi }^{\pi }F\left(x\right)cosnxdx}\\ =\frac{1}{\pi }\left[{\left.\frac{F\left(x\right)sinnx}{n}\right|}_{-\pi }^{\pi }-\frac{1}{n}{\displaystyle {\int }_{-\pi }^{\pi }{F}^{\prime }}\left(x\right)sinnxdx\right]\\ =-\frac{1}{n\pi }{\displaystyle {\int }_{-\pi }^{\pi }\left[f\left(x\right)-\frac{1}{2}{a}_0\right]sinnxdx}\\ =-\frac{1}{2\pi }{\displaystyle {\int }_{-\pi }^{\pi }f\left(x\right)sinnxdx}\\ =-\frac{b_n}{n}.\end{array}$

Analogously,

$B_n=\frac{1}{\pi }{\displaystyle {\int }_{-\pi }^{\pi }F\left(x\right)sinnxdx=\frac{a_n}{n}.}$

Therefore,

(16) $F\left(x\right)=\frac{1}{2}{A}_0+{\displaystyle \sum _{n=1}^{\infty }\frac{1}{n}\left(a_{n}sinnx-b_{n}cosnx\right).}$

To obtain the value of the remaining coefficient A ₀ we set x =   0 and obtain

(17) $0=F\left(0\right)=\frac{1}{2}{A}_0-{\displaystyle \sum _{n=1}^{\infty }\frac{b_n}{n}.}$

Hence, taking this value of A ₀ to (16),

$F\left(x\right)={\displaystyle \sum _{n=1}^{\infty }\frac{1}{2}\left[a_{n}sinnx+b_n\left(1-cosnx\right)\right]={\displaystyle \sum _{n=1}^{\infty }{\displaystyle {\int }_0^x\left(a_{n}cos\mathit{nt}+b_{n}sin\mathit{nt}\right)dt,}}}$

Q.E.D.

Theorem 2.8 is not very useful for computation because term-by-term integration of ${\displaystyle {\sum }_{n=1}^{\infty }{b}_n}$ sin nt from zero to x leads to the infinite series ${\displaystyle {\sum }_{n=1}^{\infty }{b}_n}/n,$ whose evaluation may not be possible. For computational purposes this result can be restated in the following form, in which, for convenience and without loss of generality, we set a =   0.

Theorem 2.9

Let f be as in Theorem 2.8. Then, for any x in [–π, π],

${\displaystyle {\int }_0^{x}f-\frac{1}{2}{a}_{0}x=\frac{1}{2\pi }{\displaystyle {\int }_{-\pi }^{\pi }\left({\displaystyle {\int }_0^{x}f}\right)dx+{\displaystyle \sum _{n=1}^{\infty }\frac{1}{n}\left(a_{n}sinnx-b_{n}cosnx\right),}}}$

where the right-hand side is the Fourier series on (–π, π) of the function on the left-hand side.

Proof

Use the definition of A ₀ in (16), instead of the value furnished by (17), Q.E.D.

It is possible to state and prove the equivalent of Theorems 2.8 and 2.9 for Fourier series on arbitrary intervals. This is left to Exercise 8.4.

Example 2.14

In the Fourier series expansion of Example 2.9

$x={\displaystyle \sum _{n=1}^{\infty }2\frac{{\left(-1\right)}^{n+1}}{n}sinnx,}$

valid for any x in (–π, π), we have a_n =   0 for all n ≥   0 and b_n =   2(–1) ^{n  +   1}/n for all n ≥   1. Since

$\frac{1}{2\pi }{\displaystyle {\int }_{-\pi }^{\pi }\left({\displaystyle {\int }_0^{x}tdt}\right)dx=\frac{{\pi }^2}{6},}$

Theorem 2.9 gives

$\frac{x^2}{2}=\frac{{\pi }^2}{6}+{\displaystyle \sum _{n=1}^{\infty }2\frac{{\left(-1\right)}^n}{n^2}cosnx.}$

Of course, the right-hand side is the Fourier series of x ²/2 on (–π, π) and it can be integrated again. But now a ₀  = π ²/3 while a_n =   2(–1) ⁿ /n ² and b_n =   0 for all n ≥   1. Using Theorem 2.9 again gives

$\frac{x^3}{6}-\frac{{\pi }^{2}x}{6}={\displaystyle \sum _{n=1}^{\infty }2\frac{{\left(-1\right)}^n}{n^3}sinnx.}$

This procedure can be repeated any number of times and all the expansions so obtained are valid on (–π, π).

