Frequency response of networks and complex frequency

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:

|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:

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