Find the peak amplitudes of the first three non-zero terms in the amplitude phase Fourier series for a symmetric triangular wave having peak amplitude x0=10 V and period T=2 ms .

Question

Find the peak amplitudes of the first three non-zero terms in the amplitude phase Fourier series for a symmetric triangular wave having peak amplitude [latex] x_0=10 \mathrm{~V}[/latex] and period [latex] T=2 \mathrm{~ms}[/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]

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

Accepted Answer

From Table 16.2 , the Fourier coefficients for a symmetric triangular wave are given by :

[latex] \begin{aligned} &X_0=x_{d c}=0 \ &X_k=x_0 \operatorname{sa}^2\left(\frac{k \pi}{2} ight) ; \quad k=\pm 1, \pm 2, \pm 3, \cdots \end{aligned}[/latex]

We can use a spreadsheet (e.g., Excel), a computer program (e.g., Matlab), or a pocket calculator to compute the coefficients. We obtain the values in Table 16.3 :

Table 16.3 See Example 16.8
k	0	1	2	3	4	5
[latex] X_k(\mathrm{~V}) [/latex]	0.000	4.053	0.000	0.45	0.000	0.162

The first three non-zero coefficients are [latex] X_1, X_3, X_5[/latex] . The peak amplitudes of the terms in the corresponding amplitude-phase series are

[latex] \begin{aligned} &A_1=2\left|X_1 ight|=8.106 \mathrm{~V}, \\ &A_3=2\left|X_3 ight|=900 \mathrm{mV}, \\ &A_5=2\left|X_5 ight|=324 \mathrm{mV} . \end{aligned} [/latex]

	$\begin{aligned} &X_k=\frac{x_0 \tau}{T} \operatorname{sa}\left(\frac{k \pi \tau}{T}\right) ; \quad k=0, \pm 1, \pm 2, \cdots \\ &x_{r m s}=\sqrt{\frac{x_0^2 \tau}{T}} \end{aligned}$
	$\begin{aligned} &X_k=\frac{x_0 \tau}{2 T} \mathrm{sa}^2\left(\frac{k \pi \tau}{2 T}\right) ; \quad k=0, \pm 1, \pm 2, \ldots \\ &x_{r m s}=\sqrt{\frac{x_0{ }^2 \tau}{3 T}} \end{aligned}$
	$\begin{aligned} &X_k=\frac{j x_0}{2 k \pi}\left[e^{-j k \pi \tau / T_{-}} \mathrm{sa}\left(\frac{k \pi \tau}{2 T}\right)\right] ; \quad k=\pm 1, \pm 2, \ldots \\ &X_0=\frac{x_0 \tau}{2 T}, \quad x_{r m s}=\sqrt{\frac{x_0^2 \tau}{3 T}} \end{aligned}$
	$X_k=\frac{x_0}{2 T\left(\tau_2-\tau_1\right)}\left[\tau_2^2 \mathrm{sa}^2\left(\frac{k \pi \tau_2}{2 T}\right)-\tau_1^2 \mathrm{sa}^2\left(\frac{k \pi \tau_1}{2 T}\right)\right]$ $k=0, \pm 1, \pm 2, \ldots$ $x_{r m s}=\sqrt{\frac{x_0^2}{3 T}\left(2 \tau_1+\tau_2\right)}$
	$\begin{aligned} &X_k=\frac{x_0 \tau\left(1-e^{-T / \tau}\right)}{T+j 2 k \pi \tau} ; \quad k=\pm 2, \pm 3, \pm 4 \cdots \\\\ &x_{r m s}=\sqrt{\frac{x_0^2 \tau}{2 T}\left(1-e^{-2 T / \tau}\right)} \end{aligned}$
	$\begin{aligned} &X_0=0 \\\\ &X_k=x_0 \mathrm{sa}\left(\frac{k \pi}{2}\right) ; \quad k=\pm 1, \pm 2, \pm 3, \ldots \\\\ &x_{r m s}=x_0 \end{aligned}$
	$\begin{aligned} &X_0=0 \\\\ &X_k=x_0 \mathrm{sa}^2\left(\frac{k \pi}{2}\right) ; \quad k=\pm 1, \pm 2, \pm 3, \cdots \\\\ &x_{r m s}=\sqrt{\frac{x_0^2}{3}} \end{aligned}$
	$\begin{aligned} &X_0=0 \\\\ &X_k=\frac{j x_0(-1)^k}{\pi k} ; \quad k=\pm 1, \pm 2, \pm 3, \cdots \\\\ &x_{r m s}=\sqrt{\frac{x_0^2}{3}} \end{aligned}$
	$\begin{aligned} &X_0=0 \\\\ &X_k=\frac{x_0}{T(T-2 \tau)}\left[(T-\tau)^2 \operatorname{sa}^2\left(\frac{k \pi[T-\tau]}{2 T}\right)-\tau^2 \mathrm{sa}^2\left(\frac{k \pi \tau}{T}\right)\right] \\\\ &k=\pm 1, \pm 2, \cdots \\\\ &x_{r m s}=\sqrt{\frac{x_0^2}{3 T}(T+4 \tau)} \end{aligned}$
	$\begin{aligned} &X_k=\frac{x_1 \tau}{2 T}\left\{\operatorname{sa}\left[\frac{(k+1) \pi \tau}{T}\right]+\operatorname{sa}\left[\frac{(k-1) \pi \tau}{T}\right]\right\}-\frac{\left(x_1-x_0\right) \tau}{T} \operatorname{sa}\left(\frac{k \pi \tau}{T}\right) \\\\ &k=0, \pm 1, \pm 2, \pm 3, \cdots \\\\ &\tau=\frac{T}{\pi} \cos ^{-1}\left(1-\frac{x_0}{x_1}\right) \end{aligned}$ $x_{r m s}=\frac{x_1^2}{2 \pi}\left\{\sin \left(\frac{\pi \tau}{T}\right)\left[\cos \left(\frac{\pi \tau}{T}\right)+4\left(\frac{x_0}{x_1}-1\right)\right]+\frac{\pi \tau}{T}\left[2\left(\frac{x_0}{x_1}\right)^2-4 \frac{x_0}{x_1}+3\right]\right\}$
	$\begin{aligned} \tau=& \frac{T}{\pi} \cos ^{-1}\left(\frac{x_0}{x_1}\right) \\\\ X_0 &=0 \\\\ X_{\pm 1} &=\frac{x_1}{2}-\frac{x_1 \tau}{T}\left[\mathrm{sa}\left(\frac{2 \pi \tau}{T}\right)+1\right]+\frac{2 x_0 \tau}{T} \mathrm{sa}\left(\frac{\pi \tau}{T}\right) \\\\ X_k &=\frac{x_1 \tau\left[(-1)^k-1\right]}{2 T}\left\{\mathrm{sa}\left[\frac{(k+1) \pi}{2}\right]+\mathrm{sa}\left[\frac{(k-1) \pi}{2}\right]-\frac{2 x_0}{x_1} \mathrm{sa}\left(\frac{k \pi \tau}{T}\right)\right\} \\\\ k &=\pm 2, \pm 3, \pm 4 \cdots \\\\ x_{r m s} &=\sqrt{\frac{2 x_0^2 \tau}{T}+\frac{x_1^2}{2 \pi T}\left[\pi(T-2 \tau)-T \sin \left(\frac{2 \pi \tau}{T}\right)\right]} \end{aligned}$
	$\begin{aligned} &X_k=\frac{x_0}{4}\left\{\mathrm{sa}\left[\frac{(k+1) \pi}{2}\right]+\mathrm{sa}\left[\frac{(k-1) \pi}{2}\right]\right\} ; k=0, \pm 1, \pm 2, \cdots \\\\ &x_{r m s}=\sqrt{\frac{x_0^2}{4}} \end{aligned}$
	$\begin{aligned} &X_k=\frac{x_0\left[1+(-1)^k\right]}{4}\left\{\mathrm{sa}\left[\frac{(k+1) \pi}{2}\right]+\mathrm{sa}\left[\frac{(k-1) \pi}{2}\right]\right\} \\\\ &k=0, \pm 1, \pm 2, \cdots \\\\ &x_{r m s}=\sqrt{\frac{x_0^2}{2}} \end{aligned}$

