Your Federal Quarterly Tax Payments are due April 15th Get Help Now >>

TUTORIAL QAF OR DFS DFT by 2m18Zj

VIEWS: 0 PAGES: 2

									 TUTORIAL QUESTIONS FOR DFS & DFT                                 DIGITAL SIGNAL PROCESSING


1. Find the DFS expansion of the sequence
                                        ~(n)  A cos(n )
                                        x
                                                      2


2. Compute the N-point DFT of each of the following sequences:
   (a) x1 (n)   (n)

   (b) x 2 (n)   (n  n0 ),         where 0  n0  N

   (c) x3 (n)   n             0n N

 3. Consider the sequence
                                x(n)   (n)  2 (n  2)   (n  3)


    (a) Find the four-point DFT, X (k ) , of          x(n) .

    (b) Find the finite-length sequence q(n) that has a four-point DFT

                                           Q(k )  W42 k X (k )

    (c) If    y(n) is the four-point circular convolution of x(n) with itself, find

         y(n) and the four-point DFT Y (k ) .

    (d) With h(n)   (n)   (n  1)  2 (n  3) , find the four-point circular

        convolution of x(n) with h(n) .
TUTORIAL QUESTIONS FOR DFS & DFT                                DIGITAL SIGNAL PROCESSING



SOLUTION:
                 3
  (a) X (k )   x(n)W4nk  1  2W42 k  W43k
                n 0




  (b) The sequence q(n) is formed by multiplying the DFT of x(n) by the complex
      exponential W42 k . Because this corresponds to a circular shift of x(n) by 2,

                                         q(n)  x(( n  2)) 4

   it follows that
                                 q(n)   (n  2)  2 (n)   (n  1)



  (c) Y (k )  X (k ) 2  (1  2W42 k  W43k )(1  2W42 k  W43k )

                        1  4W42 k  2W43k  4W44 k  4W45k  W46 k
                        5  4W4k  5W42 k  2W43k
      Therefore,
                         y(n)  5 (n)  4 (n  1)  5 (n  2)  2 (n  3)


  (d) With h(n)   (n)   (n  1)  2 (n  3) , the four-point circular convolution of

      x(n) with h(n) may be found using the tabular method. Because, the linear
      convolution of x(n) with h(n) is
                               y(n)  x(n)  h(n)  [1, 1, 2, 5, 1, 4, 2]

   then

              n                     0      1      2      3         4        5   6    7   8
            y(n)                    1      1      2      5         1        4   2    0   0
           y(n+4)                   1      4      2      0         0        0   0    0   0
            z(n)                    2      5      4      5        -         -    -   -       -


      or     z (n)  2 (n)  5 (n  1)  4 (n  2)  5 (n  3)

								
To top