Questions: All-Pass Filter Networks and Phase Equalization

Question 1

A signal with three frequency components at 100 Hz, 1 kHz, and 10 kHz passes through a well-designed all-pass filter. What happens to the amplitude of each component?

• A. All three amplitudes are reduced by equal amounts since all-pass filters attenuate uniformly
• B. Higher frequencies are attenuated more, similar to a low-pass filter
• C. All three amplitudes are unchanged — the all-pass filter has unity gain at every frequency
• D. The components are redistributed in amplitude to create a flat total power spectrum

Answer: (C)

An all-pass filter has |H(jω)| = 1 at every frequency — it passes all frequency components at full amplitude. The name is precise: all frequencies pass without attenuation or amplification. What changes is the phase response ∠H(jω), which varies with frequency, altering timing relationships between frequency components without changing their individual amplitudes. This is fundamentally different from low-pass, high-pass, or bandpass filters, which selectively attenuate amplitude.

Question 2

A digital communications system uses a sharp Chebyshev low-pass filter at the receiver. Engineers notice that received data pulses are smeared and overlapping in time, causing bit errors, even though no frequency components are missing. What is the most likely cause, and how could an all-pass filter help?

• A. The Chebyshev filter has too much passband ripple, which distorts pulse amplitudes; an all-pass filter cannot help with amplitude problems
• B. The Chebyshev filter's non-linear phase response delays different frequencies by different amounts, smearing pulses; cascading an all-pass equalizer flattens group delay without altering the amplitude response
• C. The filter's cutoff frequency is too low, removing high-frequency components needed for sharp pulse edges; a wider cutoff is needed
• D. The Chebyshev filter creates standing waves in the transmission line; an all-pass filter absorbs these reflections

Answer: (B)

Sharp filters like Chebyshev types have strongly non-linear phase responses — different frequency components experience different delays (non-constant group delay). When a pulse's frequency components arrive at different times, the pulse shape distorts, causing intersymbol interference. An all-pass phase equalizer can be designed with a complementary group delay profile that, when cascaded with the original filter, produces flat total group delay. Crucially, since the all-pass has unity magnitude, it does not disturb the amplitude response the Chebyshev filter was carefully designed to achieve.

Question 3

A cascade of most-pass filter sections can reduce the total system group delay if the sections are designed to subtract delay from heavily delayed frequency bands.

• T. True
• F. False

Answer: False

This is a fundamental constraint: every all-pass section adds delay — it can redistribute group delay across frequencies (make it more uniform) but cannot reduce the average total delay. Cascading all-pass stages always increases total system latency. The phase response of an all-pass section monotonically decreases from 0° to −180° (first-order) or −360° (second-order), representing added delay. A cascade can flatten group delay, but the flattened level is always at least as high as the original maximum delay.

Question 4

An all-pass filter achieves unity magnitude response because its transfer function has a right-half-plane zero that mirrors each left-half-plane pole across the imaginary axis.

• T. True
• F. False

Answer: True

This is the structural reason for flat magnitude response. For H(s) = (s − a)/(s + a) with a > 0: at any frequency s = jω, |jω − a| = √(ω² + a²) = |jω + a|, so numerator and denominator magnitudes are always equal and |H(jω)| = 1 identically. The right-half-plane zero is essential: it contributes the same magnitude as the pole but with different phase, creating a non-trivial phase response while preserving flat magnitude. Removing the right-half-plane zeros (as in a minimum-phase filter) would break this symmetry.

Question 5

Explain why an all-pass filter's pole-zero configuration guarantees unity magnitude response at all frequencies.

Model answer: For H(s) = (s − a)/(s + a), the zero at s = +a is the mirror image of the pole at s = −a across the imaginary axis. At any purely imaginary frequency s = jω, the distance from jω to the zero equals the distance from jω to the pole: both are √(ω² + a²). Since transfer function magnitude equals the product of distances to zeros divided by the product of distances to poles, these equal distances cancel and |H(jω)| = 1 for all ω. The phase differs because the angles to the zero and pole from jω are not equal, producing a non-trivial frequency-dependent phase shift.

This pole-zero mirror symmetry is the structural foundation of all-pass filter design. Higher-order sections use complex-conjugate pole pairs with mirrored zeros, preserving the equal-distance property for all frequencies. The geometry explains both why magnitude is flat (equal distances cancel) and why phase is not flat (unequal angles do not cancel) — the all-pass achieves pure phase modification with no amplitude consequence.