Wavelet Transform (Alternative to Fourier Transform for Non-Stationary Signals)

      The Wavelet Transform is a powerful tool used to analyze signals whose frequency content changes over time, called non-stationary signals. Unlike the traditional Fourier Transform (FT) which represents a signal as a sum of infinite-duration sinusoids (losing time information), wavelets allow simultaneous time and frequency analysis.    

      A wavelet is a short, oscillatory waveform localized in time. The Wavelet Transform breaks a signal into shifted and scaled versions of a mother wavelet.    

Continuous Wavelet Transform (CWT)

The CWT of a signal x(t) is defined as:

              W(a, b) = ∫−∞∞ x(t) * (1/√|a|) * ψ((t − b) / a) dt          

     
• ψ(t) is the mother wavelet.
     
• a is the scale parameter (controls frequency).
     
• b is the translation parameter (controls time).
   

      Smaller scales (|a| < 1) correspond to high-frequency components, larger scales (|a| > 1) correspond to low-frequency components.    

Advantages of Wavelet Transform

     
• Good time resolution at high frequencies, good frequency resolution at low frequencies.
     
• Effective for signals with transient features, such as spikes or discontinuities.
     
• Applicable in signal denoising, compression, and feature extraction.
   

Example:

      Consider a signal composed of a short pulse followed by a low-frequency sinusoid. FT will show both frequencies but cannot tell when the pulse occurs in time. Wavelet transform will provide both time and frequency localization, showing the pulse's time position and the ongoing sinusoid.    

Fourier Integral Theorem

      The Fourier Integral Theorem generalizes the idea of Fourier series (which applies to periodic functions) to non-periodic functions that are absolutely integrable.    

      It states that a function f(t), under certain conditions, can be reconstructed by its Fourier transform as:    

              f(t) = (1 / 2π) ∫−∞∞ F(ω) * ejωt dω          

      Where the Fourier transform F(ω) is:    

              F(ω) = ∫−∞∞ f(τ) * e−jωτ dτ          

      This theorem provides the foundation for the continuous Fourier transform. The integral allows the representation of signals as a continuous superposition of complex exponentials.    

Conditions for Validity:

     
• f(t) is absolutely integrable (integral of |f(t)| is finite).
     
• Function is piecewise continuous.
     
• Function satisfies Dirichlet conditions.
   

Significance:

      This theorem ensures that the Fourier transform exists and that a function can be perfectly reconstructed from its transform, given the conditions above.    

Generalized Fourier Transforms

      Generalized Fourier Transforms extend the idea of the classical Fourier transform to broader classes of functions or other domains, often to solve partial differential equations or analyze non-standard signals.    

      Examples include:    

     
• Fourier Sine and Cosine Transforms: Used for functions defined only on the positive real axis, useful for boundary value problems.
     
• Fractional Fourier Transform (FrFT): A generalization that represents a signal in intermediate domains between time and frequency.
     
• Fourier-Stieltjes Transform: Extends FT to functions of bounded variation, involving integrals with respect to measures.
   

Fourier Sine Transform (FST)

      Defined for functions on [0, ∞) as:    

              F_s(ω) = √(2/π) ∫0∞ f(t) * sin(ωt) dt          

      Inverse:    

              f(t) = √(2/π) ∫0∞ F_s(ω) * sin(ωt) dω          

Applications:

     
• Solving heat and wave equations with specific boundary conditions.
     
• Signal processing for one-sided signals.
   

Fractional Fourier Transform (FrFT)

      The FrFT can be thought of as a rotation in the time-frequency plane by an angle α. When α = π/2, it corresponds to the ordinary Fourier transform.    

      It is defined by an integral transform with a kernel that depends on α. This provides flexibility in analyzing signals with varying time-frequency characteristics.    

Multidimensional Fourier Transform

      The Multidimensional Fourier Transform extends the classical one-dimensional FT to functions of multiple variables, such as images (2D) or volumes (3D).    

Definition (2D case):

      For a function f(x, y), its 2D Fourier transform F(u, v) is given by:    

              F(u, v) = ∫−∞∞ ∫−∞∞ f(x, y) * e−j2π(ux + vy) dx dy          

      The inverse transform is:    

              f(x, y) = ∫−∞∞ ∫−∞∞ F(u, v) * ej2π(ux + vy) du dv          

Properties:

     
• Separability: Multidimensional FT can be computed by successive 1D FTs along each dimension.
     
• Shift and scaling properties extend naturally.
     
