105年 - 105 國立臺北大學_碩士班一般入學考試試題_通訊工程學系：通訊原理#147252

科目：【阿摩】未分類題庫 | 年份：105年 | 選擇題數：0 | 申論題數：13

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所屬科目：【阿摩】未分類題庫

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申論題 (13)

1. (15%)Please provide your answers with true or false to the following questions.

(a) The advantage of Polar nonreturn-to-zero (NRZ) signaling is that the power spectrum of the signal is lower near zero-frequency.
(b) In the delta modulation, the message is down-sampled due to limited bandwidth.
(c) A dispersive communication channel may introduce the inter-symbol interference.
(d) The raised-cosine filter is more bandwidth efficient than the rectangular filter.
(e) The matched filter is designed to maximize the signal-to-noise ratio.
2. (15%) What is the resulting Signal-to-quantization noise ratio for a signal uniformly distributed on [-1, 1] when uniform PCM with 256 levels is employed?
3. (20%)Please fine the Fourier-transform of \( x(t) \) in terms of \( X(f) \) to the following questions.

(a) \(x(t - t_0)\) , where \(t_0\) is a scalar.
(b) \(x(at)\), where \(a\) is a scalar.
4. (10%) A white Gaussian noise with two-sided power spectral density (PSD) \( N_0 / 2 \) is added to a signal having PSD given by \( S_{x}(f) = f^{4} \) for \( |f| \leq W \) and \( S_{x}(f) = 0 \) otherwise. The sum is filtered by an ideal low pass filter with gain 1 and bandwidth \( B > W \). Determine the signal-to-noise ratio (SNR) at the output. How much do we gain in SNR if we change the bandwidth to \( B = W \)?
5. (15%) Consider a phase modulated signal \( x(t) = A\cos (2\pi f_{c}t + m(t)) \). Provide a demodulator that can correctly recover the message \( m(t) \). Please explain how the demodulator works.
6. Consider a double sideband system with quadrature multiplexing (QM) which simultaneously sends two messages \( m_{1}(t) \) and \( m_{2}(t) \) through \( \cos 2\pi f_{c}t \) and \( \sin 2\pi f_{c}t \), respectively. The transmitted signal is given by

$$x(t) = A[m_1(t) \cos 2\pi f_c t + m_2(t) \sin 2\pi f_c t].$$

(a)(10%) Suppose the phase is perfectly known to the receiver and there is no noise, design a coherent demodulator that can recover both the messages. You must show that the demodulator can correctly demodulate.
(b)(10%) Now, assume that there is a phase error \(\theta\) but still no noise, i.e., the receiver believes the phase is \(2\pi f_{c}t + \theta\). Run your demodulator again and explain what you observe.
(c)(5%) Now, assume that there is no phase error but the signal is corrupted by a narrowband noise \( n_c(t) \cos 2\pi f_c t - n_s(t) \sin 2\pi f_c t \). Run your demodulator again and determine the output.