Fourier Series
Exponential Fourier Series Coefficients
c_n = \frac{1}{T} \int_{0}^{T} x(t)\, e^{-jn\omega_0 t}\, dt
| Symbol | Description | Unit |
|---|---|---|
| c_n | Complex Fourier coefficient | V (or signal units) |
| T | Fundamental period | s |
| \omega_0 | Fundamental angular frequency = 2π/T | rad/s |
Worked example
Find c_0 (DC component) of x(t) = 2 + 3cos(2πt) + sin(4πt).
Given: x(t) = 2 + 3cos(2πt) + sin(4πt)
- c_0 = (1/T) ∫₀ᵀ x(t) dt
- Average of cos and sin over full period = 0
- c_0 = (1/T) ∫₀ᵀ 2 dt = 2
Answer: c_0 = 2 (DC component)
Parseval's Theorem (Power, Fourier Series)
P = \frac{1}{T} \int_0^T |x(t)|^2 dt = \sum_{n=-\infty}^{\infty} |c_n|^2
| Symbol | Description | Unit |
|---|---|---|
| P | Average power | W |
Worked example
Find average power of x(t) = 3cos(2πt) + 4sin(4πt).
Given: Amplitudes: 3 and 4
- For cosine: c_1 = c_{-1} = 3/2, |c_±1|² = 9/4 each → contribution = 9/4 + 9/4 = 4.5
- For sine: c_2 = -j4/2 = -j2, |c_±2|² = 4 each → contribution = 4 + 4 = 8
- P = 4.5 + 8 = 12.5 W
- Check: P = (3²/2) + (4²/2) = 4.5 + 8 = 12.5 ✓
Answer: P = 12.5 W
Trigonometric Fourier Coefficients
a_n = \frac{2}{T}\int_0^T x(t)\cos(n\omega_0 t)\,dt,\quad b_n = \frac{2}{T}\int_0^T x(t)\sin(n\omega_0 t)\,dt
| Symbol | Description | Unit |
|---|---|---|
| a_n | Cosine coefficient | signal units |
| b_n | Sine coefficient | signal units |
Worked example
For a square wave x(t) with amplitude 1 and period T, find a_1.
Given: x(t) = +1 for 0 < t < T/2, -1 for T/2 < t < T
- a_1 = (2/T)[∫₀^{T/2} 1·cos(2πt/T)dt + ∫_{T/2}^T (-1)·cos(2πt/T)dt]
- = (2/T)[T/(2π)·sin(2πt/T)]₀^{T/2} - (2/T)[T/(2π)·sin(2πt/T)]_{T/2}^T
- = (1/π)[sin(π) - sin(0)] - (1/π)[sin(2π) - sin(π)]
- = (1/π)[0] - (1/π)[0] = 0
Answer: a_1 = 0 (square wave has no cosine terms — it is an odd function)
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) | Frequency-domain representation | V·s |
| \omega | Angular frequency | rad/s |
Worked example
Find FT of x(t) = e^{-2t}u(t).
Given: x(t) = e^{-2t}u(t), a = 2
- X(jω) = ∫₀^∞ e^{-2t} e^{-jωt} dt
- = ∫₀^∞ e^{-(2+jω)t} dt
- = 1/(2+jω)
Answer: X(jω) = 1/(2 + jω)
Inverse Fourier Transform
x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} X(j\omega)\,e^{j\omega t}\,d\omega
Worked example
Verify that IFT of X(jω) = 2/(1+jω) gives x(t) = 2e^{-t}u(t).
Given: X(jω) = 2/(1+jω)
- Standard pair: 1/(a+jω) ↔ e^{-at}u(t)
- 2/(1+jω) = 2 × [1/(1+jω)]
- Linearity: x(t) = 2 × e^{-t}u(t)
Answer: x(t) = 2e^{-t}u(t) ✓
FT Convolution Property
y(t) = x(t)*h(t) \xrightarrow{\mathcal{F}} Y(j\omega) = X(j\omega)\cdot H(j\omega)
| Symbol | Description | Unit |
|---|---|---|
| H(j\omega) | System frequency response | dimensionless |
Worked example
System with H(jω) = 1/(1+jω). Input x(t) = e^{-2t}u(t). Find Y(jω).
Given: X(jω) = 1/(2+jω), H(jω) = 1/(1+jω)
- Y(jω) = X(jω) × H(jω)
- = 1/(2+jω) × 1/(1+jω)
- = 1/[(1+jω)(2+jω)]
Answer: Y(jω) = 1/[(1+jω)(2+jω)]
Parseval's Theorem (Energy, CTFT)
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 | Signal energy | J |
Worked example
Find energy of x(t) = e^{-3t}u(t).
Given: x(t) = e^{-3t}u(t)
- E = ∫₀^∞ (e^{-3t})² dt = ∫₀^∞ e^{-6t} dt
- = [-1/6 · e^{-6t}]₀^∞
- = 0 - (-1/6) = 1/6
Answer: E = 1/6 J
Laplace Transform
Unilateral Laplace Transform
X(s) = \int_0^{\infty} x(t)\,e^{-st}\,dt
| Symbol | Description | Unit |
|---|---|---|
| X(s) | Laplace transform | V·s |
| s | Complex frequency s = σ + jω | rad/s |
Worked example
Find L{e^{-at}u(t)}.
