and division (stretch).

Transcription

1 Filterg AC signals Topi areas Eletrial and eletroni engeerg: AC Theory. Resistane, reatane and impedane. Potential divider an AC iruit. Low pass and high pass filters. Mathematis: etor addition. Trigonometry. Pythagoras theorem. Phasor multipliation and division (streth). Prerequisites It may be useful to look at the resoure AC phasors and fault detetion to trodue phasor diagrams and the relationship between resistane, reatane and impedane. Problem statement An alternatg urrent is not neessarily assoiated solely with a power supply. Analogue signals from sensors and struments an also be thought of as beg AC signals, although, ontrast to a power supply, the frequeny usually hanges with time. An example is the reordg and playg bak of sounds usg a mirophone and speaker. However, sometimes suh signals are ontamated with unwanted omponents, suh as high-frequeny mirophone hiss, or low frequeny mas hum, whih you want to filter to improve the sound quality. Another example ould be that when playg bak sounds, you want to selet and diret low frequeny sounds to a large bass speaker and higher frequeny sounds to smaller tweeter speakers to take advantage of the performane harateristis of eah. Yet another example is the detetion and deodg of radio ommuniation signals, where the reeiver selets a relatively narrow range of radio frequenies, then, general, filters the radio arrier signal to reover the origal sound or data signal. How an you selet signals based on their frequeny? Filterg AC signals 1

2 Ativity 1 Disussion Disuss some examples of where you might want to filter high or low frequenies Bakground The iruit below shows a potential divider iruit usg two resistors R 1 R 2 Figure 1 A potential divider The total resistane the iruit on the left is given by R tot = R 1 + R 2. Usg Ohm s law, the urrent flowg through the resistors is related to the voltage aross them,, as = IR I = IR tot R R + R = = tot 1 2 As the resistors are series, the same urrent must flow through eah. Ohm s law an be used on eah resistor to alulate the voltage drop aross it (the sum of the voltage drops aross eah resistor must equal the total voltage drop, ). The diagram shows that the voltage drop aross R 2 is denoted as, so that = IR2 = R + R 1 2 R Rearrangg to give R2 =. R + R This gives the ratio of the magnitude of the put to the put. A small value means that the put voltage is small ompared with the put voltage. A large value means that the put voltage is omparable with the put voltage. 2 Filterg AC signals

3 Ativity 2 Replag a resistor with a apaitor R C Figure 2 A potential divider for AC signals usg a resistor and a apaitor Figure 2 shows a potential divider for an AC iruit where R 2 Figure 1 has been replaed by a apaitor. The reatane of a apaitor of apaitane C for an AC signal of frequeny f, X is given by X C 1 = 2 fc 1 Fd the total impedane of the iruit Figure 2, Z, when C = 1 mf and. Ht Remember the impedane is alulated by fdg the vetor sum of the resistane and reatane, see resoure 2 Look at the expression for the bakground disussion above. Use this, and the expression you used to alulate the total impedane question 1 above to write an expression for the ratio terms of R and X. Filterg AC signals

4 Ativity Interative The resoure shows the iruit shown Figure 2 and plots its harateristis for a full range of signal frequenies from 1 Hz to 20 khz on a logarithmi frequeny sale. Additionally, you an slide the frequeny slider to obta a set of values of X, Z and for a partiular frequeny. Figure Sreen shot of resoure The lower plot shows for a range of frequenies; a value of one means the frequeny is passed, while a lower value shows reasg filterg. The table shows spot values dependg on the frequeny marker value. 1 Use the resoure to fill Table 1. Disuss how hangg the value of C affets the results. Case a Case b Case C = 0.1 mf, C = 10 mf, X Z Table 1 4 Filterg AC signals

5 2 Use the resoure to fill Table 2. Disuss how hangg the value of f affets the results. Case a Case b Case f = 10 Hz f = 1 khz X Z Table 2 Look at the results Table 2. Why ould the iruit Figure 2 be desribed as a low pass filter? The red le on the resoure shows the frequeny at whih the magnitude of the put is ab times the magnitude of the put. In fat it shows when the magnitude of the put is 1 2 times the magnitude of the put. Tehnially, this is the frequeny at whih the power the put signal is half the power of the put signal, where the power an AC signal is related to its root mean square (RMS) value (not overed this resoure). 4 A signal of frequeny greater than 2. khz is mixed with the desired signal. Fd a pair of values for R and C whih give a ut off frequeny near this value. Is the pair of values unique? 5 Disuss whether the shape of the filter behaviour plotted by the resoure (lower-left plot) is ideal? Ativity 4 A high pass filter Clik the button marked on the resoure This swaps the position of the resistor and the apaitor. The filter now passes high frequeny signals and filters low frequeny ones, as shown by the lower plot. 1 What do you notie ab the ut-off frequeny of a high-pass filter and a low-pass filter when usg the same R and C values? 2 Experiment with the values of R and C and observe how the ut off frequeny is affeted. A mas hum of 50 Hz is beg mixed with the signal. Fd a pair of values for R and C whih give a ut off frequeny near this value. Is the pair of values unique? 4 Write an expression for the ratio terms of R and X for the iruit the new onfiguration and expla why the iruit now filters low frequenies. Filterg AC signals 5

6 Bakground Band pass filter A signal that is first passed through a high pass filter with a ut off frequeny f 1 will ut off all frequenies less than f 1. Passg this filtered signal through a low pass filter with a ut off frequeny of f 2, will ut off all the remag frequenies greater than f 2. The ombed effet is to pass a frequeny band between than f 1 and f 2. Note, the graph below is a sketh only. A full analysis of iruit that ombes a high pass filter with a low pass filter would need to onsider how all the omponents terated a sgle iruit High pass Low pass Combed Figure 4 A band pass filter f1 f Frequeny (Hz) Suh a filter is used the detetion and deodg of radio ommuniation signals, where the reeiver selets a relatively narrow range of radio frequenies to proess to reover the origal sound or data signal. Streth and hallenge ativity The use of reative elements (apaitors and dutors) a iruit hanges the phase of the voltage with respet to the urrent, whih passive omponents (resistors) do not. A hange of phase is also seen filter iruits, where the put signal is of phase with the put signal. The phase angle is shown the upper plot of the resoure The amount of shift depends on the frequeny of the signal. An animated view of the effet of the phase shift an be seen by likg the button. 6 Filterg AC signals

7 Figure 5 Sreen shot of resoure 1 What do you notie ab the phase shift for a low pass and a high pass filter as the frequeny hanges? What is the phase shift at the ut-off frequeny? 1 2 The phase shift for a low pass filter is given by tan ( 2 fcr) =. erify that the above onfiguration of R = 4.7 kω, C = 0 nf and gives a phase shift of The phase shift for a high pass filter is given by = tan. erify that usg the 2 fcr 1 1 above omponents a high pass filter and signal of 100 Hz gives a phase shift of Usg phasor values for the resistane, reatane and impedane, and the expressions for you found earlier verify that the angles 2 and above are orret by performg a phasor division. Ht To multiple the phasor a b by the phasor d you multiply the magnitudes and add the angles, ( ) a b d = a b + d. To divide the phasor a b by the phasor d you divide the magnitudes and subtrat the angles, a b d a = ( b d ). Filterg AC signals 7

8 Notes and solutions Ativity 1 A number of examples have been given the trodution suh as noise removal from a mirophone, sendg high or low frequenies to appropriate speakers for optimal sound put and frequeny seletion for radio. Other examples lude Ativity 2 Broadband ternet data is transmitted on the same le as a voie land-le. To prevent terferene between the voie signal and the broadband data signal, an ADSL filter is attahed between the telephone let and the broadband rer. Hearg aids may selet a partiular range of frequenies assoiated with speeh that have been lost by a person and seletively amplify them. They may also filter other frequenies not assoiated with speeh to further larify the sound heard by the user. Ative prostheti limbs that are ontrolled by urrent sensors plaed on the sk need to respond only to the small deliberate biologial signals and must filter other signal soures suh as power les. 1 When C = 1 mf and, the reatane of the apaitor is X C = k d.p. 6 2 fc = = ( ) The total impedane of the iruit is given by the vetor sum of the resistane and apaitive reatane, see Figure 6. R X Z Figure 6 Usg resistane values kω, Z R X = + = + = = ( ) k d.p. 8 AC Filterg power AC and signals power fator

9 2 The iruit Figure 2 is similar to the iruit Figure 1, but with R 1 replaed by R R 2 replaed by C whih will replae a resistane of value R 2 by a reatane of X C. Similarly, the total 2 2 resistane of R tot = R 1 + R 2 will be replaed by a total impedane of Z = R + X. The expression for the ratio will be similarly affeted: Figure 1 iruit Figure 2 iruit R R = = R R R X X = = 2 2 C C 2 2 tot Z R + X Ativity 1 Case a Case b Case C = 0.1 mf, C = 10 mf, X kω ( d.p) kω ( d.p) Ω ( d.p) Z kω ( d.p) kω ( d.p) 1.01 kω ( d.p) ( d.p) ( d.p) Table 1 Results. Note than 0.1 mf = 100 nf Usg the entral results olumn (Case b) as a referene: When the value of the apaitor is redued by a fator 10 (gog from Case b to Case a), the apaitive reatane reases by a fator of 10. The total impedane does not rease by a fator of 10 as it is alulated as the vetor sum of the resistane and reatane. The ratio of reases from to 0.998, diatg that the magnitude of the put signal reases. When the value of the apaitor is reased by a fator 10 (gog from Case b to Case ), the apaitive reatane dereases by a fator of 10. The total impedane does not derease by a fator of 10 as it is alulated as the vetor sum of the resistane and reatane. The ratio of dereases from to 0.157, diatg that the magnitude of the put signal dereases. As a general pattern: A small value of C leads to a large apaitive reatane X C, whih domates the vetor sum used to alulate the total impedane, Z, for the R value used. In this ase the magnitude of the impedane is similar to the magnitude of the apaitive reatane. A large value of C leads to a small apaitive reatane X C, whih is domated the vetor sum used to alulate the total impedane, Z, by the resistor value, R. In this ase the magnitude of the impedane is similar to the magnitude of the resistane. Filterg AC signals 9

10 Table 2 Results 2 Case a Case b Case f = 10 Hz f = 1 khz X kω ( d.p) kω ( d.p) Ω ( d.p) Z kω ( d.p) kω ( d.p) 1.01 kω ( d.p) ( d.p) ( d.p) Usg the entral results olumn (Case b) as a referene: When the value of the frequeny is redued by a fator 10 (gog from Case b to Case a), the apaitive reatane reases by a fator of 10. The total impedane does not rease by a fator of 10 as it is alulated as the vetor sum of the resistane and reatane. The ratio of reases from to 0.998, diatg that the magnitude of the put signal reases. Note, this gives the same result as keepg the frequeny at 100 Hz and redug the Capaitane by a fator of 10, whih is not 1 surprisg when you onsider the expression for apaitive reatane is X C =, i.e. 2 fc the effets of f and C are multiplied the denomator. When the value of the frequeny is reased by a fator 10 (gog from Case b to Case ), the apaitive reatane dereases by a fator of 10. The total impedane does not derease by a fator of 10 as it is alulated as the vetor sum of the resistane and reatane. The ratio of dereases from to 0.157, diatg that the magnitude of the put signal dereases. As a general pattern: A small value of f leads to a large apaitive reatane X C, whih domates the vetor sum used to alulate the total impedane, Z, for the R value used. In this ase the magnitude of the impedane is similar to the magnitude of the apaitive reatane. A large value of f leads to a small apaitive reatane X C, whih is domated the vetor sum used to alulate the total impedane, Z, by the resistor value, R. In this ase the magnitude of the impedane is similar to the magnitude of the resistane. From the results above it an be seen that for small values of f, X will be large and will be lose to 1. For large values of f, X will be small and will be lose to zero. This behaviour gives a simple filter low frequenies will give an put that is similar magnitude to the put, i.e. they will be passed. High frequenies will give an put that is lose to zero, i.e. they will be filtered. Beause of this behaviour the iruit is alled a low pass filter. 4 R = 680 Ω, C = 100 nf gives a ut-off frequeny of 2.41 khz. However, this is not a unique pair, for example R = 1 kω and C = 68 nf gives the same ut-off frequeny. There are other suh pairs. 5 The filter behaviour isn t ideal. An ideal filter would pass all frequenies below 2.41 khz and blok all frequenies above it, so the frequeny response would be a step funtion. The resoure shows that stead of behavg like this, the iruit gradually redues how muh of the signal above the ut-off frequeny is passed. See Figure AC Filterg power AC and signals power fator

11 Atual filter Ideal filter Frequeny (Hz) Figure 7 Ideal and atual filter behaviour Ativity 4 1 For a given set of R or C values the ut-off frequeny for a high pass filter is the same as the ut-off frequeny for a low pass filter. 2 As with the low pass filter, reasg R or C redues the ut-off frequeny. Redug R or C reases the ut-off frequeny. R = 4.7 kω, C = 680 nf gives a ut-off frequeny of Hz. However, this is not a unique pair, for example R = 6.8 kω and C = 470 nf gives the same ut-off frequeny. There are other suh pairs This iruit still has the same impedane, Z = R + X. However, the put is given as the voltage drop aross the resistor rather than the apaitor so that R R = = Z 2 2 R + X. For small values of f, X will be large and will be lose to 0, i.e. will be small. For large values of f, X will be small and will be lose to 1, i.e. will be lose to. Filterg AC signals 11

12 Streth and Challenge 1 A low pass filter has a negative phase shift and the angle of shift beomes more negative as the magnitude of signal that is passed dereases, i.e. the shift is small for frequenies below the ut-off frequeny and beomes more negative for frequenies higher than the ut-off frequeny. The shift is never less than -90. This is shown on the plot as the red trae for laggg behd the blue trae for. The ut-off frequeny oides with a phase shift of -45. A high pass filter has a positive phase shift and the angle of shift beomes larger as the magnitude of signal that is passed dereases, i.e. the shift is small for frequenies above the ut-off frequeny and reases for frequenies lower than the ut-off frequeny. The shift is never more than 90. This is shown on the plot as the red trae for leadg the blue trae for. The ut-off frequeny oides with a phase shift of and Calulator exerise. 4 X C X C For the low pass filter = = 2 2 Z R + X. The phase angle for X is -90. The phase angle 1 X for Z is given by tan R. Usg the values from the image R = 4.7 kω X = 4.82 kω Z = 6.74 kω The phase angle for Z is tan = Note, the negative sign of X this expression; the phasor diagram, apaitive reatane pots down. The phasor division is now X C = = = 90 ( 45.7 ) = Z Whih gives the orret value of and the orret phase shift angle. For the high pass filter R =. The phasor division for this is Z R = = = 0 ( 45.7 ) = Z Aga givg the orret value of and the orret phase shift angle. Royal Aademy of Engeerg Pre Philip House, Carlton House Terrae, London SW1Y 5DG Tel: +44 (0) Registered harity number Filterg AC signals