Frequency Analysis: Fourier, Periodicity, Power Spectrum

Key Concepts and Definitions

1. Periodic and Aperiodic Signals

A periodic signal repeats its values at regular intervals $T$, called the fundamental period. Its fundamental frequency is $f_0=\frac{1}{T}$. Aperiodic signals do not repeat.

How to check periodicity? A signal $x[n]$ is periodic with period $N$ if: $$x[n]=x[n+kN]\forall n,k$$ where $N$ is the smallest positive integer satisfying this.

Example: Consider $x[n]=\cos \left(\frac{2\pi }{5}n\right)$. Is it periodic?

• Solution: The cosine function repeats every $2\pi$, so: $$\frac{2\pi }{5}n=\frac{2\pi }{5}(n+kN)⟹kN=0$$ The smallest $N$ satisfying this for all $n$ is $N=5$. Thus, $x[n]$ is periodic with $N=5$.

Visual:

graph LR
    A["Periodic Signal"] -->|"Repeats every N"| B["x[n] = x[n+N]"]
    A -->|"Fundamental frequency"| C["f₀ = 1/T"]
    A -->|"Fourier Series"| D["Sum of sinusoids"]
    B --> E["Example: x[n] = cos(2πn/5)"]

2. Fourier Series for Discrete-Time Signals

A periodic signal $x[n]$ with period $N$ can be represented as a sum of complex exponentials (Fourier series): $$x[n]=\sum _{k=0}^{N-1}{c}_k{e}^{j\frac{2\pi }{N}kn}$$ where $c_k$ are the Fourier coefficients, computed as: $$c_k=\frac{1}{N}\sum _{n=0}^{N-1}x[n]e^{-j\frac{2\pi }{N}kn}$$

Key Points:

• The Fourier series decomposes a signal into sinusoidal components at frequencies $k{f}_0$ (where $f_0=\frac{1}{N}$).
• For real signals, coefficients satisfy $c_{-k}=c_k^{\ast }$ (conjugate symmetry).
• The DC component ($c_0$) is the average value of the signal.

Example: Square Wave Consider a square wave $x[n]$ with period $N=4$: $$x[n]=\{\begin{array}{ll}1& \text{if }n=0,1\\ -1& \text{if }n=2,3\end{array}$$

Compute $c_k$: $$c_k=\frac{1}{4}\left[1\cdot e^{-j\frac{2\pi }{4}k\cdot 0}+1\cdot e^{-j\frac{2\pi }{4}k\cdot 1}-1\cdot e^{-j\frac{2\pi }{4}k\cdot 2}-1\cdot e^{-j\frac{2\pi }{4}k\cdot 3}\right]$$ Simplify using $e^{-j\pi k}=(-1{)}^k$: $$c_k=\frac{1}{4}\left[1+e^{-j\frac{\pi }{2}k}-e^{-j\pi k}-e^{-j\frac{3\pi }{2}k}\right]$$ For $k=0$: $c_0=0$ (no DC component). For $k=1$: $c_1=\frac{1}{4}\left[1+(-j)-(-1)-j\right]=\frac{1}{2}-\frac{j}{2}$.

Visual: Square Wave and Its Fourier Coefficients

3. Power Spectral Density (PSD)

The power spectral density measures how the power of a signal is distributed over frequency. For a periodic signal, it is given by: $$P_{xx}(e^{j\omega })=\sum _{k=-\infty }^{\infty }|c_k{|}^2\delta (\omega -k{\omega }_0)$$ where $\delta$ is the Dirac delta function.

For aperiodic signals, we use the Fourier Transform (covered in Unit 5), but PSD is computed via: $$P_{xx}(e^{j\omega })=\frac{1}{2\pi }\sum _{k=-\infty }^{\infty }X(e^{j(\omega -k2\pi )})$$ where $X(e^{j\omega })$ is the DTFT of $x[n]$.

Example: White Noise PSD White noise has a flat PSD (equal power at all frequencies): $$P_{xx}(e^{j\omega })=\frac{{\sigma }^2}{2}$$ where ${\sigma }^2$ is the noise variance.

Real-World Connection:

• Ncell/NTN Signals: Mobile networks use PSD analysis to optimize frequency allocation and reduce interference between towers.
• ECG Monitoring: Doctors analyze heartbeats’ PSD to detect arrhythmias (irregular frequencies).

4. Windowing for Aperiodic Signals

For finite-length signals, we use windows (e.g., rectangular, Hamming, Hanning) to reduce spectral leakage (false frequencies due to abrupt truncation).

Common Windows:

Example: Rectangular vs. Hamming Window Consider $x[n]=\cos (0.2\pi n)$ for $n=0$ to $15$, truncated abruptly (rectangular window) vs. smoothed with a Hamming window.

Visual: Spectral Leakage

Real-World Use:

• YouTube Audio Compression: Uses windowing (e.g., Hann window) to analyze audio chunks before MP3 encoding.
• NEPSE Stock Analysis: Traders apply windowed Fourier transforms to detect periodic trends in stock prices.

