GATE EC NAT 24: A signal is band-limited to 2 kHz. Enter the Nyquist sampling rate in kHz.
Correct Answer: 4
Explanation:
2fmax=4 kHz.
MCQ Collection
GATE EC NAT MCQs
Practice GATE EC NAT questions with answers and explanations.
GATE EC NAT 25: An ideal amplifier has voltage gain 3. Enter 20 log10(gain) in dB, rounded to two decimals.
Correct Answer: 9.54
Explanation:
20log₁₀(3)=9.54 dB.
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 1: An ideal amplifier has voltage gain 3. Enter 20 log10(gain) in dB, rounded to two decimals.
Correct Answer: 9.54
Explanation:
20log₁₀(3)=9.54 dB.
GATE EC NAT 2: A carrier has frequency 36 MHz. Enter its period in nanoseconds, rounded to three decimals.
Correct Answer: 27.778
Explanation:
T(ns)=1000/f(MHz)=27.778.
GATE EC NAT 3: A signal is band-limited to 5 kHz. Enter the Nyquist sampling rate in kHz.
Correct Answer: 10
Explanation:
2fmax=10 kHz.
GATE EC NAT 4: 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 5: A carrier has frequency 28 MHz. Enter its period in nanoseconds, rounded to three decimals.
Correct Answer: 35.714
Explanation:
T(ns)=1000/f(MHz)=35.714.