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A researcher wishes to digitally record analog sounds for testing animal hearing with frequencies of up to $100[kHz]$.

  1. What is the minimum sampling rate required to process these sounds? [Samples taken per Hertz will be $2$]
  2. If a $16 \text{ bit}$ (per sample) PCM (A/D) converter is used, what is the data rate of the resulting digital signal?
  3. Use Shannon's formula to find the minimum signal to noise ratio (in dB!) required to sustain the given data rate over a $500 \text{kHz}$ radio channel.
asked in Computer Networks by Active (3.6k points)
edited by | 72 views
This question needs additional information
Channel capacity isn't mentioned
However, this is a very popular question of information theory & coding
I'll add the full question
It is a complete ques.Channel capacity can be related to frequency.

1 Answer

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A.  Sampling rates are the number of samples per second in a sound.

      Here, Sample/Hz = $2$

      Frequency = $100 \text{ kHz}$

      ∴ $\color{maroon}{\text{Minimum sampling rate}}$ = $2 \times 100,000$ samples/sec

$\qquad \qquad \qquad \qquad = \color{purple}{200,000 \text{ samples/sec }}$

B. Now, we'll determine the maximum data rate or channel capacity

 Channel capacity: The maximum rate at which data can be transmitted over a given channel. Usually measured in bps (bits per sec)

∴ $\color{lightblue}{\text{Channel Capacity}}$ = $200,000 \text{ samples/sec} \times 16 \text{ bit/sample}$

$\qquad \qquad \qquad = 32,00,000 \text{ bps}$

$\qquad \qquad \qquad = \color{lightgreen}{3.2 \text{ Mbps}}$  $\left [ \text{1 Mbps = 1000000 bit Per second} \right]$

C. According to Shannon's Theorem

 $C = B * log_2 (1 + \dfrac{S}{N})$

Where $B$= Bandwidth

$\dfrac{S}{N}$ = signal to noise ratio, expressed in dB

$C$ = maximum channel capacity

∴ $\dfrac{C}{B} = log_2(1 + \dfrac{S}{N})$

 Or, $1+\dfrac{S}{N} = 2^{C/B}  $

Or, $\dfrac{S}{N} = 2^{C/B} -1 $

Here, Channel Capacity or C will be $3.2 \text{ Mbps}$

Bandwidth or B will be $500 kHz$

∴ $\dfrac{S}{N} = 2^{C/B} -1 $

Or, $\dfrac{S}{N} = 2^{3200/500} -1$              $\left [  \text{3.2 Mbps = 3200 Kilobit per sec } \right ]$

Or, $\dfrac{S}{N} = 2^{6.4} -1$

Or, $\dfrac{S}{N} = 84.4485063 -1$

Or, $\dfrac{S}{N} = 83.4485063$

Now, $\dfrac{S}{N} \text{ in dB} = 10 \times log_{10}(\dfrac{S}{N})$

Or, $\dfrac{S}{N} \text{ in dB} = 10 \times log_{10} \left ( 83.4485063 \right)$

Or, $\dfrac{S}{N} \text{ in dB} = 10 \times (1.9213743)$

Or, $\dfrac{S}{N} \text{ in dB} = 19.213743 \text{ dB}$

∴ $\color{green}{\text{Minimum signal to noise ratio (in dB) will be}} \color{Gold}{\text{19.21 dB}}$ $\color{green}{\text{ is required to sustain the given data rate}} \color{green}{\text{ over a 500kHz radio channel. }}$
answered by Boss (11.5k points)
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thanks a lot

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