First-order filters, Exams of Algebra

Download First-order filters and more Exams Algebra in PDF only on Docsity! EE 230 1st-order filters – 1 First-order filters The general form for the transfer function of a first order filter is: T (s) = Go ⋅ s + Zo s + Po There will always be a single pole at s = –Po. The pole must be real (there is only one, so no complex conjugates are not possible) and it must be negative (for stability). There will always be a zero, which can be at s = 0, as s → ±∞ (zero at infinity), or somewhere else, s = –Zo. (Note the zero can have a positive value.) There may be a gain factor, Go, which might be 1 or smaller (for a passive circuit with a voltage divider) or have a magnitude greater than 1 for an active circuit. The two most important cases are the zero at infinity, which is a low- pass filter and the zero at zero, which is the high-pass filter. T (s) = a1s + a0 b1s + bo However, we will typically recast this into a standard form: EE 230 1st-order filters – 2 Low-pass In the case were a1 = 0, we have a low-pass function. T (s) = Go ⋅ Po s + Po T (s) = ao b1s + bo In standard form, we write it as: σ jω x pole at –P0 s-plane zero as s → ±∞ The reason for this form will become clear as we proceed. We will ignore the gain initially and focus on sinusoidal behavior by letting s = jω. Po s + Po s=jω = Po Po + jω Re-expressing the complex value in magnitude and phase form: Po Po + jω = [ Po P2 o + ω2 ]exp (jθLP) θLP = − arctan ( ω Po ) EE 230 1st-order filters – 5 Use the standard definition for cut-off frequency, which is the frequency at which the magnitude is down by from the value in the pass-band. For our low-pass function, the pass band is at low frequencies, and the magnitude there is 1. (Again, we are “hiding” Go by assuming that it is unity. If Go ≠ 1, then everything is scaled by Go.) 2 Cut-off frequency M = 1 2 = Po P2 o + ω2 c With a bit of algebra, we find that . The cut-off frequency is defined by the pole. Tricky! Thus, in all of our equations, we could substitute ωc for Po. ωc = Po We can also calculate the phase at the cut-off frequency. θLP = − arctan ( ωc Po ) = − 45∘ The cut-off frequency points are indicated in the plots on the previous slide. EE 230 1st-order filters – 6 ս/3 = arctan ֌ ֌F 7/3 (M֌) = *R · ֌F M֌ + ֌F 7/3 (V) = *R · ֌F V+ ֌F 7/3 (V) = *R + V ֌F 7/3 (M֌) = *R + M ֌ ֌F ս/3 = arctan ֌ ֌F Again, we could just as easily use real frequency rather than angular frequency. As an exercise: re-express all of the above formulas using f instead of ω. To emphasize the importance of the corner frequency in the low-pass function, we can express all the previous results using ωc in place of Po. On the left are the functions in "standard" form. On the right, the functions are expressed in a slightly different form that is sometimes easier to use. 7/3 (M֌) = *R ֌F ֌ + ֌F <latexit sha1_base64="ZcBktQQdyEg5AHKJu2kZSXSy48k=">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</latexit><latexit 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(V) = =& =& + =5 =  V&  V& + 5 =  5& V+  5& = ֌F V+ ֌F The only real design consideration is choosing the RC product, which then sets the corner frequency. EE 230 1st-order filters – 10 Magnitude and phase plots for an active low-pass filter with R1 = 1 kΩ, R2 = 25 kΩ, and C = 6.4 nF, giving fo = 1000 Hz and Go = –25 (|Go|= 28 dB). EE 230 1st-order filters – 11 – + R C Vo(s) Vi(s) R1 R2 Low-pass filter circuits: non-inverting op amp Note: It might slightly disingenuous to treat this as if it were some new type of filter — we can readily see that it is a simple RC filter cascaded with a simple non- inverting amp. However, it is still a useful circuit. 