Quiz3_solution - Solutions to EE3008 Quiz 3 Problems...

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Solutions to EE3008 Quiz 3 Problems Problem 1: (a) Since X ( t ) is a white process, the power spectrum of X ( t ) is given by G X ( f ) = N 0 2 . The transfer function of the LTI system is obtained as H ( f ) = 1 τ , | f | ≤ τ 2 0 , otherwise. Since X ( t ) is a WSS process, Y ( t ) is also a WSS process with power spectrum given by G Y ( f ) = G X ( f ) | H ( f ) | 2 = N 0 2 τ 2 , | f | ≤ τ 2 0 , otherwise. As a result, Y ( t ) is not a white process because its power spectrum G Y ( f ) is not a constant for all f . The power of Y ( t ) is P Y = Z -∞ G X ( f ) df = Z τ/ 2 - τ/ 2 N 0 2 τ 2 df = N 0 2 τ . (b) Since Y ( t 0 ) ∼ N (0 , σ 2 0 ), we have Pr { Y ( t 0 ) > σ 0 } = Q σ 0 - 0 σ 0 = Q (1) = 0 . 15866 . 1
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Problem 2: (a) The Nyquist sampling rate is f S = 2 B s = 8 kHz. If each sample is uniformly quantized using a b -bit midriser, the bit rate is given by R b = b · f S = 8 b kbps. The required channel bandwidth of 4-ary PAM with 90% in-band power is 0 . 5 R b , which is bounded by 16 kHz. Therefore, we have R b 32 kbps 8 b 32 b 4. That is, the maximum b is 4.
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  • Spring '16
  • Signal Processing, TUtorial And Note, Bit rate, BPSK, BPSK signal, 4-ary PAM signal

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