Continuous-Time Fourier Transform (CTFT)
Fourier Transform Definition
X(j\omega) = \int_{-\infty}^{\infty} x(t)\, e^{-j\omega t}\, dt
| Symbol | Description | Unit |
|---|---|---|
| X(j\omega) | Fourier transform (frequency-domain representation) | dimensionless (or V·s) |
| \omega | Angular frequency = 2πf | rad/s |
| x(t) | Time-domain signal | V (or any consistent unit) |
Worked example
Find the Fourier transform of x(t) = e^{−3t} u(t).
Given: x(t) = e^{-3t} u(t), a=3
- X(jω) = ∫₀^∞ e^{-3t} e^{-jωt} dt = ∫₀^∞ e^{-(3+jω)t} dt
- = [e^{-(3+jω)t} / (-(3+jω))]₀^∞
- At t→∞: e^{-(3+jω)t} → 0 (since Re(3+jω)=3>0)
- X(jω) = 0 − [−1/(3+jω)] = 1/(3+jω)
Answer: X(jω) = 1/(3+jω)
Inverse Fourier Transform
x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega)\, e^{j\omega t}\, d\omega
| Symbol | Description | Unit |
|---|---|---|
| x(t) | Recovered time-domain signal | V |
| X(j\omega) | Fourier spectrum | V·s |
Worked example
Find x(t) for X(jω) = rect(ω/2W) (ideal LPF, bandwidth W rad/s).
Given: X(jω)=1 for |ω|≤W, 0 otherwise
- x(t) = (1/2π) ∫_{-W}^{W} e^{jωt} dω
- = (1/2π) [e^{jωt}/(jt)]_{-W}^{W}
- = (1/2πjt)(e^{jWt} − e^{-jWt})
- = (1/πt) sin(Wt) = (W/π) sinc(Wt/π)
Answer: x(t) = (W/π) sinc(Wt/π) — the sinc function in time
Fourier Transform of Rectangular Pulse
\Pi\!\left(\frac{t}{\tau}\right) \xrightarrow{\mathcal{F}} \tau\, \text{sinc}\!\left(\frac{\omega \tau}{2\pi}\right) = \tau\, \frac{\sin(\omega\tau/2)}{\omega\tau/2}
| Symbol | Description | Unit |
|---|---|---|
| \tau | Pulse width | s |
| \text{sinc}(x) | Normalised sinc = sin(πx)/(πx) | dimensionless |
Worked example
Find |X(jω)| at ω=2π/τ (first null) for a unit-amplitude rectangular pulse of duration τ.
Given: τ is pulse width, x(t)=1 for |t|≤τ/2
- X(jω) = τ·sin(ωτ/2)/(ωτ/2)
- At ω = 2π/τ: ωτ/2 = 2π/τ × τ/2 = π
- X(j·2π/τ) = τ·sin(π)/π = τ×0/π = 0
- First null confirms bandwidth ≈ 2π/τ rad/s = 1/τ Hz
Answer: First null at ω = 2π/τ (f = 1/τ Hz); bandwidth ≈ 1/τ Hz
Fourier Transform of Impulse
\delta(t) \xrightarrow{\mathcal{F}} 1, \quad 1 \xrightarrow{\mathcal{F}} 2\pi\,\delta(\omega)
| Symbol | Description | Unit |
|---|---|---|
| \delta(t) | Dirac delta function | s⁻¹ |
Worked example
Find X(jω) for x(t) = 5δ(t−2).
Given: x(t)=5δ(t−2)
- Apply time-shift property: FT{δ(t−t₀)} = e^{−jωt₀}
- X(jω) = 5·e^{−jω·2} = 5e^{−j2ω}
- |X(jω)| = 5 (constant magnitude — impulse has flat spectrum)
Answer: X(jω) = 5e^{−j2ω}
Fourier Transform Properties
Linearity
a\,x_1(t) + b\,x_2(t) \xrightarrow{\mathcal{F}} a\,X_1(j\omega) + b\,X_2(j\omega)
| Symbol | Description | Unit |
|---|---|---|
| a, b | Arbitrary constants | dimensionless |
Worked example
Find FT of x(t) = 3e^{−2t}u(t) − 2e^{−5t}u(t).
Given: a=3, a1=2, b=−2, a2=5
- FT{e^{-at}u(t)} = 1/(a+jω)
- X(jω) = 3/(2+jω) + (−2)/(5+jω)
- = 3/(2+jω) − 2/(5+jω)
Answer: X(jω) = 3/(2+jω) − 2/(5+jω)
Time-Shifting Property
x(t - t_0) \xrightarrow{\mathcal{F}} e^{-j\omega t_0} X(j\omega)
| Symbol | Description | Unit |
|---|---|---|
| t_0 | Time delay | s |
Worked example
x(t) is a rectangular pulse with FT X(jω)=sinc(ω/2π). Find FT of y(t)=x(t−3).
Given: X(jω)=sinc(ω/2π), t₀=3
- Y(jω) = e^{−jω·3} × X(jω)
- = e^{−j3ω} × sinc(ω/2π)
- |Y(jω)| = |X(jω)| (magnitude unchanged)
- ∠Y(jω) = ∠X(jω) − 3ω (phase shifted by −3ω)
Answer: Y(jω) = e^{−j3ω}·sinc(ω/2π)
Frequency-Shifting (Modulation) Property
e^{j\omega_0 t} x(t) \xrightarrow{\mathcal{F}} X(j(\omega - \omega_0))
| Symbol | Description | Unit |
|---|---|---|
| \omega_0 | Carrier angular frequency | rad/s |
Worked example
A message signal x(t) has FT X(jω) bandlimited to 2π×10³ rad/s. It is modulated: y(t)=x(t)cos(2π×100×10³ t). Find the bandwidth of Y(jω).
Given: Message BW = 1 kHz, carrier = 100 kHz
- cos(ω₀t) = (e^{jω₀t} + e^{-jω₀t})/2
- Y(jω) = (1/2)[X(j(ω−ω₀)) + X(j(ω+ω₀))]
- Two sidebands, each 1 kHz wide, centred at ±100 kHz
- Total bandwidth of Y = 2 × 1 kHz = 2 kHz
Answer: Bandwidth of Y(jω) = 2 kHz (DSB-SC modulation)
Convolution in Time ↔ Multiplication in Frequency
x(t) * h(t) \xrightarrow{\mathcal{F}} X(j\omega)\, H(j\omega)
| Symbol | Description | Unit |
|---|---|---|
| H(j\omega) | System frequency response | dimensionless |
Worked example
Input x(t)=e^{−t}u(t), system h(t)=e^{−2t}u(t). Find Y(jω).
Given: X(jω)=1/(1+jω), H(jω)=1/(2+jω)
- Y(jω) = X(jω)·H(jω)
- = 1/(1+jω) × 1/(2+jω)
- = 1/[(1+jω)(2+jω)]
- Partial fractions (if needed): = 1/(1+jω) − 1/(2+jω)
Answer: Y(jω) = 1/[(1+jω)(2+jω)]
Differentiation in Time
\frac{dx(t)}{dt} \xrightarrow{\mathcal{F}} j\omega\, X(j\omega)
| Symbol | Description | Unit |
|---|---|---|
| j\omega | Frequency-domain differentiation operator | rad/s |
Worked example
x(t) = te^{-3t}u(t). Find FT of dx/dt without direct integration.
Given: X(jω) = 1/(3+jω)² (standard pair for te^{-at}u(t))
- FT{dx/dt} = jω × X(jω)
- = jω/(3+jω)²
Answer: FT{dx/dt} = jω/(3+jω)²
Parseval's Theorem (Energy)
E = \int_{-\infty}^{\infty} |x(t)|^2\, dt = \frac{1}{2\pi} \int_{-\infty}^{\infty} |X(j\omega)|^2\, d\omega
| Symbol | Description | Unit |
|---|---|---|
| E | Total signal energy | J (normalised) |
| |X(j\omega)|^2 | Energy spectral density | J·s/rad |
Worked example
Find the energy of x(t)=e^{-3t}u(t) using Parseval's theorem.
Given: X(jω)=1/(3+jω)
- E = (1/2π)∫_{-∞}^{∞} |1/(3+jω)|² dω
- = (1/2π)∫_{-∞}^{∞} 1/(9+ω²) dω
- Using ∫_{-∞}^{∞} 1/(a²+ω²)dω = π/a with a=3:
- E = (1/2π)×(π/3) = 1/6
Answer: E = 1/6 (verify: ∫₀^∞ e^{-6t}dt = 1/6 ✓)
Common Fourier Transform Pairs
One-Sided Exponential
e^{-at} u(t) \xrightarrow{\mathcal{F}} \frac{1}{a + j\omega}, \quad a > 0
| Symbol | Description | Unit |
|---|---|---|
| a | Decay constant (must be positive) | s⁻¹ |
Worked example
Find the 3 dB bandwidth of x(t)=e^{-5t}u(t).
