EE299 HW1 Solution -- Winter 2008
1.
(a)
(i)
For a 4-bit word, we will have
0 = 0000
1 = 0001
2 = 0010
3 = 0011
….
….
14 = 1110
15 = 1111
Now we want to represent the range form 0 to 120 (or say [0,120])by 4-bit word. That means we have to map [0, 15] to [0,120].
That means:
0 = 0000 è 0
15 = 1111 è 120 = 15 X 8
From above, it hints every number should multiply by 8.
0 = 0000 è 0
1 = 0001 è 8
2 = 0010 è 16
3 = 0011 è 24
….
….
14 = 1110 è 112
15 = 1111 è 120
So the precision (the amount of minimum unit) of representing [0,120] by 4-bit word is 8 if you use a uniform mapping as we have done here.
(ii)
If you drive on a country road, your speed limitation should be low (like 20 miles, but we only have “16”, “24” here) and the error is roughly 20%. This representation method will not provide you enough precision. The percentage is lower at higher speeds, but still much worse than speedometer errors in most cars.
(b)
Here we use sign-magnitude form for representing negative and positive.
ð For negative or positive: 1 bit
For number 100:
è We need 7 bits.
For precision:
è It costs 4 bits.
So it is 12 bits in total.
2.
Some examples:
· Digital music players (e.g. MP3) vs. analog record players: Some people believe they can here distortion due to digitization and sampling even for CD quality, though digital music is much more compact and less likely to be damaged.
· Digital vs. analog watch: some people like the style of analog, some people think digital is easier to read, others think analog gives you a better sense of time
· Digital vs. analog cameras: Analog cameras have more resolution, so you can blow pictures up to a bigger size, though this is becoming less of an issue with new digital cameras that have more memory and more pixels per image. Digital cameras avoid the costs of developing film.
3.
Sound / Spectrogram1 / C
2 / A
3 / B
Spectrogram A illustrates a “chirp” which is essentially a sinusoid with time-varying frequency that is increasing.
Spectrogram B illustrates a sequence of tones, as evidenced by the horizontal lines.
Spectrogram C corresponds to speech with voiced sounds, since the harmonics are visible through out. The smoothly changing fundamental frequency is characteristic of speech.
4.
Storage space:
(a)
(8 × samples/second) × (12 bits/sample) × (30 seconds)
= 2.88 × bits
= 2.88 Mbits
(b)
(16 × samples/second) × (8 bits/sample) × (30 seconds)
= 3.84 × bits
= 3.84 Mbits
(c)
(44.1 × samples/second) × (16 bits/sample) × (30 seconds)
= 21.168 × bits
= 21.168 Mbits
5.
45 rpm = 45/60 rps = 0.75 rps (revolutions per second)
The sampling rate is much higher than 0.75.
So no aliasing occurs.