Transfer Functions

The single complex-valued function that tells you everything a linear circuit does to a sinusoid — at any frequency.

Circuit AnalysisElectrical Engineering Year 2Free preview
⏱️ About 16 min

Your phone's speaker plays a bass drum and a piccolo at the same volume — but the tiny amplifier inside treats them very differently. Why?

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The big idea: Every linear circuit acts as a frequency-dependent scale and shifter for sinusoidal signals. The transfer function H(jω) captures that behavior in a single complex expression — its magnitude tells you how much of each frequency passes through, and its angle tells you how much each frequency is shifted in time.
🎯 By the end, you'll be able to
  • Define the transfer function H(jω) = Vout(jω)/Vin(jω) and explain why it is complex-valued
  • Derive the transfer function of an RC low-pass filter from voltage-divider analysis in the phasor domain
  • Compute the magnitude |H(jω)| and phase ∠H(jω) at specific frequencies, including the corner frequency ωc
  • Explain the asymptotic behavior of a low-pass filter as ω approaches zero and infinity
📎 Helpful to know first
  • Impedance & Admittance

What a transfer function captures

When you feed a sinusoidal input Vin(jω) into a linear circuit at angular frequency ω, the steady-state output Vout(jω) is another sinusoid at the same frequency — but generally with a different amplitude and a shifted phase. The ratio of output to input is the transfer function:

H(jω) = Vout(jω) / Vin(jω)

Because phasor voltages are complex numbers, H(jω) is complex-valued. Its magnitude |H(jω)| is the gain (amplitude scaling) at that frequency, and its argument ∠H(jω) is the phase shift the circuit imposes. Together these two real-valued functions of ω constitute the circuit's frequency response.

\[ H(j\omega) = \frac{V_{out}(j\omega)}{V_{in}(j\omega)}, \qquad |H(j\omega)| = \text{gain}, \qquad \angle H(j\omega) = \text{phase shift} \]
The transfer function fully characterizes how a linear circuit responds to a sinusoidal input at angular frequency ω.

Deriving the RC low-pass transfer function

Consider a series resistor R and capacitor C driven by Vin across the pair, with Vout measured across the capacitor. In the phasor domain the capacitor's impedance is Z_C = 1/(jωC), so this is simply a voltage divider:

H(jω) = Z_C / (R + Z_C) = (1/(jωC)) / (R + 1/(jωC))

Multiplying numerator and denominator by jωC gives the clean form H(jω) = 1 / (1 + jωRC). Defining the corner frequency ωc = 1/(RC) lets us write the transfer function in normalized form as 1/(1 + jω/ωc), which separates the circuit's behavior from its specific component values.

\[ H(j\omega) = \frac{1}{1 + j\omega R C} = \frac{1}{1 + j\,\omega/\omega_c}, \qquad \omega_c = \frac{1}{RC} \]
Transfer function of the RC low-pass filter. The corner frequency ωc (in rad/s) is set entirely by the RC product.

Magnitude and phase at key frequencies

The magnitude and phase of H(jω) = 1/(1 + jω/ωc) are obtained by taking the modulus and argument of the denominator:

|H(jω)| = 1/√(1 + (ω/ωc)²) and ∠H(jω) = −arctan(ω/ωc)

Three regimes tell the whole story. At low frequencies (ω ≪ ωc), the filter passes the signal with unity gain and negligible phase shift. At the corner (ω = ωc), the gain drops to 1/√2 ≈ 0.707 and the phase is −45°. At high frequencies (ω ≫ ωc), the gain falls off as 1/ω and the phase approaches −90°. This is why it is called a low-pass filter: low frequencies pass through, high frequencies are attenuated.

\[ |H(j\omega)| = \frac{1}{\sqrt{1+(\omega/\omega_c)^2}}, \qquad \angle H(j\omega) = -\arctan\!\left(\frac{\omega}{\omega_c}\right) \]
Magnitude (gain) and phase of the first-order low-pass transfer function. At ω = ωc the gain is 1/√2 — the half-power point.
📝 Worked example: An RC low-pass filter has R = 1 kΩ and C = 1 μF. Find the corner frequency ωc, then evaluate the magnitude and phase at (a) ω = ωc and (b) ω = 3ωc.
  1. Compute RC = 1000 × 1×10⁻⁶ = 10⁻³ s, so ωc = 1/(RC) = 1/10⁻³ = 1000 rad/s.
  2. (a) At ω = ωc = 1000 rad/s: |H| = 1/√(1+1²) = 1/√2 ≈ 0.7071, and ∠H = −arctan(1) = −45°.
  3. (b) At ω = 3ωc = 3000 rad/s: |H| = 1/√(1+3²) = 1/√10 ≈ 0.3162, and ∠H = −arctan(3) ≈ −71.57°.
  4. Check limits: as ω → 0, H → 1 (0° phase); as ω → ∞, H → 0 (−90° phase). Consistent with low-pass behavior.
✓ (a) At ωc = 1000 rad/s: |H| ≈ 0.7071, ∠H = −45°. (b) At 3ωc = 3000 rad/s: |H| ≈ 0.3162, ∠H ≈ −71.57°.
✏️ Practice: An RC low-pass filter uses R = 500 Ω and C = 1 μF. What is the corner frequency ωc in rad/s?
rad/s
Solution
  1. RC = 500 × 1×10⁻⁶ = 5×10⁻⁴ s.
  2. ωc = 1/(RC) = 1/(5×10⁻⁴) = 2000 rad/s.

Check your understanding

1. For the RC low-pass filter H(jω) = 1/(1 + jω/ωc), what is the phase shift as ω → ∞?
As ω → ∞ the imaginary term dominates the denominator, so H(jω) approaches 1/(j∞), whose angle is −90°.
2. At the corner frequency ω = ωc, the magnitude |H| equals 1/√2. Why is this called the 'half-power point'?
Power is proportional to V². When |H| = 1/√2, the output voltage is 1/√2 of the input, so the output power is (1/√2)² = 1/2 of the input power.
🔎 Related lessons
Bode Plots
✅ Key takeaways
  • The transfer function H(jω) = Vout/Vin is a complex ratio whose magnitude gives gain and whose angle gives phase shift at each frequency.
  • An RC low-pass filter has H(jω) = 1/(1 + jω/ωc) with corner frequency ωc = 1/(RC).
  • At ω = ωc, the gain is 1/√2 ≈ 0.707 (the half-power point) and the phase is −45°.
  • As ω → 0 the filter passes signals unaltered; as ω → ∞ it attenuates them toward zero with phase approaching −90°.
➡️ Knowing the exact magnitude and phase at every frequency is powerful, but plotting them on a linear scale quickly becomes unreadable — the next lesson introduces the logarithmic Bode plot, the engineer's standard tool for visualizing frequency response.
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