Continuous and Discrete Time Signals and Systems (Mandal & 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) [] , f o r 0 5 a n d [ 6 ] [] x kk k x k x k =≤ + = . 0 0 22 . 63 K π ππ Ω= = = () [] 00 00000 55 0 2345 11 6 1 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 j n jn Dx k e k e K ee e e e nnnnn n n n n −Ω == − Ω =+ + + + Ω ∑∑ Expressing 0 5 0 1 6 k jn n k Dk e = = and using the series sum formula 12 2 0 (1 ) ) MM M k k rM r M r kr r ++ = −+ + = , we can also represent n D in a compact form as follows: 000 0 0 67 5 2 0 6 5 66 ( 1 ) jn j n j n k jn n jn k e e e = , with 0 . 3 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 . 9 K =
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Chapter 11 2 () ( ) 00 0 0 0 0 0 88 0 2345 23 2 32 11 [] 9 1 1 0.5 0.5 0.5 9 1 10 . 5 1 9 1 . 5 1 9 jn k jn k n kk jn j n j n j n j n jn j n j n jn j n jn j n Dx k e k e K ee e e e e e −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 0e l s e w h e r e . j j n je n Dj e n π −= , 7 nn DD + = The magnitude and phase spectra for 71 2 n −≤ ≤ are shown below.
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Solutions 3 (iv) Because 226 5 5/ 3 rational π Ω == = , [] x k is a periodic function with a fundamental period 26 6 3 5 o Km m =⋅ = = (for m =5). Therefore, 0 0 22 63 K ππ Ω === . The function x k can be expressed in terms of complex exponential functions as follows. 0 0 5 5 1 34 3 4 0 [] 2 2 jn k j K jk n n xk e e e De Ω ⎛⎞ + ⎜⎟ ⎝⎠ = = From the above equation, we can state that 4 25 0 0 4 n j D en n = = ≤≤ and 6 nn DD + = (since n D is periodic with period 6). Note that 0 4 n D n n = = and /4 5 0 4 n D n n = = ) . The amplitude and phase spectrums for 61 1 n are shown below. (v) [ 5 ] m x kk m δ =−∞ =− 0 0 . 5 K Ω= = 00 44 0 11 1 55 jn k jn k n D xke ke K −Ω = ∑∑ The magnitude and phase spectra for 51 0 n −≤ ≤ are shown below.
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Chapter 11 4 (vi) 10 2 2 2 2 2 [ ] cos cos cos 4 cos cos cos 35 3 5 3 5 kk xk k π ππ ⎛⎞ == = ⎜⎟ ⎝⎠ 15 15 15 11 6 14 cos cos 21 5 2 1 5 period period period = =+ ±²³²´ ±²²²²²³²²²²²´ 0 0 22 . 15 K Ω= = () 00 0 77 7 7 6 1 1 4 [ ] cos cos cos 2 cos 2 1 52 1 1 1 5 4 1 1 cos cos cos 7 cos 2 5 2 1 5 2 2 1 4 jk jk jk jn k n n k ee De Ω− Ω Ω Ω =− = + = Ω + Ω ⎡⎤ + + ⎣⎦ = where 1/4 2, 7 0 n D n otherwise = =± ± The magnitude and phase spectra for 81 4 n −≤ ≤ are shown below.
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Solutions 5 (vii) () [ ] cos 2 3 and [ +3]= [ ] 10 0.5 1,2 x kk x k x k k k π == = = . [] x k is a periodic function with fundamental period of 3. 0 0
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Continuous and Discrete Time Signals and Systems (Mandal & Asif) solutions - chap11

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