(1) The document discusses obtaining first-order low-pass, high-pass, band-reject, and band-pass filters from a first-order all-pass filter. (2) It shows how to derive the transfer functions for each filter type by adding or subtracting the input and output of one or two cascaded all-pass filter sections. (3) Key specifications of each filter like cutoff frequencies, gain, and bandwidth are calculated and verified through SPICE simulation, showing good agreement between calculated and simulated responses.

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: