Chapter 8: Problem 11
Show that the average power transmitted to a load impedance \(\breve{Z}_{i}\) is given by \(P=\frac{1}{2 Z_{0}}\left(\left|\ddot{v}_{+}\right|^{2}-\left|\tilde{v}_{-}\right|^{2}\right)\) \(=\frac{1}{2 Z_{0}}\left|{v}_{\max }\right|\left|{v}_{\operatorname{mis}}\right|\) \(-\frac{1}{2 Z_{0}} \frac{\left|\ddot{v}_{\max }\right|^{2}}{\text { VSWR }}=\frac{1}{2 Z_{0}}\left|\tilde{v}_{\operatorname{mia}}\right|^{\top} V S W R\), where \(\left|\tilde{v}_{\max }\right|\) and \(\left|\tilde{\theta}_{\min }\right|\) are the amplitudes of the voltage at maxima and minima of the standingwave pattern.
Short Answer
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Key Concepts
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