A certain triangular pulse train has period T=4 ms and pulse duration τ=1 ms . How many terms are needed in a finite Fourier series to account for at least 99.9 % of the power the pulse train would cause to be dissipated in a resistive load?

Question

A certain triangular pulse train has period [latex] T=4 \mathrm{~ms} [/latex] and pulse duration [latex] au=1 \mathrm{~ms}[/latex] . How many terms are needed in a finite Fourier series to account for at least 99.9 % of the power the pulse train would cause to be dissipated in a resistive load?

Accepted Answer

The Fourier coefficients of a triangular pulse train are given by

[latex] X_k=\frac{x_0 au}{2 T} \mathrm{sa}^2\left(\frac{k \pi au}{2 T} ight) ; \quad k=0, \pm 1, \pm 2, \cdots [/latex]

From Table 16.2, the mean-squared amplitude of the triangular pulse train is

[latex] x_{r m s}^2=\frac{x_0^2 au}{3 T}=\frac{x_0^2}{12}[/latex]

Table 16.2 Fourier coefficients for selected waveforms[latex] { }^5 [/latex]
	[latex] \begin{aligned} &X_k=\frac{x_0 au}{T} \operatorname{sa}\left(\frac{k \pi au}{T} ight) ; \quad k=0, \pm 1, \pm 2, \cdots \ &x_{r m s}=\sqrt{\frac{x_0^2 au}{T}} \end{aligned} [/latex]
	[latex] \begin{aligned} &X_k=\frac{x_0 au}{2 T} \mathrm{sa}^2\left(\frac{k \pi au}{2 T} ight) ; \quad k=0, \pm 1, \pm 2, \ldots \ &x_{r m s}=\sqrt{\frac{x_0{ }^2 au}{3 T}} \end{aligned} [/latex]
	[latex] \begin{aligned} &X_k=\frac{j x_0}{2 k \pi}\left[e^{-j k \pi au / T_{-}} \mathrm{sa}\left(\frac{k \pi au}{2 T} ight) ight] ; \quad k=\pm 1, \pm 2, \ldots \ &X_0=\frac{x_0 au}{2 T}, \quad x_{r m s}=\sqrt{\frac{x_0^2 au}{3 T}} \end{aligned} [/latex]
	[latex] X_k=\frac{x_0}{2 T\left( au_2- au_1 ight)}\left[ au_2^2 \mathrm{sa}^2\left(\frac{k \pi au_2}{2 T} ight)- au_1^2 \mathrm{sa}^2\left(\frac{k \pi au_1}{2 T} ight) ight][/latex] [latex] k=0, \pm 1, \pm 2, \ldots [/latex] [latex] x_{r m s}=\sqrt{\frac{x_0^2}{3 T}\left(2 au_1+ au_2 ight)}[/latex]
	[latex] \begin{aligned} &X_k=\frac{x_0 au\left(1-e^{-T / au} ight)}{T+j 2 k \pi au} ; \quad k=\pm 2, \pm 3, \pm 4 \cdots \\ &x_{r m s}=\sqrt{\frac{x_0^2 au}{2 T}\left(1-e^{-2 T / au} ight)} \end{aligned} [/latex]
	[latex] \begin{aligned} &X_0=0 \\ &X_k=x_0 \mathrm{sa}\left(\frac{k \pi}{2} ight) ; \quad k=\pm 1, \pm 2, \pm 3, \ldots \\ &x_{r m s}=x_0 \end{aligned} [/latex]
	[latex] \begin{aligned} &X_0=0 \\ &X_k=x_0 \mathrm{sa}^2\left(\frac{k \pi}{2} ight) ; \quad k=\pm 1, \pm 2, \pm 3, \cdots \\ &x_{r m s}=\sqrt{\frac{x_0^2}{3}} \end{aligned}[/latex]
	[latex] \begin{aligned} &X_0=0 \\ &X_k=\frac{j x_0(-1)^k}{\pi k} ; \quad k=\pm 1, \pm 2, \pm 3, \cdots \\ &x_{r m s}=\sqrt{\frac{x_0^2}{3}} \end{aligned} [/latex]
	[latex] \begin{aligned} &X_0=0 \\ &X_k=\frac{x_0}{T(T-2 au)}\left[(T- au)^2 \operatorname{sa}^2\left(\frac{k \pi[T- au]}{2 T} ight)- au^2 \mathrm{sa}^2\left(\frac{k \pi au}{T} ight) ight] \\ &k=\pm 1, \pm 2, \cdots \\ &x_{r m s}=\sqrt{\frac{x_0^2}{3 T}(T+4 au)} \end{aligned} [/latex]
	[latex] \begin{aligned} &X_k=\frac{x_1 au}{2 T}\left\{\operatorname{sa}\left[\frac{(k+1) \pi au}{T} ight]+\operatorname{sa}\left[\frac{(k-1) \pi au}{T} ight] ight\}-\frac{\left(x_1-x_0 ight) au}{T} \operatorname{sa}\left(\frac{k \pi au}{T} ight) \\ &k=0, \pm 1, \pm 2, \pm 3, \cdots \\ & au=\frac{T}{\pi} \cos ^{-1}\left(1-\frac{x_0}{x_1} ight) \end{aligned} [/latex] [latex] x_{r m s}=\frac{x_1^2}{2 \pi}\left\{\sin \left(\frac{\pi au}{T} ight)\left[\cos \left(\frac{\pi au}{T} ight)+4\left(\frac{x_0}{x_1}-1 ight) ight]+\frac{\pi au}{T}\left[2\left(\frac{x_0}{x_1} ight)^2-4 \frac{x_0}{x_1}+3 ight] ight\} [/latex]
	[latex] \begin{aligned} au=& \frac{T}{\pi} \cos ^{-1}\left(\frac{x_0}{x_1} ight) \\ X_0 &=0 \\ X_{\pm 1} &=\frac{x_1}{2}-\frac{x_1 au}{T}\left[\mathrm{sa}\left(\frac{2 \pi au}{T} ight)+1 ight]+\frac{2 x_0 au}{T} \mathrm{sa}\left(\frac{\pi au}{T} ight) \\ X_k &=\frac{x_1 au\left[(-1)^k-1 ight]}{2 T}\left\{\mathrm{sa}\left[\frac{(k+1) \pi}{2} ight]+\mathrm{sa}\left[\frac{(k-1) \pi}{2} ight]-\frac{2 x_0}{x_1} \mathrm{sa}\left(\frac{k \pi au}{T} ight) ight\} \\ k &=\pm 2, \pm 3, \pm 4 \cdots \\ x_{r m s} &=\sqrt{\frac{2 x_0^2 au}{T}+\frac{x_1^2}{2 \pi T}\left[\pi(T-2 au)-T \sin \left(\frac{2 \pi au}{T} ight) ight]} \end{aligned} [/latex]
	[latex] \begin{aligned} &X_k=\frac{x_0}{4}\left\{\mathrm{sa}\left[\frac{(k+1) \pi}{2} ight]+\mathrm{sa}\left[\frac{(k-1) \pi}{2} ight] ight\} ; k=0, \pm 1, \pm 2, \cdots \\ &x_{r m s}=\sqrt{\frac{x_0^2}{4}} \end{aligned} [/latex]
	[latex] \begin{aligned} &X_k=\frac{x_0\left[1+(-1)^k ight]}{4}\left\{\mathrm{sa}\left[\frac{(k+1) \pi}{2} ight]+\mathrm{sa}\left[\frac{(k-1) \pi}{2} ight] ight\} \\ &k=0, \pm 1, \pm 2, \cdots \\ &x_{r m s}=\sqrt{\frac{x_0^2}{2}} \end{aligned}[/latex]

