SME
Bioinstrumentation
Signal filtering
Use this for “An RC low-pass has 10 kΩ and 0.1 μF. What is its cutoff, and its gain at 500 Hz?”
Diagram
fc
=
1
2πRC
=
159
Hz
|H|
=
1
÷
√(1
+
(
f
fc
)
2
)
=
0.303,
or
−10.4
dB
This is an example. Type your own values to start a new problem.
fc (Hz)
=
1
2π
×
R (kΩ)
×
C (μF)
|H|
=
1
1
+
f (Hz)
fc (Hz)
2
G
Gain in decibels
calculated · dB
Formulas
fc
=
1
2π ×
R
×
C
159
=
1
2π × 10,000 × (1 ×
10
-7
)
|H|
=
1
÷
√(1
+
(
f
fc
)
2
)
0.303
=
1
÷
√(1
+
(
500
159
)
2
)
G
=
20
×
log₁₀(|H|)
−10.4 = 20 × log₁₀(0.303)
Formulas are worked in Ω, F. The step-by-step shows the conversions.
Assumptions
•
A first-order RC low-pass: the output is taken across C, with nothing loading it.
•
The input is a sine wave at f; the gain |H| is output amplitude over input amplitude.
•
At the cutoff fc the gain is 1/√2, or −3 dB; well above it the gain falls 20 dB a decade.
Step-by-step
WE KNOW
R = 10 kΩ, C = 0.1 μF, f = 500 Hz
FIND
cutoff frequency (fc), gain (|H|), gain in decibels (G)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to Ω, F).
R = 10 kΩ = 10,000 Ω (1 kΩ = 1000 Ω)
C = 0.1 μF = 1 × 10⁻⁷ F (1 F = 10⁶ μF)
Step 2 · Find cutoff frequency (fc)
Use fc = 1 ÷ (2π × R × C)
The
cutoff
is
where
R
equals
the
capacitor’s
reactance
1
2πfC
.
fc
=
1
2π × 10,000 ×
10
-7
fc = 159 Hz
Step 3 · Find gain (|H|)
Use |H| = 1 ÷ √(1 + (f ÷ fc)²)
R
and
C
share
the
input
as
a
divider;
the
higher
f
is,
the
less
C
keeps.
|H|
=
1
÷
√(1
+
(
500
159
)
2
)
|H|
=
1
÷
√
1 +
3.145
2
|H|
=
1
÷
√
1 + 9.891
|H|
=
1
÷
√
10.891
|H|
=
1
3.3
|H| = 0.303
Step 4 · Find gain in decibels (G)
Use G = 20 × log₁₀(|H|)
Decibels count the gain on a log scale: 20 dB for each factor of 10 in amplitude.
G
=
20
×
log₁₀(0.303)
G
=
20
×
(−0.5186)
G
=
−10.4
dB
Check
Put the numbers back in. Both sides should match.
159
=
1
2π × 10,000 ×
10
-7
✓
0.303
=
1
÷
√(1
+
(
500
159
)
2
)
✓
−10.4 = 20 × log₁₀(0.303) ✓
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