Sinusoidal steady state and the s-plane
The gain of a filter at any frequency can be measured with a ruler on a drawing: it is the distance between a point on the imaginary axis and the pole. The s-plane turns the frequency response into geometry.
01The complex frequency \( s \)
The previous topic used \( j\omega \) to turn derivatives into products. That \( j\omega \) is a particular case of something more general: the complex frequency \( s = \sigma + j\omega \). A signal of the form \( e^{st} \) covers, in a single stroke, all the elementary signals of the signals unit:
| Value of \( s \) | Signal \( e^{st} \) |
|---|---|
| \( s = 0 \) | A constant: pure DC |
| \( s = \sigma \) real and negative | A decaying exponential: a transient |
| \( s = j\omega \), purely imaginary | A sinusoid of constant amplitude: steady state |
| \( s = \sigma + j\omega \) with \( \sigma<0 \) | A damped sinusoid: an oscillatory transient |
| \( \sigma>0 \) | Something that grows without bound: an unstable circuit |
Sinusoidal steady state is, then, what happens along a single line of the plane: the imaginary axis. The rest of the \( s \)-plane describes transients, and is studied in the transient response unit.
02The transfer function, its poles and its zeros
With impedances written in \( s \) —\( Z_L = sL \), \( Z_C = 1/sC \)— the relationship between the output and the input of a linear circuit is a ratio of polynomials in \( s \): the transfer function \( H(s) \). The roots of the numerator are the zeros; those of the denominator, the poles. They are enough to know the entire circuit, up to a constant.
That evaluation has a geometric reading that is worth internalizing. The magnitude of \( H(j\omega) \) is inversely proportional to the distance between the point \( j\omega \) and the pole: when \( \omega \) is small, the point is close to the origin and the distance to the pole hardly changes; when \( \omega \) exceeds \( |p| \), the distance grows with \( \omega \) and the gain falls. The phase angle is, with its sign changed, the angle of that same vector.
On the left, the \( s \)-plane with the RC filter’s pole and the point \( j\omega \) that runs along the imaginary axis. On the right, the gain in decibels that comes from measuring that vector. You move the frequency and the two views move together.
03Decibels and Bode plots
The gain of a circuit varies over many orders of magnitude, and so does the frequency. That is why the frequency response is plotted with both axes logarithmic: the frequency in decades and the gain in decibels, \( 20\log_{10}|H| \). On that scale, each pole contributes a straight line with a slope of −20 dB per decade starting at its cutoff frequency, and each zero, +20 dB per decade.
| \( f/f_c \) | \( |H| \) | Gain | Phase |
|---|---|---|---|
| 0.1 | 0.995 | −0.04 dB | −5.7° |
| 1 | 0.707 | −3.01 dB | −45° |
| 10 | 0.0995 | −20.04 dB | −84.3° |
| 100 | 0.0100 | −40.00 dB | −89.4° |
One decade above the cutoff, the filter already attenuates ten times; two decades above, a hundred. That rule of −20 dB per decade per pole is what makes it possible to sketch by hand, in a minute, the response of a circuit with several poles: you add up straight lines.
04Power in sinusoidal steady state
With sinusoidal voltage and current out of phase by an angle \( \varphi \), the instantaneous power oscillates, and only part of it is turned into useful work. Three powers organize the matter:
At 230 V it draws 10 A with a power factor of 0.8 lagging. The apparent power is 2300 VA; the active power, 1840 W; the reactive power, 1380 var. The utility has to carry 10 A even though only 8 A “work”: that is why it charges surcharges to industrial users with a low power factor, and why power-factor-correction capacitors exist, which locally supply the reactive power the motor asks for.
05Where it shows up in electronics
| Case | What is read from the s-plane or from the Bode plot |
|---|---|
| Anti-aliasing filter before an A/D converter | How many poles are needed to attenuate enough above half the sampling frequency |
| Stability of a feedback amplifier | If any closed-loop pole crosses into the right half-plane, the amplifier oscillates |
| Op amp datasheet | The gain-bandwidth product is, literally, the location of its dominant pole |
| Switched-mode power supply with power factor correction | The PFC stage makes the mains current follow the voltage, with a \( pf \) close to 1 |
06In the lab
Build an RC with 1.6 kΩ and 100 nF, drive it with a function generator and measure the output amplitude and phase shift at 100 Hz, 1 kHz and 10 kHz. Compare with the table in section 3.
Using this page’s lab, verify that at \( f = f_c \) the vector joining the pole to \( j\omega \) makes 45° with the real axis, and that at \( f = 10f_c \) it makes almost 84°.
Calculate the parallel capacitor that takes the motor of the example from a pf of 0.8 to a pf of 0.95, at 230 V and 50 Hz.
07Common mistakes
- Confusing \( s \) with \( j\omega \). Sinusoidal steady state is \( s=j\omega \); the rest of the plane describes transients.
- Using \( 10\log \) for a voltage ratio. For voltages and currents it is \( 20\log \); \( 10\log \) is for powers.
- Forgetting the sign of a pole’s phase. A pole lags: it contributes negative phase, down to −90°.
- Adding apparent powers as if they were active. \( S \) is a complex number; its magnitudes do not add unless all the loads have the same power factor.
- Believing that reactive power is “lost.” It is not dissipated; what it costs is the extra current it makes circulate through the conductors.
08Self-assessment
On which part of the \( s \)-plane is the frequency response evaluated?
On the imaginary axis, \( s = j\omega \).
Where is the pole of an RC low-pass filter?
At \( s = -1/RC \), on the negative real half-axis.
What is the gain at the cutoff frequency?
\( 1/\sqrt2 \approx 0.707 \), that is, −3.01 dB, with −45° of phase.
What slope does each pole contribute on a Bode plot?
−20 dB per decade above its cutoff frequency.
What does the power factor measure?
The fraction of the apparent power that is active: \( \cos\varphi = P/|S| \).
What does a pole in the right half-plane indicate?
An unstable circuit: its response grows without bound.
09Further reading
- Charles K. Alexander and Matthew N. O. Sadiku. Fundamentals of Electric Circuits. 7th ed., McGraw-Hill, 2021. The chapters on AC power and frequency response, with hand-drawn Bode plots.
- Mac E. Van Valkenburg. Network Analysis. 3rd ed., Prentice Hall, 1974 (Spanish edition, “Análisis de redes”, Limusa). The classic text that builds the whole theory of networks around the s-plane, with the geometric interpretation of poles and zeros.
- Adel S. Sedra and Kenneth C. Smith. Microelectronic Circuits. 8th ed., Oxford University Press, 2020. The appendix on transfer functions and Bode plots, oriented toward what amplifier stability later demands.