A wireless digital transmission scheme is using 16-QAM over an additive white Gaussian noise channel and a maximum-likelihood receiver. Consider the information bit rate from source to be 4 × 10⁶ bits per second. The minimum transmission bandwidth (in MHz) of the modulated signal necessary for optimum recovery of information at the receiver is ________. (rounded off to two decimal places)

The correct answer is 1.00.

To find the minimum transmission bandwidth of the modulated signal, we can break the problem down into the following step-by-step calculation:

Step 1: Determine the number of bits per symbol
The modulation scheme is 16-QAM (Quadrature Amplitude Modulation). The number of constellation points is:
M = 16
The number of bits transmitted per symbol, represented by k, is calculated as:
k = log 2 ( M ) = log 2 ( 16 ) = 4 bits/symbol

Step 2: Find the symbol transmission rate (Baud rate)
The source bit rate is given as:
R b = 4 × 10 6 bits per second
The symbol rate, R_s, is the bit rate divided by the number of bits per symbol:
R s = R b k = 4 × 10 6 4 = 1 × 10 6 symbols per second (or 1 MBaud)

Step 3: Calculate the minimum transmission bandwidth
For a passband modulation scheme like QAM, the minimum transmission bandwidth (Nyquist bandwidth) required for the optimum recovery of information without intersymbol interference (ISI) is equal to the symbol rate:
B min = R s
Substituting the value of the symbol rate:
B min = 1 × 10 6 Hz = 1.00 MHz

Therefore, the minimum transmission bandwidth of the modulated signal necessary for optimum recovery is 1.00 MHz.