Exponential Signals

Exponential signals are the eigenfunctions of linear time-invariant systems, meaning an LTI system scales them but never changes their form. This property underlies every transfer function derivation in control systems and filter design.

Core Concept

The continuous-time exponential signal is x(t) = A e^(st) where s = sigma + j omega is a complex number. When s is purely real and negative (s = -a, a > 0), the signal decays toward zero. When s is purely imaginary (s = j omega), the signal becomes the complex sinusoid e^(j omega t) with constant magnitude.

The real part sigma controls growth or decay. If sigma > 0, the signal grows exponentially and is unstable in system terms. If sigma < 0, it decays and carries finite energy. If sigma = 0, the magnitude stays constant forever and the signal is a power signal, not an energy signal.

Discrete-time exponential signals take the form x[n] = A r^n where r = |r| e^(j omega) is a complex number. The magnitude |r| plays the role of e^(sigma). If |r| < 1, the sequence decays. If |r| > 1, it grows. If |r| = 1, it oscillates with constant amplitude like a complex sinusoid.

Key Formula

Continuous-time: x(t) = A e^(st), s = sigma + j omega. When s is real and negative: x(t) = A e^(-at) u(t), Laplace transform X(s) = A/(s+a), ROC: Re(s) > -a. Discrete-time: x[n] = A r^n u[n], Z-transform X(z) = Az/(z-r), ROC: |z| > |r|.

Given: x(t) = 5 e^(-2t) u(t). Find Laplace transform and classify the signal.

Formula:
  L{e^(-at) u(t)} = 1/(s + a),  ROC: Re(s) > -a

Step 1: Identify parameters
  A = 5,  a = 2

Step 2: Apply transform pair
  X(s) = 5 * 1/(s + 2) = 5/(s + 2)

Step 3: State ROC
  ROC: Re(s) > -2  (right-half plane to the right of pole at s = -2)

Step 4: Find energy
  E = integral(0 to inf) |5 e^(-2t)|^2 dt
    = 25 * integral(0 to inf) e^(-4t) dt
    = 25 * [1/4] = 25/4

Final Answer:
  X(s) = 5/(s+2),  pole at s = -2
  E = 25/4  (finite)  =>  energy signal
  P = 0  (consequence of finite energy)

Exam Tip: GATE frequently gives a pole location and asks about signal behaviour. A pole in the left-half s-plane (Re(s) < 0) means the corresponding exponential mode decays — system is BIBO stable. A pole on the imaginary axis gives sustained oscillation — marginal stability. A pole in the right-half plane gives an exponentially growing mode — unstable. For discrete-time, map these rules to |r| < 1, |r| = 1, and |r| > 1 respectively.

Properties Summary

• General form: x(t) = A e^(st); s = sigma + j omega controls decay rate and frequency.
• Decaying (sigma < 0): finite energy, energy signal, BIBO stable mode.
• Growing (sigma > 0): infinite energy and power, unstable mode.
• Pure imaginary (sigma = 0): e^(j omega t) has constant magnitude, power signal.
• Laplace pair: e^(-at) u(t) <-> 1/(s+a), ROC Re(s) > -a.
• Discrete: x[n] = r^n u[n] <-> z/(z-r) in Z-domain, ROC |z| > |r|.
• Eigenfunction property: LTI system with transfer function H(s) gives output H(s0) e^(s0 t) for input e^(s0 t).

Quick Revision

• x(t) = A e^(st); s = sigma + j omega.
• sigma < 0: decaying, energy signal; sigma > 0: growing, unstable.
• e^(-at)u(t) has Laplace transform 1/(s+a) with pole at s = -a.
• Discrete: r^n u[n] has Z-transform z/(z-r) with pole at z = r.
• Energy of e^(-at)u(t) = 1/(2a).
• Euler's formula: e^(j omega t) = cos + j sin; magnitude 1.
• Exam trap: writing the Laplace transform ROC as Re(s) > a instead of Re(s) > -a for a decaying exponential e^(-at)u(t). The pole is at s = -a, so the ROC is to the right of -a.