Why Stable Systems Have Poles Inside the Unit Circle but AR Model Roots Are Outside? Why Stable Systems Have Poles Inside the Unit Circle but AR Model Roots Are Outside? If you've studied Digital Signal Processing (DSP) or Time Series Analysis , you've probably encountered two statements that seem contradictory: A stable discrete-time system must have its poles inside the unit circle. A stable AR (Auto-Regressive) model requires its roots to lie outside the unit circle. At first glance, these statements appear to conflict. Fortunately, they are both correct. The apparent contradiction comes from the fact that DSP and statistics define the characteristic polynomial differently. This article explains why both statements are true and clears up one of the most common sources of confusion in signal processing. Stability of Continuous-Time Systems For a continuous-time system, the transfer function is H(s) = N(s) / D(s) A s...

  Question 36 A signal x(t)=100cos(24π×10 3 t) is ideally sampled with a sampling period of T s =50 µs and then passed through an ideal low-pass filter having a cutoff frequency of 15 kHz . Which frequencies are present at the filter output? 12 kHz only 8 kHz only 12 kHz and 9 kHz 12 kHz and 8 kHz Solution Step 1: Find the signal frequency Since 2πf=24π×10 3 Therefore, f=12 kHz Step 2: Find the sampling frequency f s =1/T s =1/(50×10 -6 ) =20000 Hz =20 kHz Step 3: Spectral Replicas The sampled spectrum contains frequencies |nf s ±f| For n = 0 12 kHz For n = 1 20-12=8 kHz 20+12=32 kHz For n = 2 40-12=28 kHz and so on. Step 4: Apply LPF LPF cutoff frequency 15 kHz The frequencies below 15 kHz are 8 kHz ✔ 12 kHz ✔ Therefore, both components pass through the filter. Final Output Frequencies 8 kHz and 12 kHz Correct Answer: (d) GATE EC Previous Year Papers with Soluti...

  Question 35 Consider a sampled signal y(t)=5 × 10 -6 x(t) Σ n=-∞ ∞ δ(t-nT s ) where x(t)=10cos(8π×10 3 t) and T s =100 µs When y(t) is passed through an ideal low-pass filter having a cutoff frequency of 5 kHz , the output is 5 × 10 -6 cos(8π×10 3 t) 5 × 10 -5 cos(8π×10 3 t) 5 × 10 -1 cos(8π×10 3 t) 10 cos(8π×10 3 t) Solution Step 1: Find the signal frequency Since 2πf = 8π×10 3 Therefore, f = (8π×10 3 )/(2π) =4 kHz Step 2: Find the sampling frequency f s =1/T s =1/(100×10 -6 ) =10000 Hz =10 kHz Step 3: Check Nyquist condition Nyquist Frequency=f s /2=5 kHz Since 4 kHz < 5 kHz there is no aliasing . Step 4: Reconstruction amplitude The sampled signal is y(t)=5×10 -6 x(t) Σδ(t-nT s ) The impulse train contributes a factor 1/T s =10000 Hence, 5×10 -6 ×10000 =5×10 -2 Original signal amplitude = 10 10×5×10 -2 =0.5 =5×10 -1 Final Output y(t)=5×10 -1 cos(8π×10 3 t) Correct Answer:...

Cloud Kitchen Mobile App 1 Architecture Overview The Tech Stack ✅ FastAPI – High-performance Async API ✅ React Native ✅ Firebase (for authentication) ✅ MySQL (database) Project Structure frontend/ ├── app │ └── UserDashboard.tsx └── ... ├── assets/images └── components └── constants └── hooks └── scripts └── app.json └── eslint.config.js └── expo-env.d.ts └── app.json └── package-lock.json └── package.json └── tsconfig.json └── README.md Frontend Implementation frontend/index.html import React, { useState, useEffect } from "react"; import { View, Text, TouchableOpacity, ScrollView, StyleSheet, TextInput } from "react-native"; impo...

Time-Bandwidth Product, GMSK, and Pulse Shaping: A Comprehensive Guide Understanding Time-Bandwidth Product (TBP): From Raised Cosine to GMSK Exploring the trade-off between signal duration, spectral width, and system performance. 1. What is the Time-Bandwidth Product (TBP)? The Time-Bandwidth Product (TBP) is a fundamental metric in signal processing that defines the relationship between a signal's duration ($\Delta t$) and its spectral width ($\Delta f$). It is the signal-processing equivalent of the Heisenberg Uncertainty Principle . $$TBP = B \times T$$ Where $B$ is the bandwidth and $T$ is the symbol duration (or pulse width). No signal can be simultaneously "t...

IIR Filter & Feedback Lab IIR Filter & Feedback Lab Exploring Recursive Systems: Where Output depends on Output 1 Input Signal $x[n]$ Signal Type Unit Impulse δ[n] Sine Wave Unit Step u[n] Frequency (for Sine) Notice: With an Impulse input, the output of an IIR filter can ring forever (Infinite response). 2 Feedback Coefficients Numerator $b$ (Feed-forward) Denominator $a$ (Feedback - $a_0$ is always 1) System Stable Try setting $a$ to 1.0, -1.1 to see an unstable system! ...

Ultimate FIR Filter Interactive Lab FIR Filter & LTI System Lab A visual workflow for understanding Discrete-Time Convolution 1 Define Input Signal $x[n]$ Signal Type Clean Sine Wave Sine Wave + White Noise Unit Impulse δ[n] Square Wave Frequency (Hz) This is your raw data. In an LTI system, we want to modify this signal's characteristics. 2 Define Filter $h[k]$ Filter Presets (Coefficients) Moving Average (Low Pass) Simple Differentiator (High Pass) Gaussian-like (Smooth LP) ...

  The Yule-Walker equations are fundamentally derived using the properties of a Wide-Sense Stationary (WSS) random process. Without the WSS assumption, the classical Yule-Walker equations cannot be obtained in their standard form. What is a Wide-Sense Stationary (WSS) Process? A random process \(X(t)\) is said to be Wide-Sense Stationary (WSS) if it satisfies three important conditions. 1. Constant Mean $$ E[X(t)] = \mu $$ where the mean \(\mu\) does not depend on time. 2. Constant Variance $$ Var(X(t))=\sigma^2 $$ The variance remains unchanged over time. 3. Autocovariance Depends Only on Lag Instead of depending on two different time instants, $$ C_X(t_1,t_2), $$ the covariance depends only on their difference, $$ C_X(\tau) = C_X(t_1-t_2). $$ Similarly, the autocorrelation function becomes $$ R_X(\tau) = E[X(t)X(t+\tau)]. $$ This property is the key assumption used in deriving the Yule-Walker equations. What are the Yule-Walker Equa...