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The signal $x(t)$ has FTP $T_0=1 \&$ the following Fourier coefficients, $C_k=\left\{\begin{array}{cc}\left(-\frac{1}{3}\right)^k, & k \geq 0 \\ 0, & k<0\end{array}\right.$ Find $x(t)$
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Solution:

$ \mathrm{x}(t)=\sum_{k=-\infty}^{\infty} C_k \mathrm{e}^{j k \omega_0 t}=\sum_{k=0}^{\infty}\left(-\frac{1}{3}\right)^k \mathrm{e}^{j k \omega_0 t}\\ $

$ \begin{aligned}\\ & =\sum_{k=0}^{\infty}\left(-\frac{1}{3} \mathrm{e}^{j \omega_0 t}\right)^k \\\\ & =1+\left(-\frac{1}{3} \mathrm{e}^{j \omega_0 t}\right)+\left(-\frac{1}{3} \mathrm{e}^{j \omega_0 t}\right)^2 \\\\ & =\frac{1}{\left(1+\frac{1}{3} \mathrm{e}^{j \omega_0 t}\right)} \\\\ & =\frac{1}{\left(1+\frac{1}{3} \mathrm{e}^{j 2 \pi t}\right)}\\ \end{aligned} $

$ \begin{aligned} c_n & =\frac{A}{T} \int_0^d e^{-j n \omega_0 t} d t \\\\ & =\left.\frac{A}{T} \frac{1}{-j n \omega_0} e^{-j n \omega_0 t}\right|_0 ^d \\\\ & =\frac{A}{T}\left(\frac{1}{-j n \omega_0} e^{-j n \omega_0 d}-\frac{1}{-j n \omega_0}\right)\\ \end{aligned} $

$ \begin{aligned} = & \frac{A}{T} \frac{1}{j n \omega_0}\left(1-e^{-j n \omega_0 d}\right) \\\\ = & \frac{A}{T} \frac{1}{j n \omega_0} e^{-j n \omega_0 d / 2}\left(e^{j n \omega_0 d / 2}-e^{-j n \omega_0 d / 2}\right) \\\\ = & \frac{A d}{T} \frac{\sin \left(\frac{n \pi d}{T}\right)}{\left(\frac{n \pi d}{T}\right)} e^{-j n \omega_0 d / 2} \end{aligned}\\ $

$ x(t)=\sum_{n=-\infty}^{\infty} c_n e^{j n \omega_0 t} $

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