Fourier Transform

calendar_today July 21, 2025 article 322 words access_time 2 min replay Modified July 22, 2025

Table of Contents

Introduction

FT decomposes a signal into its frequency components
Unlike the Laplace Transform (which uses $s = \sigma + j\omega$), FT is restricted to $s = j\omega$

$$ X(j\omega) = \int_{-\infty}^{\infty} x(t) e^{-j \omega t} \, \mathrm{d}t $$

$$ x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega) e^{j \omega t} \, \mathrm{d}\omega $$

Use the table of pairs and properties [Fourier transform table (course handout)] to evaluate FT/iFT

Although possible in some cases, it’s not really practical to use FT when dealing with initial conditions or systems involving transient behavior
General rule of thumb: LT for response, FT for spectral analysis (which isn’t really covered in this course)

Use when the impulse response $h(t)$ and input $x(t)$ are both known
Transform both signals:
- $H(j\omega) = \mathcal{F}\{ h(t) \}$
- $X(j\omega) = \mathcal{F}\{ x(t) \}$
Multiply in frequency domain: $$ Y(j\omega) = H(j\omega) X(j\omega) $$
Inverse transform to get output: $$ y(t) = \mathcal{F}^{-1} \left\{ Y(j\omega) \right\} $$

Replace each circuit element with its Fourier-domain impedance
- Resistor: $R$
- Inductor: $j\omega L$
- Capacitor: $\frac{1}{j\omega C}$
Take the FT of the input signal
Solve the circuit using methods of circuit analysis
Compute output in the Fourier domain
$$ Y(j\omega) = H(j\omega) X(j\omega) $$
Take the iFT to find $y(t)$
$$ y(t) = \mathcal{F}^{-1} \left\{ Y(j\omega) \right\} $$

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