Rattn_aPi 2025-03-25

\(a=\dfrac{1}{\sqrt{A}}\)

R₀ =

\(R_0=\dfrac{\left(a^2-1\right)\cdot R_b\cdot\sqrt{R_k}}{\left(a^2+1\right)\cdot\sqrt{R_k}-2\cdot a\cdot\sqrt{R_b}}\)

R₁ =

\(R_1=\dfrac{\left(a^2-1\right)\cdot\sqrt{R_b\cdot R_k}}{2\cdot a}\)

R₂ =

\(R_2=\dfrac{\left(a^2-1\right)\cdot R_k\cdot\sqrt{R_b}}{\left(a^2+1\right)\cdot\sqrt{R_b}-2\cdot a\cdot\sqrt{R_k}}\)


Rb =
Rk =
AP =
☺
			\(a=\dfrac{1}{\sqrt{A}}\)
R₀ =			\(R_0=\dfrac{\left(a^2-1\right)\cdot R_b\cdot\sqrt{R_k}}{\left(a^2+1\right)\cdot\sqrt{R_k}-2\cdot a\cdot\sqrt{R_b}}\)
R₁ =			\(R_1=\dfrac{\left(a^2-1\right)\cdot\sqrt{R_b\cdot R_k}}{2\cdot a}\)
R₂ =			\(R_2=\dfrac{\left(a^2-1\right)\cdot R_k\cdot\sqrt{R_b}}{\left(a^2+1\right)\cdot\sqrt{R_b}-2\cdot a\cdot\sqrt{R_k}}\)

document.write( SzHuEnSzSz( 'Aszimmetrikus Π csillapító', 'Asymmetric Π attenuator' ) );