ISTITUTO TECNICO INDUSTRIALE STATALE GUGLIELMO MARCONI - Forlì School Year 2009/2010 C.L.I.L. MODULE: ANALYSIS AND DESIGN OF FIRST ORDER RC FILTERS Subjects: ELECTRONICS/ENGLISH Teachers: Miria Bovino and Roberto Versari Class: 4AEL Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Outline of the CLIL module Lesson 1: RC low pass filter schematic and transfer function (with C.A.D. example) Lesson 2: frequency response graphical representation (with excel example) Lesson 3: Analysis of the CR high-pass filter (with excel + C.A.D. example) Lesson 4: An introduction to active filters and circuit simulations (with C.A.D. example) Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Lesson 1: RC low pass filter schematic and transfer function (2h) Lesson 1 Targets: Learn how to draw the schematic of an RC low-pass filter. Learn how to calculate its transfer function G(jw) Learn the filter constant time () definition Learn how to calculate the RC low-pass filter output for any sinusoidal input. Learn how to realize an RC low pass filter with a C.A.D. tool Istituto Tecnico Industriale Statale “G. Marconi” - Forlì RC low pass filter schematic The first order passive RC low pass filter is made of only two components: R VI(jw) stimulus or input sinusoidal voltage C VO(jw) output sinusoidal voltage VI and VO can be represented by the corresponding vector quantities: VI=VIej I, VO= VOej O Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Transfer function calculation (1/2) The RC low pass filter can be easily seen as a simple impedance divider with Z1=R and Z2=1/jwC Z1=R VI(jw) Z2=1/jwC VO(jw) The impedance divider formula provides: VO/VI=Z2/(Z1+Z2). VO/VI is the transfer function G(jw) of the RC filter. Hence it is: G(jw)=Z2/(Z1+Z2) Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Transfer function calculation (2/2) Let’s substitute Z1=R and Z2=1/jwC in G(jw) expression 1 1 1 1 jwC G ( jw ) 1 jwC 1 jwRC 1 R R jwC jwC R.C= [s] is defined as the time constant of the low-pass filter. Hence: 1 1 G ( jw ) 1 jw 1 j 2 f Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Output calculation (1/2) G(jw) is a complex number with: (jw ) Amplitude: G Phase: 1 2 1 ( w ) artg ( w ) G (jw ) VO=G(jw). VI can be written using complex numbers properties: V 1 G ( jw ) V V I VO Amplitude: V O I I 2 2 1 ( w ) 1 ( w ) VO Phase: artg ( w ) O G ( jw ) I I G(jw) depends only on and f (input signal frequency) Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Output calculation (2/2) RC low-pass filter 1) 2) 3) 4) 5) VI, I, f =R. C sinusoidal input G(jw) G(jw) VO, O, f sinusoidal output VI, I and f are given the filter schematic is known is calculated. f and are known G(jw) and G(jw) are calculated VO =|G(jw)| . VI is calculated O = G(jw) + I is calculated Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Output calculation example Exercise: calculate the output VO of the circuit corresponding to the given sinusoidal input VI 1) VI=1V, I=00, f=10KHz remember: w2 f 2) =RC=320*50nF=16s 3) |G(10KHz)|=1/ 1+(2**10KHz*16s)2 =0,7 G(10KHz)=-artg(2**10KHz*16s)=-450 The output voltage is attenuated 4) VO=0,7*1V=0,7V of a factor of 0,7 and delayed of 0 0 0 5) O=-45 +0 =-45 45 degrees when f=10KHz. Istituto Tecnico Industriale Statale “G. Marconi” - Forlì LTspiceIV CAD tool Download free from: http://www.linear.com/designtools/software CAD tools: schematic implementation and real time simulation needed to decrease electronic circuit time to market. English language prerequisite for most CAD tools. Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Lesson 2: RC low pass filter frequency response graphical representation (2h) Lesson 2 Targets: Learn the graphical method to calculate any filter output. Learn the low-pass filter cutoff frequency (fC) definition, its bandwidth (Bw) and the relationship between fC and . Learn how to read and draw first-order low pass filters amplitude diagrams in logarithmic scale. Learn how to design an RC low-pass filter for a given bandwidth and/or fC. Learn how to draw the amplitude and phase diagrams of first-order RC low-pass filters with MS-excel. Istituto Tecnico Industriale Statale “G. Marconi” - Forlì The graphical approach to the calculation of filter output Lesson 1 result: the filter output voltage can be derived by the knowledge of the input voltage and the analitycal calculation of the filter transfer function G(jw) – both amplitude and phase. Disadvantages of the analitycal approach: - two calculations (amplitude and phase) for each input frequency. - G(jw) analitycal expression depends on filter topology. New graphical approach: two plots, namely |G(jw)| vs. f and G(jw) vs. f are associated to each filter. The calculation of the output voltage does not need G(jw) value calculation because the values of amplitude and phase of G(jw) at the given frequency can be simply read from the two given plots of the filter. Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Amplitude and Phase diagrams G(jw) vs. f is the filter phase diagram* Amplitude diagram 1 0,9 0,8 0,7 0,6 0,5 0,4 0,3 0,2 0,1 0 Phase diagram 0 -15 -30 . G(jw) |G(jw)| |G(jw)| vs. f is the filter amplitude diagram* -45 -60 -75 -90 1000000 100000 10000 frequency 1000 100 10 1 0,1 1000000 100000 10000 1000 100 10 1 0,1 frequency Examples of amplitude and phase diagrams of an RC low pass filter with R=1K and C=1F. *=remember that: w2 f Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Output amplitude calculation with graphical method (1/2) VI, f Amplitude diagram 1 0,9 0,8 0,7 0,6 0,5 0,4 0,3 0,2 0,1 0 VO, f 1000000 100000 10000 frequency 1000 100 10 1 0,1 sinusoidal input |G(jw)| RC low-pass filter amplitude diagram sinusoidal output 1) VI, I and f are given 2) the filter amplitude diagram value at the frequency f can be read from the plot. 3) VO =|G(jw)| . VI is easily calculated, since |G(jw)| is the value read from the plot in step 2). Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Output amplitude calculation with graphical method (2/2) Case 1): VI=1,5V I=00 f=100Hz From the filter amplitude diagram at f=100Hz we read |G(jw)| =0,8. Hence the output voltage amplitude at f=100Hz is VO=0,8*1,5V=1,2V Case 2): VI=1,5V I=00 f=1KHz From the filter amplitude diagram at f=1KHz we read |G(jw)| =0,18. Hence the output voltage amplitude at f=1KHz is VO=0,18*1,5V=0,27V Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Output phase calculation with graphical method (1/2) RC low-pass filter phase diagram Phase diagram 0 -15 -30 O, f G(jw) I, f -45 -60 -75 -90 1000000 100000 10000 frequency 1000 100 10 1 0,1 sinusoidal input sinusoidal output 1) VI, I and f are given 2) the filter phase diagram value at the frequency f can be read from the plot. 3) O = G(jw)+I is easily calculated, since G(jw) is the value read from the plot in step 2). Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Output phase calculation with graphical method (2/2) Case 1): VI=1,5V I=00 f=100Hz From the filter phase diagram at f=100Hz we read G(jw)=-350. Hence the output voltage phase at f=100Hz is O=-350+0=350 Case 2): VI=1,5V I=00 f=1KHz From the filter phase diagram at f=1KHz we read G(jw)=-810. Hence the output voltage phase at f=1KHz is O=-810+0=-810 Graphical method advantages: - no need to calculate amplitude and phase: they are simply read from the plots. - it is a general approach valid for all kind of filters: it is only needed the knowledge of the amplitude and phase plots. Istituto Tecnico Industriale Statale “G. Marconi” - Forlì |G(jw)| Amplitude diagram analysis Amplitude diagram 1 0,9 0,8 0,7 0,6 0,5 0,4 0,3 0,2 0,1 0 fC=160Hz stop band pass band B frequency scale is logarithmic 1000000 100000 10000 1000 100 10 1 0,1 frequency 1) The cutoff frequency fC is defined as the frequency value at which |G(jw)|=0,7 In fact, all input signals with f< fC “pass”, i.e. we can find them at the filter output without significant attenuation. On the contrary, all input signals with f> fC are “cut”, i.e. the filter output has an amplitude significantly lower than the input. 2) The filter band B is defined as the range of input frequencies for which the filter attenuation is 0,7. In practice B=[0, fC]. The bandwidth is the width of B=fC-0=fC 3) The relationship between fC and is: 1 1 f C 2 2 R C Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Phase diagram analysis fC=160Hz Phase diagram 0 -15 -30 G(jw) frequency scale is logarithmic -45 -60 -75 -90 1000000 100000 10000 1000 100 10 1 0,1 frequency The cutoff frequency fC corresponds to the frequency value at which G(jw)=-450. The filter introduces a maximum phase shift (a delay) of –900. All input signals with frequencies within the filter pass band [0, fC] are delayed less than 45/(360*f)=0,125/f. Input signals with frequencies outside the filter pass band are delayed more than 0,125*f, with a maximum delay of 0,25/f corresponding to the maximum phase shift of –900. Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Amplitude diagram 1 0,9 0,8 0,7 0,6 0,5 0,4 0,3 0,2 0,1 0 frequency log. scale |G(jw)| |G(jw)| Why a frequency logarithmic scale? fC=160Hz frequency linear scale frequency 1) frequency logarithmic scale: the whole waveform is clearly outlined and the amplitude values at different frequencies can be read with reasonable accuracy 2) frequency linear scale: the amplitude values at different frequencies cannot be read so the graph is useless. The same considerations hold for the phase plot. 1000000 900000 800000 700000 600000 500000 400000 300000 200000 100000 0,1 1000000 100000 10000 1000 100 10 1 0,1 frequency Amplitude diagram 1 0,9 0,8 0,7 0,6 0,5 0,4 0,3 0,2 0,1 0 Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Whole logarithmic scale: the Bode plot Analitycal formula for output calculation: VO = G(jw) . VI Normalization (dimensionless): VO/1V= G(jw) . VI/1V logarithmic: 20Log(VO/1V)=20Log G(jw) +20Log(VI/1V) Output calculation in dB: VO[dBV]= G(jw) [dB]+VI[dBV] Amplitude diagram in dB (Bode Plot) |G(jw)| [dB] 0 -10 -20 -30 -40 -50 -60 fC=160Hz 20dB 1dec -70 -80 1000000 100000 10000 1000 100 10 1 0,1 frequency |G(jw)|[dB] vs. f is Bode plot It is a full log. diagram. f=fC |G(jw)|=-3dB (in fact, 20Log(0,7)=-3dB). When f>fC Bode plot slope is 20dB/dec. Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Bode plot representation advantages Very simple output calculation: VO=|G(jw)|+VI [dB] (only a sum) Very easy to draw!!! it can be approximated by: |G(jw)|=0 for f<fC horizontal line |G(jw)|=-20*Log(f/fC)straight line with slope -20dB/dec |G(jw)| [dB] 0dB -20dB -40dB -60dB fC 10fC 100fC 1000fC f (log) only approximation: |G|=-3dB and not 0 when f=fC Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Design for a given fC or B The design of a first order RC low pass filter for a given fC or B is very simple: 1 1 1) calculate the filter time constant: 2 fC 2 B 2) Choose a capacitor value among those available in the laboratory: for instance, C=1F. 3) calculate the corresponding value of R: R C 4) if R is in the [100,100K] range OK!!!, otherwise go back to point 2) choosing another C value. Example: design a low pass filter with fC=1KHz. 1 4 1) 1 , 59 10 s2) C 2 f C 1F 4 1 , 59 10 s 3) R 159 4) OK!!! 6 10 F Istituto Tecnico Industriale Statale “G. Marconi” - Forlì How to use MS-excel to draw Amplitude and Phase diagrams Electronic sheet tools are useful for automated calculations and diagram representations very useful for transfer function plot drawings. Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Lesson 3: Analysis of the CR high-pass filter (2h) Lesson 3 Targets: Learn the schematic of a high-pass CR filter and its transfer function G(jw) Learn how to draw the amplitude and phase plots of the filter Learn the relationship between , fC and B of the filter Learn how to calculate the output of the filter Learn how to design the filter Learn how to simulate the filter with CAD tools in English language. Istituto Tecnico Industriale Statale “G. Marconi” - Forlì CR high pass filter schematic First-order passive CR high-pass filter schematic: C O 1µ VI(jw) R 159 VI=VIeji Same schematic as the RC low pass filter but with R and C interchanged. VO(jw) =RC=0,16ms same formula as VO=VOejo the low pass filter Using the same impedance divider model: G(jw)=VO(jw)/VI(jw)=Z2/(Z1+ Z2), where Z1=1/jwC and Z2=R. w After a few mathematics we get: G (jw ) jw G (jw ) 1 jw 2 1 ( w ) 0 90 artg ( w ) G (jw ) Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Amplitude and Phase diagrams Amplitude diagram in dB (Bode Plot) 0 -10 |G(fC)| = -3dB -20 -30 20dB -40 -50 -60 B=[fC,] 60 fC=1KHz G(fc)=+450 45 30 fC=1KHz 15 0 1000000 100000 fC and B are defined in the same way as the RC low pass. However, B=[fC,] because the filter attenuates only the frequencies < fC, significantly. 10000 frequency 1000 100 10 1 0,1 1000000 100000 10000 1000 100 10 1 0,1 frequency Phase diagram 75 1dec -70 -80 0 90 artg ( w ) G (jw ) 90 G(jw) |G(jw)| [dB] w G ( jw ) [ dB ] 20 Log ( ) 2 1 ( w ) Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Diagrams characteristics Amplitude: B=[fC,] because signals are not attenuated when f>fC. Bw=. |G(fC)|=-3dB (corresponding to a 0,7* attenuation) when f<fC|G(f)| increases of 20dB for each decade of increase of f (i.e. for a frequency increase of a factor of 10). Phase: f<<fCmaximum phase shift of +900, corresponding to an output voltage anticipated of 0,25/f with respect to input. f=fCG=+450; output voltage anticipated of 0,125/fC. f>>fC G=00; output is in phase with input. High-pass: input high frequencies are not attenuated nor shifted “pass”. Input low frequencies are attenuated and anticipated “cut”. fC is the frequency cut threshold. Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Filter design and Bode plot drawing The filter design for a given fC can be done using the same 4 steps algorithm as the low pass filter: 1 2 fC 1) calculate the filter time constant: 2) Choose a capacitor value among those available in the laboratory: for instance, C=1F. R 3) calculate the corresponding value of R: C 4) if R [100,100K] OK!!!, otherwise go back to point 2) choosing another C value available. |G(jw)| [dB] 0,01fC 0,1fC fC 0dB -20dB -40dB -60dB 0,001fC f (log) Bode plot drawing: |G(jw)|=0 for f>fC horizontal line |G(jw)|=20*Log(f/fC)straight line with slope +20dB/dec when f<fC only approx. |G(fC)|=-3dB and not 0dB. Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Output calculation examples CR high pass filter with amplitude and phase diagrams of slide 27. VO[dBV]=|G(f)|[dB]+VI[dBV] O=G(f)+I Case 1): VI=1V=0dBV I=00 f=100Hz G(100Hz)=830. |G(100Hz)|=-20dB VO=-20dB+0dBV=-20dBV=0,1V O=830+00=830 Case 2): VI=1V=0dBV I=00 f=1000Hz=1KHz G(1KHz)=450. |G(1KHz)|=-3dB VO=-3dB+0dBV=-3dBV=0,7V O=450+00=450 Case 3): VI=1V=0dBV I=00 f=10000Hz=10KHz G(10KHz)=70. |G(10KHz)|=-0,09dB VO=-0,09dB+0dBV=-0,09dBV=0,99V O=70+00=70 Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Circuit simulation: LTspiceIV LTspiceIV download free from http://www.linear.com/designtools/software/ Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Pros and cons of simulation Simulation advantages: Fast time to market: you do not need to wait physical implementation of the circuit to know its behavior No hand calculations no human error, no interconnection errors Simulation drawbacks: If components models are not accurate the circuit can exhibit unexpected behavior Parasitic components are not included and are circuit layout dependent difficult to predict. That’s what differentiates the good from the bad designer. The Pros are exceedingly better than the Cons: that’s why simulation is ALWAYS used to prototype new designs. Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Lesson 4: An introduction to active filters and simulation examples (2h) Lesson 4 Targets: Learn the main differences between RC passive and active filters. Learn the schematics of low-pass and high-pass RC active filters. Learn the amplitude and phase diagrams of active filters. Learn the schematic and the frequency response diagrams of the pass-band second order RC active filter. Learn how to simulate active filters. Learn some practical applications of filters. Learn the difference between first-order and higher order filters. Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Passive vs. active filters Let’s consider an RC low pass filter connected to a 1K load RL: R 159 O C VI(jw) 1µ VI=VIeji RL 1K V ( jw ) Z O G ( jw ) 2 V jw ) Z Z I( 1 2 VO(jw) VO=VOejo The effect of RL is to change Z2: Z2=C//RL=RL/(1+jwCRL) 1 (jw ) Consequently G(jw) changes too and becomes: G R 1 jw Consequences: RL 1) Maximum amplitude = 1+R/RL <1 2) fC = (1+R/RL)/2 Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Limits of passive filters The main limits of passive filters are: 1) the frequency response of the filter is dependent on the RL load. In particular, the maximum amplitude of the filter when connected to a load is <1 and load dependent. fC value is load dependent too. 2) the maximum amplitude of a passive filter is =1 (no signal gain) even when not connected to any RL load. Active filters are able to provide both a frequency response independent of any RL load and a signal gain >1. They are obtained adding an active component such as an OP.AMP. (OPerational AMPlifier) to a passive filter. Istituto Tecnico Industriale Statale “G. Marconi” - Forlì First order low-pass active filter Inverting amplifier configuration R21.59K G(jw)=-Z2/Z1 where Z1=R1=159 and Z2=C//R2=R2/(1+jw2), 2=C*R2=0,159ms VCC Schematic: C100n R1 U1 VI VO 159 -VSS OP05 R 2 1 G ( jw ) R 1 1 jw 2 Advantages: G(jw) independent of any RL (>100) connected to filter output VO because of the low OPAMP output resistance -VSS VI Programmable low frequency gain = R2/R1 V1 V3 Istituto Tecnico Industriale Statale “G. Marconi” - Forlì low-pass active filter frequency response R 2 1 0 G ( jw ) [ dB ] 20 Log () 20 Log ( ) 180 artg ( w ) 2 G ( jw ) R 1 1 ( w ) 2 Amplitude diagram in dB (Bode Plot) 20 165 |G(fC)|=17dB 0 G(fC)=1350 150 G(jw) |G(jw)| [dB] 10 Phase diagram 180 135 -20 -10 fC=1KHz -30 120 105 90 -40 Amplitude: Same diagram as passive filter shifted up of 20*Log(R2/R1) Phase: Same diagram as passive filter shifted up of 1800 fC is the frequency at which |G| is –3dB with respect to its low