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# Filter dengan-op-amp

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### Filter dengan-op-amp

1. 1. Active Filters
2. 2. FiltersA filter is a system that processes a signal in some desired fashion. • A continuous-time signal or continuous signal of x(t) is a function of the continuous variable t. A continuous-time signal is often called an analog signal. • A discrete-time signal or discrete signal x(kT) is defined only at discrete instances t=kT, where k is an integer and T is the uniform spacing or period between samples
3. 3. Types of Filters There are two broad categories of filters: • An analog filter processes continuous-time signals • A digital filter processes discrete-time signals. The analog or digital filters can be subdivided into four categories: • Lowpass Filters • Highpass Filters • Bandstop Filters • Bandpass Filters
4. 4. Analog Filter ResponsesH(f) H(f) 0 f 0 f fc fc Ideal “brick wall” filter Practical filter
5. 5. Ideal Filters Lowpass Filter Highpass Filter M(ω) Stopband Passband Passband Stopband ωc ω ωc ω Bandstop Filter Bandpass Filter M(ω) Passband Stopband Passband Stopband Passband Stopband ωc ωc ω ωc ωc ω 1 2 1 2
6. 6.  There are a number of ways to build filters and of these passive and active filters are the most commonly used in voice and data communications.
7. 7. Passive filters Passive filters use resistors, capacitors, and inductors (RLC networks). To minimize distortion in the filter characteristic, it is desirable to use inductors with high quality factors (remember the model of a practical inductor includes a series resistance), however these are difficult to implement at frequencies below 1 kHz. • They are particularly non-ideal (lossy) • They are bulky and expensive
8. 8.  Active filters overcome these drawbacks and are realized using resistors, capacitors, and active devices (usually op-amps) which can all be integrated: • Active filters replace inductors using op-amp based equivalent circuits.
9. 9. Op Amp Advantages Advantages of active RC filters include: • reduced size and weight, and therefore parasitics • increased reliability and improved performance • simpler design than for passive filters and can realize a wider range of functions as well as providing voltage gain • in large quantities, the cost of an IC is less than its passive counterpart
10. 10. Op Amp Disadvantages Active RC filters also have some disadvantages: • limited bandwidth of active devices limits the highest attainable pole frequency and therefore applications above 100 kHz (passive RLCfilters can be used up to 500 MHz) • the achievable quality factor is also limited • require power supplies (unlike passive filters) • increased sensitivity to variations in circuit parameters caused by environmental changes compared to passive filters For many applications, particularly in voice and data communications, the economic and performance advantages of active RC filters far outweigh their disadvantages.
11. 11. Bode Plots Bode plots are important when considering the frequency response characteristics of amplifiers. They plot the magnitude or phase of a transfer function in dB versus frequency.
12. 12. The decibel (dB)Two levels of power can be compared using aunit of measure called the bel. P2 B = log10 P1The decibel is defined as: 1 bel = 10 decibels (dB)
13. 13. P2 dB = 10 log10 P1A common dB term is the half power pointwhich is the dB value when the P2 is one-half P1. 1 10 log10 = −3.01 dB ≈ −3 dB 2
14. 14. LogarithmsA logarithm is a linear transformation used to simplify mathematical and graphical operations. A logarithm is a one-to-one correspondence.
15. 15. Any number (N) can be represented as abase number (b) raised to a power (x). N = (b) x The value power (x) can be determined by taking the logarithm of the number (N) to base (b). x = log b N
16. 16.  Although there is no limitation on the numerical value of the base, calculators are designed to handle either base 10 (the common logarithm) or base e (the natural logarithm). Any base can be found in terms of the common logarithm by: 1 log q w = log10 w log10 q
17. 17. Properties of Logarithms The common or natural  The log of the quotient logarithm of the number of two numbers is the 1 is 0. log of the numerator The log of any number minus the less than 1 is a negative denominator. number.  The log a number The log of the product of taken to a power is two numbers is the sum equal to the product of of the logs of the the power and the log numbers. of the number.
18. 18. Poles & Zeros of the transferfunction pole—value of s where the denominator goes to zero. zero—value of s where the numerator goes to zero.
19. 19. Single-Pole Passive Filter R vout ZC 1 / sC = = vin R + Z C R + 1 / sCvin C vout 1 1 / RC = = sCR + 1 s + 1 / RC Firstorder low pass filter Cut-off frequency = 1/RC rad/s Problem : Any load (or source) impedance will change frequency response.
20. 20. Single-Pole Active Filter R vin C vout Same frequency response as passive filter. Buffer amplifier does not load RC network. Output impedance is now zero.
21. 21. Low-Pass and High-PassDesigns High Pass Low Passvout 1 1 = =vin 1 1 + sRC vout 1 / RC +1 sCR sCR = sRC s vin s + 1 / RC = = RC ( s + 1 / RC ) ( s + 1 / RC )
22. 22. To understand Bode plots, you need to use Laplace transforms! R The transfer function of the circuit is: V in (s) Vo ( s ) 1 / sC 1Av = = = Vin ( s ) R + 1 / sC sRC + 1
23. 23. Break FrequenciesReplace s with jω in the transfer function: 1 1 1 Av ( f ) = = = jωRC + 1 1 + j 2πRCf  f  1 + j  f   bwhere fc is called the break frequency, or cornerfrequency, and is given by: 1 fc = 2πRC
24. 24. Corner Frequency The significance of the break frequency is that it represents the frequency where Av(f) = 0.707∠-45°. This is where the output of the transfer function has an amplitude 3-dB below the input amplitude, and the output phase is shifted by -45° relative to the input. Therefore, fc is also known as the 3-dB frequency or the corner frequency.
25. 25. Bode plots use a logarithmic scale forfrequency. One decade 10 20 30 40 50 60 70 80 90 100 200where a decade is defined as a range offrequencies where the highest and lowestfrequencies differ by a factor of 10.
26. 26.  Consider the magnitude of the transfer function: 1 Av ( f ) = 1 + ( f / fb ) 2 Expressed in dB, the expression isAv ( f ) dB = 20 log 1 − 20 log 1 + ( f / f b ) 2 [ = −20 log 1 + ( f / f b ) = −10 log 1 + ( f / f b ) 2 2 ] = −20 log( f / f b )
27. 27.  Lookhow the previous expression changes with frequency: • at low frequencies f<< f , |Av| b dB = 0 dB • low frequency asymptote • at high frequencies f>>f , b |Av(f)|dB = -20log f/ fb • high frequency asymptote
28. 28. Note that the two asymptotes intersect at fbLow frequency asymptote Magnitude where |Av(fb )|dB = -20log f/ fb −3 dB 0 20 Actual response curve20 . log ( P ( ω ) ) 40 High frequency asymptote 60 0.1 1 10 100 ω rad sec
29. 29.  The technique for approximating a filter function based on Bode plots is useful for low order, simple filter designs More complex filter characteristics are more easily approximated by using some well-described rational functions, the roots of which have already been tabulated and are well-known.
30. 30. Real Filters The approximations to the ideal filter are the: • Butterworth filter • Chebyshev filter • Cauer (Elliptic) filter • Bessel filter
31. 31. Standard Transfer Functions Butterworth • Flat Pass-band. • 20n dB per decade roll-off. Chebyshev • Pass-band ripple. • Sharper cut-off than Butterworth. Elliptic • Pass-band and stop-band ripple. • Even sharper cut-off. Bessel • Linear phase response – i.e. no signal distortion in pass-band.
32. 32. Butterworth FilterThe Butterworth filter magnitude is defined by: 1 M (ω ) = H ( jω ) = (1 + ω ) 2 n 1/ 2 where n is the order of the filter.
33. 33. From the previous slide: M (0) = 1 1 M (1) = for all values of n 2For large ω: 1 M (ω ) ≅ n ω
34. 34. And 20 log10 M (ω ) = 20 log10 1 − 20 log10 ω n = −20n log10 ω implying the M(ω) falls off at 20n db/decade for large values of ω.
35. 35. 10 1T1 i 20 db/decadeT2 iT3 i 40 db/decade 0.1 60 db/decade 0.01 0.1 1 10 w i 1000
36. 36. To obtain the transfer function H(s) from the magnitude response, note that 2 1M (ω ) = H ( jω ) = H ( jω ) H (− jω ) = 2 1+ ω ( ) 2 n
37. 37. Because s = jω for the frequency response, we have s2 = − ω2. 1 1 H ( s) H (− s) = = ( 1+ − s ) 2 n 1 + ( − 1) s n 2n The poles of this function are given by the roots of1 + ( − 1) s − j ( 2 k −1)π n 2n = −1 = e , k = 1,2,  ,2n
38. 38. The 2n pole are: e j[(2k-1)/2n]π n even, k = 1,2,...,2n sk = e j(k/n)π n odd, k = 0,1,2,...,2n-1Note that for any n, the poles of the normalized Butterworthfilter lie on the unit circle in the s-plane. The left half-planepoles are identified with H(s). The poles associated withH(-s) are mirror images.
39. 39. Recall from complex numbers that the rectangular formof a complex can be represented as: z = x + jyRecalling that the previous equation is a phasor, we canrepresent the previous equation in polar form: z = r ( cosθ + j sin θ )where x = r cosθ and y = r sin θ
40. 40. Definition: If z = x + jy, we define e z = e x+ jy to be thecomplex number e = e (cos y + j sin y ) z xNote: When z = 0 + jy, we have e jy = (cos y + j sin y )which we can represent by symbol: jθ e
41. 41. The following equation is known as Euler’s law. jθ e = (cosθ + j sin θ )Note that cos( − θ ) = cos( − θ ) even function sin ( − θ ) = − sin (θ ) odd function
42. 42. This implies that − jθ e = (cosθ − j sin θ )This leads to two axioms: jθ − jθ e +e jθ − jθ e −ecos θ = and sin θ = 2 2j
43. 43.  Observe that e jθ represents a unit vector which makes an angle θ with the positivie x axis.
44. 44. Find the transfer function that corresponds to a third-order(n = 3) Butterworth filter.Solution:From the previous discussion: sk = e jkπ/3, k=0,1,2,3,4,5
45. 45. Therefore, s0 = e j0 jπ / 3 s1 = e j 2π / 3 s2 = e jπ s3 = e j 4π / 3 s4 = e j 5π / 3 s5 = e
46. 46. The roots are: p1 1 p6 1 p2 .5 .. 0.8668j p5 .5 0.866 j p3 .5 . . 0.866j p4 .5 0.866 j
47. 47. 2Im p i 2 0 2 2 Re p i
48. 48. Using the left half-plane poles for H(s), we get 1H ( s) = ( s + 1)( s + 1 / 2 − j 3 / 2)( s + 1 / 2 + j 3 / 2) which can be expanded to: 1 H ( s) = ( s + 1)( s + s + 1) 2
49. 49.  The factored form of the normalized Butterworth polynomials for various order n are tabulated in filter design tables.
50. 50. n Denominator of H(s) for Butterworth Filter1 s+12 s2 + 1.414s + 13 (s2 + s + 1)(s + 1)4 (s2 + 0.765 + 1)(s2 + 1.848s + 1)5 (s + 1) (s2 + 0.618s + 1)(s2 + 1.618s + 1)6 (s2 + 0.517s + 1)(s2 + 1.414s + 1 )(s2 + 1.932s + 1)7 (s + 1)(s2 + 0.445s + 1)(s2 + 1.247s + 1 )(s2 + 1.802s + 1)8 (s2 + 0.390s + 1)(s2 + 1.111s + 1 )(s2 + 1.663s + 1 )(s2 + 1.962s + 1)
51. 51. Frequency Transformations
52. 52. So far we have looked at the Butterworth filterwith a normalized cutoff frequency ω c = 1 rad / secBy means of a frequency transformation, wecan obtain a lowpass, bandpass, bandstop, orhighpass filter with specific cutoff frequencies.
53. 53. Lowpass with Cutoff Frequencyωu Transformation: sn = s / ω u
54. 54. Highpass with Cutoff Frequencyωl Transformation: sn = ω l / s
55. 55. Bandpass with CutoffFrequencies ωl and ωu Transformation: F G s2 + ω 2 ω 0 s ω 0 I J sn = = H + K 0 Bs B ω0 s where ω 0 = ω uω l B = ωu −ωl
56. 56. Bandstop with CutoffFrequencies ωl and ωu Transformation: Bs B sn = 2 s +ω0 2 = Fs + ω I ω G J H sK 0 ω 0 0
57. 57. Active Filters To achieve a gain > 1, we need to utilize an op-amp based circuit to form an active filter We can design all the common filter types using op-amp based active filters Recall for an ideal op-amp that V+ = V– I+ = I – = 0 Lect23 EEE 202 61
58. 58. Op Amp ModelNon-inverting V+ input + Rout Vo Rin + – – Inverting input A(V+ –V– ) V– Lect23 EEE 202 62
59. 59. Non-inverting Op-Amp Circuit + Vout H ( jω ) = – + Vin Z2Vin + = 1+ – Z2 Vout Z1 Z1 – Lect23 EEE 202 63
60. 60. Inverting Op-Amp Circuit Z2 Z1 – – Vin + + – Vout + Vout Z 2H ( jω ) = = Vin Z1 Lect23 EEE 202 64
61. 61. An Integrator C R – + + Vin + – Vout –Earlier in the semester, we saw this op-amp based integrator circuit. What type offilter does it create? Does this help you to understand time and frequency domaininterrelations? Lect23 EEE 202 65
62. 62. A Differentiator R C – + + Vin + – Vout –We also observed this op-amp based differentiator circuit. What type offiltering does it produce? How could we move the zero away from the origin? Lect23 EEE 202 66
63. 63. Practical Example Youare shopping for a stereo system, and the following specifications are quoted: • Frequency range: –6 dB at 65 Hz and 22 kHz • Frequency response: 75 Hz to 20 kHz ±3 dB What do these specs really mean? • Note that the human hearing range is around 20 Hz to 20 kHz (audible frequencies) Could you draw a rough Bode (magnitude) plot for the stereo system? Lect23 EEE 202 67
64. 64. MATLAB Filter Example How can we filter a real measurement signal? One method involves using a numerical algorithm called the Fast Fourier Transform (FFT) which converts a time domain signal into the frequency domain Start MATLAB, then download and run the ‘EEE202Filter.m’ file • Can you observe the reduction in the high frequency components in both the time and frequency domain plots of the output signal?68 Lect23 EEE 202
65. 65. MATLAB Filter Example (cont’d) Can we extend the knowledge acquired about transfer functions and filtering? • How can we further reduce the high frequency component magnitudes? Show me. • How can we remove more high frequencies from the output signal? Show me. Fast Freq.Input Inverse Output Fourier DomainSignal FFT Signal Transform Filtering Lect23 EEE 202 69