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.
MCQ Collection
gate-ec MCQs
Practice gate-ec questions with answers and explanations.
GATE EC NAT 28: An ideal amplifier has voltage gain 6. Enter 20 log10(gain) in dB, rounded to two decimals.
Correct Answer: 15.56
Explanation:
20log₁₀(6)=15.56 dB.
GATE EC NAT 29: A carrier has frequency 42 MHz. Enter its period in nanoseconds, rounded to three decimals.
Correct Answer: 23.81
Explanation:
T(ns)=1000/f(MHz)=23.81.
GATE EC NAT 30: A signal is band-limited to 8 kHz. Enter the Nyquist sampling rate in kHz.
Correct Answer: 16
Explanation:
2fmax=16 kHz.
GATE EC NAT 26: A carrier has frequency 16 MHz. Enter its period in nanoseconds, rounded to three decimals.
Correct Answer: 62.5
Explanation:
T(ns)=1000/f(MHz)=62.5.
GATE EC NAT 12: A signal is band-limited to 6 kHz. Enter the Nyquist sampling rate in kHz.
Correct Answer: 12
Explanation:
2fmax=12 kHz.
GATE EC NAT 13: An ideal amplifier has voltage gain 7. Enter 20 log10(gain) in dB, rounded to two decimals.
Correct Answer: 16.9
Explanation:
20log₁₀(7)=16.9 dB.
GATE EC NAT 14: A carrier has frequency 24 MHz. Enter its period in nanoseconds, rounded to three decimals.
Correct Answer: 41.667
Explanation:
T(ns)=1000/f(MHz)=41.667.
GATE EC NAT 15: A signal is band-limited to 9 kHz. Enter the Nyquist sampling rate in kHz.
Correct Answer: 18
Explanation:
2fmax=18 kHz.
GATE EC NAT 16: An ideal amplifier has voltage gain 2. Enter 20 log10(gain) in dB, rounded to two decimals.
Correct Answer: 6.02
Explanation:
20log₁₀(2)=6.02 dB.
GATE EC NAT 17: A carrier has frequency 15 MHz. Enter its period in nanoseconds, rounded to three decimals.
Correct Answer: 66.667
Explanation:
T(ns)=1000/f(MHz)=66.667.
GATE EC NAT 18: A signal is band-limited to 4 kHz. Enter the Nyquist sampling rate in kHz.
Correct Answer: 8
Explanation:
2fmax=8 kHz.