5. Common Fourier Transform Pairs

Fourier Transform pairs are pairs of functions where one is a time-domain function and the other is its corresponding frequency-domain representation (Fourier Transform). Knowing common pairs helps solve many problems in signal processing, physics, and engineering easily.

5.1. Delta Function (Dirac Delta)

Definition:

The Dirac delta function, denoted δ(t), is not a function in the traditional sense but a generalized function or distribution with these properties:

• Zero everywhere except at t = 0
• Infinite at t = 0 such that:
  ∫-∞∞ δ(t) dt = 1
• Sifting property: For any continuous function f(t),
  ∫-∞∞ f(t) δ(t - t0) dt = f(t0)

Fourier Transform of δ(t):

F(ω) = ∫-∞∞ δ(t) e-j ω t dt = e-j ω · 0 = 1

Interpretation: The Fourier transform of the delta function is a constant function 1 across all frequencies. It contains all frequencies equally — it’s the perfect "impulse" in time domain, representing all frequencies.

Inverse Fourier Transform:

f(t) = (1 / 2π) ∫-∞∞ 1 · ej ω t dω = δ(t)

This means the inverse transform of the constant function 1 is the delta function.

Example Use:

• Modeling an ideal impulse in time (e.g., an instantaneous voltage spike).
• Sampling in signal processing.

5.2 Rectangular Pulse

Definition:

A rectangular pulse of width T and height 1 centered at zero is:

rect(t / T) = { 1    if |t| ≤ T/2
        0    otherwise }

Fourier Transform:

F(ω) = ∫-T/2T/2 e-j ω t dt = (2 / ω) sin(ω T / 2) = T · sinc( (ω T) / (2π) )

Where the sinc function is defined as:

sinc(x) = sin(π x) / (π x)

Explanation:

• The rectangular pulse in time domain transforms into a sinc function in frequency domain.
• The sinc function has a main lobe and decaying oscillations.
• The width of the pulse in time is inversely related to the width of the sinc in frequency. Narrow pulses in time → wide spectrum.

Example:

If T = 1, the transform is:

F(ω) = (2 sin(ω/2)) / ω

Plotting this function shows how the energy is distributed around zero frequency with oscillations.

5.3 Sine and Cosine Functions

Definitions:

• Cosine function:
  f(t) = cos(ω0 t)
• Sine function:
  f(t) = sin(ω0 t)

Fourier Transform of Cosine:

Using Euler's formula cos θ = (ejθ + e-jθ) / 2,

F(ω) = π [ δ(ω - ω₀) + δ(ω + ω₀) ]

Fourier Transform of Sine:

Using sin θ = (ejθ - e-jθ) / (2j),

F(ω) = j π [ δ(ω + ω₀) - δ(ω - ω₀) ]

Explanation:

• Both sine and cosine transform to impulses (delta functions) at their positive and negative frequencies.
• Cosine has two symmetric impulses with equal weight.
• Sine has two impulses with opposite signs (phase shift).

Example:

A cosine wave at frequency f0 = ω0 / 2π = 1 Hz has spectral lines (impulses) at ±1 Hz.

5.4 Gaussian Function

Definition:

A Gaussian function centered at zero:

f(t) = e ^{-t2 / (2 σ2)}

Where σ controls the width of the Gaussian.

Fourier Transform:

F(ω) = σ √(2π) e ^{-σ2 ω2 / 2}

Explanation:

• The Fourier transform of a Gaussian is also a Gaussian in the frequency domain.
• Width σ in time and frequency are inversely related:
  • Narrow Gaussian in time → wide Gaussian in frequency.
  • Wide Gaussian in time → narrow in frequency.
• This is a unique function that retains its shape under Fourier transform.

Example:

If σ = 1,

f(t) = e^-t2/2

then

F(ω) = √(2π) e ^{-ω2 / 2}

which is a Gaussian centered at zero frequency.

