Jun 14, 2016 #1 J Johnny_YU Member level 2 Joined Oct 17, 2013 Messages 45 Helped 2 Reputation 4 Reaction score 2 Trophy points 8 Activity points 371 Hi,all I have a problem with the first order low pass filter. I kown how to calculate the H(s). H(s) = 1/(sRC+1), the bode diagram is below. In bode diagram , the slope is -20dB, how to calculate that slope? Thank you
Hi,all I have a problem with the first order low pass filter. I kown how to calculate the H(s). H(s) = 1/(sRC+1), the bode diagram is below. In bode diagram , the slope is -20dB, how to calculate that slope? Thank you
Jun 14, 2016 #2 S shanmei Advanced Member level 1 Joined Jul 26, 2006 Messages 430 Helped 8 Reputation 16 Reaction score 8 Trophy points 1,298 Location USA Activity points 4,496 H(s)=1/(SRC+1), when S is large, H(s)=1/SRC. H1(s)=1/10RC, when s=10; H2(s)=1/100RC, when s=100; 20log(H2(s)/H1(s))=20log(0.1)=-20dB
H(s)=1/(SRC+1), when S is large, H(s)=1/SRC. H1(s)=1/10RC, when s=10; H2(s)=1/100RC, when s=100; 20log(H2(s)/H1(s))=20log(0.1)=-20dB
Jun 14, 2016 #3 C CataM Advanced Member level 4 Joined Dec 23, 2015 Messages 1,275 Helped 314 Reputation 628 Reaction score 312 Trophy points 83 Location Madrid, Spain Activity points 8,409 H(jω)=1/(1+jω·T) T=RC for simplification. You are asking about the gain, so I will omit the phase. |H(jω)|(dB)=-20·log[√1+(ω·T)² ] if ω·T>>>1 results in -20·log(ω·T) Lets say -20·log(ω·T)=a [dB] => -20·log(10*ω·T)=-20+a [dB] i.e. every decade you subtract 20 dB.
H(jω)=1/(1+jω·T) T=RC for simplification. You are asking about the gain, so I will omit the phase. |H(jω)|(dB)=-20·log[√1+(ω·T)² ] if ω·T>>>1 results in -20·log(ω·T) Lets say -20·log(ω·T)=a [dB] => -20·log(10*ω·T)=-20+a [dB] i.e. every decade you subtract 20 dB.