3. Continuous Fourier Transform (CFT)

3.1 Definition and Basic Formula

The Continuous Fourier Transform (CFT) is a mathematical tool used to convert a time-domain signal f(t) into its frequency-domain representation F(ω).

What is it?

• The Fourier Transform decomposes any signal (even non-periodic) into a sum (integral) of complex sinusoids (waves) with different frequencies.
• It helps us analyze which frequencies exist in the original signal and how much of each frequency is present.

The formula:

F(ω) = ∫−∞∞ f(t) · e−jωt dt

• F(ω): Fourier Transform of f(t), a complex function showing amplitude and phase of frequency ω.
• f(t): Original time-domain signal.
• ω: Angular frequency (radians per second), related to frequency f by ω = 2πf.
• j: Imaginary unit, where j² = −1.
• e−jωt: Complex sinusoid (basis function).

Interpretation of the formula

For each frequency ω, multiply the original signal f(t) by e−jωt (a complex wave with frequency ω) and integrate (sum) over all time.

This operation measures how much of frequency ω is present in f(t). The result F(ω) is a complex number whose magnitude tells the amplitude (strength) of that frequency, and whose angle (phase) tells its phase shift.

3.2 Inverse Fourier Transform

To recover the original time-domain signal f(t) from its frequency-domain F(ω), we use the Inverse Fourier Transform:

f(t) = (1 / 2π) ∫−∞∞ F(ω) · ejωt dω

• This reconstructs the signal by summing all frequencies back, weighted by F(ω).
• The factor 1 / 2π is a normalization constant that depends on the Fourier transform definition.

Explanation of inverse formula:

• F(ω) contains frequency information.
• Multiplying by ejωt “rebuilds” the oscillations.
• Integrating over all frequencies sums these oscillations to recreate the original signal.

3.3 Conditions for Existence (Dirichlet Conditions)

For the Fourier Transform to exist (be defined and finite), the signal f(t) must satisfy some conditions, called Dirichlet conditions:

1. Absolute integrability:
   ∫−∞∞ |f(t)| dt < ∞
   The total area under the absolute value of f(t) must be finite (finite energy signal).
2. Finite number of discontinuities:
   f(t) should have a finite number of jumps (discontinuities) in any finite interval.
3. Finite number of maxima and minima:
   f(t) should have a finite number of peaks and valleys (extrema) within any finite interval.

Why these conditions?

They guarantee that the integral in the Fourier transform converges.
Without these, the transform might diverge or not be meaningful.

3.4 Example: Fourier Transform of a Rectangular Pulse

Define the signal:

f(t) = { 1, for |t| ≤ T/2
         0, otherwise }

A pulse of amplitude 1 lasting from −T/2 to T/2.
This is a simple non-periodic signal.

Find the Fourier Transform F(ω):

F(ω) = ∫−T/2T/2 1 · e−jωt dt

Calculate the integral:

F(ω) = [ e−jωt / (−jω) ]−T/2T/2 = (1 / −jω) [ e−jω(T/2) − ejω(T/2) ]

Simplify using Euler's formula:

Recall:
ejθ − e−jθ = 2j sin θ

So,

F(ω) = (1 / −jω) · (−2j) sin(ωT/2) = (2 sin(ωT/2)) / ω

Final form:

F(ω) = 2 · (sin(ωT/2) / ω)

Interpretation:

• F(ω) is a sinc-like function (since sin x / x is called sinc).
• The pulse in time domain corresponds to a continuous spectrum with main lobe and side lobes in frequency domain.
• As T increases, the main lobe in frequency domain gets narrower (frequency resolution increases).

Important note: The amplitude of each frequency component in F(ω) shows how much that frequency is present in the rectangular pulse.

3.5 Summary

3.6 Why is Fourier Transform important?

• Signal analysis: Understand which frequencies are present in a signal (audio, electrical signals, images).
• Filtering: Design filters that remove or enhance certain frequencies.
• Communications: Modulation and demodulation of signals.
• Physics & Engineering: Analyze wave phenomena, vibrations, quantum mechanics, optics.

20 Solved Examples on Continuous Fourier Transform (CFT)

These examples progress from simple to complex and cover key concepts and formulas from Continuous Fourier Transform. Each solution explains every symbol, why formulas are used, and how to interpret results, to help you fully master the topic.

Example 1: Fourier Transform of a Delta Function

Problem: Find the Fourier Transform of f(t) = δ(t), the Dirac delta function.

