first order all pass filter

1. First Order Active RC Sections
Problem 1:
For the first order all pass filter (1st
APF = phase equalizer) shown below:
(i) Show that first order LPF can be obtained as: 1st
LPF = 1 - 1st
APF
(ii) Show that first order HPF can be obtained as: 1st
HPF = 1+1st
APF
(iii) Show that Band reject filter (BRF) can be obtained as: BRF = 1 + two
cascaded sections of 1st
APF
(iv) Show that Band pass filter (BPF) can be obtained as: BPF = 1 - two cascaded
sections of 1st
APF.
(v) Verify (i)-(iv) using Spice simulation.

2. First Order Active RC Sections
Problem 1:
For the first order all pass filter (1st
APF) shown below:
(i) Show that first order LPF can be obtained as: 1st
LPF = 1 - 1st
APF
Solution:
The transfer function of 1st
APF is given by
𝑇(𝑆) =
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
=
𝑆𝐶 − 𝐺 𝐺
𝑆𝐶 + 𝐺 𝐺
=
𝑆 −
1
𝑅 𝐺 𝐶
𝑆 +
1
𝑅 𝐺 𝐶
= −
1 − 𝑆𝐶𝑅 𝐺
1 + 𝑆𝐶𝑅 𝐺
Using subtractor based on op-amp where two inputs of the subtractor circuit are
the input voltage and the output of the 1st
APF. The output of the subtractor is
𝑉𝑜1
𝑉𝑖𝑛
= 1 − (−
1 − 𝑆𝐶𝑅 𝐺
1 + 𝑆𝐶𝑅 𝐺
) = (
1 + 𝑆𝐶𝑅 𝐺 + 1 − 𝑆𝐶𝑅 𝐺
1 + 𝑆𝐶𝑅 𝐺
)
𝑉𝑜1
𝑉𝑖𝑛
= (
2
1 + 𝑆𝐶𝑅 𝐺
)
It is 1st
order LPF. By using C = 0.1µF and RG = R = RF = 41.42kΩ:
The Dc gain is Dc gain = 20 log(2) = 6dB
the cut-off frequency is 𝜔 𝑜 =
1
𝐶𝑅 𝐺
= 241.42rad/ sec = 2𝜋 (38.4Hz)

3. (ii) Show that first order LPF can be obtained as: 1st
HPF = 1 +1st
APF
Solution:
The transfer function of 1st
APF is given by
𝑇(𝑆) =
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
= −
1 − 𝑆𝐶𝑅 𝐺
1 + 𝑆𝐶𝑅 𝐺
Using adder based on op-amp circuit where the two inputs of the adder circuit are
the input voltage and the output of the 1st
APF. The output of the adder is
−
𝑉𝑜1
𝑉𝑖𝑛
= 1 + (−
1 − 𝑆𝐶𝑅 𝐺
1 + 𝑆𝐶𝑅 𝐺
) = (
1 + 𝑆𝐶𝑅 𝐺 − 1 + 𝑆𝐶𝑅 𝐺
1 + 𝑆𝐶𝑅 𝐺
)
−
𝑉𝑜1
𝑉𝑖𝑛
= (
2𝑆𝐶𝑅 𝐺
1 + 𝑆𝐶𝑅 𝐺
)
It is 1st
HPF. By using C = 0.1µF and RG = R = RF = 41.42kΩ:
The Dc gain at 𝑇(𝑗𝜔 = ∞) is
Dc gain = 20 log(2) = 6dB
the cut-off frequency is
𝜔 𝑜 =
1
𝐶𝑅 𝐺
= 241.42rad/sec = 2π (38.4Hz)