Given: a=5, X(jω)=1/(5+jω)
- |X(jω)|² = 1/(25+ω²)
- At ω=0: |X|²=1/25
- Half-power point: 1/(25+ω²) = 1/(2×25) → 25+ω²=50 → ω²=25 → ω=5
- f_{-3dB} = ω/(2π) = 5/(2π) ≈ 0.796 Hz
Answer: Bandwidth = 5 rad/s ≈ 0.796 Hz
Signum Function
\text{sgn}(t) \xrightarrow{\mathcal{F}} \frac{2}{j\omega}
| Symbol | Description | Unit |
|---|---|---|
| \text{sgn}(t) | +1 for t>0, −1 for t<0 | dimensionless |
Worked example
Express the Fourier transform of u(t) in terms of known pairs using u(t) = (1+sgn(t))/2.
Given: FT{1}=2πδ(ω), FT{sgn(t)}=2/(jω)
- FT{u(t)} = FT{1/2} + FT{sgn(t)/2}
- = (1/2)·2πδ(ω) + (1/2)·2/(jω)
- = πδ(ω) + 1/(jω)
Answer: U(jω) = πδ(ω) + 1/(jω)
Gaussian Pulse
e^{-\alpha t^2} \xrightarrow{\mathcal{F}} \sqrt{\frac{\pi}{\alpha}}\, e^{-\omega^2/(4\alpha)}
| Symbol | Description | Unit |
|---|---|---|
| \alpha | Gaussian width parameter | s⁻² |
Worked example
Find FT of x(t)=e^{-2t²} and identify its bandwidth.
Given: α=2
- X(jω) = √(π/2)·e^{-ω²/8}
- At ω=0: X(0) = √(π/2) ≈ 1.253
- −3 dB when e^{-ω²/8} = 1/√2 → ω²/8 = ln√2 = 0.347 → ω² = 2.77 → ω = 1.664 rad/s
- f_{-3dB} = 1.664/(2π) = 0.265 Hz
Answer: X(jω) = √(π/2)·e^{−ω²/8}; −3 dB bandwidth ≈ 1.664 rad/s
Fourier Series
Complex Fourier Series Coefficients
c_n = \frac{1}{T_0} \int_0^{T_0} x(t)\, e^{-jn\omega_0 t}\, dt, \quad \omega_0 = \frac{2\pi}{T_0}
| Symbol | Description | Unit |
|---|---|---|
| c_n | nth complex Fourier coefficient | same as x(t) |
| T_0 | Fundamental period | s |
| \omega_0 | Fundamental angular frequency | rad/s |
Worked example
Find c₀ and c₁ for a square wave: x(t)=1 for 0≤t<T₀/2, −1 for T₀/2≤t<T₀.
Given: T₀=1 s (ω₀=2π), square wave amplitude ±1
- c₀ = (1/T₀)[∫₀^{T₀/2} 1 dt + ∫_{T₀/2}^{T₀} (−1) dt] = (1/1)[0.5 − 0.5] = 0
- c₁ = (1/1)[∫₀^{0.5} e^{−j2πt}dt + ∫_{0.5}^{1}(−1)e^{−j2πt}dt]
- = [e^{−j2πt}/(−j2π)]₀^{0.5} − [e^{−j2πt}/(−j2π)]_{0.5}^{1}
- = (1/(−j2π))(e^{−jπ}−1) − (1/(−j2π))(e^{−j2π}−e^{−jπ}) = (1/(j2π))(−2)(−2)/(2) ... = 1/(jπ)
Answer: c₀ = 0 (no DC); c₁ = 1/(jπ) ≈ −j0.318
Trigonometric Fourier Series
x(t) = a_0 + \sum_{n=1}^{\infty}\left[a_n \cos(n\omega_0 t) + b_n \sin(n\omega_0 t)\right]
| Symbol | Description | Unit |
|---|---|---|
| a_0 | DC component = average value | V |
| a_n, b_n | Cosine and sine amplitudes | V |
Worked example
For a full-wave rectified sine: x(t)=|sin(πt/T)| with T=1 s. Find a₀ and a₂.
Given: x(t)=|sin(πt)|, T₀=1 (period of rectified wave = T/2... fundamental = 2/T=2 Hz)
- a₀ = (2/π) = 0.6366 (well-known result for |sin|)
- a₂ = −4/(π(4−1)) = −4/(3π) = −0.4244 (n=1 in |sin| expansion)
- Full series: x(t) = 2/π − (4/π)[cos(2ωt)/3 + cos(4ωt)/15 + ...]