• Parseval’s theorem applies in multidimensions, preserving energy.
   

Applications:

     
• Image processing: filtering, enhancement, and compression.
     
• Medical imaging (MRI, CT scans).
     
• Pattern recognition and computer vision.
   

Example:

      Suppose f(x, y) is a 2D rectangular pulse defined as 1 inside a rectangle and 0 outside. Its 2D FT results in a sinc function in both u and v directions, showing how sharp edges in space translate to spread frequency components.    

Spectrograms and Time-Frequency Representations

      The classical Fourier transform gives frequency content but loses time information. Spectrograms and other time-frequency representations solve this by showing how frequency content varies with time.    

Short-Time Fourier Transform (STFT):

      The STFT is computed by applying the Fourier transform over a short, sliding window of the signal.    

      Mathematically,    

              STFT{x}(t, ω) = ∫−∞∞ x(τ) * w(τ − t) * e−jωτ dτ          

     
• w(t) is a window function, typically of finite length (e.g., Hamming, Gaussian).
     
• t is the time shift of the window.
   

      The magnitude squared of STFT gives the spectrogram:    

              Spectrogram(t, ω) = |STFT{x}(t, ω)|²          

Trade-off: Time vs Frequency Resolution

      The window length controls the resolution:    

     
• Short window → good time resolution, poor frequency resolution.
     
• Long window → good frequency resolution, poor time resolution.
   

Example:

      For a signal that contains a short chirp (frequency increasing over a short period), the spectrogram visually shows the frequency sweep over time, which is invisible in the classic Fourier transform.    

Other Time-Frequency Representations:

     
• Wigner-Ville Distribution: Provides high-resolution time-frequency analysis but suffers from cross-term interference.
     
• Wavelet Scalogram: Time-frequency representation based on wavelet transform coefficients.
     
• Hilbert-Huang Transform: Adaptive data analysis method for non-linear and non-stationary signals.
   

Summary:

      Spectrograms and related time-frequency methods extend the classical Fourier transform's abilities by providing information about when frequencies occur, essential for analyzing real-world signals that vary in time.    

Example 1: Compute the Fourier Transform of a rectangular pulse

Problem: Find the Fourier transform of the rectangular pulse f(t) defined as:

f(t) = 1 for |t| ≤ 1, and f(t) = 0 otherwise.

Solution:

The Fourier transform is given by:

F(ω) = ∫_−∞^∞ f(t) e^−jωt dt

Since f(t) ≠ 0 only between −1 and 1:

F(ω) = ∫₋₁¹ e^−jωt dt

Integrate:

F(ω) = [ (e^−jωt) / (−jω) ]₋₁¹ = (e^−jω − e^jω) / (−jω)

Use Euler's identity: e^jθ − e^−jθ = 2j sin(θ), so:

F(ω) = (−2j sin(ω)) / (−jω) = 2 sin(ω) / ω

Interpretation: The function sinc(x) = sin(x)/x is the frequency representation of a rectangular time pulse.

Why this formula? The integral of complex exponentials gives the frequency components. Here the rectangular pulse contains a spread of frequencies weighted by the sinc function.

Example 2: Verify the inverse Fourier transform of Example 1 recovers f(t)

Problem: Show that applying inverse Fourier transform to F(ω) from Example 1 returns the rectangular pulse f(t).

Recall: Inverse Fourier transform:

f(t) = (1 / 2π) ∫_−∞^∞ F(ω) e^jωt dω

Solution: Substituting F(ω) = 2 sin(ω)/ω:

f(t) = (1 / 2π) ∫_−∞^∞ (2 sin(ω)/ω) e^jωt dω

This integral equals 1 for |t| ≤ 1 and 0 otherwise, by the Fourier integral theorem and properties of the sinc function.

Why? The sinc function in frequency domain corresponds to a rectangular pulse in time domain.

Example 3: Calculate the Continuous Wavelet Transform (CWT) of a delta function δ(t)

Problem: Compute CWT of δ(t) using mother wavelet ψ(t).

Recall: CWT formula:

W(a,b) = ∫_−∞^∞ δ(t) * (1/√|a|) * ψ((t−b)/a) dt

Solution: Since δ(t) "picks out" the value at t=0:

W(a,b) = (1/√|a|) * ψ((0−b)/a) = (1/√|a|) * ψ(−b/a)

Interpretation: The CWT of δ(t) is just a scaled and shifted version of the mother wavelet, evaluated at (−b/a).