9+ (V) = =& =& + =5 =  5& V+  5& 9R (V) = + 5 5 9+ (V) non-inverting amp simple RC Low-pass with and *R = + 5 5 ֌F =  5& 7 (V) = 9R (V) 9L (V) = + 5 5  5& V +  5& EE 230 1st-order filters – 12 Low-pass filter circuits: another RC – + + –Vi(s) ZR1 = R1 Vo(s) =  V& ZC ZR2 = R2 =3 = =5 =& = 5  V& 5 +  V& = 5 + V5& 7 (V) = 9R (V) 9L (V) = =3 =3 + =5 = 5 +V5& 5 +V5& + 5 = 5 5 + 5 + V55& = 5 5 + 5  53& V+  53& = *R · ֌F V + ֌F 53 = 5 5 *R = 5 5 + 5 Note the voltage divider. “Gain” < 1. ֌F =  53& Low-pass. The corner depends on the parallel combination. EE 230 1st-order filters – 15 M = ω P2 o + ω2 θHP = 90∘ − arctan ( ω Po ) angular frequency (rad/s) angular frequency (rad/s) angular frequency (rad/s) m ag ni tu de 0.0 0.2 0.4 0.6 0.8 1.0 1.2 10 100 1000 10000 100000 m ag ni tu de (d B ) -50 -40 -30 -20 -10 0 10 10 100 1000 10000 100000 ph as e (° ) 0 15 30 45 60 75 90 10 100 1000 10000 100000 We can use the functions to make the magnitude and phase as frequency response plots. magnitude - linear scale magnitude - Bode plot phase EE 230 1st-order filters – 16 Cut-off frequency M = 1 2 = ω P2 o + ω2 c With a bit of algebra, we find that . The same result as for low- pass response, except that pass-band is above the cut-off frequency in this case. Once again, we see the importance of the poles in determining the behavior of the transfer functions. ωc = Po We can calculate the phase at the cut-off frequency. θHP = 90∘ − arctan ( ωc Po ) = 45∘ Use the standard definition for cut-off frequency, which is the frequency at which the magnitude is down by from the value in the pass-band. For our high-pass function, the pass band is at high frequencies, and the magnitude there is 1. (Again, we are “hiding” Go by assuming that it is unity. If Go ≠ 1, then everything is scaled by Go.) 2 The cut-off frequency points are indicated in the plots on the previous slide. EE 230 1st-order filters – 17 Exercise: Re-express all of the above formulas using f instead of ω. 7+3 (V) = *R · V V+ ֌F 7+3 (M֌) = *R · M֌ M֌ + ֌F 7/3 (M֌) = *R · ֌ ֌ + ֌F ս+3 = arctan ֌  arctan ֌ ֌F =  arctan ֌ ֌F 7+3 (V) = *R + ֌F V = *R  M ֌F ֌ 7+3 (M֌) = *R + ֌F M֌ ս+3 = + arctan ֌F ֌ 7+3 (M֌) = *R + ֌F ֌  → *R  = *R · 3 ֌F + 3 The corner frequency is the value of the pole.P0 = ωc To emphasize the importance of the corner frequency in the high-pass function, we can express all the previous results using ωc in place of Po. EE 230 1st-order filters – 20 Z2 Z1 – + C R1 R2 Vi(s) Vo(s) = = 5 +  V& We see that this is also high-pass with and *R = 5 5 ֌F =  5& The same comments about the phase apply here: the –1 in the gain factor introduces an extra 180° (or –180°) of phase. High-pass filter circuits: inverting op amp 7 (V) = 9R (V) 9L (V) = = = = 5 5 +  V& = 5 5 V V +  5& EE 230 1st-order filters – 21 High-pass filter circuits: non-inverting op amp Again, this is simple RC high- pass cascaded with a non- inverting amp. 9R (V) = + 5 5 9+ (V) non-inverting amp simple RC high pass High-pass with and *R = + 5 5 ֌F =  5& – + R C Vo(s) Vi(s) R1 R2 9+ (V) = =5 =& + =5 = V V+  5& 7 (V) = 9R (V) 9L (V) = + 5 5 V V +  5& EE 230 1st-order filters – 22 High-pass filter circuits: another RC *R = 5 5 + 5 Note the voltage divider. “Gain” < 1. High-pass. The corner depends on the series combination. 7 (V) = 9R (V) 9L (V) = =5 =5 + =5 + =& = 5 5 + 5 +  V& = 5 5 + 5 V V+  (5+5)& = *R · V V + ֌F ֌F =  (5 + 5)& – + + –Vi(s) ZR2 = R2 Vo(s) ZC=  V&ZR1 = R1