Frequency Response

Learning Objectives

• Define frequency response and explain what magnitude response and phase response describe.
• Derive and use the transfer function of a simple RC circuit.
• Calculate the cutoff (-3dB) frequency of an RC filter.
• Convert a magnitude ratio to decibels (dB) and interpret the result.
• Explain how bandwidth is defined and why it matters for real circuits.
• Connect frequency response concepts to practical filter and amplifier design.

Quick Answer

Frequency response describes how a circuit's output changes — in both amplitude and phase — as you feed it input signals of different frequencies. Every circuit that contains a capacitor or inductor behaves differently at different frequencies, because reactance (X_C = 1/ωC, X_L = ωL) depends on frequency. Frequency response is captured by two curves: magnitude response (how much the circuit amplifies or attenuates each frequency) and phase response (how much it shifts the timing of each frequency). This matters enormously in practice — it's why an audio equalizer can boost bass without touching treble, why a WiFi filter passes 2.4GHz but blocks other frequencies, and why an amplifier that sounds "flat" in the midrange might roll off badly at high frequencies.

The Transfer Function

The relationship between a circuit's output and input, as a function of frequency, is captured by the transfer function H(jω):

H(jω) = V_out(jω) / V_in(jω)

where ω = 2πf is the angular frequency (f in hertz) and j is the imaginary unit.

Worked Example 1: Derive the transfer function of a simple RC low-pass filter (R in series, output taken across C).

By voltage division using impedances: V_out/V_in = Z_C / (R + Z_C)
Z_C = 1/(jωC), so:
H(jω) = [1/(jωC)] / [R + 1/(jωC)] = 1 / (1 + jωRC)

Magnitude Response

The magnitude response |H(jω)| tells you how much the output amplitude is scaled at each frequency.

For the RC low-pass filter above:

|H(jω)| = 1 / √(1 + (ωRC)²)

Worked Example 2: For R = 1kΩ and C = 1µF, find the magnitude response at f = 100Hz and at f = 10kHz.

RC = 1000 × 1×10⁻⁶ = 1×10⁻³ s (time constant)

At f = 100 Hz: ω = 2π(100) = 628.3 rad/s
ωRC = 628.3 × 0.001 = 0.6283
|H| = 1/√(1+0.6283²) = 1/√1.395 = 1/1.181 ≈ 0.847 (about 85% of input passes)

At f = 10,000 Hz: ω = 2π(10000) = 62,832 rad/s
ωRC = 62,832 × 0.001 = 62.83
|H| = 1/√(1+62.83²) = 1/√(1+3948) ≈ 1/62.84 ≈ 0.0159 (only 1.6% passes)

This confirms the filter's behavior: low frequencies pass through almost unchanged, while high frequencies are heavily attenuated — exactly what a low-pass filter is designed to do.

Phase Response

The phase response φ(ω) tells you the phase shift the circuit introduces between input and output at each frequency.

For the same RC low-pass filter:

φ(ω) = −arctan(ωRC)

Worked Example 3: Find the phase shift at f = 100Hz and f = 10kHz using the same R and C as above.

At f = 100 Hz: φ = −arctan(0.6283) ≈ −32.1°
At f = 10,000 Hz: φ = −arctan(62.83) ≈ −89.1°

At low frequencies the output barely lags the input; near and above the cutoff frequency, the lag approaches 90°, which is characteristic of a first-order RC low-pass filter.

Cutoff Frequency and Bandwidth

The cutoff frequency (or -3dB frequency) is where the output power drops to half its maximum value — equivalently, where the magnitude drops to 1/√2 ≈ 0.707 of its maximum.

f_c = 1 / (2πRC)

Worked Example 4: Find the cutoff frequency for R = 1kΩ, C = 1µF.

f_c = 1 / (2π × 1000 × 1×10⁻⁶) = 1 / (2π × 0.001) = 1 / 0.006283 ≈ 159.2 Hz

Bandwidth is the range of frequencies over which the circuit passes signals effectively (within a defined tolerance, typically -3dB of the maximum). For this single-pole low-pass filter, the bandwidth is simply from 0Hz up to f_c ≈ 159.2Hz.

