First-order filters The general form for the transfer function of a first order filter is: a s + a T (s) = 1 0 b1s + bo However, we will typically recast this into a standard form:
s + Zo T (s) = Go ⋅ 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.
EE 230 1st-order filters – 1 Low-pass jω In the case were a1 = 0, we have a low-pass function. s-plane
ao T (s) = zero as s → ±∞ b1s + bo x σ
In standard form, we write it as: pole at –P0
Po T (s) = Go ⋅ s + Po 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.
P P o = o s + Po Po + jω s=jω Re-expressing the complex value in magnitude and phase form:
Po Po ω = exp (jθLP) θLP = − arctan Po + jω [ 2 2 ] ( Po ) Po + ω EE 230 1st-order filters – 2 By looking at the magnitude expression, we can see the low-pass behavior.
Po Po For low frequencies (ω << Po): ≈ = 1. 2 2 2 Po + ω Po
Po Po Po At high frequencies (ω >> Po): ≈ = . 2 2 2 ω Po + ω ω At low frequencies, the magnitude is 1 (the output is equal to the input) and at high frequencies, the magnitude goes down inversely with frequency, consistent with the notion of a low-pass response.
We can also examine the phase at the extremes.
ω For low frequencies (ω ≈ 0): θLP = − arctan ≈ 0. ( Po )
→ ω ∘ At high frequencies (ω +∞): θLP = − arctan ≈ − 90 . ( Po )
EE 230 1st-order filters – 3 P We can use the functions to make the magnitude M = o and phase as frequency response plots. 2 2 Po + ω magnitude - linear scale magnitude - Bode plot 1.2 10
1.0 0
0.8 -10
0.6 -20
magnitude 0.4 -30 magnitude (dB) magnitude 0.2 -40
0.0 -50 10 100 1000 10000 100000 10 100 1000 10000 100000 angular frequency (rad/s) angular frequency (rad/s) phase 0
-15
-30 ω θLP = − arctan -45 ( Po ) phase(°) -60
-75
-90 10 100 1000 10000 100000 EE 230 angular frequency (rad/s) 1st-order filters – 4 Cut-off frequency Use the standard definition for cut-off frequency, which is the frequency at which the magnitude is down by 2 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.) 1 P M = = o 2 2 2 Po + ωc
With a bit of algebra, we find that ωc = Po. The cut-off frequency is defined by the pole. Tricky! Thus, in all of our equations, we could substitute ωc for Po.
We can also calculate the phase at the cut-off frequency.
ωc ∘ θLP = − arctan = − 45 ( Po ) The cut-off frequency points are indicated in the plots on the previous slide. EE 230 1st-order filters – 5 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. ( )= ( )= · + +
( )= ( )= · + + ( ) = ( ) = +
Resistor and capacitor in series — + + Z output taken across the capacitor. V (s) C Vo(s) i – = – Use a voltage divider to find the transfer function.
( )= ( ) +
( ) ( )= = = = = ( ) + + + +
Clearly, this is low-pass with Go = 1 and =
The only real design consideration is choosing the RC product, which then sets the corner frequency.
EE 230 1st-order filters – 7 Low-pass filter circuits: simple RL ZL = sL Inductor and resistor in series — + output taken across the resistor. V (s) + Z V (s) i – R o = R – Use a voltage divider to find the transfer function.
( )= ( ) +
( ) ( )= = = = = ( ) + + + +
Again, low-pass behavior with Go = 1, but now with =
As with the previous example choosing the “RL time constant” , we can define the pass-band of this low-pass filter.
EE 230 1st-order filters – 8 Low-pass filter circuits: inverting op amp C = = R + Z1 2 R 1 V (s) – i V (s) + o
( ) + ( )= = = = = ( ) + +
Clearly, this is also low-pass with = and =
Be careful with the extra negative sign in the gain: –1 = exp( j180° )
At low frequencies: |T| → R2/R1, and θT → 180° (= – 180°)
At high frequencies: |T| → (ωR1C)–1, and θT → +90° (= – 270°)
EE 230 1st-order filters – 9 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 – 10 Low-pass filter circuits: non-inverting op amp
R Note: It might slightly V (s) + i V (s) disingenuous to treat this as if C – o it were some new type of filter — we can readily see R that it is a simple RC filter R 2 1 cascaded with a simple non- inverting amp. However, it is still a useful circuit. ( )= = + + simple RC +
( )= + + ( ) non-inverting amp ( ) ( )= = + ( ) +
Low-pass with = + and = EE 230 1st-order filters – 11 Low-pass filter circuits: another RC ZR1 = R1 ( ) ( )= = ( ) + + + Z Z V (s) C R2 Vo(s) i – + – = = = R2 + + = = + + = + = = + + = + = Low-pass. · +
= Note the voltage divider. “Gain” < 1. + = The corner depends on the parallel combination. EE 230 1st-order filters – 12 High-pass jω In the case were a0 = 0, we have a high-pass function. s-plane a s T (s) = 1 b1s + bo zero at 0 x σ In standard form, we write it as: pole at –P0 s T (s) = Go ⋅ s + Po
We will ignore the gain initially (set Go = 1) and focus on sinusoidal behavior by letting s = jω.
