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electrical_engineering_1:introduction_in_alternating_current_technology [2023/12/20 09:53]
mexleadmin
electrical_engineering_1:introduction_in_alternating_current_technology [2023/12/20 09:55] (aktuell)
mexleadmin
Zeile 476: Zeile 476:
 Up to now, we used the following formula to represent alternating voltages: Up to now, we used the following formula to represent alternating voltages:
  
-$$u(t)= \sqrt{2} \hat{U\cdot \sin (\varphi)$$+$$u(t)= \sqrt{2} U \cdot \sin (\varphi)$$
  
 This is now interpreted as the instantaneous value of a complex vector $\underline{u}(t)$, which rotates given by the time-dependent angle $\varphi = \omega t + \varphi_u$. This is now interpreted as the instantaneous value of a complex vector $\underline{u}(t)$, which rotates given by the time-dependent angle $\varphi = \omega t + \varphi_u$.
Zeile 504: Zeile 504:
 Generally, from now on not only the voltage will be considered as a phasor, but also the current $\underline{I}$ and derived quantities like the impedance $\underline{X}$. \\ Generally, from now on not only the voltage will be considered as a phasor, but also the current $\underline{I}$ and derived quantities like the impedance $\underline{X}$. \\
 Therefore, the known properties of complex numbers from Mathematics 101 can be applied: Therefore, the known properties of complex numbers from Mathematics 101 can be applied:
-  * A multiplication with $j$ equals a phase shift of $+90°$ +  * A multiplication with $j\omega$ equals a phase shift of $+90°$ 
-  * A multiplication with $-j$ equals a phase shift of $-90°$+  * A multiplication with ${{1}\over{j\omega}}$ equals a phase shift of $-90°$
  
 ===== 6.5 Complex Impedance ===== ===== 6.5 Complex Impedance =====
Zeile 556: Zeile 556:
 \\ \\ \\ \\
 The relationship between ${\rm j}$ and integral calculus should be clear:  The relationship between ${\rm j}$ and integral calculus should be clear: 
-  - The derivative of a sinusoidal value - and therefore a phasor - can simply be written as "$\cdot {\rm j}$", \\ which also means a phase shift of $+90°$: \\ \begin{align*}{{\rm d}\over{{\rm d}t}} {\rm e}^{{\rm j}(\omega t + \varphi_x)} = {\rm j} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)}\end{align*} +  - The derivative of a sinusoidal value - and therefore a phasor - can simply be written as "$\cdot {\rm j}\omega$", \\ which also means a phase shift of $+90°$: \\ \begin{align*}{{\rm d}\over{{\rm d}t}} {\rm e}^{{\rm j}(\omega t + \varphi_x)} = {\rm j} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)}\end{align*} 
-  - The integral of a sinusoidal value - and therefore a phasor - can simply be written as "$\cdot (-{\rm j})$", \\ which also means a phase shift of $-90°$.((in general, here the integration constant must be considered. This is however often neglectable since only AC values (without a DC value) are considered.)) <WRAP> +  - The integral of a sinusoidal value - and therefore a phasor - can simply be written as "$\cdot (-{{1}\over{ {\rm j}\omega}})$", \\ which also means a phase shift of $-90°$.((in general, here the integration constant must be considered. This is however often neglectable since only AC values (without a DC value) are considered.)) <WRAP> 
 \begin{align*} \begin{align*}
                      \int {\rm e}^{{\rm j}(\omega t + \varphi_x)}                       \int {\rm e}^{{\rm j}(\omega t + \varphi_x)} 
-  = {{1}\over{\rm j}} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)}  +  = {{1}\over{\rm j\omega}} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)}  
-  =         - {\rm j} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)}+  = -{{\rm j}\over{\omega}} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)}
 \end{align*} \end{align*}
 </WRAP> </WRAP>