1. Nyquist Theorem or theorem of sampling says that if a signal to be imitate digitally correct than the sampling size should be 2 times the original signal.

2. Given an original signal with frequency in between 20Hz to 20 000Hz, the suitable sampling rate used to digitize the signal based on the theory of sampling is 40Hz to 40 000Hz.

3. Y is sampled at 15 000Hz

a) Play in between 0.2 to 2.0 seconds

wavplay(Y(0.2 x 15000 : 2.0 x 15 000), 15000)

wavplay(Y(3000 : 30000), 15000)

b) Sample in between 0.2 to 2.0

-> (2.0 – 0.2) x 15000 = 1.8 x 15000), 15000)

-> (2.0 x 15000) – (0.2 x 15000) = 30000 – 3000 = 27000

c) Playback effect

-> If the value of playback is lower than the signal for example 8000Hz the sound will be slower or lower pitch

-> If the value of playback is higher than the original signal for example 20 000Hz than the sound play will be faster of higher pitch

4. v = [9700 19000 23000 31700];

u = [9900 11000 20000 23200 33700 36100 3800];

-> In time (divide by 16000)

v = [0.6062 1.1875 1.4375 1.9813]

u = [0.6188 0.6875 1.2500 1.4500 2.1063 2.2563 2.3750]

So from the question, these information can be directly extracted

p = 4; // number of reference points

n = 7; // number of auto segmentation points

Calculate match at 0.05 by calculating the difference between reference points and auto segmentation points. The easiest way is by developing the difference table between auto and reference points.

0.61875 | 0.6875 | 1.25 | 1.45 | 2.10625 | 2.25625 | 2.375 | |

0.60625 | 0.0125 | 0.08125 | 0.64375 | 0.84375 | 1.5 | 1.65 | 1.76875 |

1.1875 | 0.56875 | 0.5 | 0.0625 | 0.2625 | 0.91875 | 1.06875 | 1.1875 |

1.4375 | 0.81875 | 0.75 | 0.1875 | 0.0125 | 0.66875 | 0.81875 | 0.9375 |

1.98125 | 1.3625 | 1.29375 | 0.73125 | 0.53125 | 0.125 | 0.275 | 0.39375 |

m = 2

i = n – m = 7 – 2 = 5

o = p – m = 4 – 2 = 2

p(m) = m/p x 100 = 2/4 x 100 = 50%

p(i) = i/n x 100 = 5/7 x 100 = 71.4%

p(o) = o/p x 100 = 2/4 x 100 = 50%

5. The information can be extracted from the question. Total signal. S = 8000 x 2.0 s = 16000 samples

Size of processing windows, N = 8000 x 40ms = 320 samples

Size of overlapped windows, m = 8000 x 20ms = 160 samples

Number of auto correction p=12

Number of N in the signal can be calculated based on this formula

nN = ((S –m) / N) x 2

nN = ((16000 – 160)/ 320) x 2

nN = 15840/320 x 2

nN = 49.2 x 2

nN = 99

So, the number of features vector is N x p = 99 x 12 = 1188

The compression ratio is original signal = 16000 sample value

Compressed signal = 1188 sample value

So, compression ratio (16000 – 1188) / 16000 x 100 = 92.6%

P/s: Good Luck for final exam! Please share more notes and tips hehe... =)

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