Answer: a₀ = 2/π ≈ 0.637; a₂ = −4/(3π) ≈ −0.424
Parseval's Theorem for Periodic Signals
P = \frac{1}{T_0} \int_0^{T_0} |x(t)|^2\, dt = \sum_{n=-\infty}^{\infty} |c_n|^2
| Symbol | Description | Unit |
|---|---|---|
| P | Average power | W (normalised) |
| c_n | Complex Fourier series coefficients | V |
Worked example
Verify the power of a unit square wave (±1) using Parseval's theorem.
Given: c_n = 2/(jnπ) for odd n, 0 for even n
- Direct: P = (1/T₀)∫|x|²dt = 1 (since x²=1 always)
- Via Parseval: P = Σ|c_n|² = Σ_{n odd} |2/(nπ)|² = (4/π²)(1+1/9+1/25+...) = (4/π²)(π²/8) = 1/2+1/2=1
- Sum Σ_{n odd} 1/n² = π²/8
Answer: P = 1 W (normalised) confirmed by both methods
Discrete-Time Fourier Transform (DTFT)
DTFT Definition
X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} x[n]\, e^{-j\omega n}
| Symbol | Description | Unit |
|---|---|---|
| X(e^{j\omega}) | DTFT (periodic with period 2π) | dimensionless |
| \omega | Digital frequency | rad/sample |
Worked example
Find DTFT of x[n] = (0.5)^n u[n].
Given: x[n]=(0.5)^n u[n]
- X(e^{jω}) = Σ_{n=0}^∞ (0.5)^n e^{-jωn} = Σ(0.5e^{-jω})^n
- Geometric series with r=0.5e^{-jω}, |r|=0.5<1 ∀ω
- X(e^{jω}) = 1/(1−0.5e^{-jω})
Answer: X(e^{jω}) = 1/(1−0.5e^{-jω})
Sampling Theorem — Nyquist Rate
f_s \geq 2 f_{max}, \quad f_s = \frac{1}{T_s}
| Symbol | Description | Unit |
|---|---|---|
| f_s | Sampling frequency | Hz |
| f_{max} | Highest frequency component in x(t) | Hz |
| T_s | Sampling period | s |
Worked example
An audio signal is bandlimited to 20 kHz. What is the minimum sampling rate? If sampled at 44.1 kHz, what is the digital frequency of a 5 kHz tone?
Given: f_max=20 kHz, f_s=44100 Hz, f_tone=5000 Hz
- Minimum sampling rate = 2 × 20000 = 40 kHz
- Digital frequency ω = 2π × f_tone / f_s
- = 2π × 5000/44100
- = 2π × 0.1134 = 0.7125 rad/sample
Answer: f_s(min) = 40 kHz; ω = 0.7125 rad/sample at 44.1 kHz
Spectral Analysis and Applications
Energy Spectral Density
S_{xx}(\omega) = |X(j\omega)|^2
| Symbol | Description | Unit |
|---|---|---|
| S_{xx}(\omega) | Energy spectral density | J·s/rad |
Worked example
Find the energy spectral density of x(t)=e^{-3t}u(t) and the fraction of energy below ω=3 rad/s.
Given: X(jω)=1/(3+jω), E=1/6
- S_xx(ω) = 1/(9+ω²)
- E_{|ω|≤3} = (1/2π)∫_{-3}^{3} 1/(9+ω²) dω = (1/2π)·(1/3)[arctan(ω/3)]_{-3}^{3}
- = (1/6π)[arctan(1)−arctan(−1)] = (1/6π)[π/4+π/4] = (1/6π)(π/2) = 1/12
- Fraction = (1/12)/(1/6) = 1/2 = 50%
Answer: 50% of signal energy lies within |ω| ≤ 3 rad/s
Convolution Theorem in Frequency Domain
x_1(t)\cdot x_2(t) \xrightarrow{\mathcal{F}} \frac{1}{2\pi} X_1(j\omega) * X_2(j\omega)
| Symbol | Description | Unit |
|---|---|---|
| * | Convolution in frequency domain | dimensionless |
Worked example
x₁(t)=cos(100t) (carrier) and x₂(t) is a message bandlimited to 10 rad/s. Describe the spectrum of y(t)=x₁(t)x₂(t).
