---
title: "Frequency response of networks and complex frequency"
subject: "Electrical Circuit Theory"
chapter: 5
key: "semester_3/electrical_circuit_theory:frequency-response-complex-frequency-role"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/frequency-response-complex-frequency-role
site: BCT Engineering
---

# Frequency response of networks and complex frequency

Electrical Circuit Theory · Chapter 5: Frequency Response of Network

- Asked in: 2068 Baisakh, 6 marks
- Video: [What is Complex Frequency? What is s?](https://www.youtube.com/watch?v=8QFwB0M7Ig8) — Coulomb Cloud
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/frequency-response-complex-frequency-role
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-5

## Question

What do you understand by frequency response of networks and hence highlight the role of complex frequency in studying the frequency response.

## Short answer

The frequency response is the sinusoidal steady-state output magnitude and phase as functions of frequency ω, given by evaluating the network function H(s) at s=jω. Complex frequency s=σ+jω allows a unified description: the s-plane pole-zero diagram governs stability, resonance, damping, and Bode plots by showing how H(jω) varies along the imaginary axis.

## Answer

**Frequency response** of a network is the way its steady-state output, for sinusoidal input, varies with the frequency of the input. It is described by the network function H(jω) = |H(jω)| ∠ φ(ω), as two plots against ω:

- magnitude response |H(jω)| (often in dB), which shows which frequencies pass and which are attenuated (low-pass, high-pass, band-pass, band-reject);
- phase response φ(ω), which shows the phase shift given to each frequency.

#### Complex frequency

The complex frequency is s = σ + jω.

- σ (nepers/s) gives exponential growth or decay, e^(σt).
- ω (rad/s) gives sinusoidal oscillation.
- The signal e^(st) = e^(σt)(cos ωt + j sin ωt) therefore covers d.c. (s = 0), pure sinusoids (s = jω), decaying and growing exponentials and damped sinusoids in a single expression.

#### Role in frequency response

1. A network function H(s) (ratio of polynomials in s) describes the network for every kind of excitation. The sinusoidal steady-state response is obtained simply by putting s = jω, i.e. evaluating H(s) on the imaginary axis of the s-plane.
2. H(s) is fixed by its poles and zeros (critical frequencies): H(s) = K (s − z1)(s − z2)... / [(s − p1)(s − p2)...]. For a given ω:

```text
|H(jω)| = K · Π|jω − zi| / Π|jω − pi|
φ(ω)    = Σ ∠(jω − zi) − Σ ∠(jω − pi)
```

So the response can be sketched from the pole-zero plot: as the point jω moves up the imaginary axis, the magnitude falls when it moves away from a pole or towards a zero, and it peaks near a pole that lies close to the axis (resonance).
3. Poles in the left half plane mean a stable network; the further a pole is from the jω axis, the more damped is the response and the flatter the peak.
4. It leads directly to Bode plots: each factor (s + a) contributes a break at ω = a with a slope change of ±20 dB/decade.

**Example.** H(s) = 1/(s + 1), a pole at s = −1:

| ω (rad/s) | \|H(jω)\| | Phase |
|---|---|---|
| 0.1 | 0.995 | −5.7° |
| 1 | 0.707 | −45° |
| 10 | 0.0995 | −84.3° |

So it is a low-pass response with −20 dB/decade roll-off above the corner ω = 1 rad/s.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What does the frequency response of a network describe?
   A. How the steady-state sinusoidal output (magnitude and phase) varies with input frequency
   B. How the transient decays after switching
   C. The network's response to a d.c. input only
   D. How component values change with temperature
   Answer: A. It is H(jω) shown as magnitude and phase plots against ω. (Key point: Frequency response meaning)

2. What does s = σ + jω represent?
   A. σ is the oscillation frequency and ω the decay
   B. Both are only angular frequencies
   C. σ is the phase and ω the amplitude
   D. σ gives exponential growth/decay and ω gives oscillation
   Answer: D. e^(st) = e^(σt)(cos ωt + j sin ωt) combines decay and oscillation. (Key point: Complex frequency)

3. How is frequency response obtained from H(s)?
   A. Put s = 0
   B. Put s = σ with ω = 0
   C. Put s = jω, evaluating on the imaginary axis
   D. Take the inverse Laplace transform of H(s)
   Answer: C. Sinusoidal steady state corresponds to s on the jω axis. (Key point: Getting the sinusoidal response)

4. How do poles and zeros shape |H(jω)|?
   A. Magnitude falls near poles and rises near zeros
   B. Magnitude rises near a pole close to the jω axis and falls approaching a zero
   C. Poles and zeros affect only the phase
   D. Magnitude is fixed by the zeros alone
   Answer: B. |H| = K Π|jω - zi| / Π|jω - pi|, so poles in the denominator raise it when close. (Key point: Role of poles and zeros)

## Related in Electrical Circuit Theory

- [Current response and average power, non-sinusoidal source](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Afourier-average-power-nonsinusoidal-r-l) — Electrical Circuit Theory, chapter 6
- [Poles and zeros of a network function](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Apoles-and-zeros-significance) — Electrical Circuit Theory, chapter 5

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