4. (iii) Show that Band reject filter (BRF) can be obtained as: BRF = 1 + two
cascaded sections of 1st
APF (BRF = 1 + 2nd
APF)
Solution:
The transfer function of two cascaded section of 1st
APF is given by
𝑇(𝑆) = (−
1 − 𝑆𝐶𝑅 𝐺
1 + 𝑆𝐶𝑅 𝐺
) × (−
1 − 𝑆𝐶𝑅 𝐺
1 + 𝑆𝐶𝑅 𝐺
)
𝑇(𝑆) =
1 − 2𝑆𝐶𝑅 𝐺 + (𝑆𝐶𝑅 𝐺)2
1 + 2𝑆𝐶𝑅 𝐺 + (𝑆𝐶𝑅 𝐺)2
It is a 2nd
order APF.
𝜔𝑧 = 𝜔 𝑝 =
1
𝐶𝑅 𝐺
= 241.42 rad/sec = 2π (38.4Hz)
Using adder based on op-amp circuit where the two inputs of the adder circuit are
the input voltage and the output of the 2st
order APF. The output of the adder is
−
𝑉𝑜1
𝑉𝑖𝑛
= 1 + (
1 − 2𝑆𝐶𝑅 𝐺 + (𝑆𝐶𝑅 𝐺)2
1 + 2𝑆𝐶𝑅 𝐺 + (𝑆𝐶𝑅 𝐺)2
)
−
𝑉𝑜1
𝑉𝑖𝑛
=
1 + 2𝑆𝐶𝑅 𝐺 + (𝑆𝐶𝑅 𝐺)2
+ 1 − 2𝑆𝐶𝑅 𝐺 + (𝑆𝐶𝑅 𝐺)2
1 + 2𝑆𝐶𝑅 𝐺 + (𝑆𝐶𝑅 𝐺)2
−
𝑉𝑜1
𝑉𝑖𝑛
= 2
1 + (𝑆𝐶𝑅 𝐺)2
1 + 2𝑆𝐶𝑅 𝐺 + (𝑆𝐶𝑅 𝐺)2
It is 2nd
order notch filter. By using C = 0.1µF and RG = R = RF = 41.42kΩ:
In the two passbands ( ω → 0 and ω → ∞ ) the gain is
Dc gain = 20 log(2) = 6dB
while the notch frequency is
𝜔 𝑜 =
1
𝐶𝑅 𝐺
= 241.42rad/ sec = 2𝜋 (38.4Hz)
The pole quality factor Q is obtained as
𝜔 𝑜
𝑄
=
2
𝐶𝑅 𝐺
𝑄 =
𝜔 𝑜
2/𝐶𝑅 𝐺
= 0.5

5. 𝑄 =
𝜔 𝑜
𝐵𝑊
= 0.5
𝐵𝑊 = (𝜔2 − 𝜔1) =
𝜔 𝑜
𝑄
= 482.84 rad/ sec = 2𝜋 (76.89Hz)
𝐵𝑊 = (𝜔2 − 𝜔1) =
𝜔 𝑜
𝑄
= 482.84 rad/sec
𝜔 𝑜 = √ 𝜔1 𝜔2 = 241.42 rad/sec
𝜔2 = 𝜔1 + 𝐵𝑊 = 𝜔1 + 482.84
𝜔2 =
𝜔 𝑜
2
𝜔1
=
(241.42)2
𝜔1
By substituting 𝜔2
𝜔1 + 482.84 =
(241.42)2
𝜔1
𝜔1
2
+ 482.84𝜔1 − (241.42)2
= 0
Either 𝜔1 = 100.123 rad/sec or 𝜔1 = −582.123 rad/sec (ignore)
∴ 𝜔1 = 100.123 rad/ sec = 2𝜋 (16Hz) → The lower cut-off frequency
𝜔2 = 𝜔1 + 𝐵𝑊 = 100.123 + 482.84 = 582.963 rad/ sec = 2𝜋 (93Hz)
𝜔2→ the higher cut-off frequency.
The depth of the notch in dB can be found as
|−
𝑉𝑜1
𝑉𝑖𝑛
| = |2|
|1 + (𝑗𝜔𝐶𝑅 𝐺)2|
|1 + 2𝑗𝜔𝐶𝑅 𝐺 + (𝑗𝜔𝐶𝑅 𝐺)2|
|−
𝑉𝑜1
𝑉𝑖𝑛
| = |2|
√(1 + −𝜔2 𝐶2 𝑅 𝐺
2
)
2
√(1 − 𝜔2 𝐶2 𝑅 𝐺
2
)
2
+ (2𝜔𝐶𝑅 𝐺)2
At notch frequency
𝜔 = 𝜔 𝑜 =
1
𝐶𝑅 𝐺

