Question For each of the periodic signals shown in Fig. P6.1-1, find the compact trigonometric Fourier series and sketch the amplitude and phase spectra. If either the sine or cosine terms are absent in the Fourier series, explain why. 1 п н д г. ДДДДДД math Аллилла Figure P6.1-1

a) T_{0}=4, \omega_{0}=\frac{2 \pi}{T_{0}}=\frac{2 \pi}{4}=\frac{\pi}{2}Because of quen symmetry, all the sine terms are zero.\begin{array}{l}f(t)=a_{0}+\sum_{n=1}^{\infty}\left[a_{n} \cos \left(\frac{n \pi}{2} t\right)+b_{n} \sin \left(\frac{n \pi}{2} t\right)\right] \\a_{0}=0 \quad\left(b_{y}\right. \text { inspection) } \\a_{n}=\frac{4}{4}\left[\int_{0}^{1} \cos \left(\frac{n \pi}{2} t\right) d t-\int_{1}^{2} \cos \left(\frac{n \pi}{2} t\right) d t\right]=\frac{4}{n \pi} \sin \frac{n \pi}{2}\end{array}Theredore the Fourier series for f(t) isf(t)=\frac{4}{\pi}\left(\cos \frac{\pi t}{2}-\frac{1}{3} \cos \frac{3 \pi t}{2}+\frac{1}{5} \cos \frac{5 \pi t}{2}-\frac{1}{7} \cos \frac{\pi \pi t}{2}+\cdots\right)Here b_{n}=0, and we allaw c_{n} to take negative values, The figure below shows the plat for C_{n}b) T_{\theta}=10 \pi, \omega_{0}=\frac{2 \pi}{T_{0}}=\frac{2 \pi}{10 \pi}=\frac{1}{5}Because of quen symmetry, all the sine tesms are zero.\begin{array}{l}f(t)=a_{0}+\sum_{n=1}^{\infty}\left[a_{n} \cos \left(\frac{n}{5}\right) t+b_{n} \sin \left(\frac{n}{5} t\right)\right] \\a_{0}=\frac{1}{5}\left(b_{y} \text { inspection }\right) \\a_{n}=\frac{2}{10 \pi} \int_{-\pi}^{\pi} \cos \left(\frac{n}{5}\right) t d t=\frac{1}{5 \pi} \times\left.\left(\frac{5}{n}\right) \sin \left(\frac{n}{5}\right) t\right|_{t=-\pi} ^{\pi}=\frac{2}{\pi n} \sin \left(\frac{n \pi}{5}\right)\end{array}b_{n}=\frac{2}{10 \pi} \int_{-\pi}^{\pi} \sin \left(\frac{n}{5} t\right) d t=0 (integrand is an odd function of t ) Hese b_{n}=0, and we allow C_{n} to take neqative values. Note that c_{n}=a_{n} for n=0,1,2,3 \cdots fiquse below staws the ampletude plet of C_{n} and phase spectrum is zero.c)\begin{array}{l}T_{0}=2 \pi \Rightarrow \omega_{0}=\frac{2 \pi}{T_{0}}=\frac{2 \pi}{2 \pi}=1 \\f(t)=a_{0}+\sum_{n=1}^{\infty} a_{n} \cos n t+b_{n} \sin n t\end{array}with a_{0}=0.5 (by inspection)\begin{array}{l} a_{n}=\frac{1}{\pi} \int_{0}^{2 \pi} \frac{t}{2 \pi} \cos n t d t=0, b_{n}=\frac{1}{\pi} \int_{0}^{2 \pi} \frac{t}{2 \pi} \sin n t d t=-\frac{1}{\pi n} \\\text { and } f(t)=0.5-\frac{1}{\pi}\left(\sin t+\frac{1}{2} \sin 2 t+\frac{1}{3} \sin 3 t+\frac{1}{4} \sin 4 t+\cdots\right) \\=0.5+\frac{1}{\pi}\left[\cos \left(t+\frac{\pi}{2}\right)+\frac{1}{2} \cos \left(2 t+\frac{\pi}{2}\right)+\frac{1}{3} \cos \left(3 t+\frac{\pi}{2}\right)+\cdots \cdot\right]\end{array}The reason for vanishing of the cosine terms is that when 0.5 (the d c component) is substracted from f(t), the remaining function has add symmetry. Hence, the Fovrier sezies wowld contain d c and sine terms only. Fiqure above shows the plot for c_{n} and \theta_{n}d) T_{0}=\pi, \omega_{0}=\frac{2 \pi}{T_{0}}=\frac{2 \pi}{\pi}=2 and f(t)=\frac{4}{\pi} tBy inspection a_{0}=0Because of odd symmetry a_{n}=0 \quad(n>0)Hence, the Forrier seyies wovld contain sine terms only.\begin{aligned}b_{n} & =\frac{4}{\pi} \int_{0}^{\pi / 4} \frac{4}{\pi} t \sin 2 n t d t=\frac{2}{\pi n}\left(\frac{2}{\pi n} \sin \frac{\pi n}{2}-\cos \frac{\pi n}{2}\right) \\\Rightarrow f(t) & =\frac{4}{\pi^{2}} \sin 2 t+\frac{1}{\pi} \sin 4 t-\frac{4}{9 \pi^{2}} \sin 6 t-\frac{1}{2 \pi} \sin 8 t+\cdots \\& =\frac{4}{\pi^{2}} \cos \left(2 t-\frac{\pi}{2}\right)+\frac{1}{\pi} \cos \left(4 t-\frac{\pi}{2}\right)+\frac{4}{9 \pi^{2}} \cos \left(6 t+\frac{\pi}{2}\right)+\frac{1}{2 \pi} \cos \left(8 t+\frac{\pi}{2}\right)+\cdots\end{aligned}The plot for C_{n} and \theta_{n} is shown as follows.e)\begin{array}{l}T_{0}=3, \omega_{0}=\frac{2 \pi}{T_{0}}=\frac{2 \pi}{3} \\a_{0}=\frac{1}{3} \int_{0}^{1} t d t=\frac{1}{6} \\a_{n}=\frac{2}{3} \int_{0}^{1} t \cos \frac{2 \pi \pi}{3} t d t=\frac{3}{2 \pi^{2} n^{2}}\left[\cos \frac{2 \pi n}{3}+\frac{2 \pi n}{3} \sin \frac{2 \pi n}{3}-1\right] \\b_{n}=\frac{2}{3} \int_{0}^{1} t \sin \frac{2 \pi \pi}{3} t d t=\frac{3}{2 \pi^{2} n^{2}}\left[\sin \frac{2 \pi n}{3}-\frac{2 \pi n}{3} \cos \frac{2 \pi n}{3}\right]\end{array}Therefore C_{0}=\frac{1}{6}, C_{n}=\frac{3}{2 \pi^{2} n^{2}}\left[\sqrt{2+\frac{4 \pi^{2} n^{2}}{9}-2 \cos \frac{2 \pi n}{3}-\frac{4 \pi n}{3} \sin \frac{2 \pi n}{3}}\right] and \theta_{n}=\tan ^{-1}\left(\frac{\frac{2 \pi n}{3} \cos \frac{2 \pi n}{3}-\sin \frac{2 \pi n}{3}}{\cos \frac{2 \pi n}{3}+\frac{2 \pi n}{3} \sin \frac{2 \pi n}{3}-1}\right)T_{0}=6, w_{0}=\pi / 3, a_{0}=0.5 (by inspection) Even symmetry; b_{n}=0\begin{aligned}a_{n} & =\frac{4}{6} \int_{0}^{3} f(t) \cos \frac{n \pi}{3} d t=\frac{2}{3}\left[\int_{0}^{1} \cos \frac{n \pi}{3} d t+\int_{1}^{2}(2-t) \cos \left(\frac{n \pi}{3}\right) t d t\right] \\& =\frac{6}{\pi^{2} \pi^{2}}\left[\cos \frac{n \pi}{3}-\cos \frac{2 n \pi}{3}\right] \\f(t) & =0.5+\frac{6}{\pi^{2}}\left(\cos \frac{\pi}{3} t-\frac{2}{9} \cos \pi t+\frac{1}{25} \cos \frac{5 \pi}{3} t+\frac{1}{49} \cos \frac{7 \pi}{3} t+\cdots\right)\end{aligned}obsesve that even harmonics vanish. The reason is that if the d c(0.5) is substracted from f(t), the resulting function has half-wave symmetry. Figure shows the plot for Cn ...