The frequency response 𝐻(𝑓) of a linear time-invariant system has magnitude as shown in the figure.   Statement I: The system is necessarily a pure delay system for inputs which are bandlimited to −𝛼≤𝑓≤𝛼. Statement II : For any wide-sense stationary input process with power spectral density 𝑆H(𝑓), the output power spectral density 𝑆x(𝑓) obeys 𝑆y(𝑓)=𝑆x(𝑓) for −𝛼≤𝑓≤𝛼. Which one of the following combinations is true?  Answer: Option 1. Both Statements are correct Solution: System has magnitude 1 and bandlimited, so statement 1 is true. 𝑆y(𝑓)=𝑆x(𝑓) is true as System has magnitude 1 GATE EC Previous Year Papers with Solutions → Electronics and Communication Study Material →

Advanced Jakes Model Simulator Interactive Jakes Model Simulator Time Envelope, Rayleigh Statistics, and Doppler Bathtub Spectrum As per the Central Limit Theorem , as the number of NLOS multi-paths increase, the resulting complex envelope becomes a Gaussian process, and its magnitude follows a Rayleigh distribution . Mathematical Foundation The Jakes model approximates the fading process as a sum of $N$ complex sinusoids: \[ Z(t) = X(t) + jY(t) = \sqrt{\frac{2}{N}} \sum_{n=1}^{N} e^{j(2\pi f_d t \cos\alpha_n + \phi_n)} \] 1. Rayleigh Distribution (Envelope) \[ p(r) = \frac{r}{\sigma^2} e^{-r^2 / 2\sigma^2}, \quad r \ge 0 \] As $N \to \infty$, the magnitude $|Z(t)|$ follows this curve. 2. Doppler PSD (Bathtub ...

Jakes Model & Sum of Sinusoids The Philosophy of "Sum of Sinusoids" In wireless communication, the received signal is the sum of many sinusoids with different amplitudes, phases, and frequencies. As per the Central Limit Theorem , as $N$ increases, the magnitude follows a Rayleigh distribution . Simulator: Convergence to Rayleigh Parameters Number of Paths ($N$): 8 Performance: AWGN vs Rayleigh The Sum of Sinusoids model helps visualize why fading causes a massive drop in Bit Error Rate (BER) performance compared to a static (AWGN) channel. Characteristics of the Jakes Model: Oscillator Count: Jakes showed that using $N=8$ to $10$ oscillators is sufficient to accurately model the statistical properties of Raylei...

Q-Function & Modulation in Rayleigh Fading Q-Function's Role in Rayleigh Fading Analyzing AWGN vs. Rayleigh Fading using the General Error Rule 1. The "General" Rule for the Q-Function Regardless of the modulation, the process for Rayleigh fading always follows this template: Identify the AWGN Error Probability: \( P_e(\gamma) = A \cdot Q(\sqrt{B\gamma}) \) Average it over Rayleigh: \[ \int_{0}^{\infty} P_e(\gamma) \cdot p_{Rayleigh}(\gamma) d\gamma \] BPSK Analysis The Rayleigh closed-form solution used in the code: \[ P_{b, Rayleigh} = \frac{1}{2} \left( 1 - \sqrt{\frac{\bar{\gamma}}{1 + \bar{\gamma}}} \right) \] is the exact analytical result of integrating \( Q(\sqrt{2\gamma}) \) over the Rayleigh distribution. BER Ana...

  Since five Root Locus branches originate from the same point, the system must have five coincident open-loop poles at that location. This is because Root Locus branches always start from open-loop poles, and the number of Root Locus branches equals the number of open-loop poles (or zeros, whichever is greater). Hence, the system has five repeated real poles , which corresponds to Option (A):  1/(s+1)^5 GATE EC Previous Year Papers with Solutions → Electronics and Communication Study Material →

The point of interest is -1/K = -0.5 on the real axis. Counting the encirclements: Outer Loop: Clockwise, enclosing -0.5. (Count = +1) Left Inner Loop: Also clockwise, enclosing -0.5. (Count = +1) Right Inner Loop: Counter-clockwise, but -0.5 lies outside. (Count = 0) Total Encirclements (N) = 2 Using the stability relation: Z = N + P = 2 + 0 = 2 Hence, the closed-loop transfer function has 2 poles in the closed right-half plane. Correct Option: (C) 2 Nyquist Stability Analysis 1. The Fundamental Principle The number of closed-loop poles in the right-half plane (RHP), denoted as Z , is determined by the Nyquist Stability Criterion: Z = N + P N : Net number of clockwise (CW) encirclements of the critical point. P : Number of open-loop poles of G(s) in the RHP. 2. Identify Given Parameters ...

Advanced Bode Plot & Logarithmic Frequency Simulator SYSTEM DYNAMICS: BODE MAGNITUDE & PHASE ANALYSIS Precision Frequency Domain Simulation with Exact Transfer Functions System Transfer Function $H(s)$ Low-Pass Filter (1 / (1 + s/ωc)ⁿ) High-Pass Filter ((s/ωc)ⁿ / (1 + s/ωc)ⁿ) Pure Integrator (1 / (s/ωc)ⁿ) Filter Order / Poles ($n$) n = 1 n = 2 n = 3 Corner Frequency ($f_c$): 1000 Hz Display Overlays Show Asymptotic Lines Show Actual Physical Curve THEORETICAL SLOPES: Decade Rate: -20.0 dB/dec Octave Rate: -6.02 dB/oct At $2f_c$ (+1 Octave): -3.01 dB actual At $10f_c$ (+1 Decade): -10.04 dB...

  The ideal long channel nMOSFET and pMOSFET devices shown in the circuits have threshold voltages of 1 V and −1 V, respectively. The MOSFET substrates are connected to their respective sources. Ignore leakage currents and assume that the capacitors are initially discharged. For the applied voltages as shown, the steady state voltages are _________________. Answer: for nMOSFET, Vgs ≥ Vth  Or, (Vg - Vs) ≥ Vth So, the maximum output voltage is limited to Vg - Vth or 5-1 = 4V For nMOSFET, Vgs  less than or equal to Vth  Or, (Vg - Vs) <= Vth If you were using a pMOS to pull an output down to ground (passing a "0"), it would be limited. It would turn off once the output reached: Vout,min =Vg + |Vth| or, 5V GATE EC Previous Year Papers with Solutions → Electronics and Communication Study Material →