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Lecture 23 Filters Hung-yi Lee Filter Types Lowpass filter Bandpass filter wco : cutoff frequency Bandwidth B = wu - wl Highpass filter Notch filter Real World Ideal filter Transfer Function – Rules • Filter is characterized by its transfer function H s = N s D s =K s z1 s z 2 s z m s p1 s p 2 s p n The poles should be at the left half of the s-plane. We only consider stable filter. Given a complex pole or zero, its complex conjugate is also pole or zero. Transfer Function – Rules • Filter is characterized by its transfer function H s = N s D s =K s z1 s z 2 s z m s p1 s p 2 s p n n m :improper filter As the frequency increase, the output will become infinity. n m :proper filter We only consider proper filer. The filters consider have more poles than zeros. Filter Order H s = N s D s =K s z1 s z 2 s z m s p1 s p 2 s p n Order = n The order of the denominator is the order of the filter. order=1 order=4 order=4 Outline • Textbook: Chapter 11.2 Second-order Filter First-order Filters Lowpass Filter Highpass Filter Lowpass Filter Highpass Filter Bandpss Filter Notch Filter First-order Filters Firsr-order Filters H s = N s zero or first order first order D s 0 or 1 zero 1 pole Case 1: Case 2: 1 pole, 0 zero 1 pole, 1 zero H s = K 1 s p H s = K sz s p Firsr-order Filters - Case 1 Lowpass filter H s = K w 1 s p As ω increases Magnitude decrease Phase decrease Pole p is on the negative real axis Firsr-order Filters - Case 1 • Amplitude of the transfer function of the first-order low pass filter Ideal Lowpass filter First-order Lowpass filter Firsr-order Filters - Case 1 • Find cut-off frequency ωco of the first-order low pass filter w At DC Lowpass filter H s = K 0 = K 1 s p 1 | p| Find cut-off frequency ωco such that | p| w co = K 1 | p| 1 2 w co = | p | Firsr-order Filters - Case 2 H s = K sz s p w Zero can be positive or negative |z| | p| Case 2-1: Absolute value of zero is smaller than pole Magnitude is proportional to the length of green line divided by the length of the blue line Low frequency ≈ |z|/|p| Because |z|<|p| The low frequency signal will be attenuated If z=0, the low frequency can be completely block Not a low pass Firsr-order Filters - Case 2 H s = K sz s p w Case 2-1: Absolute value of zero is smaller than pole Magnitude is proportional to the length of green line divided by the length of the blue line High frequency The high frequency signal will pass |z| | p| If z=0 (completely block low frequency) H s = K s s p High pass First-order Filters - Case 2 H s = K s s p • Find cut-off frequency ωco of the first-order high pass filter w = K | jw | w co | jw p | w |p| 2 2 co = K w co = K 2 = 1 2 w co = | p | | j w co | | j w co - p | = K 2 (the same as low pass filter) First-order Filters - Case 2 H s = K sz s p w Case 2-2: Absolute value of zero is larger than pole Low frequency ≈ |z|/|p| Because |z|>|p| The low frequency signal will be enhanced. High frequency: magnitude is 1 The high frequency signal will pass. Neither high pass nor low pass First-order Filters vh 1 H lp s = 1 sC R = 1 1 sRC 1 sC 1 vl 1 H lp j w = w = RC jw 1 Lowpass filter If vh is output Highpass filter 0 = 1 = 0 RC 1 2 w RC RC If vl is output RC 1 s RC 1 sC Consider vin as input = w co = 1 2 (pole) RC s H hp s = 1 H lp s = s 1 RC 0 = 0 = 1 w co = 1 RC (pole) First-order Filters H lp s = vh R R sL R = L s R L R R H lp j w = vl L jw R w = L R w L 2 L 0 = 1 = 0 w co = H hp s = 1 H lp s = 2 s s R L R L (pole) Cascading Two Lowpass Filters V out Vx V in H lp 1 s = H lp s = Vx V in H lp 2 s = H lp 1 s V out Vx H lp 2 s Cascading Two Lowpass Filters V out Vx V in 1 1 H lp s = = sC 1 R1 1 sC 1 sC 2 R2 = 1 1 sC 1 R1 1 sC 2 1 1 s C 1 R 1 C 2 R 2 s C 1 C 2 R 1 R 2 2 1 sC 2 R 2 1 Cascading Two Lowpass Filters V out Vx V in Z eq s The first low pass filter is influenced by the second low pass filter! H lp 1 s = Vx V in = Z eq s Z eq s R1 Cascading Two Lowpass Filters V out Vx V in 1 Z eq s = || R 2 sC 1 sC 2 1 1 R2 sC 1 sC 2 = 1 1 R 2 sC 1 sC 2 Z eq s 1 2 s C 1C 2 2 s C 1C 2 = sC 2 R 2 1 s C 1 C 2 s C 1 C 2 R 2 2 Cascading Two Lowpass Filters V out Vx V in Z eq s H lp s = H lp 1 s H lp 2 s = = 1 Z eq s Z eq s R1 sC 2 1 sC 2 1 1 s R 1C 1 C 2 R 2 R 1C 2 s C 1C 2 R 1 R 2 2 R2 Second-order Filters Second-order Filter H s = N s 0, 1 or 2 zeros D s Second order Must having two poles 2 poles Case 1: No zeros Case 2: One zeros Case 3: Two zeros Second-order Filter – Case 1 Case 1-1 Case 1-2 w w p1 p1 p2 p2 Second-order Filter – Case 1 Case 1-1 Real Poles w The magnitude is l1 The magnitude monotonically decreases. l1 p1 p2 l1l 2 As ω increases l2 l2 K Decrease faster than first order low pass Second-order Filter – Case 1 Case 1-2 Complex Poles w The magnitude is l1 As ω increases, l2 p1 l1 l2 p2 K l1l 2 l1 decrease first and then increase. l2 always decrease What will happen to magnitude? 1. Increase 2. Decrease 3. Increase, then decrease 4. Decrease, then increase Second-order Filter – Case 1 Case 1-2 Complex Poles w If ω > ωd l1 l2 p1 wd wd p2 l1 and l2 both increase. The magnitude must decrease. What will happen to magnitude? 1. Increase 2. Decrease 3. Increase, then decrease 4. Decrease, then increase Second-order Filter – Case 1 Case 1-2 Complex Poles l1 p1 wd wd p2 When ω < ωd w w l2 l1 = w d w l2 = w d w 2 2 2 2 Maximize the magnitude K l1l 2 Minimize l1l 2 = 2 wd w 2 2 wd w 2 Second-order Filter – Case 1 Minimize l l = w w w w Minimize f w = w w w w df w = 2 w w 1 w w 2 w w w dw w w 1 w w w w w w = 0 2 2 1 2 2 2 d d 2 2 2 2 d d 2 2 d 2 2 2 w w d w d w 2 2 w d = 0 w wd w 2 = wd w 2 =0 d w w d w 2w = 0 2 w 2 2 d w wd w d 2 d 2 w w d d 2 d 2 d 2 w wd 2 = 0 w d w w d w = 0 2 w = wd 2 2 (maximize) Second-order Filter – Case 1 w = wd 2 2 Lead to maximum The maxima exists when w d w Peaking w p1 p1 wd wd wd wd p2 wd p2 No Peaking wd Peaking Second-order Filter – Case 1 w = wd 2 2 Lead to maximum The maxima exists when w d wd l1 w wd p2 w = w p1 =K l2 Peaking K l1l 2 1 w d w 2 2 w d w 2 2 Assume w = w d2 2 w = K 1 2w 0 = K d 1 wd 2 2 Second-order Filter – Case 1 K H s = s 2 w0 Q sw 2 0 w0 Q p1 , p 2 = w0 Q 2 4 w 02 2 For complex poles p1 w0 wd wd 4 w 02 0 = w0 wd = w0 2 Q wd = w0 1 2Q w0 p2 w0 Q 2 2 1 2 1 4Q 2 Second-order Filter – Case 1 H s = s 2 w0 Q sw jw 2 0 w = w0 = Q times p1 K H jw = K w w0 wd wd w0 Q K w 2 jw w 0 2 j w0 w Q K H jw 0 = w = w0 w0 2 j w0 p2 2 0 2 Q w 0 = K w 2 0 Q 0 = K w0 2 Q times of DC gain Second-order Filter – Case 1 w = w 2 d 2 K H s = s 2 w0 w = K Lead to maximum sw Q For complex poles 2 0 w0 Q = p1 , p 2 = w0 Q w0 Q 2 2 w0 2Q wd = w0 1 2w 4 w 02 2 4 w 02 0 1 1 Q 2 1 4Q 2 d Second-order Filter – Case 1 w = w 2 d 2 H s = s 2 w0 = s w0 2 Q w = w0 1 w = K Lead to maximum K 1 2Q 2 w0 wd = w0 1 2Q 1 2w d 1 4Q 2 Lead to maximum The maximum value is K The maximum exist when Q 1 2 1 2w = 0 . 707 = K d 1 Q w0 2 1 1 4Q 2 Second-order Filter – Case 1 Case 1-1 Real Poles Case 1-2 Complex Poles w w p1 p1 p2 wd wd Q 0 .5 (No Peaking) p2 0 . 707 Q 0 . 5 Which one is considered as closer to ideal low pass filter? K H s = s 2 w0 Q Q 0 .5 s w0 2 Q = 1 Complex poles = 0 . 