modified 9 years ago

RC dolní propust

00:42:52

    Xc=1/(6,28fC)
    Z=(R^2 +  Xc^2)^0,5
    uc(t)=u1(t) x (Xc/Z)
-----------------------------------------------------
  a) Výpočet výstup.napětí při f=10 kHz

    Xc=1/6.28x10^-5
    Xc=15,923.57 Ohm

    Z=(10x10^9 + 253,560,081.5)^0.5
    Z=101,259.9 Ohm

    uc(t)=10 x (15,923.57/101259.9)
    uc(t)=1.57 V
-----------------------------------------------------
  b) Výpočet mezní frekvence a výstup.napětí pro pokles -3dB 

   f0=1/(6.28RC)
   pro -3dB platí:.       uc(t)=0.707 x u1(t)
-----------------------------------------------------
    f0=1/[6.28 x 100 000 x (1x10^-9)]
    f0=1,592.4 Hz
  
    uc(t)=0.707 x 10
    uc(t)=7.07 V

published 9 years ago

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