Jan 21, 2011 #1 E ericyeoh Member level 2 Joined Jun 3, 2009 Messages 46 Helped 1 Reputation 2 Reaction score 0 Trophy points 1,286 Location malaysia Activity points 1,635 Guys i still dun understand the band pass design in below attrached circuit. the 1st IC1A and IC1B opamp hv , FL and FH at 1.59hz. FH = (1/[2(pie) 10mic x 10K]) = 1.59hz FL = (1/[2(pie) 0.1mic x 1M]) = 1.59hz how come both cut off frequency at 1.59hz? if this is the case, the bandwidth =0 hz? Please help me pls, i'm lost Attachments PIR forum.png 24.5 KB · Views: 111
Guys i still dun understand the band pass design in below attrached circuit. the 1st IC1A and IC1B opamp hv , FL and FH at 1.59hz. FH = (1/[2(pie) 10mic x 10K]) = 1.59hz FL = (1/[2(pie) 0.1mic x 1M]) = 1.59hz how come both cut off frequency at 1.59hz? if this is the case, the bandwidth =0 hz? Please help me pls, i'm lost
Jan 21, 2011 #2 keith1200rs Super Moderator Staff member Joined Oct 9, 2009 Messages 10,865 Helped 2,065 Reputation 4,130 Reaction score 1,596 Trophy points 1,403 Location Yorkshire, UK Activity points 57,270 Re: PIR sensor need help Your calculations are correct but that doesn't mean you will have zero bandwidth. The cut off frequencies are where the signal has dropped by 3dB. So, the peak response will be around 1.6Hz and it will roll off on either side. Keith. Last edited: Jan 22, 2011
Re: PIR sensor need help Your calculations are correct but that doesn't mean you will have zero bandwidth. The cut off frequencies are where the signal has dropped by 3dB. So, the peak response will be around 1.6Hz and it will roll off on either side. Keith.