Equation (17) clearly shows that, if f is as in Theorem 2.8, then the series

${\displaystyle \sum _{n=1}^{\infty }\frac{b_n}{n}}$

converges. This result is sometimes useful to show that a certain trigonometric series cannot be a Fourier series (as in Definition 2.1 or 2.5). The following example was given by Pierre Fatou (1878–1929) in 1906.

Example 2.15

The series

${\displaystyle \sum _{n=2}^{\infty }\frac{1}{logn}sinnx}$

is not a Fourier series because

${\displaystyle \sum _{n=1}^{\infty }\frac{b_n}{n}={\displaystyle \sum _{n=2}^{\infty }\frac{1}{nlogn}}}$

diverges.

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Signal Analysis in the Frequency Domain

John Semmlow , in Circuits, Signals and Systems for Bioengineers (Third Edition), 2018

3.4.4 The Inverse Discrete Fourier Transform

The inverse Fourier transform performs the reverse Fourier transform and constructs a waveform from its Fourier coefficients. It implements the discrete version of the Fourier synthesis equation, Equation 3.38. It would not be difficult to code Equation 3.38 using a slightly modified version of Example 3.6. But, as usual, it is easier to use MATLAB's inverse Fourier transform routine, ifft. An opportunity to code a direct implementation of Equation 3.38 and show you are as good as MATLAB is provided in one of the problems.

The format of MATLAB's ifft routine is:

  x = ifft(Xf,N);       % Inverse Fourier transform

where Xf is the Fourier coefficients and N is an optional argument limiting the number of coefficients to use. A few other less common options are described in the associated help file.

Example 3.13

The final example uses the Morlet waveform used in Example 3.11, with the waveform initially on the left side of the signal array. We take the Fourier transform, and then modify the resulting spectrum before reconstructing a time domain signal using the inverse Fourier transform. Specifically, we multiply the phase spectrum by 4. In Example 3.11, we saw that a time shift alters only the phase; specifically, a shift that increases time increases the downward slope of the phase spectrum. Since the phase curve in Example 3.11 is linear, multiplying the phase curve by 4 increases the downward slope of the phase curve by 2 and should shift the Morlet waveform to the right, but not change its shape.

Solution: The first part of the program is identical to Example 3.11 with a slight change in the construction of the signal: the Morlet waveform is placed at the beginning of the signal array. This is done just for variety and the visual effect when the waveform is time shifted. The DFT is taken and the magnitude and phase are determined. The phase is left in radians and not unwrapped since we want to convert back using the inverse DFT after modification. The phase spectrum is multiplied by 4 and the complex spectrum is constructed from the combination of modified phase spectrum and unmodified magnitude spectrum.

MATLAB's inverse Fourier transform routine, iff(Y), expects the input, Y, to be in the complex form:

$X_m=a_m+j{b}_m;$

where: a _m   = C _m cos(θ_m ) and b _m   =   −C _m sin(θ_m ) where C_m is the magnitude and θ_m is the modified phase (see Equation E-1, Appendix E).

% Example 3.13 Take the Fourier transform of the Morlet waveform used in Example 3.11 .

%   Modify the phase spectrum then reconstruct the time signal using the inverse

%   Fourier transform.

%

fs =150;               % Assumed sample frequency. Same as Example 3.11 .

N1 = 512;               % Next 4 lines, same as Example 3.11

wo1 = pi∗sqrt(2/log2(2));                   % Const. for wavelet time scale

t1 = ((0:(N1/2)-1)/fs)∗2;                   % Time vector for Morlet wavelet

mor = (exp(-t1.ˆ2).∗ cos(wo1∗t1));         % Generate Morlet wavelet

x = [fliplr(mor), mor, zeros(1,2∗N1)];     % Construct signal

%

N = length(x);                   % Signal number of points

t = (1:N)/fs;                     % Time vector for plotting

f = (0:N-1)∗fs/N;               % Frequency vector for plotting

%

subplot(1,2,1);    

plot(t,x);                         % Plot original signal

  ........labels and axis.......

%

X = fft(x);                       % Signal complex spectrum

x1 = ifft(X);                     % Unmodified inverse Fourier transform

Mag = abs(X);                     % Get magnitude

Phase = angle(X);               % Get phase in radians not unwrapped

Phase1 = 4∗Phase;               % Modify phase spectrum

%

am = Mag.∗cos(Phase1);         % Calculate new real coefficients

bm = Mag.∗sin(Phase1);         % Calculate new imaginary coefficients

Y = am + j∗bm;                   % New complex spectrum

y = ifft(Y);                     % Take modified inverse Fourier transform.

%

subplot(1,2,2); hold on;         % Plot modified and original signal

plot(t,x1);                       % Signal from unmodified spectrum

plot(t,y,':');                   % Signal from modified spectrum

  ........labels and axis.......

Results: The original signal, Figure 3.27A, is the same as that used in Example 3.11 except that the Morlet wavelet occurs at the beginning of the signal. The signal reconstructed from the unmodified complex signal is identical to the original, Figure 3.27B (solid line). The signal constructed from the modified phase spectrum has the same shape as the original, Figure 3.27B (dotted line), but has been shifted in time approximately 5   s. This is expected because increasing the slope of the phase trajectory is equivalent to a time shift.

Figure 3.27. (A) The original signal used in Example 3.13 is similar to that used in Example 3.11 except for the position of the wavelet waveform. (B) The signal reconstructed from the unmodified Fourier transform of the signal in A is the same. The signal reconstructed from the Fourier transform after the phase shift has been modified by increasing its slope is the same but shifted in time.

There are many signal processing algorithms that follow the strategy used in this example; they move signals to the frequency domain, process the resulting spectral curves, and then move the signals back into the time domain. Although this may seem convoluted, the speed of the FFT and inverse Fourier transform makes this approach attractive and often faster than operations in the time domain.

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BIOSIGNAL PROCESSING

Monty A. Escabí PhD , in Introduction to Biomedical Engineering (Second Edition), 2005

Solution

As for Problem 10.6, the Fourier coefficients are obtained by integrating from −1 to 1. Because a single cycle of the square wave signal has nonzero values between −1/2 to +1/2, the integral can be simplified by evaluating it between these limits:

$c_m=\frac{1}{T}{\int }_{T}x(t)e^{-jm{\omega }_{o}t}dt=\frac{1}{2}{\int }_{-1/2}^{1/2}5e^{-jm\pi t}dt=\frac{5}{2}\cdot \frac{e^{-jm\pi t}}{-jm\pi }|\begin{array}{l}1/2\hfill \\ -1/2\hfill \end{array}$

$\frac{5}{2}\cdot \frac{e^{+m\pi /2}-e^{-jm\pi /2}}{jm\pi }=\frac{5}{2}\cdot \frac{sin(m\pi /2)}{m\pi /2}$

Therefore,

$x(t)=\sum _{m=-\infty }^{+\infty }{c}_m{e}^{ik\omega o^t}=\sum _{m=-\infty }^{\infty }\frac{5}{2}\cdot \frac{\text{sin}(m\pi /2)}{m\pi /2}\cdot e^{jm\pi t}$

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Powder X-Ray Diffraction, Applications

Daniel Louër , in Encyclopedia of Spectroscopy and Spectrometry, 1999

List of symbols

A _n , B _n = Fourier coefficients; A ^D = distortion coefficient; A ^S = size coefficient; B = isotropic atomic displacement; d = interlayer spacing; D = crystallite diameter; e = microstrain; f = atomic scattering factor; F = structure factor; h,k,l = Miller indices; I = intensity; Lp = angle-dependent factor; m = Pearson exponent; M ₂₀ = de Wolff figure of merit; m _k = reflection multiplicity; N = site occupation factor; R = goniometer circle radius; R _p = profile factor; s = scale factor; W = weight fraction; Z = number of molecules per unit cell; β = integral breadth; δ = specimen displacement; ɛ = apparent crystallite size; η = pseudo-Voigt mixing factor; θ = diffraction angle; λ = wavelength; ϕ = shape factor; Φ = line-profile function.

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Thermal Engineering of Steel Alloy Systems

T. Lübben , ... H.-W. Zoch , in Comprehensive Materials Processing, 2014

12.09.5.1.1 Connection between Fourier Coefficients of Roundness Plots and Strains

In a first step, it is necessary to have relations between a measured roundness plot and the curvature radius. To distinguish roundness plots and curvature radius, the first are marked with a bar (like $\overline{R}$ ). Furthermore, the radius of the median fiber of the ring will be used for further calculations, which is the average value of outer and inner radius:

[13] $\overline{R}\left(\alpha \right)=\frac{\overline{R}{\left(\alpha \right)}_{\text{outer}}+\overline{R}{\left(\alpha \right)}_{\text{inner}}}{2}$

The average radius makes it possible to describe the overall behavior of the ring by this one variable, whereas the local behavior can be described by the wall thickness W(α):

