SME
Measurements & Instrumentation · Problem type
RC low-pass filter
Use this for “An RC filter has R = 10 kΩ and C = 0.1 μF. Find the cutoff and the gain at 500 Hz.”
Diagram
Corner at fc = 159.2 Hz.
At f = 500 Hz: |H| = 0.303 = −10.4 dB, phase −72.3°.
This is an example. Type your own values to start a new problem.
fc (Hz)
=
1
2π
×
R (kΩ)
×
C (μF)
G
=
1
1
+
(
f (Hz)
÷
fc (Hz)
)
2
GdB
Gain in decibels
calculated · dB
Formulas
fc
=
1
2π ×
R
×
C
159.1549
=
1
2π × 10,000 × (1 ×
10
-7
)
G
=
1
÷
√(1
+
(
f
fc
)
2
)
0.3033
=
1
÷
√(1
+
(
500
159.1549
)
2
)
GdB
=
20
×
log₁₀(
G
)
−10.3621 = 20 × log₁₀(0.3033)
Formulas are worked in Ω, F. The step-by-step shows the conversions.
Assumptions
•
First order: one resistor, then one capacitor, the output taken across the capacitor.
•
The load draws no current (a buffered output) and the source is ideal.
•
A sine input; at fc the gain is 0.707 (−3 dB), and it falls 20 dB a decade above.
Step-by-step
WE KNOW
R = 10 kΩ, C = 0.1 μF, f = 500 Hz
FIND
cutoff frequency (fc), gain ratio (G), gain in decibels (GdB)
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)
At
the
cutoff
the
capacitor’s
reactance
1
2πfC
equals
R
.
fc
=
1
2π × 10,000 ×
10
-7
fc = 159.1549 Hz
Step 3 · Find gain ratio (G)
Use G = 1 ÷ √(1 + (f ÷ fc)²)
R
and
the
capacitor
share
the
input:
the
higher
f
is,
the
less
of
it
reaches
C
.
G
=
1
÷
√(1
+
(
500
159.1549
)
2
)
G
=
1
÷
√
1 +
3.14159
2
G
=
1
÷
√
10.8696
G
=
1
3.29691
G
=
0.3033145
Step 4 · Find gain in decibels (GdB)
Use GdB = 20 × log₁₀(G)
A ratio of voltages in decibels is 20 log₁₀ of it.
GdB = 20 × log₁₀(0.3033145)
GdB = 20 × (−0.5181)
GdB = −10.3621 dB
Check
Put the numbers back in. Both sides should match.
159.1549
=
1
2π × 10,000 ×
10
-7
✓
0.3033145
=
1
÷
√(1
+
(
500
159.1549
)
2
)
✓
−10.3621 = 20 × log₁₀(0.3033145) ✓
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