Understanding Poles, Zeros and Transfer Functions in Passive Circuits

[@isacdaimary9908]

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

[@pablomarco5118]

amazinge

[@coffeecuppepsi]

thank you Mark Ruffalo for explaining this so well

[@karastom2304]

is there more lectures for LTspice?

[@MrMisi6]

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

[@rachelstokes1340]

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]

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

[@leiaz6762]

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]

@@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