[14] $W\left(\alpha \right)=\overline{R}{\left(\alpha \right)}_{\text{outer}}-\overline{R}{\left(\alpha \right)}_{\text{inner}}$

The Fourier series of the median fiber can be written as

[15] $\overline{R}\left(\alpha \right)={\overline{R}}_0+\sum _{k=2}^{\mathrm{\infty }}{a}_k·\cos \left(k·\alpha +{\alpha }_k\right)$

where the coefficients a _k and α _k , $k\in N$ are typical ≠ 0. This description is equivalent to eqn [4]. The coefficient of the order k = 1 can be neglected because it describes only a shift of the complete ring. The corresponding curvature radius R(α) can be calculated in polar coordinates by (8)

[16] $\frac{{\overline{R}}^2+2·{\left(\frac{\partial \overline{R}}{\partial \alpha }\right)}^2-\overline{R}·\frac{{\partial }^2\overline{R}}{\partial {\alpha }^2}}{{\left({\overline{R}}^2+{\left(\frac{\partial \overline{R}}{\partial \alpha }\right)}^2\right)}^{3/2}}=\frac{1}{R\left(\alpha \right)}$

In general, $\overline{R}\left(\alpha \right)$ is not the local curvature radius R(α). This is only true for an ideal ring.

The calculation of the curvature in general gives quite complex equations. For simplification, the infinite sum in eqn [15] should be reduced to the constant term ${\overline{R}}_0$ (mean radius) and one additional order k ≥ 2 of the Fourier series. In this case, without any loss of generality α _k can be set to zero. The evaluation of eqn [16] yields

[17] $\frac{1}{R\left(\alpha \right)}=\frac{{\left({\overline{R}}_0+a_k·\cos \left(k\alpha \right)\right)}^2+2·k^2·a_k^2·{\sin }^2\left(k\alpha \right)+k^2·a_k·\cos \left(k\alpha \right)·\left({\overline{R}}_0+a_k·\cos \left(k\alpha \right)\right)}{{\left({\left({\overline{R}}_0+a_k·\cos \left(k\alpha \right)\right)}^2+\left(k^2·a_k^2·{\sin }^2\left(k\alpha \right)\right)\right)}^{3/2}}$

Because α _k is set to zero, $\overline{R}\left(\alpha \right)$ becomes maximal at α = 0 (see eqn [15]). At that position, eqn [17] becomes

[18] $\frac{{\overline{R}}_0+a_k\left(1+k^2\right)}{{\left({\overline{R}}_0+a_k\right)}^2}=\frac{1}{R\left(0\right)}$

Equation [18] can be solved to a _k (only the negative square root of the solution has a physical relevance):

[19] $a_k=\frac{1}{2}R\left(0\right)·\left(1+k^2\right)-{\overline{R}}_0-\sqrt{\frac{1}{4}R{\left(0\right)}^2{\left(1+k^2\right)}^2-{\overline{R}}_0·R\left(0\right)·k^2}$

By this equation, the unknown Fourier coefficient could be calculated if the average radius of the ring and the curvature radius at the position of the maximum radius ( α = 0°) were known. The first value is easily measureable, but the second value is normally unknown. For the calculation of this parameter, two basic types of deformation of a ring have to be considered:

•

bending and

•

isotropic expansion.

12.09.5.1.1.1 Calculation of the Fourier coefficients in the case of pure bending

In case of a bending of a circular segment (Figure 38), an additional deformation that leads to a modified circuit occurs. R ₀ and R represent the curvature radius of the neutral fiber ds ₀ before and after the additional deformation. In this case, R ₀ equals the average radius ${\overline{R}}_0$ . The length of ds ₀ is given by

Figure 38. Change of curvature radius by an additional bending of a circular segment .

Reproduced from Frerichs, F.; Lübben, Th.; Hoffmann, F.; Zoch, H.-W. Ring Geometry as an Important Part of Distortion Potential. Mat.-wiss. u. Werkstofftech. 2012, 43 (1-2), 16–22.