Example 4: Use Fourier Sine Transform to find the transform of f(t) = e^−at, t ≥ 0, a > 0

Recall Fourier Sine Transform (FST):

F_s(ω) = √(2/π) ∫₀^∞ f(t) sin(ωt) dt

Solution:

F_s(ω) = √(2/π) ∫₀^∞ e^−at sin(ωt) dt

Using integral formula:

∫₀^∞ e^−pt sin(qt) dt = q / (p² + q²), for p > 0

Here, p = a, q = ω, so:

F_s(ω) = √(2/π) * (ω / (a² + ω²))

Meaning: The FST transforms an exponentially decaying signal into a function revealing its frequency content weighted by ω/(a²+ω²).

Example 5: Compute 2D Fourier transform of f(x,y) = e^{−(x² + y²)}

Problem: Find F(u,v) where:

F(u,v) = ∫∫ e^{−(x² + y²)} e^{−j2π(ux + vy)} dx dy

Solution:

This is a product of two 1D Gaussian functions, so the 2D FT separates:

F(u,v) = ∫ e^−x² e^−j2πux dx * ∫ e^−y² e^−j2πvy dy

1D FT of Gaussian f(t) = e^−t² is:

√π * e^−(πω)²

So,

F(u,v) = (√π * e^−(πu)²) * (√π * e^−(πv)²) = π * e^{−π²(u² + v²)}

Interpretation: The Fourier transform of a Gaussian is another Gaussian, showing the self-similarity of Gaussian shapes in both domains.

Example 6: Explain the trade-off in Short-Time Fourier Transform (STFT) window size

Problem: Why does increasing the STFT window length improve frequency resolution but reduce time resolution?

Explanation:

• Window length defines how much signal we analyze at once.
• Long window covers more data → better frequency resolution because FT sees more oscillations to distinguish frequencies.
• Long window blurs when frequency changes happen → poor time resolution.
• Short window is localized → detects quick changes → good time resolution.
• Short window sees fewer cycles → frequency resolution worsens.

Formula relation: The time-bandwidth product (uncertainty principle) limits simultaneous time and frequency resolution.

Example 7: Compute Fourier Transform of δ(t − t₀)

Recall: δ(t − t₀) is the delta function shifted to t₀.

F(ω) = ∫_−∞^∞ δ(t − t₀) e^−jωt dt

Solution:

By sampling property of δ:

F(ω) = e^−jωt₀

Interpretation: The Fourier transform of a shifted impulse is a complex exponential with linear phase corresponding to shift t₀.

Example 8: Explain why Wavelet Transform is better than FT for signals with transients

Answer:

• Fourier Transform uses infinite duration sine/cosine basis functions → loses time info.
• Wavelets use short-time, localized functions → captures both frequency and exact time of transient events.
• This allows detection of sudden changes, spikes, and short-duration features.

Example 9: Use Parseval’s theorem to verify energy preservation in FT

Parseval’s theorem:

∫ |f(t)|² dt = (1/2π) ∫ |F(ω)|² dω

Given: f(t) = e^−t²

We know F(ω) = √π e^−(ω²)/4

Calculate both integrals:

• Energy in time domain = ∫ e^−2t² dt = √(π/2)
• Energy in frequency domain = (1/2π) ∫ π e^−(ω²)/2 dω = √(π/2)

Both energies match, confirming energy preservation by FT.

Example 10: Compute the Fractional Fourier Transform (FrFT) of order α = π/2 of f(t)

Answer: FrFT with α=π/2 equals the classical Fourier Transform.

Therefore, FrFT_π/2{f(t)} = FT{f(t)}.

This property helps interpret FrFT as a rotation in time-frequency plane.

Example 11: Calculate DFT of x[n] = {1, 2, 3, 4}

DFT formula:

X[k] = Σ_n=0^N-1 x[n] e^−j2πkn/N, k=0,1,...,N−1

Given: N=4

Calculate:

• X[0] = 1+2+3+4 = 10
• X[1] = 1 + 2e^−jπ/2 + 3e^−jπ + 4e^−j3π/2 = 1 + 2(−j) + 3(−1) + 4j = (1−3) + j(−2+4) = −2 + 2j
• X[2] = 1 + 2e^−jπ + 3e^−j2π + 4e^−j3π = 1 − 2 + 3 − 4 = −2
• X[3] = 1 + 2e^−j3π/2 + 3e^−j3π + 4e^−j9π/2 = 1 + 2j + 3(−1) + 4(−j) = (1−3) + j(2−4) = −2 − 2j

Example 12: Explain the role of the mother wavelet in Wavelet Transform

The mother wavelet ψ(t) is the prototype function from which all wavelets are derived by scaling and translation.

