Lecture 17
First order filters
Objectives:
1. Study different types of first order filters
2. Develop the Bode plot of their frequency response
3. Present their passive realization
4. Analyze op amp RC realization
At the end of this class you will be able to:
1. Recognize first-order filter transfer functions
2. Recognize different types of first-order filters
4. Determine the different filter parameters
5. Design the filter to meet desired specifications
6. Plot the transfer function of any first-order filter
* High order-filters can be realized by cascading first and second order sections. These
filters can be implemented by passive RC or LC circuit. However, using active-RC
filters based on op amps provides several advantages:
(a) Gain which can be set to a desired value
(b) Independent of some of the filter parameters without affecting others
(c) The output impedance is very low (ideally zero) allowing cascading
(i.e. any high-order filter function can be realized using first-order and secondorder filters)
* In fact, active-RC filters are suitable for discrete implementation. But they are not
suitable for high-frequency operation
* In standard form the denominator of the first-order transfer function is written as s + ω o
where ωo is the pole frequency. Based on the value of the coefficient of s1 and s 0 in the
numerator, first order filters are classified as lowpass, highpass, general, and allpass as
follows:
1. Low-pass filter:
* Its transfer function is given by:
ao
T (s) =
s + ωo
T(s) has zero at infinity.
* The bode plot of the lowpass filter can be developed as shown in Fig. 1 (a). For
frequencies less than the pole frequency ωo, the transfer function is constant and
equals to 20log(ao/ ωo). After ωo the frequency response decreases by 20dB/decade.
Fig. 1: First-order LPF (a) Bode plot (b) passive realization (c) active RC realization
* This transfer characteristics can be implemented with simple series RC circuit
shown in Fig. 1(b). The TF can directly be found by VDR:
1/ sC
1
Vi ( s ) =
Vi ( s)
R + 1/ sC
sCR + 1
V ( s)
1/ CR
⇒ o
=
Vi ( s ) s + 1/ CR
Vo ( s ) =
Thus, the 3dB corner frequency will be at ωo =1/RC and the lowpass gain is unity.
Example 1: Find the TF of the active RC filter shown in Fig.1 (c).
Solution: the TF can be obtained by writing KCL equation at v-=0 and then solving
for Vo(s) as follows:
0 − Vi 0 − Vo
+
+ (0 − Vo ) sC = 0
R1
R2
Multiplying by R1R2 and solving for Vo results in:
Vo ( s )
1/ CR1
R2
=−
=−
Vi ( s )
sCR1 R2 + R1
s + 1/ CR2
Therefore, ωo=1/CR2 and dc gain of –R2/R1. This gain can be controlled via R1
without changing the corner frequency.
2. Fist order highpass filter (HPF):
* The TF of first-order HPF is given by:
as
T (s) = 1
s + ωo
T(s) has zero at dc frequency. The Bode plot of the magnitude response of T(s) can
be `developed as shown in Fig. 2(a)
Fig. 2: First-order HPF (a) Bode plot (b) passive realization (c) active RC realization
* This transfer function has a corner frequency of ωo. The HP gain is obtained by
letting s → ∞ resulting in gain of a1.
* The passive realization of HPF is obtained by taken the voltage a cross the resistor
of the RC series connection shown in Fig. 2(b) resulting in ωo =1/RC and the HP
gain of unity
Example 2: Find the TF of the op amp based first-order HPF shown in Fig. (c).
Solution: Writing a single KCL at v-=0 and solving for Vo yields:
0 − Vi
0 − Vo
+
=0
R1 + 1/ sC
R2
⇒ Vo = −
R2
Vi
R1 + 1/ sC
Vo( s )
sCR2
R /Rs
=−
=− 2 1
Vi ( s )
sCR1 + 1
s + 1/ CR1
It can be seen that the corner frequency is ωo =1/CR1 and the high frequency gain is –
R2/R1.
3. First-order general filter
* The standard form of the general first order filter is given by:
a s + a0
(1)
T (s) = 1
s + ωo
T(s) has zero frequency at –a0/a1. The Bode plot of the magnitude response of T(s)
can be developed as shown in Fig. 3(a).
Fig. 3: First-order general filter (a) Bode plot (b) passive realization (c) active RC
realization
* At low frequencies ω is less than ωo, the dc terms in the denominator and numerator
are dominant thus the transfer function will be approximately a0/ ωo, Whereas at
high frequencies where ω is more than a0 / a1 , the s terms become dominant and
the transfer function becomes approximately a1 . For frequencies between ωo and
a0 / a1 the function decreases by 20dB/decade.
* This general first-order filter can be realized by the passive circuit shown in
Fig. 3(b) or the active RC circuit of Fig. 3(c).
Example 3: find the transfer function of the circuit of Fig. 3(c).
Solution: the TF can be found by writing KCL equation at v-=0.
0 − Vi
0 − Vo
+ (0 − Vi ) sC1 +
+ (0 − Vo ) sC2 = 0
R1
R2
T
hen, solving for Vo results in:
Vo ( s )
sC + 1/ R1
sC / C + 1/( R1C2 )
=− 1
=− 1 2
Vi ( s )
sC2 + 1/ R2
s + 1/( R2C2 )
(2)
Comparing equations (1) and (2) suggests that: ω 0 =
1
1
, a0 / a1 =
, dc gain = –
C2 R2
C1 R1
R2/R1 and high frequency gain= -C1/C2.
4. First-order Allpass filter :
* The first-order Allpass function can be written in standard form as:
s − ω0
T ( s ) = − a1
s + ωo
The magnitude of this filter can be obtained as follows:
( jω ) + ( −ω o )
2
2
( jω ) + (ω o ) o
2
T ( jω ) = a1
2
= a1
Thus the allpass filter has a constant gain for all frequencies.
* However, its phase response shows frequency dependent characteristics. The
phase of T(s) can b obtained as follows.
φ ( jω ) = tan −1 ( −ω / ω o ) − tan −1 (ω / ω o ) = −2 tan −1 (ω / ω o )
Therefore, for low frequencies the phase will be approximately zero. At ωo the
phase will be -90. Finally, at high frequencies the phase approaches -180 as
shown in Fig. 3(b).
* The first-order allpass filter can be used to adjust the phase of an input signal by
any value between zero and -180.
* An allpass first order filter can be implemented by the passive circuit of Fig. 4(c)
or the active circuit of Fig. 4(d).
Example 4: Find the transfer function of the circuit of Fig. 4(d).
Solution: the TF can be found as follows.
1/ sC
1
Vi
Vi =
R + 1/ sC
sCR + 1
v −V v −V
1
By KCL: − i + − o = 0 ⇒ v− = (Vi + Vo )
R1
R1
2
1
1
1
But v+ = v− ⇒
Vi = Vi + Vo
sCR + 1
2
2
Vo ( s ) − sCR + 1
sCR − 1
s − 1 / CR
⇒
=
=−
=−
Vi ( s )
sCR + 1
sCR + 1
s + 1/ CR
By VDR: v+ =
Thus, ωo=1/RC and Flat gain=a1=1
Fig. 4: