**Unformatted text preview: **bandwidth needed for practical audio is 44 , 100 × 16 / 2 = 352 , 800 Hz. 3. (a) Since g 1 ( t ) and g 2 ( t ) have inﬁnite bandwidths, the bandwidths of y 1 ( t ) and y 2 ( t ) are 10 kHz and 15 kHz, respectively. Since Y ( f ) = Y 1 ( f ) * Y 2 ( f ), the bandwidth of y ( t ) is 25 kHz and the Nyquist rate is 50 kHz (50,000 samples/sec). (b) (i) The smallest setting of the local oscillator is 9 . 2 + . 455 = 9 . 655 MHz and the largest setting is 9 . 9 + . 455 = 10 . 355 MHz. (ii) Since the range of the broadcast band is 700 kHz and 2 f IF = 910 kHz, it is not possible for a desired station and its image station to be in the same band....

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- Spring '09
- Yee
- Frequency, Hertz, Signal Processing, 47%, 30k, 110K