First Order System Response

When a control system contains a single energy-storing element, whether a capacitor, an inductor, or a thermal mass, its dynamic behavior is described by a first-order differential equation. The first-order system is the simplest dynamic model in control theory, yet it is foundational because it introduces the concept of the time constant τ, which governs how quickly a system responds to an input. Understanding its step response quantitatively is essential for GATE and for designing real control loops.

Core Concept: The Time Constant

The transfer function of a standard first-order system is G(s) = K / (τs + 1), where K is the DC gain and τ (tau) is the time constant in seconds. This transfer function has a single pole at s = -1/τ on the real axis of the s-plane. The system is always stable since the pole lies in the left half of the s-plane for any positive τ. There is no oscillation, and the response to any bounded input grows smoothly without any ringing.

When a unit step is applied, the output in the time domain is c(t) = K(1 - e^(-t/τ)) for t ≥ 0. This expression shows exponential approach to the final value K. The time constant τ is the time at which the response reaches exactly 63.2% of its final value. This follows directly from substituting t = τ: c(τ) = K(1 - e^-1) = K(1 - 0.368) = 0.632K. This 63.2% figure is a universal and frequently tested result in GATE examinations.

Physically, τ represents the inertia of the system against change. A large τ means the system is sluggish and takes a long time to respond, much like a large thermal mass that heats slowly. A small τ means the system is fast and responsive. In electrical terms, an RC circuit with τ = RC is a perfect first-order system, and the voltage across the capacitor follows exactly c(t) = V(1 - e^(-t/RC)) after a step input.

Time Domain Specifications

Three key performance specifications are defined for a first-order step response. The rise time Tr is defined as the time taken to go from 10% to 90% of the final value. For a first-order system, Tr = 2.2τ. This is derived by solving K(1 - e^(-Tr1/τ)) = 0.1K and K(1 - e^(-Tr2/τ)) = 0.9K and subtracting: Tr = τ·ln(9) ≈ 2.2τ.

The settling time Ts is the time after which the output stays within a specified percentage band of the final value. For a 2% band: Ts = 4τ. For a 5% band: Ts = 3τ. These follow from solving 1 - e^(-Ts/τ) = 0.98 and 0.95 respectively. A first-order system has no overshoot and no peak time, since the response is monotonically non-decreasing. These two facts distinguish it fundamentally from second-order underdamped systems.

Practical Understanding

First-order systems appear extensively in thermodynamics, electrical circuits, and fluid flow. An RC low-pass filter, a liquid level tank with a drain, a thermal system with one heat exchange surface, and a simple motor with only back-EMF (no inertia) are all first-order. The response never oscillates and always settles smoothly. In a feedback control system, adding an integrator to a first-order plant creates a second-order closed-loop, which introduces the possibility of oscillation and overshoot.

Given:
A first-order system G(s) = 5 / (2s + 1)
K = 5, τ = 2 seconds
Input: Unit step R(s) = 1/s

Why this formula applies:
Step response of first-order system is c(t) = K(1 - e^(-t/τ))
Time specs use: Tr = 2.2τ, Ts(2%) = 4τ

Formula:
c(t) = K(1 - e^(-t/τ))
Tr = 2.2τ
Ts = 4τ

Substitution:
c(t) = 5(1 - e^(-t/2))
Tr = 2.2 × 2
Ts = 4 × 2

Calculation:
c(τ) = c(2) = 5(1 - e^-1) = 5 × 0.632 = 3.16 (63.2% of 5)
Tr = 4.4 seconds
Ts = 8 seconds (2% settling)

Final Answer:
Final output = 5 units, Rise time = 4.4 s, Settling time = 8 s, No overshoot

Exam Tip: For a first-order system, memorize the triplet: 63.2% at t=τ, Tr=2.2τ, Ts=4τ (2%). The slope of the response at t=0 is K/τ. The tangent at t=0 meets the final value exactly at t=τ. These three facts appear repeatedly in GATE fill-in-the-blank questions.

Quick Revision

• Transfer function: G(s) = K/(τs+1). Single pole at s = -1/τ. Always stable for τ > 0.
• Step response: c(t) = K(1 - e^(-t/τ)). Output reaches 63.2% of final value at exactly t = τ.
• Rise time Tr = 2.2τ (10% to 90% of final value).
• Settling time: Ts = 3τ for 5% band, Ts = 4τ for 2% band.
• No overshoot, no oscillation, no peak time for first-order systems. Monotonic response only.
• Initial slope of response = K/τ. The tangent at origin hits the steady-state value exactly at t = τ.
• Trap: Do not apply overshoot or peak time formulas to first-order systems. These apply only to underdamped second-order systems.