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Continuous and Discrete Time Signals and Systems (Mandal &amp; Asif) solutions - chap09

# Continuous and Discrete Time Signals and Systems (Mandal & Asif) solutions - chap09

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Chapter 9: Sampling and Quantization Problem 9.1 (a) ( ) 1 1 200 200 ( ) 5sinc 200 sinc 40 t x t t π π π π × = = . Therefore ( ) 1 400 1/ 40 200 1 ( )= rect 0 200 . 40 X ω π ω π ω ω π = > The maximum frequency is given by 100 Hz. Based on the Nyquist theorem, the maximum sampling period is 1 5 200 s T s ms = = . (b) ( ) ( ) 2 max 100 max 50 max 100 ( ) 5sinc 200 8sin 100 freq Hz freq Hz freq Hz x t t t π = = = = + ±²³²´ ±²²³²²´ ±²²²²³²²²²´ . Therefore, maximum sampling period is given by 1 5 200 s T s ms = = . Fig. S9.1: A ( f ) denotes the spectrum of sinc(200 t ) and B ( f ) denotes the spectrum of sin(100 π t ). X 3 ( f ) is the sum of two shifted replicas of A ( f ), and is non-zero within the band [ 150,150] Hz. (c) Since x 3 ( t ) is a product of sinc(200 t ) and sin(100 π t ), the spectrum of x 3 ( t ) can be obtained by convolving the spectrums of sinc(200 t ) and sin(100 π t ). From the theory of CT convolution, it can easily be seen that (see Fig. S9.1), the maximum frequency present in x 3 ( t ) is 150 Hz. Therefore, the maximum sampling period is given by

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