Understanding Poles, Zeros and Transfer Functions in Passive Circuits

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  • เผยแพร่เมื่อ 28 พ.ย. 2024

ความคิดเห็น • 9

  • @isacdaimary9908
    @isacdaimary9908 2 ปีที่แล้ว

    It helped like anything else, I can't thank you enough with words,, God bless you

  • @pablomarco5118
    @pablomarco5118 8 หลายเดือนก่อน

    amazinge

  • @coffeecuppepsi
    @coffeecuppepsi 4 ปีที่แล้ว +1

    thank you Mark Ruffalo for explaining this so well

  • @karastom2304
    @karastom2304 8 ปีที่แล้ว

    is there more lectures for LTspice?

  • @MrMisi6
    @MrMisi6 3 ปีที่แล้ว

    Could you please explain me What is S? it is 2 times pi times F?

  • @rachelstokes1340
    @rachelstokes1340 5 ปีที่แล้ว

    why are we looking at when the impedances of R and C are the same at 5:22? Why is the result -20db down? Why do we fix the values and call it a pole?

  • @rachelstokes1340
    @rachelstokes1340 5 ปีที่แล้ว

    can you please help me so I can understand this video, otherwise its very confusing

    • @leiaz6762
      @leiaz6762  5 ปีที่แล้ว +1

      The C's impedance is 1/sC or 1/2pifC. when 1/sC=R, a transition point, also called -3dB point in power, since the output is exactly 50% of input voltage. For power, dB=10log(Pout/Pin). For V, I or R, dB=20log(Vout/Vin). So for Vout/Vin, 1/sC=R is -6dB on the vertical scale or 20log(0.5Vin/Vin)=-6dB. Why -20dB, decade in frequency in a factor of 10. If f_pole=1kHz, one decade is 10kHz. Simulate RC circuit in LTSpice with R=10k, C=0.01u, f_pole=1/2piRC=1.591kHz. If you check the dB value at 15.91kHz, you will see the Vout/Vin in dB will be measured at -20dB, hence the output ratio in voltage is falling at a rate of -20dB/dec. At 159.1kHz, the dB value will be at -40dB, because it is 2 decade awa from f_pole. Hope this helps.

    • @rachelstokes1340
      @rachelstokes1340 5 ปีที่แล้ว

      @@leiaz6762 I'm so sorry, I thought I responded to this right after, this was very helpful thank you so much for explaining, I really appreciate it