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응용전자 회로 생체의공학과 2010103786 박기택.

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Presentation on theme: "응용전자 회로 생체의공학과 2010103786 박기택."— Presentation transcript:

1 응용전자 회로 생체의공학과 박기택

2 Active Filter OpAmp를 사용하고 이득이 있는 filter Passive filter Active filter
LPF HPF

3 Active Filter Lowpass Filter Highpass Filter Bandpass Filter
실제 필터 Lowpass Filter Highpass Filter pass band stop band 𝑊 𝑐 𝑊 𝑐 𝑊 𝑐 Transmit band Bandpass Filter Bandstop Filter(Notch Filter) 𝑊 𝑐 :𝑐𝑢𝑡𝑜𝑓𝑓 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑊 𝑐 𝑊 𝐿 𝑊 𝑐 𝑊 𝐿

4 Active Filter V(s)=R(s) 𝑉(𝑗𝑤) 𝐼(𝑗𝑤) =𝑗𝑤𝐿 𝑉(𝑠) 𝐼(𝑠) =𝐿 ,
+ V(s)=R(s) 𝑉 𝑡 =𝐿 𝑑𝑖(𝑡) 𝑑𝑡 i V - 𝑉 𝑠 =𝑠𝐿𝐼 𝑠 𝑉 𝑗𝑤 =𝑗𝑤𝐿𝐼 𝑗𝑤 𝑠=𝜎+𝑗𝑤 + i V 𝑉(𝑠) 𝐼(𝑠) =𝐿 𝑉(𝑗𝑤) 𝐼(𝑗𝑤) =𝑗𝑤𝐿 𝑖 𝑡 =𝐶 𝑑𝑉(𝑡) 𝑑𝑡 , - 𝐼 𝑠 =𝑠𝐶𝑉 𝑠 𝐼 𝑗𝑤 =𝑗𝑤𝐶𝑉 𝑗𝑤 𝑠=𝜎+𝑗𝑤 𝑉(𝑠) 𝐼(𝑠) = 1 𝑠𝑐 𝑉(𝑗𝑤) 𝐼(𝑗𝑤) = 1 𝑗𝑤𝑐 ,

5 Active Filter H Vi(t) Vo(t)
H(S) = Vo(S) Vi(S) = 𝑏 𝑚 𝑆 𝑚 + ⋯ 𝑏 0 𝑎 𝑛 𝑆 𝑛 + ⋯ + 𝑎 0 Transfer function = N(S) D(S) 𝐻 𝑠 = 𝐻 0 𝑠− 𝑍 1 𝑠− 𝑍 2 ⋯(𝑠− 𝑍 𝑚 ) 𝑠− 𝑃 1 𝑠− 𝑃 2 ⋯(𝑠− 𝑃 𝑛 ) 𝑍 𝑖 :𝑧𝑒𝑟𝑜 𝑠= 𝑍 𝑖 𝐻 𝑠 =0 𝑃 𝑖 :𝑝𝑜𝑙𝑒 𝑠= 𝑃 𝑖 𝐻 𝑠 =∞

6 Pole – Zero Diagram 𝑗𝑤 𝑠=𝜎+𝑗𝑤 𝜎 Pole zero diagram을 그렸을때 나타나는 특성을 통해 전달함수의 특성을 알 수 있다. *보통 통신에 사용 되는 전달함수는 𝜎가 0이다 Left half Plame Right half Plame

7 Active Filter 𝐻 𝑠 = 𝐴 1 𝑠− 𝑃 1 + 𝐴 2 𝑠− 𝑃 2 +⋯+ 𝐴 𝑛 𝑠− 𝑃 𝑛
𝐻 𝑠 = 𝐴 1 𝑠− 𝑃 𝐴 2 𝑠− 𝑃 2 +⋯+ 𝐴 𝑛 𝑠− 𝑃 𝑛 CASE 1 𝑃 𝑘 = 𝜎 𝑘 (실근) 𝜎 𝑘 >0 Diverge 𝐿 −1 𝐴 𝑘 𝑠− 𝜎 𝑘 = 𝐴 𝑘 𝑒 𝜎 𝑘 𝑡 𝑢(𝑡) 𝜎 𝑘 <0 Die out * BIBO(Bounded Input Bounded Output) Stability 𝜎 𝑘 <0

8 Active Filter CASE 2 𝐿 −1 𝐴 𝑘 𝑠− (𝜎 𝑘 +𝑗 𝑤 𝑘 ) + 𝐴 𝑘 ′ 𝑠− (𝜎 𝑘 −𝑗 𝑤 𝑘 ) = 2𝐴 𝑘 𝑒 𝜎 𝑘 𝑡 cos⁡( 𝑤 𝑘 𝑡+ 𝐴 𝑘 ) 𝜎 𝑘 =0 𝜎 𝑘 >0 Diverge Osillation 𝜎 𝑘 <0 𝜎 𝑘 <0 : 모든 pole이 LHP에 있음 Natural Response “die out” Transient Response “homogeneous” Steady-stata Frequency Response S=jw 인 H(s) , w축에서의 변화만 분석 Die out

9 Transfer Function Vi(jw) Vo(jw) H Vi(t) Vo(t)
+ 3 - 2 V+ 7 V- 4 OUT 6 OS1 1 OS2 5 R1 C R2 Transfer Function Vi(jw) Vo(jw) H Vi(t) Vo(t) 𝑉 𝑖 𝑡 = 𝑉 𝑖𝑚 cos⁡(𝑤𝑡+ 𝜃 𝑖 ) H(jw)= 𝑉 𝑜 (jw) 𝑉 𝑖 (jw) 𝑉 𝑜 𝑡 = 𝑉 𝑜𝑚 cos⁡(𝑤𝑡+ 𝜃 𝑜 ) Lowpass filter H jw = 𝑉 𝑜 (jw) 𝑉 𝑖 (jw) =− 𝑅 2 1+𝑗𝑤 𝑅 2 𝑐 𝑅 1 =− 𝑅 2 𝑅 𝑗𝑤 𝑅 2 𝑐 𝐻 3dB log 1 𝑅 2 𝐶 log 𝑤

10 Transfer Function Highpass filter
20log 𝑅 2 𝑅 1 3dB 𝑅 𝑗𝑤𝑐 𝑤 𝑐 = 1 𝑅 1 𝐶 1 log 𝑤 Bandpass filter 𝑅 2 1+𝑗𝑤 𝑅 2 𝑐 H= 𝑉 𝑜 (𝑗𝑤) 𝑉 𝑖 (𝑗𝑤) =− 𝑅 2 1+𝑗𝑤 𝑅 2 𝑐 𝑅 𝑗𝑤𝑐 =− 𝑅 2 𝑅 1 𝑗𝑤 𝑅 2 𝐶 2 (1+𝑗𝑤 𝑅 1 𝐶 1 )(1+𝑗𝑤 𝑅 2 𝐶 2 ) 20log 𝑅 2 𝑅 1 𝑅 𝑗𝑤𝑐 𝑤 𝐿 = 1 𝑅 1 𝐶 1 𝑤 𝐻 = 1 𝑅 2 𝐶 2 log 𝑤

11 First order Active Filter
Lowpass filter H jw = 𝑉 𝑜 (jw) 𝑉 𝑖 (jw) =− 𝑅 2 1+𝑗𝑤 𝑅 2 𝑐 𝑅 1 =− 𝑅 2 𝑅 𝑗𝑤 𝑅 2 𝑐 = 𝐻 𝑜 1 1+𝑗 𝑤 𝑤 𝑜 𝐻 𝑜 =− 𝑅 2 𝑅 1 𝑤 𝑜 = 1 𝑅 2 𝑐 Dc gain(low frequency gain) 𝑤=𝑤 𝑜 = 1 𝑅 2 𝑐 일때 H= 𝐻 𝑜 1+𝑗 = 1 2 𝐻 H= 𝐻 𝑜 −3[𝑑𝐵] 3dB -20dB/dec 𝑓 𝑜 = 𝑤 𝑜 2𝜋 = 1 2𝜋 𝑅 2 𝑐 log 1 𝑅 2 𝐶 log 𝑤 3dB cut off frequency

12 First order Active Filter
Highpass filter H jw = 𝑉 𝑜 (jw) 𝑉 𝑖 (jw) = 𝐻 𝑜 𝑗 𝑤 𝑤 𝑜 1+𝑗 𝑤 𝑤 𝑜 𝐻 𝑜 =− 𝑅 2 𝑅 1 𝑤 𝑜 = 1 𝑅 1 𝑐 High frequency gain 𝑓 𝑜 = 𝑤 𝑜 2𝜋 = 1 2𝜋 𝑅 1 𝑐 3dB cut off frequency 20log 𝑅 2 𝑅 1 3dB 𝑤=𝑤 𝑜 = 1 𝑅 1 𝑐 1 일때 20db/dec H= 𝐻 𝑜 −3[𝑑𝐵] 𝑤 𝑐 = 1 𝑅 1 𝐶 1 log 𝑤

