We are given the problem of communicating a series of symbols, each of which can take on one of 512 values at a rate of f_s =〖10〗^5 symbols/s.
A) What is the required bandwidth if PCM is used with m=1?
B) Repeat for PAM with m=2, if each pulse can take on 8 amplitude levels.
C) Repeat for PPM with m=2, if each pulse can take on 8 different time slots.
A) For PCM with m = 1, the required bandwidth can be calculated using the Nyquist formula:
Bandwidth = 2 * m * f_s
Given f_s = 10^5 symbols/s and m = 1, we can substitute these values into the formula:
Bandwidth = 2 * 1 * 10^5 = 2 * 10^5 = 200 kHz
Therefore, the required bandwidth for PCM with m = 1 is 200 kHz.
B) For PAM with m = 2 and each pulse being able to take on 8 amplitude levels, the required bandwidth can be calculated using the Nyquist formula:
Bandwidth = 2 * m * f_s * log2(L)
where L is the number of amplitude levels.
Given f_s = 10^5 symbols/s, m = 2, and L = 8, we can substitute these values into the formula:
Bandwidth = 2 * 2 * 10^5 * log2(8) = 4 * 10^5 * log2(8) = 4 * 10^5 * 3 = 12 * 10^5
Therefore, the required bandwidth for PAM with m = 2 and 8 amplitude levels is 12 MHz.
C) For PPM with m = 2 and each pulse being able to take on 8 different time slots, the required bandwidth can be calculated using the Nyquist formula:
Bandwidth = 2 * m * f_s * log2(L)
where L is the number of time slots.
Given f_s = 10^5 symbols/s, m = 2, and L = 8, we can substitute these values into the formula:
Bandwidth = 2 * 2 * 10^5 * log2(8) = 4 * 10^5 * log2(8) = 4 * 10^5 * 3 = 12 * 10^5
Therefore, the required bandwidth for PPM with m = 2 and 8 time slots is 12 MHz.
To find the required bandwidth for each modulation technique, we need to understand the concepts underlying each modulation.
A) PCM (Pulse Code Modulation):
PCM is a method of transmitting analog signals in digital form. In PCM, the continuous analog signal is sampled and quantized into discrete digital values. The bandwidth requirement for PCM can be calculated using the Nyquist formula:
Bandwidth = 2 * (sampling rate) * (number of bits per sample)
In this case, the sampling rate (f_s) is given as 10^5 symbols/s, and m = 1 bit/sample. Therefore:
Bandwidth = 2 * 10^5 symbols/s * 1 bit/sample = 2 * 10^5 Hz = 200 kHz
So, the required bandwidth for PCM with m=1 would be 200 kHz.
B) PAM (Pulse Amplitude Modulation):
PAM is a technique in which the amplitude of the pulse is varied to represent different values. The number of amplitude levels (m) determines how many discrete values can be represented. For m=2, where each pulse can take on 8 amplitude levels, the number of bits per symbol can be calculated as:
Number of bits per symbol = log2(m)
Number of bits per symbol = log2(8) = 3 bits/symbol
To find the bandwidth for PAM, we can use the Nyquist formula:
Bandwidth = 2 * (symbol rate) * (number of bits per symbol)
In this case, the symbol rate (f_s) is given as 10^5 symbols/s and the number of bits per symbol is 3. Therefore:
Bandwidth = 2 * 10^5 symbols/s * 3 bits/symbol = 6 * 10^5 Hz = 600 kHz
So, the required bandwidth for PAM with m=2 would be 600 kHz.
C) PPM (Pulse Position Modulation):
PPM is a technique where the position or time slot of the pulse is varied to represent different values. For m=2, where each pulse can take on 8 different time slots, the number of bits per symbol can be calculated as:
Number of bits per symbol = log2(m)
Number of bits per symbol = log2(8) = 3 bits/symbol
To find the bandwidth for PPM, again we can use the Nyquist formula:
Bandwidth = 2 * (symbol rate) * (number of bits per symbol)
In this case, the symbol rate (f_s) is given as 10^5 symbols/s and the number of bits per symbol is 3. Therefore:
Bandwidth = 2 * 10^5 symbols/s * 3 bits/symbol = 6 * 10^5 Hz = 600 kHz
So, the required bandwidth for PPM with m=2 would also be 600 kHz.