Question 16.8: Find the peak amplitudes of the first three non-zero terms i...

Related Answered Questions

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? ...

Verified Answer:

Obtain the Fourier coefficients for the periodic signal (rectangular pulse train) defined in Fig. 16.1. ...

Verified Answer:

The Fourier coefficients for a certain Fourier series are given by Xk=1/1+jk V;k=0,±1,±2,⋯. Obtain the quadrature form of the series. ...

Verified Answer:

Obtain the Fourier coefficients for a Fourier series expressed in amplitude phase form as x(t)=5+10cos(2πf0t)+5cos(4πf0t+π/4)V ...

Verified Answer:

The Fourier coefficients for a certain Fourier series are given by Xk=1/1+jk V;k=0,±1,±2,⋯ Obtain expressions for the peak amplitudes and relative phases of the amplitude-phase form of the series. ...

Verified Answer:

Choose an appropriate interval of expansion for the signal defined graphically by Fig. 16.3. ...

Verified Answer:

Verified Answer:

Obtain and plot the twosided amplitude and phase spectra for the rectangular pulse train of Example 16.13 k=0,±1,±2,⋯,±8 . ...

Verified Answer:

Obtain and plot the first 12 values of the amplitude spectrum and the phase spectrum for a symmetric rectangular pulse train (see Table 16.2) having pulse duration τ=250μs , period T=1 ms and amplitude V0=500mV ...

Verified Answer:

Refer to Fig. 16.9, where the op amp is ideal and the excitation \nu_SνS is a halfwave rectified 800kHz sinusoid having peak amplitude x0=200mV. (a) Obtain an expression for the Fourier coefficients for the response νL. (b) Calculate the peak amplitudes of the first four non-zero terms in the ...

Verified Answer: