关于FM信号解调描述正确的是
A.宽带FM采用非相干解调
B.窄带FM只能采用相干解调
C.窄带FM可采用相干解调,也可采用非相干解调
D.宽带FM可采用相干解调
- · 有3位网友选择 D,占比30%
- · 有3位网友选择 B,占比30%
- · 有2位网友选择 C,占比20%
- · 有2位网友选择 A,占比20%
A.宽带FM采用非相干解调
B.窄带FM只能采用相干解调
C.窄带FM可采用相干解调,也可采用非相干解调
D.宽带FM可采用相干解调
A、T=300M, W=30M
B、T=3000M, W=1000M
C、T=3000M, W=300M
D、T=1000M, W=300M
A、1.25
B、2.5
C、5
D、10
A、f=8000, P=25, R=8
B、f=16000, P=25/2, R=7
C、f=8000, P=25/3, R=9
D、f=16000, P=25/3, R=8
A、L=60, R=5, T=0.000833
B、L=30, R=5, T=0.000833
C、L=60, R=5, T=0.001667
D、L=60, R=6, T=0.000833
A compact disk (CD) is used to store music. Suppose that both the two independent channels of the true stereo music with the highest frequency 22.05 kHz are sampled at the Nyquist sampling rate, then the sampling rate for each channel is f (kHz). The encoded PCM is to have an average SNR of at least 96 dB. Then, the minimum number of the uniform quantization of the sampled data should be R bits. If the Beethoven's Symphony No. 9 with 74 minutes in PCM data can be stored in CD, the minimum storage capacity of the CD should be T (MB, Megabyte). (Hint:).
A、f=44.1, R=16, T=783
B、f=22.05, R=16, T=392
C、f=44.1, R=15, T=700
D、f=44.1, R=16, T=392
A、T=111.11
B、T=55.56
C、T=27.78
D、T=13.89
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