[20] $d{s}_0=R_0·d{\alpha }_0={\overline{R}}_0·d{\alpha }_0=R·d\alpha$

Before the additional deformation occurs, the length of a fiber with a distance of r ′ from ds ₀ is given by

[21] $ds\left({\overline{R}}_0+r^{\prime },d{\alpha }_0\right)=\left({\overline{R}}_0+r^{\prime }\right)·d{\alpha }_0$

After the deformation this fiber has the length

[22] $ds(R+r^{\prime },d\alpha )=(R+r^{\prime })·d\alpha$

The tangential strain of the neutral fiber ds ₀ is 0. In a distance $r^{\prime }$ from the neutral fiber, the tangential strain can be calculated by use of eqns [21] and [22]:

[23] ${\epsilon }_{\tan \text{g}}(R+r^{\prime })=\frac{\left(R+r^{\prime }\right)·d\alpha -\left({\overline{R}}_0+r^{\prime }\right)·d{\alpha }_0}{\left({\overline{R}}_0+r^{\prime }\right)·d{\alpha }_0}$

Using the side condition $d{s}_0={\overline{R}}_0·d{\alpha }_0=R·d\alpha$ , the unknown curvature radius can be calculated from eqn [23]:

[24] $R=\frac{{\overline{R}}_0·r^{\prime }}{({\overline{R}}_0+r^{\prime })·\left({\epsilon }_{\tan \text{g}}\left(R+r^{\prime }\right)+1-\frac{{\overline{R}}_0}{{\overline{R}}_0+r^{\prime }}\right)}$

Further on in this contribution, the considered strains are in all cases tangential strains. Therefore, it will be no longer indicated with the index 'tang.' $\epsilon \left(R+r^{\prime }\right)$ in eqn [24] can be approximated by a Taylor series. Because of ε(R) = 0 eqn [24] can be rewritten as

[25] $R=\frac{{\overline{R}}_0}{\left({\overline{R}}_0+r^{\prime }\right)·\left(\frac{\partial \epsilon }{\partial r^{\prime }}\left(R\right)+\frac{1}{2}\frac{{\partial }^2\epsilon }{\partial r^{\prime 2}}\left(R\right)·r^{\prime }+\mathrm{\dots }+\left(\frac{1}{{\overline{R}}_0+r^{\prime }}\right)\right)}$

In case of ${\lim }_{r^{\text{'}}\to 0}$ eqn [5] becomes

[26] $R=\frac{1}{\left(\frac{\partial \epsilon }{\partial r^{\prime }}\left(R\right)+\frac{1}{{\overline{R}}_0}\right)}$

With this equation for the curvature radius in eqn [19] , the unknown Fourier coefficient can be calculated:

[27] $a_k=\frac{1}{2}\frac{\left(1+k^2\right)}{\left(\frac{\partial {\epsilon }_{0k}}{\partial r^{\prime }}+\frac{1}{{\overline{R}}_0}\right)}-{\overline{R}}_0-\sqrt{\frac{1}{4}\frac{{\left(1+k^2\right)}^2}{{\left(\frac{\partial {\epsilon }_{0k}}{\partial r^{\prime }}+\frac{1}{{\overline{R}}_0}\right)}^2}-\frac{{\overline{R}}_0·k^2}{\left(\frac{\partial {\epsilon }_{0k}}{\partial r^{\prime }}+\frac{1}{{\overline{R}}_0}\right)}}$

Please note that eqn [19] is only valid at α = 0° (see Figure 39 (64), left picture). Therefore, $\frac{\partial {\epsilon }_{0k}}{\partial r^{\prime }}$ is the gradient of the tangential strain at this special position. If $\frac{\partial {\epsilon }_{0k}}{\partial r^{\prime }}·{\overline{R}}_0<0.01$ eqn [27] can be approximated in very good agreement (difference < 1%) with

Figure 39. Strain distribution and local curvature radius R(α) for k = 2. Left: bending; right: isotropic expansion.

Reproduced from Frerichs, F.; Lübben, Th.; Hoffmann, F.; Zoch, H.-W. Distortion of Conical Formed Bearing Rings made of SAE 52100. Mat.-wiss. u. Werkstofftech. 2009, 40 (5-6), 402–407.

[28] $a_k\approx \frac{\frac{\partial {\epsilon }_{0k}}{\partial r^{\prime }}·{\overline{R}}_0^2}{k^2-1}$

By use of eqn [28], the shape of a former ideal cylindrical ring with radius ${\overline{R}}_0$ in the case of pure bending can be calculated by

[29] $\overline{R}\left(\alpha \right)={\overline{R}}_0+\sum _{k=2}^{\mathrm{\infty }}\frac{\frac{\partial {\epsilon }_{0k}}{\partial r^{\prime }}·{\overline{R}}_0^2}{k^2-1}·\cos \left(k·\alpha +{\alpha }_k\right)$

The corresponding distribution of strain gradients is given with very good approximation by

[30] $\frac{\partial \epsilon }{\partial r^{\prime }}\left(\alpha \right)=\sum _{k=0}^{\mathrm{\infty }}\frac{\partial {\epsilon }_{0k}}{\partial r^{\prime }}·\cos \left(k·\alpha +{\alpha }_k\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{a_k·\left(k^2-1\right)}{{\overline{R}}_0^2}·\cos \left(k·\alpha +{\alpha }_k\right)$

The last two equations give the possibility to calculate roundness plots if the strain gradients are known respectively vice versa to calculate the strain gradients if the Fourier coefficients of the roundness plots are known.