Solution:

Recall the Fourier Transform formula:

F(ω) = ∫−∞∞ f(t) · e−jωt dt

Substitute f(t) = δ(t):

F(ω) = ∫−∞∞ δ(t) · e−jωt dt

By the sifting property of delta:

F(ω) = e−jω·0 = 1

Explanation: The delta function picks out the value of e−jωt at t=0. The transform is constant 1 for all frequencies, meaning the delta contains all frequencies equally.

Example 2: Fourier Transform of a Constant Signal

Problem: Find the Fourier Transform of f(t) = 1.

Solution:

F(ω) = ∫−∞∞ 1 · e−jωt dt

This integral does not converge in the usual sense because the signal is infinite in duration and constant.

Interpretation: This signal does not satisfy the Dirichlet conditions (finite energy), so the Fourier Transform does not exist in the ordinary sense. We use generalized functions and get F(ω) = 2π δ(ω), a delta function in frequency domain.

Example 3: Fourier Transform of an Exponential Decay

Problem: Compute the Fourier Transform of f(t) = e−at u(t), where a > 0 and u(t) is the unit step function.

Solution:

Fourier Transform formula:

F(ω) = ∫0∞ e−at · e−jωt dt

Combine exponents:

F(ω) = ∫0∞ e−(a + jω)t dt

Calculate integral:

F(ω) = [−1 / (a + jω)] e−(a + jω)t |0∞ = 1 / (a + jω)

Because Re(a) > 0, the exponential vanishes at infinity.

Interpretation: The frequency response decreases for large frequencies, showing the exponential decay has low-pass characteristics.

Example 4: Fourier Transform of a Cosine Wave

Problem: Find the Fourier Transform of f(t) = cos(ω₀ t).

Solution:

Recall Euler's formula: cos(ω₀ t) = (ejω₀ t + e−jω₀ t)/2

By linearity, transform is:

F(ω) = (1/2) ∫ e−jω t ejω₀ t dt + (1/2) ∫ e−jω t e−jω₀ t dt

Simplify exponents:

F(ω) = (1/2) ∫ e−j(ω − ω₀)t dt + (1/2) ∫ e−j(ω + ω₀)t dt

Using the transform of delta, the result is:

F(ω) = π [δ(ω − ω₀) + δ(ω + ω₀)]

Interpretation: A cosine wave corresponds to two spikes (delta functions) in the frequency domain at frequencies ±ω₀.

Example 5: Fourier Transform of a Sine Wave

Problem: Find the Fourier Transform of f(t) = sin(ω₀ t).

Solution:

Using Euler's formula: sin(ω₀ t) = (ejω₀ t − e−jω₀ t) / (2j)

Fourier Transform becomes:

F(ω) = (1/(2j)) ∫ e−jω t (ejω₀ t − e−jω₀ t) dt

Splitting integral:

F(ω) = (1/(2j)) [∫ e−j(ω−ω₀)t dt − ∫ e−j(ω+ω₀)t dt]

Using delta transform, we get:

F(ω) = jπ [δ(ω + ω₀) − δ(ω − ω₀)]

Interpretation: Sine wave has two delta spikes in frequency domain, with a phase difference indicated by the imaginary unit.

Example 6: Fourier Transform of a Rectangular Pulse (Revisited)

Problem: Find the Fourier Transform of a rectangular pulse:

f(t) = 1, for |t| ≤ T/2; 0 otherwise

Solution:

Formula:

F(ω) = ∫−T/2T/2 e−jω t dt

Integrate:

F(ω) = [e−jω t / (−jω)]−T/2T/2 = (1 / −jω) [e−jω (T/2) − ejω (T/2)]

Simplify using ejθ − e−jθ = 2j sin θ:

F(ω) = (2 sin(ω T/2)) / ω

Interpretation: The transform shows a sinc pattern, illustrating how time-domain truncation spreads energy over frequencies.

Example 7: Effect of Time Shift on Fourier Transform

Problem: If f(t) has transform F(ω), find the transform of g(t) = f(t − t₀).

Solution:

Using the time-shift property:

G(ω) = e−jω t₀ F(ω)

Explanation: Shifting a signal in time adds a phase factor in the frequency domain but does not change magnitude.

Example 8: Effect of Frequency Shift on Fourier Transform

Problem: If f(t) has transform F(ω), find the transform of g(t) = ejω₀ t f(t).

Solution:

Using frequency shift property:

G(ω) = F(ω − ω₀)

Explanation: Multiplying by a complex exponential in time shifts the spectrum by ω₀ in frequency domain.