5.5 Exponential Decay Functions

Definition:

A one-sided exponential decay:

f(t) = e^{-a t} u(t)

where u(t) is the unit step function (0 for t < 0, 1 for t ≥ 0), and a > 0.

Fourier Transform:

F(ω) = ∫₀^∞ e^{-a t} e^{-j ω t} dt = 1 / (a + j ω)

Explanation:

• The function represents a causal exponential decay starting at t=0.
• The transform is a complex function showing how amplitude and phase vary with frequency.
• The magnitude decreases with frequency ω, the phase shifts depending on frequency.

Magnitude and Phase:

Magnitude:
|F(ω)| = 1 / √(a² + ω²)

Phase:
∠ F(ω) = - arctan(ω / a)

Example:

If a = 1, then

F(ω) = 1 / (1 + j ω)

which decays as ω grows, showing that high frequencies are attenuated more.

Summary Table of These Pairs

Bonus: Understanding the Formula Components

• t: Time domain variable (can represent seconds, samples, etc.)
• ω: Angular frequency in radians per second (ω = 2π f)
• j: Imaginary unit (j² = -1)
• e-j ω t: Kernel of Fourier transform — represents rotation in complex plane
• δ(t): Dirac delta — ideal impulse function
• u(t): Unit step function — zero before zero time, one after
• sinc(x): Defined as sin(π x) / (π x), important in signal reconstruction

Next steps?

If you want, I can continue with:

• Detailed proofs and derivations of each Fourier transform pair
• Visual plots of these functions and their transforms
• More transform pairs and applications
• Properties of Fourier transform (linearity, shifting, convolution, etc.)

Detailed Proofs and Derivations of Common Fourier Transform Pairs

Recall the Fourier Transform (FT) definition for a function f(t):

F(ω) = ∫-∞∞ f(t) · e-jωt dt

1. Delta Function δ(t)

Goal:

Find F(ω) for f(t) = δ(t).

Step-by-step derivation:

F(ω) = ∫_-∞^∞ δ(t) · e^-jωt dt

By the sifting property of delta:

∫_-∞^∞ f(t) · δ(t - t₀) dt = f(t₀)

Here, f(t) = e^-jωt and t₀ = 0.

Thus:

F(ω) = e-jω·0 = 1

Interpretation:

The delta function "picks out" the value of the kernel e^-jωt at t=0, which is 1.
So, the Fourier Transform of δ(t) is a constant 1 for all ω.

2. Rectangular Pulse

Function:

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

Goal:

Find

F(ω) = ∫_-∞^∞ f(t) · e^-jωt dt = ∫_-T/2^T/2 e^-jωt dt

Step-by-step:

1. Evaluate the integral:
   F(ω) = ∫_-T/2^T/2 e^-jωt dt
2. Integral of e^ax is (1/a) e^ax, so:
   F(ω) = [e^-jωt / (-jω)]_{t = -T/2}^T/2 = (1 / -jω)(e^-jωT/2 - e^jωT/2)
3. Use the identity:
   e^jθ - e^-jθ = 2j sin(θ)
4. Thus:
   F(ω) = (1 / -jω) (-2j sin(ωT/2)) = (2 sin(ωT/2)) / ω
5. Rearranged using sinc:
   F(ω) = T · sinc((ωT) / (2π))
   where sinc(x) = sin(πx) / (πx)

Interpretation:

The rectangular pulse in time corresponds to a sinc function in frequency.
Wider pulse → narrower frequency spectrum, and vice versa.