s jω = s + Po Po + jω s=jω Re-expressing the complex value in magnitude and phase form:
jω ω ∘ ω = exp (jθHP) θHP = 90 − arctan Po + jω [ 2 2 ] ( Po ) Po + ω EE 230 1st-order filters – 13 By looking at the magnitude expression, we can see the high-pass behavior. ω ω ω For low frequencies (ω << Po): ≈ = . 2 2 2 Po Po + ω Po ω ω At high frequencies (ω >> Po): ≈ = 1. 2 2 2 Po + ω ω At low frequencies, the magnitude is increasing with frequency, and at high frequencies, the magnitude is 1 (the output is equal to the input). This behavior is consistent with a high-pass response.
We can also examine the phase at the extremes.
∘ ω ∘ For low frequencies (ω << Po): θHP = 90 − arctan ≈ 90 . ( Po )
∘ ω ∘ At high frequencies (ω >> Po): θHP = 90 − arctan ≈ 0 . ( Po )
EE 230 1st-order filters – 14 ω We can use the functions to make the magnitude M = 2 2 and phase as frequency response plots. Po + ω magnitude - linear scale magnitude - Bode plot 1.2 10
1.0 0
0.8 -10
0.6 -20
magnitude 0.4 -30 magnitude (dB) magnitude 0.2 -40
0.0 -50 10 100 1000 10000 100000 10 100 1000 10000 100000 angular frequency (rad/s) angular frequency (rad/s) phase 90
75
60 ∘ ω θHP = 90 − arctan 45 ( Po ) phase(°) 30
15
0 10 100 1000 10000 100000 EE 230 angular frequency (rad/s) 1st-order filters – 15 Cut-off frequency Use the standard definition for cut-off frequency, which is the frequency at which the magnitude is down by 2 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.) 1 ω M = = 2 2 2 Po + ωc
With a bit of algebra, we find that ωc = Po. 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. We can calculate the phase at the cut-off frequency.
∘ ωc ∘ θHP = 90 − arctan = 45 ( Po ) The cut-off frequency points are indicated in the plots on the previous slide. EE 230 1st-order filters – 16 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.
The corner frequency is the = · → P0 = ωc value of the pole. + ( )= ( )= · + +
( )= ( )= · + + ( ) = = · + = arctan arctan ( ) = + = arctan = + arctan Exercise: Re-express all of the above formulas using f instead of ω. EE 230 1st-order filters – 17 High-pass filter circuits: simple RC ZC = Capacitor and resistor in series — output taken across the resistor. + V (s) + Z V (s) i – R o Use a voltage divider to find the transfer =R – function.
( )= ( ) +
( ) ( )= = = = = ( ) + + + +
Clearly, this is high-pass with G = 1 and = o
EE 230 1st-order filters – 18 High-pass filter circuits: simple RL
ZR = R + V (s) + Z V (s) i – L o = sL –
( )= ( ) +
( ) ( )= = = = = ( ) + + + +
Again, high-pass with G = 1 but with = o
EE 230 1st-order filters – 19 High-pass filter circuits: inverting op amp
Z2
Z1 R2 C R1 V (s) – i V (s) + o = +
( ) ( )= = = = ( ) + +
We see that this is also high-pass with = and =
The same comments about the phase apply here: the –1 in the gain factor introduces an extra 180° (or –180°) of phase.
EE 230 1st-order filters – 20 High-pass filter circuits: non-inverting op amp
C Vi(s) + Again, this is simple RC high- Vo(s) R – pass cascaded with a non- inverting amp.
R2 R1
+ ( )= = simple RC high pass + +
( )= + + ( ) non-inverting amp
( ) ( )= = + ( ) +
High-pass with = + and = EE 230 1st-order filters – 21 High-pass filter circuits: another RC = Z = R ZC ( ) R1 1 ( )= = ( ) + + + V (s) + ZR2 V (s) i – o – = R2 = + +
= + + ( + ) = High-pass. · +
= Note the voltage divider. “Gain” < 1. +
= The corner depends on the series combination. ( + )
EE 230 1st-order filters – 22 Case study: inductors can be trouble Cheap inductors can have a relatively large series resistance. This parasitic resistance can cause trouble in certain circumstances.
R ideal
+ V (s) + L V (s) ( )= = i – o + + –
As usual, zero at s = 0,
pole at s = –ωc. R real +
L + + + V (s) + ( )= = + = i + + + + – Vo(s) Rs Pole and zero – are both shifted!
EE 230 1st-order filters – 23 + + ( )= = + + = arctan arctan Effect of inductor parasitic resistance on high-pass filter:
L = 0.027 H, R1 = 1 kΩ, and Rs = 60 Ω. The frequency responses for both magnitude and phase are quite different.
EE 230 1st-order filters – 24