Nomogram for the Analysis of Periodic Curves
Abstract Generated abstract
The paper presents a graphical method for determining low-order Fourier components of periodic curves by approximating a given curve with a polygonal line and decomposing it into triangular elements. It derives expressions for the sine and cosine harmonic coefficients of a triangular periodic segment and transforms the required coefficient calculation into a form suitable for a nomogram using the variables nb, nc, and b/c. The method is extended to symmetric and nonsymmetric curves, odd and even harmonics, constant components, higher-order components through subdivision, and estimation of approximation error. Worked examples show that the first few components can be obtained independently and relatively quickly with acceptable accuracy for simple curves.
Full Text
NOMOGRAM FOR THE ANALYSIS OF PERIODIC CURVES
R. A. Voronov
In the analysis of periodic curves, especially in those cases where there is no need to find harmonic components of higher orders (for example, above the 5th or 7th), one may use the method given below, which considerably shortens the entire work.
Suppose that a periodic curve is given which is a triangle having one vertex at the origin and repeating after periods \(2\pi\). One period of such a curve is shown in Fig. 1.
Fig. 1.
Denoting the height of the triangle by \(a\), and the abscissae of the vertices by \(b\) and \(c\), we can express the sides \(OA\) and \(AC\) by the equations:
\[ y_1=\frac{a}{b}x;\qquad y_2=a\frac{c-x}{c-b}. \tag{1} \]
This curve will have components of the \(n\)-th order:
\[ A_n=\frac{1}{\pi}\int_0^{2\pi} y\cdot \sin nx\, dx, \]
\[ B_n=\frac{1}{\pi}\int_0^{2\pi} y\cdot \cos nx\, dx; \tag{2} \]
substituting into the first of these the value of \(y\) from equations (1), we obtain:
\[ A_n=\frac{1}{\pi}\int_0^b \frac{a}{b}x\sin nx\, dx +\frac{1}{\pi}\int_b^c a\frac{c-x}{c-b}\sin nx\, dx, \]
which, after integration and simplification, gives
\[ A_n=\frac{a}{\pi n^2}\cdot\frac{c\sin nb-b\sin nc}{b(c-b)}, \]
or, denoting
\[ \alpha=\frac{1}{\pi n}\cdot\frac{c\sin nb-b\sin nc}{b(c-b)}, \]
we shall have
\[ A_n=\alpha\frac{a}{n}. \tag{3} \]
The coefficient \(\alpha\) may be represented in the form
\[ \alpha=0.318\,\frac{nc\cdot\sin nb-nb\cdot\sin nc}{nb(nc-nb)}, \tag{4} \]
if the values \(b\) and \(c\) are taken in radians, or
\[ \alpha=18.23\,\frac{nc\cdot\sin nb-nb\cdot\sin nc}{nb(nc-nb)}, \tag{4a} \]
if \(b\) and \(c\) are taken in degrees.
From the expressions for the coefficient \(\alpha\) it is clear that it depends exclusively on \(nb\) and \(nc\). In two cases these equations yield an indeterminate form, readily resolved by the general rules. For \(b=c\) we obtain:
\[ \alpha=0.318\cos nc-D\frac{\sin nc}{nc}, \tag{4b} \]
and for \(b=0\)
\[ \alpha=0.318-D\frac{\sin nc}{nc}, \tag{4c} \]
where \(D=0.318\) when the values of \(c\) are in radians, and \(D=18.23\) when the values of \(c\) are in degrees.
If there are several curves and it is required to find the components for the curve representing their sum, then, according to the general rules, it is necessary to find the components for the individual curves and then add the results obtained.
From what has been stated above there follows a method for analyzing curves. The given curve (Fig. 2) is replaced by a broken line, as close to it as possible, and all the obtained points of inflection are connected with the origin of coordinates. From each point vertical straight lines are drawn to the lines belonging to
Fig. 2.
to the next neighboring points and measure their lengths, taking into account the direction (positive downward and negative upward). With such a construction the entire curve is broken up into a series of triangles which, if lowered onto the axis, will have the form shown below (Fig. 2). These triangles can easily be decomposed into components according to equations (3 and 4), after which the components for the entire curve can be found as the sum of the corresponding components of the individual triangles. It is easy to see that there is no need to transfer the triangles; it is sufficient merely to measure the segments of the vertical straight lines (the values of \(a\)) and the projections of the sides onto the axis (the values of \(b\) and \(c\)).