Given: x(t) = e^{-at}u(t)
- X(s) = ∫₀^∞ e^{-at} e^{-st} dt = ∫₀^∞ e^{-(s+a)t} dt
- = [e^{-(s+a)t} / (-(s+a))]₀^∞
- = 0 - (-1/(s+a)) = 1/(s+a)
Answer: L{e^{-at}u(t)} = 1/(s+a), ROC: Re(s) > -a
Initial Value Theorem
x(0^+) = \lim_{s \to \infty} s\,X(s)
Worked example
Find x(0+) from X(s) = (s+3)/[(s+1)(s+2)].
Given: X(s) = (s+3)/[(s+1)(s+2)]
- x(0+) = lim_{s→∞} s·(s+3)/[(s+1)(s+2)]
- = lim_{s→∞} (s²+3s)/(s²+3s+2)
- = lim_{s→∞} (1 + 3/s)/(1 + 3/s + 2/s²)
- = 1/1 = 1
Answer: x(0+) = 1
Final Value Theorem
x(\infty) = \lim_{s \to 0} s\,X(s)
Worked example
Find x(∞) from X(s) = (s+3)/[(s+1)(s+2)].
Given: X(s) = (s+3)/[(s+1)(s+2)]
- Check: all poles (s = -1, s = -2) in LHP — FVT applies
- x(∞) = lim_{s→0} s·(s+3)/[(s+1)(s+2)]
- = 0·(3)/(1·2) = 0
Answer: x(∞) = 0
Transfer Function
H(s) = \frac{Y(s)}{X(s)} = \frac{b_m s^m + \ldots + b_0}{a_n s^n + \ldots + a_0}
| Symbol | Description | Unit |
|---|---|---|
| H(s) | Transfer function | dimensionless |
| Y(s), X(s) | Output and input Laplace transforms | V·s |
Worked example
System: dy/dt + 3y = 2x. Find H(s) and pole location.
Given: ODE: dy/dt + 3y = 2x
- Take Laplace: sY(s) + 3Y(s) = 2X(s)
- Y(s)(s+3) = 2X(s)
- H(s) = Y(s)/X(s) = 2/(s+3)
- Pole at s = -3 (LHP, so stable)
Answer: H(s) = 2/(s+3), pole at s = -3
Z-Transform
Z-Transform Definition
X(z) = \sum_{n=-\infty}^{\infty} x[n]\, z^{-n}
| Symbol | Description | Unit |
|---|---|---|
| X(z) | Z-transform | dimensionless |
| z | Complex variable | dimensionless |
Worked example
Find Z{a^n u[n]}.
Given: x[n] = a^n u[n]
- X(z) = Σ_{n=0}^∞ a^n z^{-n} = Σ_{n=0}^∞ (a/z)^n
- Geometric series converges for |a/z| < 1, i.e. |z| > |a|
- X(z) = 1/(1 - a·z^{-1}) = z/(z-a)
Answer: X(z) = z/(z-a), ROC: |z| > |a|
Z-Transform Time Shifting
x[n-k] \xrightarrow{\mathcal{Z}} z^{-k}X(z)
| Symbol | Description | Unit |
|---|---|---|
| k | Delay in samples | samples |
Worked example
If X(z) = z/(z-0.5), find Z{x[n-2]}.
Given: X(z) = z/(z-0.5), delay k = 2
- Z{x[n-2]} = z^{-2} × X(z)
- = z^{-2} × z/(z-0.5)
- = z^{-1} / (z-0.5)
- = 1 / [z(z-0.5)]
Answer: Z{x[n-2]} = 1/[z(z-0.5)]
Inverse Z-Transform (Partial Fractions)
X(z)/z = \sum_k \frac{A_k}{z - p_k} \Rightarrow x[n] = \sum_k A_k p_k^n u[n]
| Symbol | Description | Unit |
|---|---|---|
| p_k | Pole of X(z) | dimensionless |
| A_k | Residue at pole k | dimensionless |
Worked example
Find inverse Z-transform of X(z) = z/[(z-1)(z-0.5)], ROC: |z|>1.
Given: X(z) = z/[(z-1)(z-0.5)]
- X(z)/z = 1/[(z-1)(z-0.5)]
- Partial fractions: A/(z-1) + B/(z-0.5)
- A = 1/(1-0.5) = 2; B = 1/(0.5-1) = -2
- X(z) = 2z/(z-1) - 2z/(z-0.5)
- x[n] = 2(1)^n u[n] - 2(0.5)^n u[n]
Answer: x[n] = [2 - 2(0.5)^n] u[n]
Sampling and Reconstruction
Nyquist Sampling Theorem
f_s \geq 2 f_{max} = 2B
| Symbol | Description | Unit |
|---|---|---|
| f_s | Sampling frequency | Hz |
| f_{max} | Maximum signal frequency (bandwidth B) | Hz |
Worked example
Find minimum sampling rate for a signal with bandwidth 4 kHz.
