GATE EC NAT 27: A signal is band-limited to 5 kHz. Enter the Nyquist sampling rate in kHz.
Correct Answer: 10
Explanation:
2fmax=10 kHz.
Practice GATE EC NAT questions with answers and explanations.
2fmax=10 kHz.
20log₁₀(6)=15.56 dB.
T(ns)=1000/f(MHz)=23.81.
2fmax=16 kHz.
20log₁₀(2)=6.02 dB.
T(ns)=1000/f(MHz)=66.667.
2fmax=8 kHz.
20log₁₀(5)=13.98 dB.
T(ns)=1000/f(MHz)=23.81.
2fmax=14 kHz.
20log₁₀(8)=18.06 dB.
T(ns)=1000/f(MHz)=12.346.