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Arrange the following logic families in increasing order of their noise margin

24. Arrange the following logic families in increasing order of their noise margin TTL (Transistor - Transistor Logic) ECL (Emitter Coupled Logic) MOS (Metal Oxide Semiconductor Logic) I 2 L (Integrated Injection Logic) A. (b), (d), (a), (c) B. (a), (c), (b), (d) C. (c), (b), (d), (a) D. (d), (c), (a), (b) Answer: Option A Previous yr Question papers with Full Explanations →

The output of a control system is c(t) = 1 + 0.25e ...

22. The output of a control system is c(t) = 1 + 0.25e -50t - 1.25e -10t For an unit step input, its undamped natural frequency and damping ratio will be: A. 50 rad/s & 1.34 B. 500 rad/s & 0.5 C. 22.36 rad/s & 0.35 D. 22.36 rad/s & 1.34 Answer: Option D Solution: Solution: Given output response: c(t) = 1 + 0.25e -50t - 1.25e -10t Comparing with the standard overdamped second-order system response: c(t) = 1 + A e s₁t + B e s₂t The poles are: s₁ = -50 s₂ = -10 For a second-order system: Characteristic equation: s² + 2ζω n s + ω n 2 = 0 Sum of roots: s₁ + s₂ = -2ζω n (-50) + (-10) = -2ζω n...

A 6-bit DAC has a step size of 50 mV. The full-scale output voltage and percentage resolution for this will be:

21. A 6-bit DAC has a step size of 50 mV. The full-scale output voltage and percentage resolution for this will be: A. 5 V, 1.587% B. 3.15 V, 98.413% C. 5 V, 33.33% D. 3.15 V, 1.587% Answer: Option D Solution: Given: Number of bits, n = 6 Step size = 50 mV = 0.05 V For a DAC: Step Size = V FS / (2 n - 1) Substituting the values: 0.05 = V FS / (2 6 - 1) 0.05 = V FS / 63 V FS = 0.05 × 63 V FS = 3.15 V Percentage Resolution: % Resolution = [1 / (2 n - 1)] × 100 = (1 / 63) × 100 = 1.587% Correct Answer: D. 3.15 V, 1.587% Previous yr Question papers with Full Explanations →

BPSK Demodulation Simulation (with AWGN + Rayleigh)

Virtual Labs Instructions for BASK Simulation (with AWGN + Rayleigh) Note: Use the input fields to enter the SNR (in dB), number of multi-paths, excess delay (in ms), bit rate (in bps), and the symbol duration (in ms) Step 1: Click 'Generate BASK Message' to generate the input BASK modulated signal Step 2 & 3: Click 'Generate Multi-path for Rayleigh' and 'ADD AWGN Noise' to simulate fading and add noise Step 4: Click 'LTI Output' to generate the output after convolution with the channel Step 5: Click 'Plot BER vs SNR' to compare BER with and without Rayleigh fading BASK Simulation (with AWGN + Rayleigh) Enter SNR value in dB: Enter the number of Multi-paths: Excess Delay (in ms): Bit Rate (bps): Symbol duration (ms): ...

BPSK Online Simulator (with AWGN + Rayleigh)

BPSK Simulation (with AWGN + Rayleigh) Number of Bits: Enter SNR value in dB: Enter the number of Multi-paths: Excess Delay: Choose time unit: Centiseconds (cs) Milliseconds (ms) Bit Rate (bps): Symbol duration (ms): Carrier Frequency (Hz): BER vs SNR (AWGN + Rayleigh) SNR (dB) BER BER vs SNR (AWGN only)(without Rayleigh ...

Compared with the GaAs and InP based HBTs, the Si/SiGe HBT has a lower cutoff frequency because of:

18. Compared with the GaAs and InP based HBTs, the Si/SiGe HBT has a lower cutoff frequency because of: Lower mobilities in Si Higher mobilities in Si Lower mobilities in Ge Higher mobilities in Ge Answer: Option A Solution: In case of mobilty GaAs > Ge> Si  the Si/SiGe HBT has a lower cutoff frequency because Si has lower cutoff mobilities (i.e., lower electron and hole mobilities) Previous yr Question papers with Full Explanations →

When a plane electromagnetic wave from medium 1 (ε1) is incident normally on the surface of a perfect dielectric (i.e. medium 2 with ε2), ...

16. When a plane electromagnetic wave from medium 1 (ε 1 ) is incident normally on the surface of a perfect dielectric (i.e. medium 2 with ε 2 ), then what is the relationship of the electric field strength of the reflected wave (E r ) to the electric field strength of the incident wave, (assuming that permeabilities (μ) of known insulators do not differ appreciably from free space i.e. μ 1 = μ 2 = μ 0 ) A. E r E i = √ε 2 - √ε 1 √ε 1 + √ε 2 B. E r E i = 2√ε 1 √ε 1 + √ε 2 C. E r E i = 2√ε 2 √ε 1 + √ε 2 ...

Notch Filter Interactive Online Simulator

Notch Filter Interactive Simulator (Band-Stop) Upload CSV or Audio to remove specific frequency interference Generate Signal with 50Hz Hum Notch Frequency (Hz): Q Factor (Sharpness): Sample Rate (fs): Download Filtered Signal (CSV) Note: A Notch Filter is used to remove a specific narrow frequency (like 50Hz or 60Hz power line hum) while leaving the rest of the signal intact. Q Factor: Higher values create a narrower, sharper "notch" (removes less surrounding frequencies). Lower values create a wider "notch". Return to Home → Other...

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