Question
If 2 K bits/second is bit rate, what is the minimum PCM
bandwidth required for successful transmission?ÂSolution
In Pulse Code Modulation (PCM), the minimum bandwidth required is equal to half of the bit rate (Nyquist criterion). Bit rate = 2 Kbits/s
 Minimum bandwidth = Bit rate / 2 = 2000 / 2 = 1000 Hz = 1 kHz
1, 27, ?, 343, 729, 1331
29    51    95    183    ?     711
5 14 56 220 1125 6786
...55 46 60 41 65 36 ?
20 25 54 165 662 ?
...140Â Â Â Â 300Â Â Â Â 380Â Â Â Â Â 420Â Â Â Â Â 440Â Â Â Â Â ?
...If 4 10 16 26 x 50
Then find the value of (x² - 1) + ( x² + 1).
...720    ?     240     180      144     120
...If  2,  7, 20,  x ,  110,  235,
Then,  7 x + 7√x = ?
6000Â Â Â Â Â Â 3191Â Â Â Â Â Â 5216Â Â Â Â Â Â 3847Â Â Â Â Â Â 4688Â Â Â Â Â Â 4247
6125       a             b  �...
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