Question

The discrete-time signal x(n)=6.35cos(π/10)n is quantized with a resolution (a) Δ= 0.1 or (b) Δ=0.02. How many bits are required in the A/D converter in each case?

Answer 1

To quantize a discrete-time signal with resolutions of 0.1 and 0.02, the A/D converter requires 7 bits and 10 bits, respectively.

Step-by-step explanation:

To determine how many bits are required in the A/D converter for quantizing the discrete-time signal x(n) = 6.35cos(π/10)n with two different resolutions, we need to calculate the number of quantization levels first and then find the number of bits needed to represent these levels.

For resolution Δ = 0.1, the peak-to-peak amplitude of the cosine function, which is twice the amplitude (2 * 6.35 = 12.7), must be divided by the resolution to find the number of quantization levels (L):

1. L = (peak-to-peak amplitude) / Δ = 12.7 / 0.1 = 127 levels.
2. The number of bits (b) is found by the equation 2b ≥ L. Solving for b results in b = 7 bits because 27 = 128 which is the smallest power of 2 that is greater than or equal to 127.

For resolution Δ = 0.02, we follow the same steps:

1. L = (peak-to-peak amplitude) / Δ = 12.7 / 0.02 = 635 levels.
2. Here, b = 10 bits because 210 = 1024 which is the smallest power of 2 that is greater than or equal to 635.

In summary, for resolution Δ = 0.1, the A/D converter requires 7 bits, and for resolution Δ = 0.02, it requires 10 bits.