13 First order Active Filter
Bandpass filter H= 𝑉 𝑜 (𝑗𝑤) 𝑉 𝑖 (𝑗𝑤) =− 𝑅 2 1+𝑗𝑤 𝑅 2 𝑐 𝑅 𝑗𝑤𝑐 = 𝐻 𝑜 𝑗 𝑤 𝑤 𝐿 (1+𝑗 𝑤 𝑤 𝐿 )(1+𝑗 𝑤 𝑤 𝐻 ) 𝐻 𝑜 =− 𝑅 2 𝑅 1 :𝑝𝑎𝑠𝑠𝑏𝑎𝑛𝑑 𝑔𝑎𝑖𝑛(𝑚𝑖𝑑 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑔𝑎𝑖𝑛) 𝑤 𝐿 = 1 𝑅 1 𝑐 1 :𝐿𝑜𝑤 3𝑑𝐵 𝑐𝑢𝑡𝑜𝑓𝑓 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑤 𝐻 = 1 𝑅 2 𝑐 2 :𝐻𝑖𝑔ℎ 3𝑑𝐵 𝑐𝑢𝑡𝑜𝑓𝑓 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑤=𝑤 𝐿 𝑜𝑟 𝑤= 𝑤 𝐻 일때 20log 𝑅 2 𝑅 1 H= 𝐻 𝑜 −3[𝑑𝐵] 𝑤 𝐿 = 1 𝑅 1 𝐶 1 𝑤 𝐻 = 1 𝑅 2 𝐶 2 log 𝑤

14 First order Active Filter
Phase - Shifter 𝑉 𝑝 = 1 𝑠𝑐 𝑅+ 1 𝑠𝑐 𝑉 𝑖 = 1 1+𝑠𝑅𝑐 𝑉 𝑖 𝑉 𝑜 =− 𝑉 𝑖 +2 𝑉 𝑝 = 1−𝑠𝑅𝑐 1+𝑠𝑅𝑐 𝑉 𝑖 20log 𝑅 2 𝑅 1 H= 𝑉 𝑜 (𝑗𝑤) 𝑉 𝑖 (𝑗𝑤) = 1−𝑗𝑤𝑅𝐶 1+𝑗𝑤𝑅𝐶 = −2tan −1 𝑤𝑅𝑐 log 𝑤 H log 𝑤

15 Second order Active Filter
H s = 𝑁(𝑠) { 𝑠 𝑤 𝑜 } 2 +2𝜉 𝑠 𝑤 𝑜 +1 𝑁 𝑠 :𝑠의 다항식(m<=2) 𝑤 𝑜 : undamped natural frequency 𝜉 :damping ratio 𝑃 1.2 :(−𝜉 ± 𝜉 2 −1 ) 𝑤 𝑜 1. 𝜉>1 (서로 다른 실근 2개) Over damped Natural response는 2개의 지수함수로 이루어짐 lim 𝑡→∞ 𝐴 1 𝑒 − 𝑡 𝜏 𝐴 2 𝑒 − 𝑡 𝜏 2. 0<𝜉<1(𝑐𝑜𝑚𝑝𝑙𝑒𝑥 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒) 𝑃 1.2 : −𝜉 𝑤 𝑜 ±𝑗 𝑤 𝑜 1− 𝜉 2 Natural response : damped sine 2𝐴 𝑘 𝑒 − 𝜉𝑤 0 𝑡 cos⁡( 𝑤 0 1− 𝜉 𝐴 𝑘 )

16 Second order Active Filter
3. 𝜉=0 : undamped Natural response 2𝐴 𝑘 cos⁡( 𝑤 0 𝑡+ 𝐴 𝑘 ) 4. 𝜉<0 stable Diverge (unstable) 𝜉<0 𝜉=0 𝜉<0 𝜉>1 𝜉>1 𝜉=0 𝜉<0

17 Second order Active Filter
H jw = 𝑁(𝑗𝑤) 1−{ 𝑤 𝑤 𝑜 } 2 + 𝑗 𝑤 𝑤 𝑜 𝑄 Damping Q osilation 1. LPF H jw = 𝐻 𝑂𝐿𝑝 ∗ 𝐻 𝐿𝑝 (𝑗𝑤) 𝐻 𝑂𝐿𝑝 :dc gain (low freq gain) 소자값에 따라 결정 Q값에 따라서 overshoot, undershoot 발생 = 𝐻 𝑂𝐿𝑝 1 1−{ 𝑤 𝑤 𝑜 } 2 + 𝑗 𝑤 𝑤 𝑜 𝑄 1. 𝑤 𝑤 𝑜 ≪1 3. 𝑤 𝑤 𝑜 =1 𝐻 𝑗𝑤 = 𝐻 𝑂𝐿𝑝 𝐻 𝐿𝑝 𝑗𝑤 =−𝑗𝑄 -40dB/dec 2. 𝑤 𝑤 𝑜 ≫1 𝐻= 𝐻 𝑂𝐿𝑝 +𝑄[𝑑𝐵] log 𝑤 𝐻 𝑗𝑤 = 𝐻 𝑂𝐿𝑝 ∗ −1 { 𝑤 𝑤 𝑜 } 2 𝐻 𝐿𝑝 =−40 log 𝑤 𝑤 𝑜 [dB]


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