707 2 (Butterworth filter) Q 0 . 707 Peaking Butterworth – Cut-off Frequency K H s = s 2 w0 s w0 H jw = = 0 = K s 2 = 0 . 707 2 2 Q = Q = 1 2w 0 s w 0 2 K w0 2 w co = K j w 2 w 2w 0 jw w 0 2 K 2 0 w 2 j 2w 0 w 1 K 2 w0 2 w co = w 0 ω0 is the cut-off frequency for the second-order lowpass butterworth filter (Go to the next lecture first) Second-order Filter – Case 2 Case 2: 2 poles and 1 zero Case 2-1: 2 real poles and 1 zero w z1 p 2 p1 w p2 z1 p1 w p 2 p1 z1 Second-order Filter – Case 2 Case 2: 2 poles and 1 zero Case 2-1: 2 real poles and 1 zero w p 2 p1 z1 flat Bandpass Filter Second-order Filter – Case 2 Case 2-2: 2 complex poles and 1 zero w Two Complex Poles p1 -40dB w0 w0 + z1 Zero +20dB p2 | z1 | Second-order Filter – Case 2 Case 2-2: 2 complex poles and 1 zero | z 1 | w 0 -40dB -20dB Two Complex Poles w 0 | z1 | -40dB w0 + +20dB -20dB Zero +20dB | z 1 | w 0 | z1 | w0 | z1 | Second-order Filter – Case 2 Case 2-2: 2 complex poles and 1 zero Highly Selective -20dB +20dB | z 1 |= 0 Two Complex Poles -40dB w0 + w0 Zero +20dB Bandpass Filter | z1 | Bandpass Filter • Bandpass filter: 2 poles and zero at original point H bp ( s ) = K w w 0 2 s w 0 Q s Q s w 0 2 bandpass filter K K/ 2 B wl w0 Bandwidth B = ωr - ωl wr Bandpass Filter • Bandpass filter: 2 poles and zero at original point H bp ( s ) = K H bp ( j w ) = = w 0 2 s w 0 Q s Q s w 0 2 K w 0 Q j w jw 2 w 0 Q j w w K jw Q w0 1 Find the frequency for the maximum amplitude w0Q jw = 2 0 = K j w 2 w 0 Q j w K jw Q w0 1 jw 0 Q w = w0 2 1 w 0 Q jw K w w 1 jQ 0 w w0 Bandpass Filter • Transfer function of bandpass filter H bp ( j w ) = a bp (w ) = K w w0 1 jQ w w 0 a bp ( ) = 0 a bp (w ) is maximized K w0 2 w 1 Q w w 0 a bp ( 0 ) = 0 2 when w = w 0 ω0 is center frequency Bandpass Filter - Bandwidth B K a bp (w ) = = K 2 2 w w 2 1 Q 0 w w0 2 2 w w0 1 w w0 2 =2 1 Q w w = Q2 w 0 w0 w0 1 w 1 2 2 w w 0w w 0 = 0 = Q w0 w Q 2 w = 1 Q w0 1 2 w 0 4 w 0 Q 2 Four answers? Pick the two positive ones as ωl or ωr Bandpass Filter - Bandwidth B 2 w = 1 Q w0 1 2 w 0 4 w 0 Q 2 2 1 wu = Q w0 1 2 w 0 4 w 0 Q 2 B = wu wl = 2 wl = 1 Q w0 1 2 w 0 4 w 0 Q 2 w0 Q measure the narrowness of the pass band Q Q is called quality factor Second-order Filter – Case 3 w p1 p2 Second-order Filter – Case 3 w p1 p2 Second-order Filter – Case 3 w p1 p2 Second-order Filter – Case 3 w p1 p2 Thank you! • Higher order filter • Buttorworth • Notch filter for humming • Different kinds of filter: active, passive …… Radio Amplifier • P1562 Suppose this band-stop filter were to suddenly start acting as a high-pass filter. Identify a single component failure that could cause this problem to occur: If resistor R3 failed open, it would cause this problem. However, this is not the only failure that could cause the same type of problem! • Giutar capacitor https://www.youtube.com/watch?v=3I62Xfhts 9k Algorithmic implementation • wiki High pass • They are used as part of an audio crossover to direct high frequencies to a tweeter while attenuating bass signals which could interfere with, or damage, the speaker. When such a filter is built into a loudspeaker cabinet it is normally a passive filter that also includes a low-pass filter for the woofer and so often employs both a capacitor and inductor (although very simple high-pass filters for tweeters can consist of a series capacitor and nothing else). Any second-order bandpass filter may be described by K w 0 Q s H bp ( s ) = s w 0 Q s w 0 2 2 Q = Where quality factor: w0 2 : damping coefficient the network is