SME
Circuit Analysis II
Frequency response and filters
Use this for “Find the cutoff frequency of an RC low-pass filter with R = 1 kΩ and C = 0.1 μF, and its gain at 10 kHz.”
Diagram
Corner at fc = 1,592 Hz.
At f = 10,000 Hz: |H| = 0.157 = −16.1 dB, phase −81.0°.
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 (dB)
=
20
log₁₀(
|H|
)
φ
Phase
calculated · °
Formulas
fc
=
1
2π ×
R
×
C
1,591.5494
=
1
2π × 1,000 × (1 ×
10
-7
)
|H|
=
1
÷
√(1
+
(
f
fc
)
2
)
0.1572
=
1
÷
√(1
+
(
10,000
1,591.5494
)
2
)
G
=
20
×
log₁₀(|H|)
−16.0722 = 20 × log₁₀(0.1572)
φ
=
−tan⁻¹(
f
fc
)
−80.9569
=
−tan⁻¹(
10,000
1,591.5494
)
Formulas are worked in Ω, F. The step-by-step shows the conversions.
Assumptions
•
The output is taken across C with no load on it, from an ideal source: a divider of R and the capacitor’s reactance.
•
At fc the gain is −3.01 dB and the phase −45°; far above fc the gain falls 20 dB per decade.
•
The phase φ is the output’s angle minus the input’s, so a negative φ means the output lags.
Step-by-step
WE KNOW
R = 1 kΩ, C = 0.1 μF, f = 10,000 Hz
FIND
cutoff frequency (fc), gain ratio (|H|), gain in decibels (G), phase (φ)
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 = 1 kΩ = 1,000 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 × 1,000 ×
10
-7
fc = 1,591.549 Hz
Step 3 · Find gain ratio (|H|)
Use |H| = 1 ÷ √(1 + (f ÷ fc)²)
The
divider
of
R
and
1
2πfC
:
the
higher
f
is,
the
less
of
the
input
reaches
C
.
|H|
=
1
÷
√(1
+
(
10,000
1,591.549
)
2
)
|H|
=
1
÷
√
1 +
6.283
2
|H|
=
1
÷
√
1 + 39.48
|H|
=
1
÷
√
40.48
|H|
=
1
6.362
|H| = 0.157177
Step 4 · Find gain in decibels (G)
Use G = 20 × log₁₀(|H|)
A gain ratio of voltages in decibels is 20 log₁₀ of it.
G
=
20
×
log₁₀(0.157177)
G
=
20
×
(−0.8036)
G
=
−16.0722
dB
Step 5 · Find phase (φ)
Use φ = −tan⁻¹(f ÷ fc)
The output lags: 0° far below fc, −45° at fc, toward −90° far above.
φ
=
−tan⁻¹(
10,000
1,591.549
)
φ
=
−80.9569°
Check
Put the numbers back in. Both sides should match.
1,591.549
=
1
2π × 1,000 ×
10
-7
✓
0.157177
=
1
÷
√(1
+
(
10,000
1,591.549
)
2
)
✓
−16.0722 = 20 × log₁₀(0.157177) ✓
−80.9569
=
−tan⁻¹(
10,000
1,591.549
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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