Communication Systems MCQ Questions & Answers | ECE

A. 24 MHz

B. 12 MHz

C. 9 MHz

D. 15 MHz

A. The upper sideband frequency is 1001000

B. The lower sideband frequency is 999000

C. The carrier amplitude is 5

D. x(t) is a DSB-SC signal

A. 1000 + 20cos(2π100t)

B. 2000 + 10cos(2π100t)

C. 1000 +10sin(2π100t)

D. 2cos(2π100t)

A. a-1, b-3, c-4, d-5

B. a-5, b-3, c-2, d-1

C. a-5, b-3, c-2, d-4

D. a-1, b-5, c-3, d-2

A. Less available channel space

B. More stable transmitter gives

C. More power to transmit same signal

D. Signal is less resistant to noise

A. Increase the transmission capacity

B. Improve noise performance

C. Incorporate error control coding

D. Eliminate dc component in the transmitted signal

A. carrier frequency problem

B. intermediate amplitude problem

C. Gain control problem

D. image frequency problem

A. 175 Hz

B. 525 Hz

C. 58.33 Hz

D. 3 Hz

A. $$\int\limits_{ - \infty }^\infty {{f_x}\left( x \right)dx = 1} $$

B. $$\frac{{d{f_x}}}{{dx}}\left( x \right) = {F_x}\left( x \right)$$

C. $$0 \leqslant \int_x^{} {\left( x \right) \leqslant 1} $$

D. $$\int_x^{} {\left( x \right) = \int\limits_{ - \infty }^\infty {{F_x}\left( x \right)dx} } $$

A. R(τ) = B × N₀ ℓn(2Bτ)

B. R(τ) = B × N₀ × τ × sin(2B)

C. R(τ) = B × N₀ log(2Bτ)

D. R(τ) = B × N₀ sin c(2Bτ)