Good Day Experts,
I am studying single phase uncontrolled rectifier bridge. One thing that I am confused about is whether the fundamental component of voltage and current (as well as non-fundamental component) are both an inherent property of the ac source, certain connected load, or both the ac source and the load? Is it true that in my duplex residential receptacle (120Vrms ~ 170Vpeak) there is also this presence of fundamental component and non-fundamental component? In my understanding, the source is a sinusoidal function which can be broken down into sum of infinitely many terms. The greater quantity of terms you sum, the better the approximation to the original sinusoidal function. Please confirm if my calculation of fundamental component of source current is correct. If it is, what does f(x) represent in equation below?
\[f(x)=\sum_{n=1}^{inf}b_{n}sin(nx)dx\]
\[b_{1}=\frac{2}{\pi}\int_{0}^{\pi}f(x)sin(nx)dx\]
\[b_{1}=\frac{2}{\pi}\int_{0}^{\pi}I_{0}sin(\omega t)d\omega t= -\frac{2I_{0}}{\pi}cos(\omega t )\mid _{0}^{\pi}=\frac{4I_{0}}{\pi}\]
I am studying single phase uncontrolled rectifier bridge. One thing that I am confused about is whether the fundamental component of voltage and current (as well as non-fundamental component) are both an inherent property of the ac source, certain connected load, or both the ac source and the load? Is it true that in my duplex residential receptacle (120Vrms ~ 170Vpeak) there is also this presence of fundamental component and non-fundamental component? In my understanding, the source is a sinusoidal function which can be broken down into sum of infinitely many terms. The greater quantity of terms you sum, the better the approximation to the original sinusoidal function. Please confirm if my calculation of fundamental component of source current is correct. If it is, what does f(x) represent in equation below?
\[f(x)=\sum_{n=1}^{inf}b_{n}sin(nx)dx\]
\[b_{1}=\frac{2}{\pi}\int_{0}^{\pi}f(x)sin(nx)dx\]
\[b_{1}=\frac{2}{\pi}\int_{0}^{\pi}I_{0}sin(\omega t)d\omega t= -\frac{2I_{0}}{\pi}cos(\omega t )\mid _{0}^{\pi}=\frac{4I_{0}}{\pi}\]