Questions: First-Order Systems and Frequency Response

Question 1

You apply a sinusoidal input at exactly the corner frequency ω_n to a first-order low-pass system. What do you observe in the steady-state output?

• A. The output amplitude is the same as the input and there is no phase shift — the corner frequency is where the system passes signals without distortion
• B. The output amplitude is 1/√2 times the input amplitude (−3 dB) and lags the input by exactly 45°
• C. The output amplitude is zero because the corner frequency is the cutoff where the system stops passing signals
• D. The output amplitude is half the input (−6 dB) and lags by 90°

Answer: (B)

At ω = ω_n, the magnitude is |H(jω_n)| = ω_n/√(ω_n² + ω_n²) = 1/√2 ≈ 0.707, which is −3 dB. The phase is −arctan(ω_n/ω_n) = −arctan(1) = −45°. These two values — −3 dB magnitude and −45° phase — are the defining characteristics of the corner frequency in a first-order system. The −45° phase measurement is particularly useful diagnostically: apply a sinusoid, sweep frequency until you observe exactly 45° lag, and you have located the pole. Option A describes DC behavior (ω << ω_n). Option C is wrong — the first-order system asymptotically approaches zero at high frequency but never reaches it at the corner. Option D describes the high-frequency asymptote, not the corner.

Question 2

A first-order system's pole is moved from s = −10 to s = −100 rad/s (farther from the origin). How does the step response change?

• A. The system responds more slowly because moving the pole farther left increases the time constant
• B. The system responds faster because ω_n = 100 > 10, meaning τ = 1/100 = 10 ms is shorter than τ = 1/10 = 100 ms
• C. The step response shape changes but the settling time remains the same because it depends only on damping
• D. The system becomes unstable because poles in the left half-plane always cause instability when moved farther from the origin

Answer: (B)

The pole at s = −ω_n means τ = 1/ω_n. Moving the pole from −10 to −100 increases ω_n from 10 to 100, which *decreases* the time constant from τ = 0.1 s to τ = 0.01 s. The system becomes faster: it reaches 63% of its final value in 10 ms instead of 100 ms. The geometric interpretation is that distance from the origin equals ω_n equals 1/τ — farther left means faster. Option A reverses this relationship. Option C conflates second-order damping concepts with first-order behavior (first-order systems have no damping ratio). Option D is incorrect — left-half-plane poles indicate stability.

Question 3

The time constant τ and the corner frequency ω_n of a first-order system carry independent information — knowing one does not tell you the other.

• T. True
• F. False

Answer: False

Time constant and corner frequency are reciprocals of each other: τ = 1/ω_n, equivalently ω_n = 1/τ. They are two descriptions of exactly the same physical fact — the location of the single pole. Knowing the time constant from a step response (the time to reach 63% of steady state) immediately gives you the corner frequency (where the magnitude is −3 dB and the phase is −45°), and vice versa. First-order systems have only one degree of freedom in their dynamics: the single pole location. Every characterization — time constant, corner frequency, pole location, 63% rise time, −3 dB frequency, −45° phase frequency, −20 dB/decade rolloff onset — is a different perspective on that one number.

Question 4

At frequencies well above the corner frequency ω_n, a first-order low-pass system's output magnitude continues to decrease at exactly −20 dB per decade indefinitely.

• T. True
• F. False

Answer: True

For ω >> ω_n, the magnitude simplifies to |H(jω)| ≈ ω_n/ω, which decreases as 1/ω — exactly −20 dB per decade (a factor of 10 decrease in magnitude for every decade increase in frequency). This is the asymptotic behavior of a single-pole system. It holds indefinitely in the ideal first-order model; in physical systems, additional poles at higher frequencies eventually change the rolloff slope. The −20 dB/decade slope is the frequency-domain signature of a single real pole, just as the −3 dB point identifies the corner and the −45° phase identifies ω_n.

Question 5

A technician measures the step response of an unknown first-order system and finds it reaches 63% of its final value after 5 ms. What can you determine about the system's frequency response without any additional measurements?

Model answer: The 63% rise time in a first-order step response equals the time constant τ. Therefore τ = 5 ms = 0.005 s. Since ω_n = 1/τ, the pole is at s = −200 rad/s and the corner frequency is ω_n = 200 rad/s (approximately 31.8 Hz). The magnitude response will be flat (0 dB) for frequencies well below 200 rad/s, will be −3 dB at exactly 200 rad/s, will have a −45° phase shift at 200 rad/s, and will roll off at −20 dB/decade above 200 rad/s. At any frequency ω, the magnitude is 200/√(ω² + 200²) and the phase is −arctan(ω/200).

This is the key insight about first-order systems: the single pole characterization is complete. One measurement (the 63% rise time) determines the time constant, which determines the pole location, which fully specifies every time-domain and frequency-domain characteristic. There is no additional information to uncover — the system has only one degree of freedom. This is why mastering the first-order case is foundational: it demonstrates the deep unity between time-domain and frequency-domain representations.