Given: X₁(jω)=π[δ(ω-100)+δ(ω+100)], X₂(jω) bandlimited to |ω|≤10
- Y(jω) = (1/2π)[X₁(jω)*X₂(jω)]
- Convolving X₂ with each impulse shifts it to ω=±100
- Y(jω) = (1/2)[X₂(j(ω-100)) + X₂(j(ω+100))]
- Two sidebands centred at ±100 rad/s, each 20 rad/s wide (BW=10 rad/s each side)
Answer: Y(jω) consists of two sidebands at ω = ±100 rad/s, each 20 rad/s wide
Symmetry Properties
Even and Odd Decomposition
x_e(t) = \frac{x(t)+x(-t)}{2} \xrightarrow{\mathcal{F}} \text{Re}\{X(j\omega)\}, \quad x_o(t) \xrightarrow{\mathcal{F}} j\,\text{Im}\{X(j\omega)\}
| Symbol | Description | Unit |
|---|---|---|
| x_e(t), x_o(t) | Even and odd components of x(t) | V |
Worked example
For x(t)=e^{-3t}u(t), find Re{X(jω)} using the even-part approach.
Given: x_e(t)=(e^{-3t}u(t)+e^{3t}u(-t))/2 = (1/2)e^{-3|t|}
- FT{(1/2)e^{-3|t|}} = (1/2)·2·3/(9+ω²) = 3/(9+ω²)
- X(jω) = 1/(3+jω) = 3/(9+ω²) − jω/(9+ω²)
- Re{X(jω)} = 3/(9+ω²) ✓
Answer: Re{X(jω)} = 3/(9+ω²)
Duality Property
\text{If } x(t) \xrightarrow{\mathcal{F}} X(j\omega), \text{ then } X(jt) \xrightarrow{\mathcal{F}} 2\pi\, x(-\omega)
| Symbol | Description | Unit |
|---|---|---|
| X(jt) | Spectrum evaluated at t instead of ω | dimensionless |
Worked example
Use duality to find FT of sinc(Wt/π).
Given: Known pair: rect(t/τ) ↔ τsinc(ωτ/2π)
- From the pair: τ·sinc(ωτ/2π) is the FT of rect(t/τ)
- By duality: FT{sinc(Wt/π)} gives a rectangular spectrum
- Setting W=τ/2: FT{sinc(Wt/π)} = (π/W)·rect(ω/2W)
- Ideal LPF with bandwidth W rad/s has sinc impulse response
Answer: FT{sinc(Wt/π)} = (π/W)·rect(ω/2W) — confirms sinc ↔ rect duality
Quick reference
| Formula | Expression |
|---|---|
| CTFT Definition | X(j\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t}dt |
| Inverse CTFT | x(t) = (1/2\pi)\int X(j\omega)e^{j\omega t}d\omega |
| One-Sided Exponential | e^{-at}u(t) \leftrightarrow 1/(a+j\omega) |
| Impulse Pair | \delta(t) \leftrightarrow 1 |
| Rect Pulse Pair | \Pi(t/\tau) \leftrightarrow \tau\,\text{sinc}(\omega\tau/2\pi) |
| Time Shift | x(t-t_0) \leftrightarrow e^{-j\omega t_0}X(j\omega) |
| Frequency Shift | e^{j\omega_0 t}x(t) \leftrightarrow X(j(\omega-\omega_0)) |
| Differentiation | dx/dt \leftrightarrow j\omega X(j\omega) |
| Convolution ↔ Multiplication | x*h \leftrightarrow XH |
| Parseval (energy) | E = (1/2\pi)\int|X(j\omega)|^2 d\omega |
| Parseval (periodic) | P = \sum_n |c_n|^2 |
| Nyquist Rate | f_s \geq 2f_{max} |
| DTFT | X(e^{j\omega}) = \sum_n x[n]e^{-j\omega n} |
| Energy Spectral Density | S_{xx}(\omega) = |X(j\omega)|^2 |
| Duality | X(jt) \leftrightarrow 2\pi x(-\omega) |
Exam tips
- GATE Fourier problems frequently ask you to use Parseval's theorem to evaluate definite integrals — recognise the pattern ∫|X|²dω and convert it immediately to signal energy.
- The rect–sinc duality pair appears in nearly every exam; know both directions: rect in time gives sinc in frequency, and a sinc in time means an ideal rectangular filter in frequency.
- For modulation problems, examiners check whether you correctly identify the resulting bandwidth as 2×message bandwidth for DSB-SC — a common mistake is writing the answer as the message bandwidth itself.
- When applying the time-shift property, the magnitude spectrum is unchanged and only the phase changes by −ωt₀; examiners specifically test this by asking for |X(jω)| after a delay.
- Symmetry questions often require you to state whether a signal has half-wave symmetry (no even harmonics), even symmetry (no sine terms), or odd symmetry (no cosine or DC terms) before computing coefficients.
- The Nyquist theorem assumes ideal sampling; always state that practical anti-aliasing filters require f_s > 2f_max (strictly greater than) to account for non-ideal filter roll-off.