3. Cosine Function cos(ω₀ t)

Goal:

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

Step-by-step:

1. Use Euler's formula:
   cos(θ) = (e^jθ + e^-jθ) / 2
   So, f(t) = (e^jω₀t + e^-jω₀t) / 2
2. Fourier Transform is linear, so:
   F(ω) = (1/2) ∫ e^jω₀t e^-jωt dt + (1/2) ∫ e^-jω₀t e^-jωt dt
3. Rewrite integrands:
   = (1/2) ∫ e^{-j(ω - ω₀)t} dt + (1/2) ∫ e^{-j(ω + ω₀)t} dt
4. Recall:
   ∫ e^{-j a t} dt = 2π δ(a)
5. Therefore:
   F(ω) = (1/2)·2π δ(ω - ω₀) + (1/2)·2π δ(ω + ω₀) = π [δ(ω - ω₀) + δ(ω + ω₀)]

Interpretation:

Cosine transforms to two impulses at ±ω₀, reflecting two complex exponentials.

4. Sine Function sin(ω₀ t)

Goal:

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

Step-by-step:

1. Use Euler's formula:
   sin(θ) = (e^jθ - e^-jθ) / (2j)
   So, f(t) = (e^jω₀t - e^-jω₀t) / (2j)
2. Fourier Transform:
   F(ω) = (1/2j) ∫ e^jω₀t e^-jωt dt - (1/2j) ∫ e^-jω₀t e^-jωt dt
3. Rewrite:
   = (1/2j) ∫ e^{-j(ω - ω₀)t} dt - (1/2j) ∫ e^{-j(ω + ω₀)t} dt
4. Use integral of complex exponentials:
   = (1/2j)(2π δ(ω - ω₀)) - (1/2j)(2π δ(ω + ω₀)) = (π / j) [δ(ω - ω₀) - δ(ω + ω₀)]
5. Since 1/j = -j:
   F(ω) = j π [δ(ω + ω₀) - δ(ω - ω₀)]

Interpretation:

Sine transforms to two impulses at ±ω₀ with opposite signs/phases.

5. Gaussian Function

Function:

f(t) = e^{-t² / (2σ²)}

Goal:

Find F(ω) = ∫_-∞^∞ e^-t²/(2σ²) e^-jωt dt

Step-by-step derivation:

1. Rewrite the exponent:
   -t²/(2σ²) - jωt = -(1/(2σ²)) (t² + 2jωσ² t)
2. Complete the square:
   t² + 2jωσ² t = (t + jωσ²)² - (jωσ²)² = (t + jωσ²)² + ω² σ⁴
3. Substitute back:
   -t²/(2σ²) - jωt = - (t + jωσ²)² / (2σ²) - (ω² σ²)/2
4. So, integral becomes:
   F(ω) = e^{-ω² σ² / 2} ∫ e^{-(t + jωσ²)² / (2σ²)} dt
5. Make substitution:
   z = t + jωσ², so dt = dz
6. Integral is Gaussian integral:
   ∫_-∞^∞ e^-z²/(2σ²) dz = √(2π) σ
7. Thus:
   F(ω) = √(2π) σ · e^{-ω² σ² / 2}

Interpretation:

The Gaussian transforms into another Gaussian in frequency domain. Its width in time and frequency are inversely related.

6. Exponential Decay Function

Function:

f(t) = e^{-a t} u(t), where a > 0 and u(t) is unit step function.

Goal:

Find F(ω) = ∫_-∞^∞ e^{-a t} u(t) e^{-jω t} dt = ∫₀^∞ e^{-(a + jω) t} dt

Step-by-step:

1. Integral of e^{-λ t} from 0 to ∞ is 1/λ if Re(λ) > 0
2. Here, λ = a + jω, Re(λ) = a > 0 so integral converges
3. Therefore:
   F(ω) = 1 / (a + jω)

Interpretation:

This shows the frequency response of a causal exponential decay with magnitude and phase depending on ω.

Summary Table of Fourier Transform Pairs

Common Fourier Transform Pairs with Visual Examples

Learn time-domain signals and their frequency-domain transforms visually. All formulas are shown in plain text for easy understanding.