Given: f_max = 4 kHz
- f_s(min) = 2 × f_max
- = 2 × 4000
- = 8000 Hz
Answer: Minimum sampling rate = 8 kHz
Aliasing Condition
f_{alias} = |f_s \cdot k - f|, \quad k \in \mathbb{Z}
| Symbol | Description | Unit |
|---|---|---|
| f_{alias} | Alias frequency | Hz |
| f | Original frequency | Hz |
| k | Integer | dimensionless |
Worked example
A 1.5 kHz sinusoid is sampled at 2 kHz. Find the alias frequency.
Given: f = 1500 Hz, f_s = 2000 Hz
- Alias = |f_s - f| = |2000 - 1500|
- = 500 Hz
Answer: Alias frequency = 500 Hz
Convolution and Correlation
Continuous-Time Convolution
y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau
| Symbol | Description | Unit |
|---|---|---|
| h(t) | Impulse response of LTI system | 1/s or dimensionless |
| y(t) | Output signal | V |
Worked example
Find y(t) = u(t) * u(t) for t ≥ 0.
Given: x(t) = u(t), h(t) = u(t)
- y(t) = ∫₀^t 1·1 dτ (both u(τ) and u(t-τ) are 1 for 0 ≤ τ ≤ t)
- y(t) = ∫₀^t dτ = t for t ≥ 0
Answer: y(t) = t·u(t) (ramp function)
Cross-Correlation
R_{xy}(\tau) = \int_{-\infty}^{\infty} x(t)\,y(t+\tau)\,dt
| Symbol | Description | Unit |
|---|---|---|
| R_{xy} | Cross-correlation function | signal units squared × s |
| \tau | Time lag | s |
Worked example
Find R_xx(0) for x(t) = e^{-2t}u(t). What does this equal?
Given: x(t) = e^{-2t}u(t)
- R_xx(0) = ∫_{-∞}^{∞} x(t)·x(t) dt = ∫₀^∞ e^{-4t} dt
- = [-1/4 · e^{-4t}]₀^∞ = 1/4
Answer: R_xx(0) = 1/4 = signal energy (auto-correlation at zero lag equals energy)
Discrete-Time Convolution
y[n] = x[n] * h[n] = \sum_{k=-\infty}^{\infty} x[k]\,h[n-k]
Worked example
Convolve x[n] = {1,2,3} with h[n] = {1,1}.
Given: x = [1,2,3], h = [1,1]
- y[0] = x[0]·h[0] = 1·1 = 1
- y[1] = x[0]·h[1] + x[1]·h[0] = 1·1 + 2·1 = 3
- y[2] = x[1]·h[1] + x[2]·h[0] = 2·1 + 3·1 = 5
- y[3] = x[2]·h[1] = 3·1 = 3
Answer: y[n] = {1, 3, 5, 3}
Quick reference
| Formula | Expression |
|---|---|
| Fourier Transform | X(jω) = ∫ x(t) e^{-jωt} dt |
| Inverse FT | x(t) = (1/2π) ∫ X(jω) e^{jωt} dω |
| Parseval (Energy) | E = ∫|x(t)|² dt = (1/2π) ∫|X(jω)|² dω |
| Parseval (Power, FS) | P = (1/T) ∫|x(t)|² dt = Σ|c_n|² |
| Convolution (CT) | y(t) = ∫ x(τ) h(t-τ) dτ |
| Convolution (DT) | y[n] = Σ x[k] h[n-k] |
| Laplace Transform | X(s) = ∫₀^∞ x(t) e^{-st} dt |
| Initial Value Theorem | x(0+) = lim_{s→∞} sX(s) |
| Final Value Theorem | x(∞) = lim_{s→0} sX(s) |
| Z-Transform of a^n u[n] | X(z) = z/(z-a), |z|>|a| |
| Z Time Shift | x[n-k] → z^{-k} X(z) |
| Nyquist Rate | f_s ≥ 2 f_max |
| Alias Frequency | f_alias = |f_s·k - f| |
| FT Convolution Property | x(t)*h(t) ↔ X(jω)·H(jω) |
| Transfer Function | H(s) = Y(s)/X(s) |
Exam tips
- GATE frequently uses the duality property of FT — if x(t) ↔ X(jω), then X(jt) ↔ 2π x(-ω) — verify the sign convention in the question before applying.
- For ROC questions on Z-transforms, right-sided sequences have ROC of the form |z| > r; left-sided sequences have |z| < r; always state the ROC alongside the transform.
- Final Value Theorem is only valid when all poles of sX(s) are in the open left-half s-plane — applying it when a pole is at s = 0 or in the RHP gives wrong results.
- In convolution length questions, if x has M samples and h has N samples, the output y has M+N-1 samples — examiners test both the result and the length.
- GATE energy vs power classification: finite energy signals have zero average power; periodic signals have infinite energy but finite power.
- For sampling questions, always check whether the question asks for Nyquist rate (2B) or Nyquist interval (1/2B) — they are reciprocals and confusion costs marks.