FT of Exponential Signal

The Fourier transform of a one-sided decaying exponential gives a Lorentzian spectrum in the frequency domain, a result used in modelling RC circuit responses and the spectral shape of resonant systems. Understanding this transform explains why higher decay rates produce broader frequency responses.

Core Concept

The one-sided decaying exponential x(t) = e^(-at)u(t) is zero for t < 0 and decays with rate a for t >= 0. For the Fourier transform to exist, the signal must be absolutely integrable, which requires a > 0. If a <= 0, the signal grows without bound and the standard FT does not converge.

Because x(t) is zero for t < 0, the FT integral runs only from 0 to infinity. The integrand combines two decaying exponentials: one from the signal (e^(-at)) and one from the kernel (e^(-j2*pi*f*t)). Together they form e^(-(a + j2*pi*f)t), which decays as long as a > 0.

The result 1/(a + j2*pi*f) is a complex-valued function. Its magnitude |X(f)| = 1/sqrt(a^2 + (2*pi*f)^2) is the Lorentzian shape. Larger a means faster decay in time and broader spread in frequency, demonstrating the inverse time-bandwidth relationship.

Key Formula

FT pair: e^(-at)u(t) <-> 1/(a + j2*pi*f), valid for a > 0. Magnitude: |X(f)| = 1/sqrt(a^2 + 4*pi^2*f^2). Phase: angle(X(f)) = -arctan(2*pi*f/a). The two-sided exponential pair is: e^(-a|t|) <-> 2a/(a^2 + (2*pi*f)^2), which is a real and even Lorentzian. In angular frequency: e^(-at)u(t) <-> 1/(a + j*omega).

Given: x(t) = e^(-5t) u(t)
Find: X(f), |X(f)|, and angle(X(f))

Step 1: Write FT integral
  X(f) = integral from 0 to inf of e^(-5t) e^(-j2*pi*f*t) dt

Step 2: Combine exponents
  X(f) = integral from 0 to inf of e^(-(5 + j2*pi*f)t) dt

Step 3: Evaluate
  X(f) = [-1/(5 + j2*pi*f)] * e^(-(5+j2*pi*f)t) | from 0 to inf
  At inf: exponent has positive real part 5, so term goes to 0
  At 0: term is 1
  X(f) = 1/(5 + j2*pi*f)

Step 4: Magnitude
  |X(f)| = 1/sqrt(25 + 4*pi^2*f^2)

Step 5: Phase
  angle(X(f)) = -arctan(2*pi*f/5)

Final Answer:
  X(f) = 1/(5 + j2*pi*f)
  |X(f)| = 1/sqrt(25 + 4*pi^2*f^2)
  Phase  = -arctan(2*pi*f/5)

Exam Tip: The 3 dB bandwidth of the Lorentzian spectrum of e^(-at)u(t) is f_3dB = a/(2*pi) Hz. At this frequency, |X(f)| drops to 1/sqrt(2) of its peak. Larger a means wider bandwidth. Also note that the FT of e^(+at)u(t) does not exist (a > 0 case grows); for that signal, you need the Laplace transform with appropriate ROC.

Properties Summary

• One-sided pair: e^(-at)u(t) <-> 1/(a + j2*pi*f), requires a > 0 for FT to exist.
• Two-sided pair: e^(-a|t|) <-> 2a/(a^2 + 4*pi^2*f^2), real and symmetric spectrum.
• Growing exponential: e^(at)u(t) with a > 0 has no Fourier transform; use Laplace instead.
• Magnitude at f=0: |X(0)| = 1/a, the DC gain equals the integral of x(t) from 0 to infinity.
• 3 dB frequency: |X(f)| falls to 1/sqrt(2) at f = a/(2*pi).
• Phase: linear-like for small f, saturating toward -90 degrees as f -> infinity.

Quick Revision

• e^(-at)u(t) <-> 1/(a + j2*pi*f) for a > 0.
• |X(f)| = 1/sqrt(a^2 + 4*pi^2*f^2), a Lorentzian bell curve.
• Larger a: faster decay in time, broader and flatter spectrum.
• X(0) = 1/a (area under the exponential curve).
• Two-sided: e^(-a|t|) <-> 2a/(a^2 + 4*pi^2*f^2).
• In omega form: e^(-at)u(t) <-> 1/(a + j*omega).
• Exam trap: applying the Fourier transform to e^(at)u(-t) with a > 0. This causal signal going backward in time also has a FT: 1/(a - j2*pi*f). Do not confuse it with the forward decaying case.