Common Fourier Transform Pairs and Applications

5. Common Fourier Transform Pairs

5.1 Delta Function (Dirac Delta)

Definition:

The Dirac delta function, denoted δ(t), satisfies:

• Zero everywhere except at t = 0
• Infinite at t = 0 such that
  ∫-∞∞ δ(t) dt = 1
• Sifting property: For any continuous function f(t),
  ∫-∞∞ f(t) δ(t - t₀) dt = f(t₀)

Fourier Transform of δ(t):

F(ω) = ∫_-∞^∞ δ(t) e^-jωt dt = e^-jω·0 = 1

Interpretation: Delta contains all frequencies equally.

Inverse Fourier Transform:

f(t) = (1/2π) ∫_-∞^∞ 1 · e^jωt dω = δ(t)

5.2 Rectangular Pulse

Definition:

rect(t/T) =
1, for |t| ≤ T/2
0, otherwise

Fourier Transform:

F(ω) = ∫_-T/2^T/2 e^-jωt dt = (2/ω) sin(ωT/2) = T · sinc((ωT)/(2π))

sinc(x) = sin(πx) / (πx)

Note: Narrow pulse in time means wide spread in frequency.

5.3 Sine and Cosine Functions

Cosine: f(t) = cos(ω₀ t)

Fourier Transform:

F(ω) = π [δ(ω - ω₀) + δ(ω + ω₀)]

Sine: f(t) = sin(ω₀ t)

Fourier Transform:

F(ω) = j π [δ(ω + ω₀) - δ(ω - ω₀)]

Interpretation: Sine and cosine transform into impulses at ±ω₀.

5.4 Gaussian Function

Definition:

f(t) = e^-t²/(2σ²)

Fourier Transform:

F(ω) = σ√(2π) e^{-(σ² ω²)/2}

Note: Gaussian remains Gaussian after transform.

5.5 Exponential Decay Functions

Definition:

f(t) = e^{-a t} u(t), where u(t) is the unit step, a > 0

Fourier Transform:

F(ω) = 1 / (a + jω)

Magnitude and Phase:

• |F(ω)| = 1 / √(a² + ω²)
• ∠F(ω) = -arctan(ω / a)

6. More Fourier Transform Pairs and Applications

6.1 Impulse Train (Comb Function)

Time Domain:

f(t) = Σ_n=-∞^∞ δ(t - nT)

Fourier Transform:

F(ω) = (2π / T) Σ_k=-∞^∞ δ(ω - 2π k / T)

Application: Models sampling of signals at regular intervals.

6.2 Exponential Function (Complex Frequency Shift)

Time Domain:

f(t) = e^{j ω₀ t}

Fourier Transform:

F(ω) = 2π δ(ω - ω₀)

Interpretation: Single frequency spike at ω₀.

6.3 Triangular Pulse

Time Domain:

f(t) = 1 - |t|/T for |t| ≤ T, 0 otherwise

Fourier Transform:

F(ω) = (sin(ωT/2) / (ω/2))² = T · sinc²((ωT) / (2π))

6.4 Unit Step Function u(t)

Time Domain:

u(t) = 0 for t < 0, 1 for t ≥ 0

Fourier Transform:

F(ω) = π δ(ω) + 1 / (j ω)

6.5 Ramp Function

Time Domain:

f(t) = t u(t)

Fourier Transform:

F(ω) = 1 / (jω)² + π δ'(ω)

(δ'(ω) is the derivative of delta function)

6.6 Time Shift Property

If

f(t - t₀) ↔ F(ω), then

f(t - t₀) ↔ e^{-j ω t₀} F(ω)

Time shift adds phase shift in frequency domain.

6.7 Frequency Shift Property

If

e^{j ω₀ t} f(t) ↔ F(ω - ω₀)

Multiplying by complex exponential shifts the spectrum.

7. Simple Applications of Fourier Transform

Application 1: Signal Filtering

Goal: Remove high-frequency noise, keep low-frequency components.