frequency value (same definition used for amplifiers): |G(fC)|=|G(0)|-3dB. When f= fC G(fC) = G(0) – 450 1000000 100000 frequency 10000 1000 100 10 1 1000000 100000 10000 1000 100 10 1 frequency Istituto Tecnico Industriale Statale “G. Marconi” - Forlì First order high-pass active filter Schematic: R1 VI C U1 VO OP05 -VSS 159 C 1µ VCC R21.59K G(jw)=-Z2/Z1 where Z2=R2=1,59K and Z1=R1+1/jwC=(1+jw1)/jwC, 1=C*R1=0,159ms 2=C*R2=1,59ms jw 2 G (jw ) 1 jw 1 High frequency gain =2/1=R2/R1 -VSS V1 1 VI fC V3 21 .tran 4m Independent of load RL (if RL>100) Istituto Tecnico Industriale Statale “G. Marconi” - Forlì high-pass active filter frequency response 0 270 artg ( w ) G ( jw ) 1 Amplitude diagram in dB (Bode Plot) 20 10 0 -10 -20 -30 -40 -50 -60 Phase diagram 270 255 240 G(jw) |G(fC)|=17dB |G(jw)| [dB] w R 2 1 G ( jw ) [ dB ] 20 Log () 20 Log ( ) 2 R 1 1 ( w ) 1 fC=1KHz G(fC)=2250 225 210 195 180 Amplitude: Same diagram of passive filter shifted up of 20*Log(R2/R1) Phase: Same diagram of passive filter shifted up of 1800 fC is the frequency at which |G| is –3dB with respect to its high frequency value: |G(fC)|=|G(f=)|-3dB. When f= fC G(fC) = G(0) – 450 1000000 100000 10000 frequency 1000 100 10 1 0,1 1000000 100000 10000 1000 100 10 1 0,1 frequency Istituto Tecnico Industriale Statale “G. Marconi” - Forlì What’s a pass-band active filter? Low-pass filter: only low frequency inputs are not attenuated. High-pass filter: only high frequency inputs are not attenuated. Pass-bandonly frequency inputs [f1,f2] are not attenuated. B=[f1,f2] is the pass-band of the filter. Simpler pass-band filter: cascade of low-pass + high-pass filters: fC1 fC2 OUTPUT INPUT low-pass high-pass SIGNAL SIGNAL with fC=fC1 with fC=fC2 frequencies above fC1 are attenuated frequencies above fC1 and below fC2 are attenuated Works only if fC2>fC1; It doesn’t matter if low-pass is first or second in the cascade if they are active filters. Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Second order pass-band active filter C130n VI C VCC Schematic: R1 G(jw)=-Z2/Z1 where Z2=R2//C2=R2/(1+ jw2) R21.59K Z1=R1+1/jwC1=(1+jw1)/jwC1, 1=C1*R1=30s 2=C2*R2=9.54s 12=C1*R2=47,7s U1 C2 6n VO 1K -VSS j w 12 G ( jw ) ( 1 j w 2 ) ( 1 j w 1 ) OP05 Filter pass-band B=[fC1,fC2], with fC1<fC2: 1 1 f 16 , 7 KHz f 5 , 31 KHz C 2 C 1 2 2 1 2 -VSS 1 VI V1 V3 f max 2 9 , 4 K 1 2 Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Pass-band active filter frequency response w 0 R 2 1 270 artg ( w 1 ) art ( w 2 ) G ( jw ) [ dB ] 20 Log ( ) 20 Log ( 1 ) 20 Log ( ) G ( jw ) 2 2 R 1 1 ( w ) 1 ( w ) 1 2 Amplitude diagram in dB (Bode Plot) 10 |G(jw)| [dB] 0 G(jw) -10 -20 -30 fC1=5,3K -40 fC2=16,7K -50 G(fC1)=2050 G(fC2)=1550 The Bode plot of the pass-band active filter is exactly the sum of the Bode plots of the low-pass and high-pass active filters with the same 2 and 1, respectively. The phase diagram of the pass-band active filter is the sum decreased of 1800 of the phase diagrams of the low-pass and high-pass active filters with the same 2 and 1, respectively. 1000000 100000 frequency 10000 1000 100 10 1 1E+07 1000000 100000 10000 1000 100 10 frequency Phase diagram 270 255 240 225 210 195 180 165 150 135 120 105 90 Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Filter simulation: passive vs. active C1 C2 The RL load does not affect active filter response while it affects both amplitude and phase response of the passive filter. Without RL the two responses are the same. Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Filter simulation: pass-band C1 C2 Average value of input square waveform “cut”. High frequency components “cut” too. Output is almost sinusoidal (depends on filter selectivity). Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Filter applications: ADSL modem 6th order low-pass You can phone while browsing the Internet!!! The phone signal doesn’t disturb the ADSL signal because of the high-pass filter embedded in the ADSL module. The ADSL signal doesn’t disturb the phone signal because of the low-pass filter inserted just before the phone receiver. Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Filter applications: audio systems audio signal [0,20KHz] cross-over low-pass [0,3KHz] woofer (bass) cross-over high-pass [3K,20K] tweeter (high) A Hi-Fi (High Fidelity) audio system is composed of at least two different speakers: a woofer to reproduce bass sounds and a tweeter for the high sounds. A cross-over filter is placed before each speaker so that the speaker can be reached only by the frequency components of the input audio signal it is able to reproduce. Cross-over filters are typically 2nd or 3rd order low-pass or high-pass filters. Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Higher order filters. In practical applications 2nd or higher order filters are typically used because of their higher selectivity: |G(jw)| [dB] 0dB Note: the order of a filter is equal to the number of reactive -20dB components in its schematic (i.e. the -40dB number of poles of its -60dB transfer function). fC 10fC 100fC 1000fC 1st f (log) order: -20dB/dec 2nd order: -40dB/dec 3rd order: -60dB/dec Low-pass filter attenuation at f=10*fC is: 1st order 0.1 2nd order 0.01 3rd order 0.001 The higher the order of the filter the better its ability to select only the frequencies within its pass band B. Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Bibliography (1/3) The following list of references can be used for an in-depth investigation of the subject of the presentation: 1. The Analysis and Design of Linear Circuits, 6th Edition, R. E. Thomas, A. J. Rosa, G. J. Toussaint, ISBN: 978-0-470-38330-8, 2009. (basic textbook on linear circuits analysis with a complete chapter on AC analysis and Bode plots of analog filters). 2. http://www.animations.physics.unsw.edu.au/jw/RCfilters.html (basic analysis of RC filters with animations). 3. http://www.play-hookey.com/ac_theory/filter_basics.html (basic theory of active and passive filters). 4. http://www.educypedia.be/electronics/analogfil.htm (link to several resources on analog filters, both passive and active). Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Bibliography (2/3) 5. http://www.usna.edu/Users/ee/kintzley/ee303sp09/lectures/EE3 03Sp09_L05_Filters.pdf (lecture on basic filter topologies and solutions). 6. http://dev.emcelettronica.com/performing-ac-analysis-ltspice (method to perform an AC analysis of an RC filter with Ltspice) 7. http://www.swarthmore.edu/NatSci/echeeve1/Ref/DataSheet/Int roToFilters.pdf (basic introduction to analog filters from National Semiconductor Corp., with examples of real analog filters IC's and criteria to select among filter topologies). 8. Active Filter Cookbook, D. Lancaster, 2nd ed. Thatcher, Synergetics Press, 1995 (a comprehensive textbook on active filter design and analysis) Istituto Tecnico Industriale Statale “G. Marconi” - Forlì Bibliography (3/3) 9. http://www.wisc-online.com/objects/index_tj.asp?objID=ACE2803 (analysis of the bode plots of an RC low pass filter). 10. http://www.che.ttu.edu/pcoc/software/ppt/Chap09.ppt (chapter on Bode plot analysis of linear systems with stability criteria). 11. http://eprints.iisc.ernet.in/13500/1/lec_5_web.pdf (lecture on Bode plots and second order linear systems)
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