12.09.5.1.1.2 Calculation of the Fourier coefficients in the case of isotropic expansion

In the case of an isotropic expansion or tension, the angle dα ₀ remains unchanged (Figure 40). Again the curvature radius before the expansion equals the average radius. Therefore, the tangential strain of the neutral fiber can be calculated by

Figure 40. Change of curvature radius due to isotropic expansion.

Reproduced from Frerichs, F.; Lübben, Th.; Hoffmann, F.; Zoch, H.-W. Ring Geometry as an Important Part of Distortion Potential. Mat.-wiss. u. Werkstofftech. 2012, 43 (1-2), 16–22.

[31] $\epsilon =\frac{ds\left(R\right)-ds\left(R_0\right)}{ds\left(R_0\right)}=\frac{R·d{\alpha }_0-R_0·d{\alpha }_0}{R_0·d{\alpha }_0}=\frac{R-R_0}{R_0}=\frac{R-{\overline{R}}_0}{{\overline{R}}_0}$

The new curvature radius is given by

[32] $R=\left(\epsilon +1\right)·{\overline{R}}_0$

Using this in eqn [19] and keeping in mind that this equation is only valid at α = 0°, the Fourier coefficient a _k can be calculated by

[33] $a_k={\overline{R}}_0\left(\frac{1}{2}\left(1+k^2\right)\left(1+{\epsilon }_{0k}\right)-1-\sqrt{\frac{1}{4}{\left(1+k^2\right)}^2{\left(1+{\epsilon }_{0k}\right)}^2-\left(1+{\epsilon }_{0k}\right)k^2}\right)$

ε _0k is the strain at α = 0° (see Figure 39, right picture). If $\left|{\epsilon }_{0k}\right|$ < 0.005, eqn [33] can be approximated in very good agreement (difference <1%) with

[34] $a_k\approx -\frac{{\epsilon }_{0k}·{\overline{R}}_0}{k^2-1}$

By the use of eqn [34], the shape of a former ideal cylindrical ring with radius ${\overline{R}}_0$ in the case of a pure isotropic strain can be calculated by

[35] $\overline{R}\left(\alpha \right)={\overline{R}}_0+\sum _{k=2}^{\mathrm{\infty }}\frac{{\epsilon }_{0k}·{\overline{R}}_0}{k^2-1}·\cos \left(k·\alpha +{\alpha }_k\right)$

The corresponding distribution of strain is given by

[36] $\epsilon \left(\alpha \right)=\sum _{k=0}^{\mathrm{\infty }}{\epsilon }_{0k}·\cos \left(k·\alpha +{\alpha }_k\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{a_k·\left(k^2-1\right)}{{\overline{R}}_0}·\cos \left(k·\alpha +{\alpha }_k\right)$

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Application of the theory of the electromagnetic field for calculating diffractive gratings

In Computer Design of Diffractive Optics, 2013

6.3.4 Binary dielectric gratings (TE-polarisation)

For the binary grating, the Fourier coefficients c _n ⁽¹⁾ are independent of the variable y. In this case, the field in the modulation zone is described by a system of differential equations of the second order with constant coefficients

(6.125) $\frac{d^2}{d{y}^2}\mathbf{E}\left(y\right)+{\mathbf{A}}^{\mathit{TE}}\cdot \mathbf{E}\left(y\right)=0$

where E (y) is the vector from the functions $E_p\left(y\right),p=-\overline{N,N},A^{\mathrm{TE}}$ is the matrix of the system:

(6.126) ${\mathtt{A}}_{i,j}^{\mathit{TE}}=-k_0^2{\alpha }_i^2{\mathrm{\delta }}_{i-j}+c_{i-j}^{\left(1\right)},i,j=-\overline{N,N}$

The solution of the system of differential equations(6.125) for the boundary condition with the number m in (6.113) in the matrix form [13] is:

(6.127) ${\mathbf{E}}_m^{-}\left(y\right)=cos\left(\sqrt{{\mathbf{A}}^{\mathit{TE}}}\cdot y\right){\mathbf{E}}_m^{-}\left(0\right)+\frac{sin\left(\sqrt{{\mathbf{A}}^{\mathit{TE}}}\cdot y\right)}{\sqrt{{\mathbf{A}}^{\mathit{TE}}}}\frac{\partial {\mathbf{E}}_m^{-}\left(0\right)}{\partial y}$

