Question

Television channels are \(6 \mathrm{MHz}\) wide. How many bits/sec can be sent if four-level digital signals are used? Assume a noiseless channel.

Short answer

The channel can transmit 24 million bits per second (24 Mbps).

Step-by-step solution

Understanding Channel Bandwidth

The problem gives us the bandwidth of the television channels, which is specified as 6 MHz. This means the channel can handle a frequency range of up to 6 million cycles per second.

Using the Noiseless Channel Capacity Formula

For a noiseless channel, the maximum data rate (bit rate) is calculated using the formula: \[C = 2B \, \log_2 M\]Where \(C\) is the channel capacity in bits per second (bps), \(B\) is the bandwidth in hertz (Hz), and \(M\) is the number of discrete signal levels used.

Substitute Given Values

We are given that the bandwidth \(B\) is 6 MHz (or 6 million Hz) and that the four-level digital signals imply \(M = 4\).Substitute these values into the formula:\[C = 2 imes 6,000,000 \, \log_2 4\]

Calculate Logarithm Base 2

\(\log_2 4\) asks us to calculate how many times 2 must be raised to give 4. It's calculated as follows:\[\log_2 4 = 2\]This is because \(2^2 = 4\).

Calculate the Maximum Bit Rate

Now substitute \(\log_2 4 = 2\) into the formula:\[C = 2 imes 6,000,000 imes 2\]Calculate the final answer:\[C = 24,000,000 \, \text{bps}\]

Conclusion

Therefore, the maximum number of bits per second that can be transmitted in this channel is 24 million bps.

Key concepts

Bandwidth

Bandwidth refers to the range of frequencies that a communication channel can handle. It is typically measured in Hertz (Hz). In the context of digital communication, bandwidth is crucial as it determines the rate at which data can be transmitted.

• Bandwidth is the difference between the highest and lowest frequencies that a channel can transmit.
• This "width" of frequencies directly influences how much data can be sent over the channel during a given time period.
• Larger bandwidths allow more data to be transmitted, potentially increasing the data rate.

For example, a channel with a bandwidth of 6 MHz, like in our exercise, means it can handle signals up to 6 million Hertz. Greater bandwidth typically means more capacity for sending data, which is why it's a fundamental aspect in digital communications.

Digital Signal Levels

In digital communications, signal levels refer to the distinct quantitative values that a digital signal can take. When we talk about four-level digital signals, it means that the signal can have one of four different discrete states at any given time.

• The number of levels (denoted as \( M \)) defines the potential combinations of data that can be represented with each "bit" or unit.
• More signal levels can represent more information without needing to increase the channel's frequency bandwidth.
• For example, two-level signals are binary (0 or 1), but four-level signals can be viewed as quaternary (00, 01, 10, 11).

Thus, having more levels increases the data efficiency since more data is packed into each symbol transmitted. This is a key consideration in maximizing channel capacity within the given bandwidth.

Bits per Second

Bits per second (bps) is the unit used to measure data transfer rates. It represents the number of bits (binary digits) that can be transmitted per second.

• The channel capacity in bps is indicative of how fast data can be communicated over a channel.
• Higher bps values mean faster data transmission and reception.
• The formula \(C = 2B \log_2 M \) is used to calculate the maximum bps for a noiseless channel, where \(B\) is the bandwidth and \(M\) is the number of signal levels.

In our example, the calculated channel capacity was 24 million bps, meaning the channel can transmit 24 million bits of information every second, assuming a noiseless condition and four-level signaling.

Noiseless Channel

A noiseless channel is a theoretical model where no external noise interferes with the transmission, meaning data can be sent without any errors. It's an ideal scenario that allows us to compute an upper bound on channel capacity.

• Noiseless channels provide us with "perfect" communication, which is often not the case in real-world scenarios.
• The concept helps in understanding the theoretical maximum data rate that a channel can achieve.
• The formula \( C = 2B \log_2 M \) is derived under the assumption of a noiseless channel, highlighting its application in calculating maximum channel capacity.

While real-life channels often experience noise, and thus lower effective bandwidth, assessing the potential of a noiseless channel gives insight into possible maximum efficiencies in perfect conditions.