6. |−
𝑉𝑜1
𝑉𝑖𝑛
| = |2|
√(1 + −𝜔2 𝐶2 𝑅 𝐺
2
)
2
√(1 − 𝜔 𝑜
2 𝐶2 𝑅 𝐺
2
)
2
+ (2𝜔 𝑜 𝐶𝑅 𝐺)2
= 0
But practically we need to assume that there exists mismatch between simulated
cut-off frequency ωo,sim and calculated one 1/CRG due to their tolerances
(inaccurate component values) and may include PVT mismatches (the assumed
error value between calculated ωo and simulated ωo is about 1.1%):
𝑒𝑟𝑟𝑜𝑟 =
(𝜔 𝑜,𝑐𝑎𝑙 − 𝜔 𝑜,𝑠𝑖𝑚)
𝜔 𝑜,𝑐𝑎𝑙
× 100%
1.1% =
(241.42 − 𝜔 𝑜,𝑠𝑖𝑚)
241.42
× 100%
𝜔 𝑜,𝑠𝑖𝑚 = 238.75rad/sec
Now, the depth of the notch is
|−
𝑉𝑜1
𝑉𝑖𝑛
| = 2
√(1 − (238.75 × 0.004142)2)2
√(1 − (238.75 × 0.004142)2)2 + (2 × 238.75 × 0.004142)2
|−
𝑉𝑜1
𝑉𝑖𝑛
| = 2
0.02207
√(0.02207)2 + (1.9778)2
|−
𝑉𝑜1
𝑉𝑖𝑛
| = 2
0.02207
1.9779
= 0.02232
The notch depth in dB is
20log |−
Vo1
Vin
| = 20 log(0.02232) = −33dB
The notch rejection (the notch attenuation) is around 33dB.
If we use ω = ωo = 241.42rad/sec (approximated calculated value), then
|−
𝑉𝑜1
𝑉𝑖𝑛
| = 2
√(1 − (241.42 × 0.004142)2)2
√(1 − (241.42 × 0.004142)2)2 + (2 × 241.42 × 0.004142)2
|−
𝑉𝑜1
𝑉𝑖𝑛
| = 2
0.000076719
√(1 − 0.9999)2 + (1.9999)2
= 7.6719 × 10−5
The notch depth in dB is

7. 20log |−
Vo1
Vin
| = 20 log(7.6719 × 10−5) = −82dB
The gain at the lower cut-off frequency ω1 is
|−
𝑉𝑜1
𝑉𝑖𝑛
| = 2
√(1 − (100.123 × 0.004142)2)2
√(1 − (100.123 × 0.004142)2)2 + (2 × 100.123 × 0.004142)2
|−
𝑉𝑜1
𝑉𝑖𝑛
| = 2
0.8280
√(0.8280)2 + (0.8294)2
= 1.413 𝑉/𝑉
The gain at the higher cut-off frequency ω2 is
|−
𝑉𝑜1
𝑉𝑖𝑛
| = 2
√(1 − (583.123 × 0.004142)2)2
√(1 − (583.123 × 0.004142)2)2 + (2 × 583.123 × 0.004142)2
|−
𝑉𝑜1
𝑉𝑖𝑛
| = 2
4.834
√(4.834)2 + (4.831)2
= 1.415 𝑉/𝑉
The gain in dB at both frequency is 3dB.

