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Homework_8_Solutions - Homework 8 Solutions 8.1-2(b 0.01J...

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Homework 8 Solutions 8.1-2: (b) 0.01ݏ݅݊ܿ ሺ100ߨݐሻ֞ 0.0001∆ሺ ସ଴଴గ The bandwidth of this signal is 200π rad/s or 100 Hz. Therefore the Nyquist rate is 200 Hz (samples/sec). (c) ݏ݅݊ܿሺ100ߨݐሻ+3ݏ݅݊ܿ ሺ60ߨݐሻ֞0.01ݎ݁ܿݐቀ ଶ଴଴గ ቁ+ ଶ଴ ∆ሺ ସ଴଴గ The bandwidth of ݎ݁ܿݐሺ ଶ଴଴గ is 50 Hz and that of ∆ሺ ଶସ଴గ is 60 Hz. The bandwidth of the sum is the higher of the two, that is, 60 Hz. The Nyquist sampling rate is 120 Hz. 8.1-7 (a) ܺሺݓሻ= ∆ቀ ଶ଴గ ቁ+ߨ[ߜሺݓ+20ߨሻ+ߜሺݓ−20ߨሻ] The bandwidth is 10 Hz. There is an impulse at 10 Hz, as seen from X(w). The Nyquist rate is 20 Hz. Hence, f s = 10 Hz will not permit reconstruction of x(t). This is verified from the sampled signal spectrum, shown as a function of f in Hz. (b) The Nyquist rate is 20 Hz. Hence the sampling rate f s = 20 Hz is adequate despite the fact that x(t) contains an impulse at the highest frequency 10 Hz. This is because the impulse component is cos(20πt). To reconstruct x(t) from the spectrum shown below, we need an ideal LPF of cutoff frequency 10 Hz and gain T = 1/20. Because the rect function value is 0.5 at the edge (cutoff), the LPF gain at the cutoff frequency 10 Hz is 0.5 × ଶ଴ = ସ଴ . Hence for the input of an impulse of strength 40π at ±10 Hz, the output will be an impulse of strength 50π at ݂= ±10 Hz. Hence, the filter output is the spectrum: ܺሺݓሻ= ∆ቀ ଶ଴గ ቁ+ ߨ[ߜሺݓ+20ߨሻ+ߜሺݓ−20ߨሻ] and
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ݔሺݐሻ= 5ݏ݅݊ܿ
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