SME
Circuit Analysis II · Problem type
RC high-pass filter: gain and phase
Use this for “An RC high-pass filter has R = 1 kΩ and C = 0.1 μF. Find its cutoff, and its gain and phase at 159.2 Hz.”
Diagram
Corner at fc = 1,592 Hz.
At f = 159.2 Hz: |H| = 0.0995 = −20.0 dB, phase 84.3°.
This is an example. Type your own values to start a new problem.
fc (Hz)
=
1
2π
×
R (kΩ)
×
C (μF)
|H|
=
f (Hz)
fc (Hz)
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|
=
(
f
fc
)
÷
√(1
+
(
f
fc
)
2
)
0.09953
=
(
159.2
1,591.5494
)
÷
√(1
+
(
159.2
1,591.5494
)
2
)
G
=
20
×
log₁₀(|H|)
−20.0408 = 20 × log₁₀(0.09953)
φ
=
90
−
tan⁻¹(
f
fc
)
84.2878
=
90
−
tan⁻¹(
159.2
1,591.5494
)
Formulas are worked in Ω, F. The step-by-step shows the conversions.
Assumptions
•
The output is taken across R with no load on it, from an ideal source; C blocks low frequencies.
•
At fc the gain is −3.01 dB and the phase +45°; far below fc the gain falls 20 dB per decade.
•
The phase φ is the output’s angle minus the input’s, so a positive φ means the output leads.
Step-by-step
WE KNOW
R = 1 kΩ, C = 0.1 μF, f = 159.2 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| = (f ÷ fc) ÷ √(1 + (f ÷ fc)²)
The
divider
of
1
2πfC
and
R
:
the
lower
f
is,
the
more
of
the
input
C
takes.
|H|
=
(
159.2
1,591.549
)
÷
√(1
+
(
159.2
1,591.549
)
2
)
|H|
=
159.2
1,591.549
÷
√
1 +
0.10003
2
|H|
=
159.2
1,591.549
÷
√
1 + 0.010006
|H|
=
159.2
1,591.549
÷
√
1.01
|H|
=
159.2
1,591.549
÷
1.005
|H|
=
0.10003
1.005
|H| = 0.0995316
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(0.0995316)
G
=
20
×
(−1.002)
G
=
−20.0408
dB
Step 5 · Find phase (φ)
Use φ = 90 − tan⁻¹(f ÷ fc)
The output leads: 90° far below fc, 45° at fc, toward 0° far above.
φ
=
90
−
tan⁻¹(
159.2
1,591.549
)
φ
=
84.2878°
Check
Put the numbers back in. Both sides should match.
1,591.549
=
1
2π × 1,000 ×
10
-7
✓
0.0995316
=
(
159.2
1,591.549
)
÷
√(1
+
(
159.2
1,591.549
)
2
)
✓
−20.0408 = 20 × log₁₀(0.0995316) ✓
84.2878
=
90
−
tan⁻¹(
159.2
1,591.549
)
✓
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