8. (iv) Show that Band pass filter (BPF) can be obtained as: BPF = 1 - two cascaded
sections of 1st
APF (BBF = 1 - 2nd
APF)
Solution:
The transfer function of two cascaded section of 1st
APF is given by
𝑇(𝑆) =
1 − 2𝑆𝐶𝑅 𝐺 + (𝑆𝐶𝑅 𝐺)2
1 + 2𝑆𝐶𝑅 𝐺 + (𝑆𝐶𝑅 𝐺)2
It is a 2nd
order APF.
𝜔𝑧 = 𝜔 𝑝 =
1
𝐶𝑅 𝐺
= 241.42 rad/sec = 2π (38.4Hz)
Using subtractor based on op-amp circuit where the two inputs of the subtractor
circuit are the input voltage and the output of the 2st
order APF. The output of the
subtractor is
𝑉𝑜1
𝑉𝑖𝑛
= 1 − (
1 − 2𝑆𝐶𝑅 𝐺 + (𝑆𝐶𝑅 𝐺)2
1 + 2𝑆𝐶𝑅 𝐺 + (𝑆𝐶𝑅 𝐺)2
)
𝑉𝑜1
𝑉𝑖𝑛
=
1 + 2𝑆𝐶𝑅 𝐺 + (𝑆𝐶𝑅 𝐺)2
− 1 + 2𝑆𝐶𝑅 𝐺 − (𝑆𝐶𝑅 𝐺)2
1 + 2𝑆𝐶𝑅 𝐺 + (𝑆𝐶𝑅 𝐺)2
𝑉𝑜1
𝑉𝑖𝑛
=
4𝑆𝐶𝑅 𝐺
1 + 2𝑆𝐶𝑅 𝐺 + (𝑆𝐶𝑅 𝐺)2
It is 2nd
order BPF. By using C = 0.1µF and RG = R = RF = 41.42kΩ:
In the two passbands ( ω → 0 and ω → ∞) the gain is 0dB, while the center
frequency is
𝜔 𝑜 =
1
𝐶𝑅 𝐺
= 241.42 rad/sec = 2π (38.4Hz)
The pole quality factor Q is obtained as
𝜔 𝑜
𝑄
=
2
𝐶𝑅 𝐺
𝑄 =
𝜔 𝑜
2/𝐶𝑅 𝐺
= 0.5
𝑄 =
𝜔 𝑜
𝐵𝑊
= 0.5
𝐵𝑊 = (𝜔2 − 𝜔1) =
𝜔 𝑜
𝑄
= 482.84 rad/ sec = 2𝜋 (76.89Hz)

9. 𝐵𝑊 = (𝜔2 − 𝜔1) =
𝜔 𝑜
𝑄
= 482.84 rad/sec
𝜔 𝑜 = √ 𝜔1 𝜔2 = 241.42 rad/sec
𝜔2 = 𝜔1 + 𝐵𝑊 = 𝜔1 + 482.84
𝜔2 =
𝜔 𝑜
2
𝜔1
=
(241.42)2
𝜔1
By substituting 𝜔2
𝜔1 + 482.84 =
(241.42)2
𝜔1
𝜔1
2
+ 482.84𝜔1 − (241.42)2
= 0
Either 𝜔1 = 100.123 rad/sec or 𝜔1 = −582.123 rad/sec (ignore)
∴ 𝜔1 = 100.123 rad/ sec = 2𝜋 (16Hz) → The lower cut-off frequency
𝜔2 = 𝜔1 + 𝐵𝑊 = 100.123 + 482.84 = 582.963 rad/ sec = 2𝜋 (93Hz)
𝜔2→ the upper cut-off frequency.
The gain at the center frequency 𝜔 𝑜 =
1
𝐶𝑅 𝐺
can be found as
|−
𝑉𝑜1
𝑉𝑖𝑛
| = 4
|𝑗𝜔𝐶𝑅 𝐺|
|1 + 2𝑗𝜔𝐶𝑅 𝐺 + (𝑗𝜔𝐶𝑅 𝐺)2|
|−
𝑉𝑜1
𝑉𝑖𝑛
| = 4
√(𝜔𝐶𝑅 𝐺)2
√(1 − 𝜔2 𝐶2 𝑅 𝐺
2
)
2
+ (2𝜔𝐶𝑅 𝐺)2
|−
𝑉𝑜1
𝑉𝑖𝑛
| = 4
√(1)2
√(0)2 + (2)2
|−
𝑉𝑜1
𝑉𝑖𝑛
| = 4
1
2
= 2 𝑉/𝑉
The gain in dB is
20log |−
Vo1
Vin
| = 20 log(2) = 6dB

10. (v) Verify (i)-(iv) using Spice simulation.
LTspice simulation for (i) 1st
LPF:

11. LTspice simulation for (ii) 1st
HPF:

12. LTspice simulation for (iii) 2st
BRF:

14. LTspice simulation for (iv) 2st
BPF: