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Continuous and Discrete Time Signals and Systems (Mandal &amp; Asif) solutions - chap11

# Continuous and Discrete Time Signals and Systems (Mandal & Asif) solutions - chap11

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Chapter 11: Discrete-Time Fourier Series and Transform Problem 11.1: (i) [ ] , for 0 5 and [ 6] [ ] x k k k x k x k = + = . 0 0 2 2 . 6 3 K π π π Ω = = = ( ) [ ] [ ] 0 0 0 0 0 0 0 5 5 0 0 0 2 3 4 5 0 0 0 0 0 0 0 0 0 0 1 1 [ ] 6 1 2 3 4 5 6 1 cos( ) 2cos(2 ) 3cos(3 ) 4cos(4 ) 5cos(5 ) 6 1 sin( ) 2sin(2 ) 3sin(3 ) 4sin(4 ) 5sin(5 ) 6 jn k jn k n k k jn j n j n j n j n D x k e ke K e e e e e n n n n n j n n n n n Ω Ω = = Ω Ω Ω Ω Ω = = = + + + + = Ω + Ω + Ω + Ω + Ω Ω + Ω + Ω + Ω + Ω Expressing ( ) 0 5 0 1 6 k jn n k D k e Ω = = and using the series sum formula 1 2 2 0 ( 1) (1 ) M M M k k r M r Mr kr r + + = + + = , we can also represent n D in a compact form as follows: ( ) 0 0 0 0 0 6 7 5 2 0 1 1 6 5 6 6 (1 ) jn j n j n k jn n jn k e e e D k e e Ω Ω Ω Ω Ω = + = = , with 0 . 3 π Ω = The magnitude and phase spectra for 10 10 n are shown below. (ii) ( ) ( ) ( ) ] [ ] 9 [ and 8 6 0 5 3 5 . 0 2 0 1 ] [ k x k x k k k k x = + = 0 0 2 2 . 9 K π π Ω = =

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Chapter 11 2 ( ) ( ) ( )( ) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 8 8 0 0 0 2 3 4 5 2 3 2 3 2 1 1 [ ] 9 1 1 0.5 0.5 0.5 9 1 1 0.5 1 9 1 1 0.5 1 9 jn k jn k n k k jn j n j n j n j n jn j n j n jn j n j n jn j n D x k e ke K e e e e e e e e e e e e e Ω Ω = = Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω = = = + + + + + = + + + + + = + + + The magnitude and phase spectra for 10 10 n are shown below. (iii) As solved in Example 11.5, the DTFS coefficients are obtained as (see Eq. 11.19): 4 4 3 2 3 2 for 1 for 1 0 elsewhere. j j n j e n D j e n π π = = = − , 7 n n D D + = The magnitude and phase spectra for 7 12 n are shown below.