Choosing different ψ(t) changes the analysis properties: compact support, smoothness, symmetry.

Example mother wavelets: Haar, Morlet, Daubechies.

Example 13: Calculate the STFT of a simple sinusoid x(t) = cos(2πf₀t) with a rectangular window

Window length: T seconds

STFT formula:

STFT(t, ω) = ∫ x(τ) w(τ − t) e^−jωτ dτ

With rectangular window of length T:

w(t) = 1 for |t| ≤ T/2, else 0

STFT centered at time t becomes:

STFT(t, ω) = ∫_t−T/2^t+T/2 cos(2πf₀τ) e^−jωτ dτ

This integral results in two sinc-shaped peaks in frequency at ω = ±2πf₀, showing energy concentrated at those frequencies within window around t.

Example 14: Use Fourier Integral Theorem to express f(t) in terms of its FT

Statement:

f(t) = (1/2π) ∫_−∞^∞ F(ω) e^jωt dω

Where F(ω) is the FT of f(t):

F(ω) = ∫_−∞^∞ f(τ) e^−jωτ dτ

This allows perfect reconstruction of f(t) from its frequency components, assuming conditions are met.

Example 15: Compute 1D Fourier Transform of Gaussian-modulated sinusoid

f(t) = e^−t² cos(2πf₀t)

Solution:

Using Euler’s formula, rewrite cosine as sum of exponentials:

f(t) = (1/2) e^−t² (e^j2πf₀t + e^−j2πf₀t)

Fourier transform is linear, so:

F(ω) = (1/2) [ ∫ e^−t² e^j2πf₀t e^−jωt dt + ∫ e^−t² e^−j2πf₀t e^−jωt dt ]

= (1/2) [ ∫ e^−t² e^{−j(ω−2πf₀)t} dt + ∫ e^−t² e^{−j(ω+2πf₀)t} dt ]

Each integral is Fourier transform of Gaussian shifted in frequency:

∫ e^−t² e^{−jθ t} dt = √π e^−θ²/4

So,

F(ω) = (√π/2) [ e^{−(ω−2πf₀)² / 4} + e^{−(ω+2πf₀)² / 4} ]

Interpretation: Frequency content is two Gaussians centered at ±2πf₀, showing modulation by f₀.

Example 16: Compute Fourier Transform of f(t) = t e^−at u(t), a > 0

Here, u(t) is unit step function.

Solution:

FT{f(t)} = ∫₀^∞ t e^−at e^−jωt dt = ∫₀^∞ t e^{−(a + jω)t} dt

Use integral formula:

∫₀^∞ t e^−pt dt = 1/p², for Re(p) > 0

So here p = a + jω, thus

F(ω) = 1 / (a + jω)²

Meaning: The presence of t multiplies the spectrum by a term making it decay faster.

Example 17: Calculate the 2D Fourier transform of an image with impulse at (x₀, y₀)

Problem: Let f(x, y) = δ(x − x₀, y − y₀).

Solution:

F(u, v) = ∫∫ δ(x − x₀, y − y₀) e^{−j2π(ux + vy)} dx dy

Using sampling property, this becomes:

F(u, v) = e^{−j2π(ux₀ + vy₀)}

Interpretation: The impulse's FT is a complex exponential with phase related to position.

Example 18: Use spectrogram to analyze a signal with two sine waves turning on at different times

Signal: x(t) = sin(2π 5 t) for t < 1s, and sin(2π 10 t) for t ≥ 1s

Approach: Use STFT with window size 0.5s sliding along t.

Result: Spectrogram shows 5 Hz energy before 1s and 10 Hz energy after 1s, clearly displaying frequency change over time.

Example 19: Show that convolution in time corresponds to multiplication in frequency domain

Theorem:

If y(t) = x(t) * h(t) (convolution), then Y(ω) = X(ω) H(ω)

Proof Sketch:

Using Fourier Transform properties:

FT{y(t)} = FT{∫ x(τ) h(t − τ) dτ} = X(ω) H(ω)

Why important? This property allows filtering operations in frequency domain efficiently.

Example 20: Explain why the Fourier Transform is not suitable for analyzing non-stationary signals

Answer:

• Fourier Transform analyzes global frequency content but no time localization.
• Signals with changing frequencies over time (non-stationary) lose temporal info.
• Wavelet and STFT provide time-frequency localization, overcoming this limitation.