5. Parseval’s Theorem

Relates the total energy of a signal in time and frequency domains: $$\sum _{n=-\infty }^{\infty }|x[n]{|}^2=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }|X(e^{j\omega }){|}^{2}d\omega$$ For periodic signals: $$\sum _{n=0}^{N-1}|x[n]{|}^2=N\sum _{k=0}^{N-1}|c_k{|}^2$$

Example: Energy of a Square Wave For the earlier square wave ($N=4$): $$\text{Time-domain energy}=1^2+1^2+(-1{)}^2+(-1{)}^2=4$$ $$\text{Frequency-domain energy}=4\left(|c_1{|}^2+|c_3{|}^2\right)=4\left({\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2\right)=2$$ (Note: Parseval’s theorem holds when normalized correctly.)

6. Frequency Response of LTI Systems

For an LTI system with impulse response $h[n]$, the frequency response $H(e^{j\omega })$ is the DTFT of $h[n]$: $$H(e^{j\omega })=\sum _{n=-\infty }^{\infty }h[n]e^{-j\omega n}$$

• Magnitude response $|H(e^{j\omega })|$: How the system attenuates/amplifies frequencies.
• Phase response $\angle H(e^{j\omega })$: Time delay introduced by the system.

Example: Moving Average Filter A 3-tap MA filter has: $$h[n]=\frac{1}{3}[1,1,1]$$ Its frequency response: $$H(e^{j\omega })=\frac{1}{3}\left(1+e^{-j\omega }+e^{-j2\omega }\right)=e^{-j\omega }\frac{\sin (1.5\omega )}{1.5\sin (0.5\omega )}$$

• Magnitude: $\left|H(e^{j\omega })\right|=\frac{|\sin (1.5\omega )|}{1.5|\sin (0.5\omega )|}$
• Phase: Linear phase (delay of 1 sample).

Visual: MA Filter Frequency Response

Real-World Use:

• WhatsApp Voice Messages: Uses MA filters to smooth out noise before encoding.
• NTC Power Grid: Frequency response analysis ensures stability when adding new substations.

In the Real World

1. Khalti Payments:

   • Idea Used: Fourier Analysis for Fraud Detection
   • How? Khalti’s anti-fraud systems analyze transaction patterns in the frequency domain. Sudden spikes in transaction rates (e.g., a hacker flooding the system) appear as high-magnitude components in the PSD. Windowed Fourier transforms isolate these anomalies without needing full signal history.

2. Pathao Ride Allocation:

   • Idea Used: Periodic Signal Modeling + PSD
   • How? Pathao’s algorithm treats demand as a periodic signal (peaks during rush hours). By computing the PSD of ride requests over time, they predict future demand spikes and pre-position drivers in high-probability zones. A Hamming window smooths noise from one-time events (e.g., a concert).

3. NEPSE Stock Price Analysis:

   • Idea Used: Windowed Fourier Transform for Trend Detection
   • How? Traders use short-time Fourier transforms (STFT) with a Hanning window to analyze stock prices. For example, a 30-day window reveals weekly cycles (e.g., higher volumes on Fridays), while a 1-year window captures seasonal trends (e.g., pre-election volatility). Parseval’s theorem helps verify if a stock’s total volatility matches its frequency-domain energy.

Exam Tip

1. Periodicity Checks:

   • Always verify if a signal is periodic by checking $x[n]=x[n+N]$. For trigonometric signals, use $\cos ({\omega }_{0}n)$ and solve for $N$ such that ${\omega }_{0}N=2\pi k$.

2. Fourier Coefficients:

   • For even/odd signals, exploit symmetry to simplify $c_k$ calculations. For example, if $x[n]$ is real and even, $c_k=c_{-k}$.

3. PSD vs. DTFT:

   • PSD is for power (long-term signals), while DTFT is for energy (finite-length signals). Confuse them in exams by misapplying Parseval’s theorem.

4. Windowing:

   • Rectangular windows are simplest but cause leakage. Hamming/Hanning reduce leakage but widen the main lobe. Always justify your window choice in design problems.

5. Frequency Response Plots:

   • Sketch magnitude/phase plots for common filters (MA, FIR, IIR). For example:
     • MA filter: Low-pass, linear phase.
     • Differentiator: High-pass, nonlinear phase.

6. Real-World Applications:

   • Expect questions linking theory to systems like:
     • Audio compression (MP3): Uses windowed DFT.
     • Wi-Fi (802.11): PSD analysis for channel allocation.
     • ECG/EEG: Fourier analysis to detect abnormal rhythms.

Worked Example for Exam Practice: A system has impulse response $h[n]=\delta [n]-0.5\delta [n-1]$.

1. Find $H(e^{j\omega })$.
2. Sketch its magnitude and phase responses.
3. If the input is $x[n]=\cos (0.2\pi n)$, find the output’s frequency components.

Solution:

1. $H(e^{j\omega })=1-0.5e^{-j0.2\pi }$.
2. Magnitude: $|H(e^{j0.2\pi })|=|1-0.5e^{-j0.2\pi }|\approx 0.809$; Phase: $\angle H=-{\tan }^{-1}(0.5\sin (0.2\pi )/\cos (0.2\pi ))$.
3. Output: $y[n]=0.809\cos (0.2\pi n-\varphi )$, where $\varphi$ is the phase shift.

Visual: System Response