Example 9: Scaling Property of Fourier Transform

Problem: If f(t) has transform F(ω), find the transform of g(t) = f(a t) for real a ≠ 0.

Solution:

Scaling property:

G(ω) = (1/|a|) F(ω / a)

Explanation: Compressing the signal in time domain stretches the spectrum in frequency domain and vice versa, with an amplitude scaling factor.

Example 10: Fourier Transform of Gaussian Function

Problem: Find the Fourier Transform of f(t) = e−α t², where α > 0.

Solution:

The Gaussian transform is another Gaussian:

F(ω) = √(π / α) · e−ω² / (4α)

Explanation: Gaussian functions remain Gaussian under Fourier transform, illustrating minimal uncertainty in time-frequency.

Example 11: Convolution Theorem

Problem: If y(t) = x(t) * h(t) (convolution), show the Fourier Transform relationship.

Solution:

Fourier transform of convolution:

Y(ω) = X(ω) · H(ω)

Explanation: Convolution in time domain corresponds to multiplication in frequency domain, simplifying system analysis.

Example 12: Parseval’s Theorem

Problem: Verify energy conservation between time and frequency domains.

Statement:

∫ |f(t)|² dt = (1 / 2π) ∫ |F(ω)|² dω

Interpretation: Total energy in time domain equals energy in frequency domain (up to constant factor).

Example 13: Fourier Transform of a Triangular Pulse

Problem: Find the Fourier Transform of a triangular pulse of width T.

Solution:

The transform is:

F(ω) = T · sinc²(ω T / 2)

Explanation: The triangular pulse is convolution of two rectangular pulses, so its transform is the square of the sinc.

Example 14: Fourier Transform of a Rectangular Pulse with Amplitude A

Problem: Modify Example 6 with amplitude A.

Solution:

F(ω) = A · 2 sin(ω T/2) / ω

Interpretation: Amplitude scales the magnitude of frequency components linearly.

Example 15: Fourier Transform of a Signal with Time Delay and Scaling

Problem: Find transform of g(t) = A · f(a (t − t₀)).

Solution:

Apply scaling and time-shift:

G(ω) = (A / |a|) e−jω t₀ F(ω / a)

Explanation: Combined effects of amplitude, scaling in time, and delay in time domain translate to amplitude scaling, frequency scaling, and phase shift in frequency domain.

Example 16: Fourier Transform of Rectangular Pulse Using Sinc Function Definition

Problem: Express Example 6 transform in terms of normalized sinc sinc(x) = sin(π x) / (π x).

Solution:

F(ω) = T · sinc(ω T / (2π))

Interpretation: The normalized sinc shows how pulse width inversely controls bandwidth.

Example 17: Bandwidth of a Signal

Problem: Using rectangular pulse from Example 6, estimate main lobe bandwidth.

Solution:

Main lobe zeros occur when sin(ω T/2) = 0, i.e., at ω = ±2π / T.

Bandwidth ≈ 4π / T (width of main lobe in radians per second).

Interpretation: Narrower pulses have wider bandwidth; this is the time-frequency tradeoff.

Example 18: Fourier Transform of a Periodic Signal Using Fourier Series

Problem: Show that the Fourier Transform of a periodic signal is a series of delta functions at harmonic frequencies.

Explanation:

A periodic signal x(t) with period T can be represented by Fourier series:

x(t) = Σ cₙ ej n ω₀ t, where ω₀ = 2π / T.

Fourier Transform is:

X(ω) = 2π Σ cₙ δ(ω − n ω₀)

Interpretation: Energy is concentrated at discrete harmonic frequencies.

Example 19: Windowed Fourier Transform Concept

Problem: Explain why normal Fourier Transform loses time information.

Answer:

Fourier Transform integrates over entire time, so it only shows frequency content globally, not where frequencies appear in time.

Windowed Fourier Transform (Short-Time Fourier Transform) multiplies the signal by a sliding window function w(t−τ) before transforming:

STFT(τ, ω) = ∫ f(t) w(t−τ) e−j ω t dt

This localizes frequency information in time around τ.

Example 20: Using Fourier Transform to Solve Differential Equations

Problem: Solve df(t)/dt + a f(t) = g(t) using Fourier Transform.

Solution:

Take Fourier Transform:

jω F(ω) + a F(ω) = G(ω)

Solve for F(ω):

F(ω) = G(ω) / (a + jω)

Use inverse transform to get f(t).

Interpretation: Fourier transform converts differential equations into algebraic equations in frequency domain, making them easier to solve.