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

(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?

(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.

(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.