Quantization and Noise 1The bandwidth of a music system has

Quantization and Noise

1.The bandwidth of a music system has been measured and found to be 20 kHz. What is the Nyquist frequency?

2.What is the data rate of a sampled analogue signal which has a bandwidth of 50 kHz and uses 8 bit quantization?

3.An analogue signal has a 15 kHz bandwidth, and is quantized into 32 levels. What is the data rate of the system?

4.What is the bandwidth of the analogue signal that has been sampled quantized into 65,536 levels and has a data rate of 100 kbps

5.A sharp spike of energy of 0.01 s duration appears in a line transmitting digital data with the rate of 2400 bps. How many ASCII characters could be affected?

6.A telephone line has an input power of 120 mW and an output power of 45 mW. What is the loss of the line in dB?

7.Calculate the maximum capacity C, in bps for a transmission channel with a bandwidth of 10 kHz and a signal to noise ratio of 1000

8.Calculate the maximum capacity C, in bps for a transmission channel with a bandwidth of 500 Hz and a signal to noise ratio of 10 dB

9.Data is to be transmitted over a LAN at 100 Mbps. Measurements on the system show that the S/N ratio is 90 dB. Calculate the Bandwidth required to transmit the data at the required rate.

10.The maximum through put on a digital radio channel of 5 kHz bandwidth is found to be 7 kbps. Calculate the signal to noise ratio in dB.

Solution

Question 1:

The highest frequency of the channel determines the bandwidth of the channel.

Here, the bandwidth of maximum frequency of the channel is 20kHz.

Therefore the Nyquist frequency = maximum frequency of the channel * 2 = 20 * 2 kHz = 40kHz.

Question 2:

The bandwidth of the channel is 50kHz.

Therefore the Nyquist frequency = 2 * 50 kHz = 100kHz.

Here, quantization N = 8

Therefore data rate = N * Nyquist frequency = 8 * 100kHz = 800 kbps.

Question - 3:

Here, quantization level Q = 32

As we know that Q = 2^N = 32

Therefore, N = 5 bit.

The bandwidth of the signal = 15 kHz.

The Nyquist frequency = 2 * bandwidth = 2 * 15 kHz = 30 kHz.

The data rate = N * Nyquist frequency = 5 * 30 kHz = 150 kbps.

Question-4:

Here, the quantization level Q = 65536 = 2^N

Therefore, N = 16 bit

Data rate is 100kbps

Therefore quantization frequency = 100 / 16 kHz

Therefore the bandwidth of the analog signal = 100 / (16 * 2) = 3.125kHz

Question-5:

The data rate is 2400 bps. In other words, 2400 bits of data is transmitted in 1 second.

The sharp spike was generated for 0.01 second.

For 0.01 second number of bits transmitted = 2400 * 0.01 = 2400 /100 = 24.

Therefore, 24 bits are affected. As each ASCII characters are of size 7-bit, therefore number of ASCII characters are affected = 24 / 7 = 3 (nearly)

Question- 6:

Loss of the line is measured by calculating how much energy is decreased during transmission.

Here 120 - 45 mW = 75 mW power is decreased.

75mW = (10 log₁₀ 75 / 145 ) * 1000000 = -2.04 * 1000000 dB.

Therefore the loss of power in dB = 2.04 * 10⁶ dB.

Question-7:

The capacity of a channel is measured by,

Capacity, C = B * log₂ (1 + SNR) where B = bandwidth.

Therefore, C = 10 * 1000 log₂ (1 + 1000) = 10000 log₂ 1001 = 10000 * 9.97 = 99700 bps.

Question-8:

SNR_dB (in decibel) = 10 log₁₀ SNR .

Therefore SNR = 10 ^SNR_dB / ¹⁰

Here, SNR = 10^{10 / 10} = 10

Therefore maximum capacity C = B * log₂ (1 + SNR) = 500 * log₂ 11 = 500 * 3.46 bps = 1730 bps.

Question-9:

Here C = 100Mbps = 100 * 10⁶ bps.

SNR_dB = 90 dB

Therefore SNR = 10 ^SNR_dB / ¹⁰ = 10^{90 / 10} = 10⁹

Therefore the bandwidth = 100 * 10⁶ / log₂ (1 + SNR) = 100 * 10⁶ / log₂ ( 1 + 10⁹) = 10 * 10⁶ /33.22 = 0.301 Mbps.

Question-10:

10.The maximum through put on a digital radio channel of 5 kHz bandwidth is found to be 7 kbps. Calculate the signal to noise ratio in dB.

Here, C = 7 kbps

B = 5 kHz

Therefore SNR = 2^{C / B} – 1 = 2^1.4 – 1 = 2.64 – 1 = 1.64dB.