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Signals and Systems · 144 h · Topic 27 of 34

Frequency response

An LTI system does not mix frequencies: each harmonic comes out scaled and phase-shifted, unaware of the others. But looking only at the magnitude is not enough, and an all-pass shows why.

Bode Asymptotes Group delay Minimum phase Phase margin

01What the system does to each frequency

If a complex exponential enters an LTI system and comes out multiplied by \( H(s) \), then evaluating on the imaginary axis gives what happens to each real frequency:

\[ Y(j\omega) = H(j\omega)\,X(j\omega) \] The output spectrum is the input spectrum, multiplied frequency by frequency. Convolution in time has become a product.

This means that an LTI system does not mix frequencies: each harmonic of the input comes out scaled by \( |H| \) and shifted by \( \angle H \), unaware of the others. Analyzing a circuit comes down to reading two curves.

Lab · the filter, harmonic by harmonic

On top, the input and the output in time. Below, the input spectrum in gray, the filter response in yellow and what is left at the output in green.

02Why it is plotted in decibels and on a logarithmic scale

The Bode plot is not an arbitrary convention. Taking the logarithm of a product of factors, multiplication becomes addition:

\[ 20\log|H| = 20\log|K| + \sum_i 20\log|j\omega - z_i| - \sum_k 20\log|j\omega - p_k| \] Each pole and each zero contributes its own curve, and the total is the sum. That is why it can be drawn by hand, by adding asymptotes.

And with the frequency axis on a logarithmic scale, those asymptotes are straight lines: each simple pole contributes −20 dB per decade starting at its frequency, each zero +20. With two rules you can draw the Bode plot of any system:

ElementMagnitudePhase
Simple pole at \( \omega_p \)−20 dB/dec from \( \omega_p \); −3 dB at \( \omega_p \)0 to −90°, with −45° at \( \omega_p \)
Simple zero at \( \omega_z \)+20 dB/dec from \( \omega_z \)0 to +90°
Pole at the origin−20 dB/dec throughout−90° constant
Pair of poles with \( Q \)−40 dB/dec, with a peak of \( 20\log Q \) if \( Q \gt 0.707 \)0 to −180°

The phase transition of a simple pole is not abrupt: it starts a decade before \( \omega_p \) and ends a decade after. That is why, when compensating a feedback loop, a pole that seems far away still affects the phase margin.

03Phase distorts too

It is easy to stick to the magnitude alone. But if the harmonics of a signal arrive with different delays, the waveform is distorted even though its amplitudes have been preserved. What measures that is the group delay:

\[ \tau_g(\omega) = -\frac{d\,\angle H(j\omega)}{d\omega} \] If the phase is a straight line, the group delay is constant and all the harmonics are delayed by the same amount: the signal comes out unchanged, only later.

The extreme case is in the lab: an all-pass filter has a perfectly flat magnitude —it does not attenuate any frequency— and yet it distorts the signal, because its phase is not linear. It is the demonstration that looking at the magnitude alone is not enough.

Where it matters and where it does not

The ear is fairly insensitive to phase, so in audio it is tolerated. In video and in data transmission, it is not: an uneven group delay shifts some harmonics relative to others, widens the pulses and makes one symbol spill into the place of the next. That is called intersymbol interference, and it is the reason linear-phase filters are used in communications even though they cost more order.

04Minimum phase

A system is minimum phase if all of its poles and all of its zeros are in the left half-plane. The property that makes it interesting is this: among all systems with the same magnitude, the minimum-phase one is the one with the least phase shift and the smallest group delay.

And it has a very practical consequence: in a minimum-phase system, the magnitude determines the phase. Just by measuring the amplitude curve you can deduce the phase shift, which is what makes it possible to estimate the phase margin of a loop from the slope of the Bode plot. The rule that comes out of this —crossing zero with −20 dB/decade— is the design criterion for any feedback system.

Slope at the 0 dB crossingApproximate phaseMarginBehavior
−20 dB/dec−90°90°Stable and without overshoot
−30 dB/dec−135°45°Acceptable, with some overshoot
−40 dB/dec−180°0°At the limit: it oscillates

A zero in the right half-plane breaks all of this: it contributes the magnitude of a zero but the phase of a pole, so it worsens the margin while giving nothing in return. It is what complicates the compensation of boost converters.

05In the lab

Activity 1 · Bode plot by hand and measured

Draw by asymptotes the Bode plot of \( H(s) = \dfrac{100}{(1+s/10)(1+s/1000)} \) and then measure it with the generator and the oscilloscope at ten frequencies. Compare the two on the same sheet.

Activity 2 · The all-pass distorts

Build a first-order all-pass filter, inject a square wave and compare input and output on the oscilloscope. Verify with the analyzer that the magnitude is flat, and explain why the shape changed anyway.

Activity 3 · Phase margin

Measure the open-loop Bode plot of a feedback amplifier, find the 0 dB crossover frequency and read the phase there. Contrast the margin obtained with the overshoot shown by the step response.

06Common mistakes

  • Looking only at the magnitude. Phase distorts, and an all-pass proves it.
  • Believing that a distant pole does not affect the phase. Its transition starts a decade earlier.
  • Confusing phase delay with group delay. The latter is the derivative of the phase and is the one that tells you whether the shape is preserved.
  • Assuming a system with right-half-plane zeros is minimum phase. There the magnitude no longer determines the phase.
  • Designing a loop that crosses 0 dB at −40 dB/decade. The phase margin ends up at zero.

07Self-assessment

What is the relationship between the input and output spectra?

\( Y = H\,X \): they are multiplied frequency by frequency. An LTI system does not mix frequencies.

Why does the Bode plot use decibels and a logarithmic scale?

Because that way the product of factors becomes a sum and the asymptotes are straight, which makes it possible to draw it by hand.

What does a simple pole contribute at its own frequency?

−3 dB of magnitude and −45° of phase.

What is group delay?

Minus the derivative of the phase with respect to frequency. If it is constant, the shape of the signal is preserved.

How can a filter with flat magnitude distort?

With nonlinear phase: the harmonics arrive shifted relative to one another. It is what an all-pass does.

At what slope should the loop cross 0 dB?

At −20 dB/decade, which leaves about 90° of phase margin.

08Further reading

  • Katsuhiko Ogata. Modern Control Engineering. 5th ed. (Spanish edition, “Ingeniería de control moderna”), Prentice Hall, 2010. Bode plotting by asymptotes and phase and gain margins, with complete examples.
  • Adel S. Sedra and Kenneth C. Smith. Microelectronic Circuits. 7th ed. (Spanish edition, “Circuitos microelectrónicos”), Oxford University Press, 2015. Frequency response of amplifiers and compensation of feedback loops.
  • Alan V. Oppenheim and Alan S. Willsky. Signals and Systems. 2nd ed. (Spanish edition, “Señales y sistemas”), Prentice Hall, 1997. Group delay, linear phase and minimum-phase systems, with their magnitude-phase relationship.
Development of the topic “Frequency response” of Signals and Systems (Level 2), based on the curriculum of the UTN Electronic Engineering program, 2023 curriculum — Ordinance No. 1849 of the UTN Higher Council. Back to the Topic Map · catto.ar