, calculate the maximum error value and the number of bits to be transmitted per second

Question

Assume that you have an analog signal with a maximum and minimum amplitude of +5 and -5 and a bandwidth of 1000 Hz. This signal is desired to be sampled in practice. If quantization level 256 is taken, calculate the maximum error value and the number of bits to be transmitted per second.

Answer 1

The quantization step size, Δcap delta

Δ

, is calculated by dividing the peak-to-peak amplitude by the number of quantization levels: Δ=AppL=10256≈0.0390625cap delta equals the fraction with numerator cap A sub p p end-sub and denominator cap L end-fraction equals 10 over 256 end-fraction is approximately equal to 0.0390625

Δ=𝐴𝑝𝑝𝐿=10256≈0.0390625

.The number of bits per sample, nn

𝑛

, is found by taking the base-

22

2

logarithm of the number of quantization levels: n=log2(L)=log2(256)=8 bits/samplen equals log base 2 of open paren cap L close paren equals log base 2 of 256 equals 8 bits/sample

𝑛=log2(𝐿)=log2(256)=8 bits/sample

.

The Nyquist sampling rate, fsf sub s

𝑓𝑠

, is determined by multiplying the bandwidth by 22

2

: fs=2×Bandwidth=2×1000 Hz=2000 samples/secondf sub s equals 2 cross Bandwidth equals 2 cross 1000 Hz equals 2000 samples/second

𝑓𝑠=2×Bandwidth=2×1000 Hz=2000 samples/second

.

The number of bits to be transmitted per second, Rbcap R sub b

𝑅𝑏

, is calculated by multiplying the number of bits per sample by the sampling rate: Rb=n×fs=8 bits/sample×2000 samples/second=16000 bits/secondcap R sub b equals n cross f sub s equals 8 bits/sample cross 2000 samples/second equals 16000 bits/second

𝑅𝑏=𝑛×𝑓𝑠=8 bits/sample×2000 samples/second=16000 bits/second

.