The matrix representation (6.127) can be used to express the matrices E ₀₁ and E ₁₁ in (6.123). (6.124) by the matrix of the system (6.126) and the boundary conditions(6.113):

(6.128) $\left\{\begin{array}{l}{\mathbf{E}}_{01}=cos\left(\sqrt{{\mathbf{A}}^{\mathit{TE}}}\cdot a\right)\cdot \mathbf{E}+\frac{sin\left(\sqrt{{\mathbf{A}}^{\mathit{TE}}}\cdot a\right)}{{\sqrt{\mathbf{A}}}^{\mathit{TE}}}.{\mathbf{D}}_{\beta },\\ {\mathbf{E}}_{11}=-\sqrt{{\mathbf{A}}^{\mathit{TE}}}sin\left(\sqrt{{\mathbf{A}}^{\mathit{TE}}}\cdot a\right)\cdot \mathbf{E}+cos\left(\sqrt{{\mathbf{A}}^{\mathit{TE}}}\cdot a\right)\cdot {\mathbf{D}}_{\beta },\end{array}\right\}$

where E, D _β is the unit and diagonal matrix with the elements $-i{k}_0{\mathrm{\beta }}_j,j=-\overline{N,N}$ .

The matrices E ₀₁ and E ₁₁ can also be expressed by the matrix exponent in the form identical with the representation (6.101) for TM-polarisation. In fact. the system of 2 N  +   1 second order differential equations(6.125) is replaced by the equivalent system of 4 N +   2 first order differential equations and we obtain

(6.129) $\left(\begin{array}{l}{\mathbf{E}}_{01}\\ {\mathbf{E}}_{11}\end{array}\right)=exp\left({\mathbf{A}}_1^{\mathit{TE}}\cdot a\right)\left(\begin{array}{l}\mathbf{E}\\ \mathbf{DE}\end{array}\right)$

where ${\mathbf{A}}_1^{\mathit{TE}}=\left(\begin{array}{cc}\hfill \mathbf{0}\hfill & \hfill \mathbf{E}\hfill \\ \hfill -{\mathbf{A}}^{\mathit{TE}}\hfill & \hfill \mathbf{0}\hfill \end{array}\right)$ , E is the unit matrix.

As with TM-polarisation. the matrix representations (6.128). (6.129) for the binary grating can be used to solve the diffraction problem on the grating with the continuous profile [5,6]. In fact, approximating the continuous profile by the set of N binary layers and calculating diffraction on the layer using the relationship (6.129) we obtain the coefficients R_n, T_n in the form (6.123). (6.124). The matrices E ₀₁  = E ₀₁ (a) and E ₁₁  = E ₁₁ (a) in (6.123). (6.124) are calculated. as in the case of TM-polarisation. by the multiplicative integral.

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Power spectrum and filtering

Andreas Skiadopoulos , Nick Stergiou , in Biomechanics and Gait Analysis, 2020

5.4 Fourier series

For a given period T , the Fourier coefficients ( α _k and b _k ) of the kth harmonic (k>0) of any waveform x(t), with amplitude A _k and phase θ _k , are calculated using the following Fourier series analysis equations [Eqs. (5.5–5.7)] (Winter, 2009):

(5.5) $a_k=\frac{2}{T}{\int }_0^{T}x\left(t\right)\cos \left(k{\omega }_{0}t\right)dt$

(5.6) $b_k=\frac{2}{T}{\int }_0^{T}x\left(t\right)\sin \left(k{\omega }_{0}t\right)dt$

and for k=0 the 0th harmonic is obtained by the following equation:

(5.7) $a_0=\frac{1}{T}{\int }_0^{T}x\left(t\right)dt$

For reconstruction of the x(t) waveform the following equations [Eqs. (5.8 and 5.9)] are used (Winter, 2009):

(5.8) $A_k=\sqrt{a_k^2+b_k^2}$

(5.9) ${\theta }_k={\tan }^{-1}\left(\frac{b_k}{a_k}\right)$

There is no particular reason why both a cosine and a sine wave are used to synthesize the waveform x(t), and not only a sine wave. Because of the trigonometric property $\cos \left(\theta \right)=\sin \left(\theta +(\pi /2)\right)$ , the sinusoids of Eq. (5.4) could be a set of sines only. If the phase of a sine function is shifted by $\pi /2$ , then it would match that of a cosine function. Thus, the synthesis Eq. (5.4), by means of simple trigonometric manipulations and the help of Eqs. (5.7–5.9) can also be written as follows:

(5.10) $x\left(t\right)=a_0+\underset{k=1}{\overset{\infty }{\sum A_k}}\sin \left(k{\omega }_{0}t+{\theta }_k\right)$

Any given periodic waveform x(t) can also be expressed as a series of exponential functions (Gaydecki, 2005; Peters & Williams, 1998). There is a mathematical equivalence between trigonometric and exponential functions. By using Euler's/De Moivre's formulae (Gaydecki, 2005; Peters & Williams, 1998):

(5.11) $\cos \left(k{\omega }_{0}t\right)\pm j\sin \left(k{\omega }_{0}t\right)={\left(\cos \left({\omega }_{0}t\right)\pm j\sin \left({\omega }_{0}t\right)\right)}^k=e^{\pm \mathit{jk}{\omega }_{0}t}$

that holds for any real number ${\omega }_{0}t$ and integer k, the following relationships are derived, which relate the sine and cosine waves to exponential functions:

(5.12) $\begin{array}{c}\cos \left(k{\omega }_{0}t\right)=\frac{e^{jk{\omega }_{0}t}+e^{-jk{\omega }_{0}t}}{2}\\ \sin \left(k{\omega }_{0}t\right)=\frac{-j\left(e^{jk{\omega }_{0}t}-e^{-jk{\omega }_{0}t}\right)}{2}\end{array}$

where $j=\sqrt{-1}\Rightarrow j^2=-1$ is the imaginary unit. These exponential functions are complex numbers, which are made from "real" and "imaginary" parts. The application of the complex mode is simple as soon as one becomes familiar with the imaginary unit $j$ . Thus, the sine and cosine functions of Eq. (5.4) can be described in the complex mode, as well. The complex exponential expression of the trigonometric form of the Fourier series synthesis given by Eq. (5.4) is presented by the following equation (Gaydecki, 2005):

(5.13) $x\left(t\right)=a_0+\sum _{k=1}^{\infty }{a}_k\left(\frac{e^{jk{\omega }_{0}t}+e^{-jk{\omega }_{0}t}}{2}\right)-j{b}_k\left(\frac{e^{jk{\omega }_{0}t}-e^{-jk{\omega }_{0}t}}{2}\right)\Rightarrow ···\Rightarrow x\left(t\right)=\sum _{k=-\infty }^{\infty }X\left[k\right]e^{jk{\omega }_{0}t}$

It is interesting to note that in the complex mode there are "real" and "imaginary" Fourier coefficients and both negative and positive frequencies. The "real" coefficients give the cosine terms, and the "imaginary" coefficients give the sine terms (Peters & Williams, 1998). The magnitude spectrum of the negative frequencies is the mirrored magnitude spectrum of the positive frequencies, while the coefficients of the "real" component exhibits even symmetry, and the coefficients of the "imaginary" component exhibits odd symmetry respecting the frequency axis. The negative frequencies are a mathematical artifact without any physical meaning that are necessary to compute the Fourier series when the complex form is used (Gaydecki, 2005). Going back to Eq. (5.13), the following complex quantity can be defined (Gaydecki, 2005):

(5.14) $X\left[k\right]=\{\begin{array}{c}\frac{1}{2}\left(a_k-{\mathit{jb}}_k\right),k>0\\ \frac{1}{2}\left(a_k+{\mathit{jb}}_k\right),k<0\\ a_{0,}k=0\end{array}$

where the following relationship holds true:

(5.15) $\left|\frac{1}{2}\left(a_k-{\mathit{jb}}_k\right)\right|=\left|\frac{1}{2}\left(a_k+{\mathit{jb}}_k\right)\right|=\frac{A_k}{2}$

and by substituting Eqs. (5.5) and (5.6) into a _k and b _k in Eq. (5.14, for k>0), and applying Euler's/De Moivre's formulae, Eq. (5.16) can be derived, which corresponds to the exponential formula for the harmonic analysis (Gaydecki, 2005):

(5.16) $X\left[k\right]=\frac{1}{T}{\int }_0^{T}x\left(t\right)e^{-jk{\omega }_{0}t}dt$

The outcome of Eq. (5.16) represents the two-sided frequency spectrum, where the frequency content of a real periodic signal is expressed in multiples of the fundamental frequency f ₀ $={\omega }_0/2\pi$ (Winter, 2009).

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