Gain in Decibels

Gain is often expressed in decibels (dB) because it compresses a huge range of ratios into manageable numbers and makes cascaded stages easy to combine (by simple addition instead of multiplication):

Gain (dB) = 20 × log₁₀(|H(jω)|)

Worked Example 5: Convert the magnitude response found at f=100Hz (|H|≈0.847) and f=10kHz (|H|≈0.0159) into dB.

At 100 Hz: Gain = 20 × log10(0.847) = 20 × (−0.0722) ≈ −1.44 dB
At 10 kHz: Gain = 20 × log10(0.0159) = 20 × (−1.799) ≈ −36.0 dB

At the cutoff frequency itself, |H| = 0.707 always gives exactly −3.01dB — this is where the "-3dB point" name comes from, and it's the standard reference point engineers use to define a filter's bandwidth.

Key Terms

Term	Definition
Frequency response	How a circuit's output magnitude and phase vary with input frequency
Transfer function H(jω)	The ratio V_out/V_in as a function of frequency, expressed as a complex number
Magnitude response	|H(jω)|, the amplitude scaling factor at each frequency
Phase response	The phase shift (in degrees or radians) introduced between input and output
Cutoff frequency (f_c)	The frequency where output power drops to half (magnitude drops to 0.707 of max)
Bandwidth	The range of frequencies over which a circuit passes signals effectively
Decibel (dB)	A logarithmic unit for expressing gain: dB = 20log10(magnitude ratio)

Common Mistakes

1. Misconception: "The cutoff frequency is where the signal is completely blocked." Why it's wrong: At f_c the output isn't zero — it's still 70.7% of the input amplitude (half the power). Signal continues to pass, just attenuated, beyond the cutoff. Correct: The cutoff frequency marks the -3dB point (half power), not a hard stop; attenuation increases gradually as frequency moves further past f_c.

2. Misconception: "-3dB means the signal has lost 3% of its strength." Why it's wrong: dB is logarithmic, not a percentage. A -3dB drop corresponds to the power halving (and the voltage/amplitude dropping to about 70.7%), which is a much bigger change than "3%." Correct: Use Gain(dB) = 20log10(V_out/V_in) for voltage ratios; -3dB always corresponds to a magnitude ratio of 1/√2, i.e., half power.

3. Misconception: "Phase response doesn't matter, only magnitude response affects sound/signal quality." Why it's wrong: Phase shifts can distort the timing relationships between different frequency components of a complex signal, causing audible or measurable distortion even when magnitude response is flat. Correct: Both magnitude and phase response together fully describe a circuit's effect on a signal; audio and communication engineers care about both, especially in systems combining multiple filter stages.

Comparison and Connections

Concept	Magnitude Response	Phase Response
What it shows	Amplitude scaling vs. frequency	Timing shift vs. frequency
RC low-pass formula	1/√(1+(ωRC)²)	−arctan(ωRC)
At cutoff frequency	Equals 0.707 (−3dB)	Equals −45°
Practical concern	Loudness/signal strength changes	Waveform timing/shape distortion

Practice Questions

Recall 1: What is the defining condition of the cutoff frequency in terms of magnitude and power? Answer guidance: Magnitude drops to 1/√2 ≈ 0.707 of maximum; power drops to half (−3dB point).

Recall 2: Write the formula for gain in decibels given a magnitude ratio |H|. Answer guidance: Gain(dB) = 20 × log10(|H|).

Understanding 1: Why is the phase shift exactly −45° at the cutoff frequency of a simple RC low-pass filter? Answer guidance: At f_c, ωRC = 1 (by definition of f_c = 1/2πRC), so φ = −arctan(1) = −45°.

Understanding 2: Explain why engineers prefer decibels over raw ratios for expressing gain. Answer guidance: dB compresses very large or small ratios into manageable numbers and turns cascaded multiplicative gains into simple additions, making multi-stage system analysis much easier.

Application 1: An RC low-pass filter has R = 2kΩ and C = 0.5µF. Find its cutoff frequency. Answer guidance: f_c = 1/(2π×2000×0.5×10⁻⁶) = 1/(2π×0.001) ≈ 159.2 Hz.

Application 2: A filter has a magnitude response of |H| = 0.1 at a certain frequency. Express this as a gain in dB. Answer guidance: Gain = 20×log10(0.1) = 20×(−1) = −20 dB.

Analysis 1: Two RC low-pass filters have cutoff frequencies of 100Hz and 10,000Hz. Compare their behavior at 1kHz — which filter passes more signal, and why? Answer guidance: The 10kHz-cutoff filter passes far more signal at 1kHz because 1kHz is well below its cutoff (still in the "flat" passband), whereas 1kHz is a full decade above the 100Hz filter's cutoff, where attenuation is already significant (roughly −20dB per decade beyond cutoff for a first-order filter).

Analysis 2: A student measures a circuit's output amplitude as exactly half the input amplitude and concludes they've found the cutoff frequency. Evaluate this reasoning. Answer guidance: Incorrect — the cutoff frequency is defined by a magnitude ratio of 0.707 (voltage), not 0.5. A voltage ratio of 0.5 corresponds to about −6dB, not −3dB, so this point is actually past the true cutoff frequency, further into the attenuation region.

FAQ

Q1: What's the difference between a transfer function and frequency response? The transfer function H(jω) is the complete mathematical description (a complex number at every frequency); frequency response usually refers to the two separate real-valued plots derived from it — magnitude response and phase response — often shown together as a Bode plot.

Q2: Why is the cutoff frequency also called the "half-power point"? Because at f_c, the output power (proportional to V²) is exactly half the maximum power, even though the voltage magnitude is 0.707 (not 0.5) of maximum — power scales with the square of voltage.

Q3: Does every circuit have just one cutoff frequency? No — simple first-order RC or RL circuits have one cutoff frequency, but more complex filters (second-order and higher) can have multiple cutoff frequencies, defining a passband with both a lower and upper limit (bandpass filters).

Q4: How does frequency response relate to filter design? Filters are specifically designed circuits whose frequency response selectively passes or blocks certain frequency ranges — low-pass, high-pass, bandpass, and band-stop filters are all just different shapes of frequency response.

Q5: Can frequency response be measured directly, or only calculated? Both — in the lab, you sweep a signal generator across frequencies while measuring output amplitude and phase with an oscilloscope or network analyzer; this measured response should match the theoretical transfer function for a correctly built circuit.

Quick Revision

• Frequency response = how output magnitude and phase change with input frequency.
• Transfer function: H(jω) = V_out/V_in, a complex number.
• RC low-pass magnitude: |H(jω)| = 1/√(1+(ωRC)²).
• RC low-pass phase: φ(ω) = −arctan(ωRC).
• Cutoff frequency: f_c = 1/(2πRC), where magnitude = 0.707 and power = half of maximum.
• At cutoff frequency, phase shift is exactly −45° for a first-order RC low-pass.
• Gain in dB: 20×log10(|H|); −3dB corresponds to the cutoff (half-power) point.
• Bandwidth is the range of frequencies passed effectively, usually bounded by -3dB points.
• Both magnitude and phase response matter — phase shifts distort signal timing even when magnitude is unaffected.
• Frequency response underlies all filter design: low-pass, high-pass, bandpass, band-stop.

Related Topics

Prerequisites: AC and DC Circuits, complex impedance, basic trigonometry (arctan).

Related Topics: Filters, Network Theorems, Signal and Noise Analysis.

Next Topics: Filters, Signal and Noise Analysis.