where [latex] x_0[/latex] is the amplitude of each pulse. A finite Fourier series for the triangular pulse train is

[latex] \begin{aligned} \hat{x}(t) &=\frac{x_0 au}{2 T} \sum_{k=-K}^K \mathrm{sa}^2\left(\frac{k \pi au}{2 T} ight) e^{j 2 \pi k f_0 t} \\ &=\frac{x_0}{8} \sum_{k=-K}^K \operatorname{sa}^2\left(\frac{k \pi au}{2 T} ight) e^{j 2 \pi k f_0 t} \end{aligned}[/latex]

From (16.37), the fraction of the meansquared amplitude of the pulse train that is accounted for by the first K harmonics is given by

[latex] \frac{\hat{x}_{r m s}^2}{x_{r m s}^2}=\frac{\sum_{k=-K}^K\left|X_k ight|^2}{\sum_{k=-\infty}^{\infty}\left|X_k ight|^2}=\frac{1}{x_{r m s}^2} \sum_{k=-K}^K\left|X_k ight|^2[/latex] (16.37)

[latex]\begin{aligned} \frac{\hat{x}_{r m s}^2}{x_{r m s}^2} &=\frac{12}{x_0^2} \frac{x_0^2}{64}\left[1+2 \sum_{k=1}^K \mathrm{sa}^4\left(\frac{k \pi}{8} ight) ight] \\ &=\frac{3}{16}\left[1+2 \sum_{k=1}^K \mathrm{sa}^4\left(\frac{k \pi}{8} ight) ight] \end{aligned}[/latex]

A computer calculation yields the values given in Table 16.4 , which shows that the mean-squared amplitude of a finite Fourier series including harmonics through the 12th equals 99.9 % of the mean-squared amplitude of the pulse train.

Table 16.4 Quantities calculated in Example 16.10
k	0	1	2	3	4	5	6	7	8	9	10	11	12
[latex]\left(\hat{x}_{r m s} / x_{r m s} ight)^2 [/latex]	0.188	0.526	0.772	0.914	0.975	0.994	0.997	0.997	0.997	0.997	0.998	0.998	0.999

Figure 16.7 shows graphs of one period of the triangular pulse train and the finite (12harmonic) Fourier series approximation.

[latex] { }^5 ext { In the table, symmetric means even symmetry about the vertical axis. } [/latex]

Question 16.10: A certain triangular pulse train has period T=4 ms and pul...

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