Question

Let the message signal be:

m(t)=sinc(2500t)
and the carrier is
c(t)=cos(2π×10⁶t) m(t) is VSB-modulated with filter bandwidth of 1.5 W (W is the message bandwidth). What is the bandwidth (Bᵤ) of the transmitted signal u(t) ?
a.Bᵤ=1667 Hz
b.Bᵤ=2500 Hz
c.Bᵤ=7500 Hz
d.Bᵤ=5000 Hz
e.Bᵤ=3750 Hz
f.Bᵤ=1875 Hz
​

Answer 1

The bandwidth of the transmitted signal u(t) in VSB modulation can be determined by adding the bandwidth of the message signal and the sidebands. For a sinc function, the bandwidth is determined by the first null of the sinc function. In VSB modulation, the bandwidth of each sideband is equal to the message bandwidth. Therefore, the total bandwidth of the transmitted signal can be calculated as 4 times the message bandwidth.

Step-by-step explanation:

To find the bandwidth of the transmitted signal u(t), we need to consider the bandwidth of the message signal and the modulation technique used. In VSB modulation, the upper sideband (USB) and lower sideband (LSB) are used to transmit the message signal. The bandwidth of the transmitted signal is determined by the sum of the bandwidths of the message signal and the sidebands.

For a sinc function, the bandwidth is determined by the first null of the sinc function. In this case, the first null occurs at the period of the sinc function, which can be calculated as Bm = 1/Tm, where Tm is the period. Given that the message bandwidth is 1.5 W, we can calculate the period Tm = 1/Bm.

Next, we need to calculate the bandwidth of the sidebands. In VSB modulation, the bandwidth of each sideband is equal to the message bandwidth. Therefore, the total bandwidth of the transmitted signal u(t) can be calculated as Bu = 2*Bm + 2*Bm = 4*Bm.

Substituting the values, we have Bm = 1/(1.5) = 2/3 Hz. Therefore, the bandwidth of the transmitted signal Bu = 4*(2/3) = 8/3 Hz, which is approximately 2.67 Hz.