For symmetric curves, i.e., those that have harmonic components only of odd orders, it is sufficient to perform the division into triangles for only one half-period, doubling the result obtained. Using, for odd values of \(n\), the relation
\[ \sin n(\pi-b)=\sin nb, \]
one may connect the points of inflection both with the beginning of the half-period and with its end. It is most convenient to divide the half-period into two equal parts and analyze each quarter of the period separately (see the example).
With such a division, one of the triangles necessarily has \(b=\dfrac{\pi}{2}\) and \(c=\pi\), to which correspond the following values of the coefficient
\[ \begin{aligned} n&=1 \qquad &\alpha_1&=0.405\\ n&=3 \qquad &\alpha_3&=0.135\\ n&=5 \qquad &\alpha_5&=0.081\\ n&=7 \qquad &\alpha_7&=0.058 \end{aligned} \]
The calculation of the coefficient for all other cases, carried out by equations (4), requires a considerable expenditure of time. To facilitate the determination of the values of \(\alpha\), Fig. 3 gives a nomogram constructed
Fig. 3.
necessary for this equation. Since a nomogram for the direct determination of the values of \(a\) from the values of \(nb\) and \(nc\) is obtained with a binary field and is very inconvenient to use, the equation was transformed into the following form
\[ \frac{\alpha}{18.23}\left(1-\frac{b}{c}\right)=k=\frac{\sin nb}{nb}-\frac{\sin nc}{nc} \]
and the nomogram was constructed with an intermediate blind scale \(R\) and an additional scale for the ratios \(\frac{b}{c}\). The scheme for using the nomogram is as follows:
\[ nc \longrightarrow nb \longrightarrow k;\qquad \frac{b}{c}\longrightarrow k \longrightarrow \alpha \]
In some cases the point on the blind scale lies below zero, which corresponds to negative values of \(\alpha\). To find this value it is necessary to interchange \(nb\) and \(nc\), as is done on the nomogram for the example given below.
In the nomogram presented, the values for \(nb\) and \(nc\) are given only up to \(450^\circ\), which makes it possible to carry out the analysis always up to the 5th order, and in most cases even up to the 7th. If it is necessary to obtain components of higher orders, these divisions may be extended to any limit, or the method indicated below of additional subdivision of the period may be used.
Fig. 4.
By the same method and on the same nomogram, the values of the second parts of the components \(B_n\) may be found. Since they are the multipliers in the terms with cosines \(B_n\cos nx\), and for the values
\[ n=1;5;9\ldots \]
there is the relation
\[ B_n\cos nx=B_n\sin n\left(x+\frac{\pi}{2}\right) \]
and for the values \(n=3;7\ldots\)
\[ B_n\cos nx=-B_n\sin n\left(x+\frac{\pi}{2}\right), \]
then for them the entire construction and determination of the coefficients may be repeated, but the origin of coordinates must be shifted back by a quarter of a period to the point \(O_1\) (Fig. 4). The connection of the points of the curve should be made with the points \(O_1\) and \(O_2\), the beginning and end of the new half-period. It is easy to see that, owing to the symmetry of the curves, the segment \(O_1O\) may be replaced by the segment \(O_2I\), but with a change in the sign of the ordinates \(\alpha\). This makes it possible to carry out the entire analysis for the principal half-period, but with the values for the second quarter changed to the opposite sign.
For an example, below is the analysis of a curve whose half-period is shown in Figure 5.
All the values found and the results of the calculations are brought together in the following table.
| a | $\dfrac{b}{c}$ | c | b | $\alpha_1$ | $a\alpha_1$ | 3c | 3b | $\alpha_3$ | $a\alpha_3$ | 5c | 5b | $\alpha_5$ | $a\alpha_5$ | |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| $a_0$ . . . . . | 47 | 0,500 | 180 | 90 | 0,405 | 19,0 | 540 | 270 | −0,135 | −6,5 | 900 | 450 | 0,081 | 3,8 |
| $a_1$ . . . . . | 15 | 0,444 | 90 | 40 | 0,162 | 2,4 | 270 | 120 | 0,358 | 5,4 | 450 | 200 | −0,128 | −1,9 |
| $a_2$ . . . . . | 8 | 0,235 | 40 | 17 | 0,036 | 0,3 | 120 | 51 | 0,254 | 2,0 | 200 | 85 | 0,424 | 3,4 |
| $a_3$ . . . . . | 26 | 0,744 | 90 | 67 | 0,187 | 4,9 | 270 | 201 | 0,205 | 5,3 | 450 | 335 | −0,247 | −6,4 |
| $a_4$ . . . . . | 43 | 0,716 | 67 | 48 | 0,112 | 4,7 | 201 | 144 | 0,376 | 16,2 | 335 | 240 | −0,150 | −6,5 |
| $a_5$ . . . . . | 23 | 0,713 | 48 | 35 | 0,061 | 1,4 | 144 | 105 | 0,344 | 7,9 | 240 | 175 | 0,275 | 6,3 |
| $a_6$ . . . . . | 10 | 0,600 | 35 | 21 | 0,031 | 0,3 | 105 | 63 | 0,225 | 2,2 | 175 | 105 | 0,396 | 4,0 |
| $\sum a\alpha_n$ . . . . . | — | — | — | — | — | 33,0 | — | — | — | 32,5 | — | — | — | 2,7 |
From the sums obtained, $\sum a\alpha_n$, we find the required components
\[ A_1=\frac{2\sum a\alpha_n}{n}=\frac{2\cdot 33,0}{1}=66,0; \]
\[ A_3=\frac{2\cdot 32,5}{3}=21,7;\qquad A_5=\frac{2\cdot 2,7}{5}=1.1 \]
Fig. 5.
The determination of the coefficients $\alpha_1$, $\alpha_3$, and $\alpha_5$ for the second row of the table is shown on the nomogram; moreover, for the fifth order it was necessary to interchange $nb$ and $nc$, which gave a negative value of $\alpha_5$.
To find the components $B_n$, a division into triangles was made in Fig. 6; all the results are likewise brought together in the following table:
Fig. 6.
| \(a\) | \(\dfrac{b}{c}\) | \(c\) | \(b\) | \(\alpha\) | \(a\alpha_1\) | \(3c\) | \(3b\) | \(\alpha_3\) | \(a\alpha_3\) | \(5c\) | \(5b\) | \(\alpha_5\) | \(a\alpha_5\) | |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(a_0\) . . . | 47 | 0 | 50 | 0 | 0,039 | 1,8 | 150 | 0 | 0,256 | 12,0 | 250 | 0 | 0,388 | 18,2 |
| \(a_1\) . . . | 20 | 0,685 | 73 | 50 | 0,128 | 2,6 | 219 | 150 | 0,360 | 7,2 | 365 | 250 | −0,230 | −4,6 |
| \(a_2\) . . . | 23 | 0,727 | 90 | 73 | 0,192 | 4,4 | 270 | 219 | 0,080 | 1,8 | 450 | 365 | −0,192 | −4,4 |
| \(a_\circ\) . . . | −47 | 0 | 23 | 0 | 0,009 | −0,4 | 69 | 0 | 0,071 | −3,3 | 115 | 0 | 0,175 | −8,2 |
| \(a_3\) . . . | −23 | 0,548 | 42 | 23 | 0,042 | −1,0 | 126 | 69 | 0,143 | −3,3 | 210 | 115 | 0,413 | −9,5 |
| \(a_4\) . . . | −19 | 0,764 | 55 | 42 | 0,080 | −1,4 | 165 | 126 | 0,374 | −7,1 | 275 | 210 | 0,095 | −1,8 |
| \(a_5\) . . . | −37 | 0,797 | 69 | 55 | 0,122 | −4,5 | 207 | 165 | 0,337 | −12,5 | 345 | 275 | −0,260 | 9,6 |
| \(a_6\) . . . | −62 | 0,767 | 90 | 69 | 0,188 | −11,7 | 270 | 207 | 0,120 | −7,4 | 450 | 345 | −0,230 | 14,2 |
| \(\Sigma a\alpha_n\) . . . | — | — | — | — | −10,2 | — | — | — | −12,6 | — | — | — | 13,5 |
whence we find
\[ B_n=-\frac{-10,2\cdot 2}{1}=-20,4;\qquad B_3=-\frac{-12,6\cdot 2}{3}=8,6; \]
\[ B_5=\frac{13,5\cdot 2}{5}=5.4 \]
Negative values for \(a\) correspond to the shifted part of the half-period.
This same method and this same nomogram are also suitable for nonsymmetric curves. For the components \(A_n\) and \(B_n\) of odd orders it is necessary to analyze both half-periods separately, then subtracting the obtained results algebraically, since in this case
\[ \sin n(b+\pi)=-\sin nb. \]
For the components \(A_n\) of even orders it is likewise necessary to analyze the whole period, but when summing the results for the second and fourth quarters of the period the sign must be changed to the opposite. For initial triangles having \(c=\pi\) and \(b=\dfrac{\pi}{2}\), the coefficient \(\alpha\) assumes a value equal to zero.
For the components \(B_n\) of even orders, the analysis is carried out in the same way, but with the origin shifted backward by the amount \(\dfrac{m}{n}\cdot\dfrac{\pi}{2}\), where \(m\) is an arbitrary odd integer. If \(m\) is equal to 3 or 7, then the sign of the result obtained must be changed to the opposite.
To find the constant component \(A_0\), it is necessary to compute the area of the entire curve over a whole period, which can be done from the available triangles. The value of this component will be equal to
\[ A_0^\circ=\frac{1}{360}\sum \frac{a(c-b)}{2}=0,00139\sum a(c-b), \]
if \(c\) and \(b\) are taken in degrees.
To find components of higher orders for which the nomogram has no corresponding divisions, one may divide the whole curve into parts containing an integer number of periods of the given component and analyze each such segment separately, then adding the obtain-
... results. Thus, for the 13th component the period may be divided into three parts, for example
\[ 0-\frac{10}{13}\pi;\quad \frac{10}{13}\pi-\frac{20}{13}\pi;\quad \frac{20}{13}\pi-2\pi, \]
which contain 5, 5, and 3 of its periods.
Using the nomogram, one can determine the error resulting from replacing the curve by a broken line. If, for a certain segment of the period (Fig. 7), the area \(defd\) arises in this way, then it may be regarded as the difference of the triangles \(olf\) and \(odl\), having heights \(ll_1\)
Fig. 7.
and \(dd_1\). Finding for them, from the nomogram, the values of the coefficients \(\alpha\), we compute the resulting error from the given segment
\[ \Delta A_n=\alpha_n'(ll_1)-\alpha_n''(dd_1). \]
By summing the errors of all the segments, one can find the total error. Usually this error turns out to be insignificant.
The method described is convenient in that it gives the value of any one of the components entirely independently of the others. For curves of simple form, and when it is necessary to have only the first few components, the results are obtained very quickly and with good accuracy.
Table 1
Tables for constructing the diagram
| b/c | \(x_{\mathrm{mm}}\) | \(y_{\mathrm{mm}}\) | b/c | \(x_{\mathrm{mm}}\) | \(y_{\mathrm{mm}}\) |
|---|---|---|---|---|---|
| 0,00 | 219,0 | 480,0 | 0,50 | 109,5 | 120,0 |
| 05 | 208,1 | 433,2 | 55 | 98,6 | 97,2 |
| 10 | 197,1 | 388,8 | 60 | 87,6 | 76,8 |
| 15 | 186,2 | 346,8 | 65 | 76,7 | 58,8 |
| 20 | 175,2 | 307,2 | 70 | 65,7 | 43,2 |
| 25 | 164,3 | 270,0 | 75 | 54,8 | 30,0 |
| 30 | 153,3 | 235,2 | 80 | 43,8 | 19,2 |
| 35 | 142,4 | 202,8 | 85 | 32,9 | 10,8 |
| 40 | 131,4 | 172,8 | 90 | 21,9 | 4,8 |
| 45 | 120,5 | 145,2 | 95 | 11,0 | 1,2 |
| 0,50 | 109,5 | 120,0 | 1,00 | 0,0 | 0,0 |
Table 2
| $\alpha$ | $X_{\mathrm{mm}}$ | $y_{\mathrm{mm}}$ | $\alpha$ | $X_{\mathrm{mm}}$ | $y_{\mathrm{mm}}$ |
|---|---|---|---|---|---|
| 0,00 | 0 | 0 | 0,20 | 100 | 100 |
| 02 | 10 | 1 | 22 | 110 | 121 |
| 04 | 20 | 4 | 24 | 120 | 144 |
| 06 | 30 | 9 | 26 | 130 | 169 |
| 08 | 40 | 16 | 28 | 140 | 196 |
| 0,10 | 50 | 25 | 0,30 | 150 | 225 |
| 12 | 60 | 36 | 32 | 160 | 256 |
| 14 | 70 | 49 | 34 | 170 | 289 |
| 16 | 80 | 64 | 36 | 180 | 324 |
| 18 | 90 | 81 | 38 | 190 | 361 |
| 0,20 | 100 | 100 | 0,40 | 200 | 400 |
| 42 | 210 | 441 | |||
| 0,44 | 220 | 484 |
Table 3.
| $c^\circ$ | $y_{\mathrm{mm}}$ | $c^\circ$ | $y_{\mathrm{mm}}$ | $c^\circ$ | $y_{\mathrm{mm}}$ | Note |
|---|---|---|---|---|---|---|
| 0 | 349,1 | 180 | 0,0 | 315 | —57,8 | For $b$, all $y$ values are 2 times smaller |
| 15 | 345,3 | 195 | —26,6 | 330 | —30,4 | For $b$, all $y$ values are 2 times smaller |
| 30 | 333,3 | 210 | —47,6 | 345 | —15,0 | For $b$, all $y$ values are 2 times smaller |
| 45 | 314,2 | 225 | —62,8 | 360 | 0,0 | For $b$, all $y$ values are 2 times smaller |
| 60 | 288,7 | 240 | —72,2 | 375 | 13,8 | For $b$, all $y$ values are 2 times smaller |
| 75 | 257,6 | 250 | —75,0 | 390 | 25,6 | For $b$, all $y$ values are 2 times smaller |
| 90 | 222,2 | 255 | —75,8 | 405 | 34,8 | For $b$, all $y$ values are 2 times smaller |
| 105 | 184,0 | 260 | —75,8 | 420 | 41,2 | For $b$, all $y$ values are 2 times smaller |
| 120 | 144,3 | 265 | —75,0 | 435 | 44,4 | For $b$, all $y$ values are 2 times smaller |
| 135 | 104,7 | 270 | —74,2 | 440 | 44,8 | For $b$, all $y$ values are 2 times smaller |
| 150 | 66,7 | 285 | —67,8 | 445 | 44,8 | For $b$, all $y$ values are 2 times smaller |
| 165 | 31,4 | 300 | —57,8 | 450 | 44,4 | For $b$, all $y$ values are 2 times smaller |
Equations by which these tables were computed
For $b/c$
\[ X = 219\left(1-\frac{b}{c}\right) \]
\[ Y = 480\left(1-\frac{b}{c}\right)^2. \]
For $\alpha$
\[ X = 500\alpha \]
\[ Y = 2500\alpha^2 \]
For $c$
\[ X = K \]
\[ Y = 20000\,\frac{\sin c}{c} \]
For $b$
\[ X = 0{,}5K \]
\[ Y = 10000\,\frac{\sin b}{b} \]
$K$—an arbitrary quantity, taken equal to 400 m.
The zero points of all $y$’s must lie on one straight line. In the figure all dimensions have been reduced by a factor of 6.2.