1. Take Fourier Transform of signal x(t): X(ω)
2. Multiply X(ω) by low-pass filter H(ω), where H(ω) = 1 for |ω| < ω_c, else 0
3. Apply inverse Fourier Transform to get filtered signal y(t) from Y(ω) = X(ω) · H(ω)

Application 2: Convolution Using Fourier Transform

Given y(t) = x(t) * h(t) (convolution in time)

Use property: Y(ω) = X(ω) · H(ω)

1. Compute X(ω) and H(ω)
2. Multiply them: Y(ω) = X(ω)·H(ω)
3. Inverse Fourier Transform of Y(ω) gives y(t)

Application 3: Signal Analysis

Analyze frequency content of signals (music, images) by transforming to frequency domain.

Application 4: Modulation and Demodulation

Shift frequencies by multiplying by complex exponentials (frequency translation).

8. Intuition and Why Fourier Transform Works

Any signal can be represented as a sum of sinusoids of different frequencies.

Fourier Transform breaks complex signals into simple frequency components for easier analysis.

9. Quick Tips to Remember

• Delta in time domain transforms to constant in frequency domain.
• Short pulse in time means wide frequency spectrum.
• Long, smooth signals have narrow frequency spectra.
• Time shift adds phase shift in frequency domain.
• Frequency shift shifts spectrum location.

Fourier Transform: Step-by-Step Derivations and Graphs

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

Definition:
F(ω) = ∫ from -∞ to ∞ f(t) * e^(-jωt) dt

Since f(t) = 1 for |t| ≤ T/2, integral limits reduce:
F(ω) = ∫ from -T/2 to T/2 e^(-jωt) dt

Calculate integral:
F(ω) = [ e^(-jωt) / (-jω) ] from -T/2 to T/2
     = (1 / -jω) * ( e^(-jω * T/2) - e^(jω * T/2) )

Simplify numerator:
e^(-jα) - e^(jα) = -2j sin(α), where α = ωT/2

So,
F(ω) = (1 / -jω) * (-2j sin(ωT/2))
     = (2 sin(ωT/2)) / ω

Rewrite as:
F(ω) = T * sinc( (ω T) / (2π) ), 
where sinc(x) = sin(πx) / (πx)

This is the Fourier Transform of the rectangular pulse.
  

f(t) = exp( - t² / (2σ²) )
  

Definition:
F(ω) = ∫ from -∞ to ∞ e^(-t² / (2σ²)) * e^(-jωt) dt

Combine exponents:
= ∫ e^{ - [ t²/(2σ²) + jωt ] } dt

Complete the square inside exponent:

Let A = 1/(2σ²)
Exponent = - A t² - jω t
          = -A [ t² + (jω/A) t ]

Complete square term:
t² + (jω/A) t = (t + jω/(2A))² - (ω² / (4A²))

Therefore,
Exponent = -A (t + jω/(2A))² + ω² / (4A)

So integral becomes:
F(ω) = e^{ ω² / (4A) } ∫ e^{-A (t + jω/(2A))² } dt

Since integral of Gaussian over all t is sqrt(pi / A), and shifting by a complex number doesn't change value:

F(ω) = e^{ ω² / (4A) } * sqrt(pi / A)

Substitute A:
A = 1/(2σ²)
So,
F(ω) = e^{ -σ² ω² / 2 } * σ * sqrt(2π)

Hence,
Fourier Transform of Gaussian is also Gaussian!
  

f(t) = e^{-a t} * u(t),  where a > 0 and u(t) is unit step function
  

Definition:
F(ω) = ∫ from 0 to ∞ e^{-a t} * e^{-jω t} dt  (limits from 0 because of u(t))

Combine exponent:
F(ω) = ∫ from 0 to ∞ e^{ -(a + jω) t } dt

Calculate integral:
F(ω) = [ e^{-(a + jω) t} / -(a + jω) ] from 0 to ∞

Evaluate limits:
At ∞: e^{-(a + jω) ∞} → 0 because a > 0 (exponential decay)
At 0: e^{0} = 1

So,
F(ω) = (0 - 1) / -(a + jω) = 1 / (a + jω)

This is the Fourier Transform of the causal exponential decay function.