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HW_Set_11_Solutions_v2 - Introduction to Signal Processing...

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Introduction to Signal Processing Copyright © by Hayder Radha - 1 - ECE 366 Homework Set #11 Due Friday, April 24, 2009 (In class) ECE 366 – Introduction to Signal Processing Spring 2009 Michigan State University Department of Electrical and Computer Engineering Please remember to follow the rules and policies outlined in the ECE 366 Syllabus [1] Problem 8.1-1 The highest frequency for ( ) 1 x t is: 5 1 10 B = Hz. Therefore, the Nyquist rate is: 1 5 1 2 2 10 s f B = = × samples/second. The highest frequency for ( ) 1 x t is: 5 2 1.5 10 B = × Hz. Therefore, the Nyquist rate is: 2 5 2 2 3 10 s f B = = × samples/second. For ( ) 2 1 x t , this is equivalent to multiplying ( ) 1 x t by itself. Since: (1) Multiplication in the time domain is equivalent to convolution in the frequency domain. (2) The width of the convolution of two functions is the sum of the widths, Then the highest bandwidth of ( ) 2 1 x t is: ( ) 2 1 5 2 10 x t B = × Hz. Therefore, the Nyquist rate is: ( ) ( ) 2 2 1 1 5 2 4 10 x t x t f B = = × samples/second. Similarly, for ( ) 3 2 x t : The highest bandwidth of ( ) 3 2 x t is: ( ) 2 2 5 5 5 5 1.5 10 1.5 10 1.5 10 4.5 10 x t B = × + × + × = × Hz. Therefore, the Nyquist rate is: ( ) ( ) 2 2 2 2 5 2 9 10 x t x t f B = = × samples/second.
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Introduction to Signal Processing Copyright © by Hayder Radha - 2 - ECE 366 For ( ) ( ) 1 2 x t x t : The highest bandwidth is: 5 5 5 1 2 10 1.5 10 2.5 10 B B + = + × = × Hz. Therefore, the Nyquist rate is: ( ) ( ) 1 2 5 5 2 2.5 10 5 10 x t x t f = × × = × samples/second.
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Introduction to Signal Processing Copyright © by Hayder Radha - 3 - ECE 366 [2] Problem 8.1-5 As can be seen from the following two figures, The minimum sampling frequency required is 22 kHz.
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