Question 16.8: Find the peak amplitudes of the first three non-zero terms i...
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 x_0=10 \mathrm{~V} and period T=2 \mathrm{~ms} .
Table 16.2 Fourier coefficients for selected waveforms { }^5
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\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} |
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\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} |
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\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} |
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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)} |
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\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} |
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\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} |
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\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} |
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\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} |
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\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} |
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\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\} |
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\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} |
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\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} |
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\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} |
From Table 16.2 , the Fourier coefficients for a symmetric triangular wave are given by :
\begin{aligned} &X_0=x_{d c}=0 \\ &X_k=x_0 \operatorname{sa}^2\left(\frac{k \pi}{2}\right) ; \quad k=\pm 1, \pm 2, \pm 3, \cdots \end{aligned}
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 |
| X_k(\mathrm{~V}) | 0.000 | 4.053 | 0.000 | 0.45 | 0.000 | 0.162 |
The first three non-zero coefficients are X_1, X_3, X_5 . The peak amplitudes of the terms in the corresponding amplitude-phase series are
\begin{aligned} &A_1=2\left|X_1\right|=8.106 \mathrm{~V}, \\\\ &A_3=2\left|X_3\right|=900 \mathrm{mV}, \\\\ &A_5=2\left|X_5\right|=324 \mathrm{mV} . \end{aligned}