underdamped when < w0 or Q > ½ p1 , p 2 = w0 jw 0 2Q 2 p1 = p 2 = 4Q 2 1 w0 1 2 w 0 1 2 4Q 2Q 1 p 2 = 180 cos 1 = w 0 1 2Q p 1 = 180 cos 1 1 2 Q 64 The transfer function of a second-order notch filter is H no ( s ) = 2 s w 0 Q s w 2 0 K s 2s w0 2 2 w0 2 The notch effect comes from the quadratic numerator z1 , z 2 = j w 0 2 2 jw 0 The notch width is B = wO / Q 65 Table 11.3 Simple Filter Type Lowpass Highpass Bandpass Notch Transfer Function H (s) = H (s) = H (s) = H (s) = Properties a (0 ) = K K w co s w co a (w co ) = K a ( ) = K Ks s w co a (w co ) = K K w 0 Q s 2 K s 2s w0 2 2 2 B = w0 Q a (w 0 ) = KQ w 0 s w 0 Q s w 0 2 2 a (w 0 ) = K s w 0 Q s w 0 2 2 B = w0 Q 66 Example 11.6 Design of a Bandpass Filter bandpass filter: L = 1 mH, Rw = 1.2 W, C = ?, R = ? frequency: 20kHz ± 250Hz Q par = w 0 C R // R par = B = 2 2 250 Hz w 0 = 2 20 kHz Q = w0 = C = w0 L R par = = 40 = 5 . 03 k 500 1 2 = L CR w 1 2 = w0L R // R par = Q w 0 L = 40 2 20 k 10 20 k B R // R par 20 k 1 10 2 10 3 = 63 . 3 nF 3 63 . 3 10 9 1 .2 = 13 . 2 k W R // R par = R = R R par R R par 13.2k 5.03k 13 . 2 k 5 . 03 k R = 3 R par R // R par R par R // R par = 8 . 13 k W 67 From Wiki • Butterworth filter – maximally flat in passband and stopband for the given order • Chebyshev filter (Type I) – maximally flat in stopband, sharper cutoff than Butterworth of same order • Chebyshev filter (Type II) – maximally flat in passband, sharper cutoff than Butterworth of same order • Bessel filter – best pulse response for a given order because it has no group delay ripple • Elliptic filter – sharpest cutoff (narrowest transition between pass band and stop band) for the given order • Gaussian filter – minimum group delay; gives no overshoot to a step function. Filter • A filter is a circuit that is designed to pass signals with desired frequencies and reject the others. X Filter Y H s Magnitude ratio Only input signal at these frequencies can pass Loudspeaker for home usage with three types of dynamic drivers 1. Mid-range driver 2. Tweeter Second Order Lowpass Filter H lp s = H lp j w = 1 1 s R 1C 1 C 2 R 2 R 1C 2 s C 1C 2 R 1 R 2 2 1 1 j w R 1C 1 C 2 R 2 R 1C 2 w C 1C 2 R 1 R 2 2 Compare with 1st order low pass filter H lp j w = w co j w w co As the frequency increases, the amplitude ratio drops faster than 1st order low pass filter Firsr-order Filters - Case 1 w Lowpass filter H s = K 1 s p |p| is the cut-off frequency w Larger cut-off frequency Smaller cut-off frequency Firsr-order Transfer Function - Case 2 H s = K sz s p Highpass? Lowpass? Both possible? Absolute value of pole is equal to zero Magnitude is the length of green line divided by the length of the blue line Positive zero can cause phase shift Phase is the angle of the green minus the negative one w All pass filter First Order Lowpass Filter K lp w = w K 2 2 Maximum: lp 0 = 1 1 H jw = R RC jw 1 H jw = jw RC H lp j w = K jw K Cut-off Frequency ωco: L R L K = H lp j w = 1 , RC K R L w co j w w co 1 lp w co = w 2 co K 2 1 = 2 w co = K = 2 1 RC , R L Cut off frequency 0 = K 1 K w d w co 2 w w 4 2 2 d 2 co 2 2 w 2 d w 2 co wd 2 1 = w d w co 2 2 1 2 K 2 1 wd 2 2 2 2 w d 2 w co w d w co 2 w d w co = 2 4 4 2 w d2 2 w d4 4 2 2 2 2 4 4 2 2 2 w co w co 2 w d w co = 2 w d w d 2 w 4 co w 2 4 2 2 w w 2 co 2 = w 2 d 2 d 2 co 2 4 2 w 2 2 2 d 2 4 w co = 2 =0 w 2 w 2 co = w 2 2 d 2 co 2 = w 2 2 4 w 2 2 2 wd 2 d 2 wd 2 2 4 wd 2 2 2 w co 2 w d w co 4 2 2 2 60Hz Hum First-order Filters vh vl The highpass and lowpass filters have the same cutoff frequency. w co = Are there anything wrong? 1 RC