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I don't see the how the author derived that $\frac { dv }{ dt } \leftrightarrow\overrightarrow { V } j\omega$.

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closed as too localized by Qmechanic Apr 24 '13 at 10:42

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1 Answer 1

(1) $v(t) = V_m\cos(\omega t + \phi) = Re(V_me^{j\omega t}e^{j\phi})$

(2) $\dfrac{dv}{dt}= \omega V_m\cos(\omega t + \phi + \frac{\pi}{2})=Re(\omega V_me^{j\omega t}e^{j\phi}e^{j\frac{\pi}{2}})= Re(j\omega V_me^{j\omega t}e^{j\phi})$

By inspection, comparing the last term in (1) and (2), the phasor corresponding to $v(t)$ is $V_me^{j\phi} = \vec V$ while the phasor corresponding to $\frac{dv}{dt}$ is $j\omega V_me^{j\phi} = j\omega \vec V$

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