Chapter 2: Problem Solutions
                   Discrete Time Processing of Continuous Time Signals
            Sampling
            à Problem 2.1.
            Problem:
             Consider a sinusoidal signal     xt  3cos1000t0.1
            and let us sample it at a frequency F  2kHz.
                                                 s
            a)  Determine and expression for the sampled sequence xn  xnT  and determine its Discrete 
            Time Fourier Transform X  DTFTxn;                              s
            b) Determine XF  FTxt;
            c) Recompute X from the XF and verify that you obtain the same expression as in a).
            Solution:
            a) xn  xt            3cos0.5n0.1. Equivalently, using complex exponentials,
                                 tnT
                                     s
                                                   j0.1  j0.5n         j0.1 j0.5n
                                    xn  1.5e          e        1.5e          e
            Therefore its DTFT becomes
                                                          j0.1                   j0.1         
                         X  DTFTxn  3e              3e             
                                                                          2                           2
            with 
                            j2F t
            b) Since FTe      0   FF  then
                                                 0
            2                                                                                      Solutions_Chapter2[1].nb
                                                j0.1                         j0.1
             for all F.         XF  1.5e          F  5001.5e              F  500
             c) Recall that X  DTFTxn and XF  FTxt are related as
                                                            
                                             X  F XFkF 
                                                         s                 s    FF 2
                                                          k                      s
             with F  the sampling frequency. In this case there is no aliasing, since all frequencies are contained 
                   s
             within F 2  1kHz. Therefore, in the interval  we can write
                      s                            X  F XF
                                                               s          FF 2
                                                                              s
             with F  2000Hz.  Substitute for XF from part b) to obtain
                   s
                                          j0.1                                 j0.1             
                 X  20001.5e             2000   500  1.5e            2000   500
                                                             2                                       2
             Now recall the property of the "delta" function: for any constant a  0,
                                                                    1       t
                                                       at  
             Therefore we can write                                a      a
                                                  j0.1                    j0.1          
                                   X  3e          3e              
                                                                   2                            2
             same as in b).
             à Problem 2.2.
             Problem
              Repeat Problem 1 when the continuous time signal is
             Solution                               xt  3cos3000t
             Following the same steps:
             a) xn  3cos1.5n. Notice that now we have aliasing, since 
                                 F
                                  s
             F 1500Hz   1000Hz. Therefore, as shown in the figure below, there is an aliasing at 
              0                  2
             F F 20001500Hz500Hz. Therefore after sampling we have the same signal as in 
              s     0
             Problem 1.1, and everything follows.
                       Solutions_Chapter2[1].nb                                                                                                                                                                                     3
                                                                                                                   X(F)
                                                                               1.5                                               1.5                   F(kHz)
                                                                                                                       F X(FkF)
                                                                                                                          s                                   s
                                                                                                                               k
                                                                                  1.5              0.5 0.5                         1.5                    F(kHz)
                                                                                 F                                           F
                                                                             s  1.0                                         s   1.0
                                                                                  2                                           2
                        à Problem 2.3. 
                        Problem
                        For each XF  FTxt shown, determine X  DTFTxn, where 
                        xn  xnT  is the sampled sequence. The Sampling frequency F  is given for each case.
                                                      s                                                                                                             s
                        a) XF F1000,   F  3000Hz;
                                                                                      s
                        b) XF F500F500,  F  1200Hz
                                                                                                                   s
                                                                         F
                        c) XF  3rect,   F  2000Hz;
                                                                      1000                 s
                                                                         F
                        d) XF  3rect,  F  1000Hz;
                                                                      1000                s
                                                                 F3000                                  F3000
                        e) XF  rect  rect,   F                                                   3000Hz;
                                                                   1000                                    1000                   s
                        Solution
                        For all these problems use the relation
                                                                                      X  F   X F 2kF 
                                                                                                            s                         s                          s
                                                                                                               k
                                                                                   3000                                                                                     2
                        a)  X  3000   1000  k3000  2    k2;
                                                              k                     2                                                           k                         3
            4                                                                                 Solutions_Chapter2[1].nb
                                        1200                             1200
            b) X  1200   500  k1200    500  k1200
                               k        2                              2
                            
                     2   k2 k2
                           k          0                    0
               2500 1200 1.2;
              0
                                              20002      2000                        k2
            c) X  20003rect  k   6000rect shown 
                                  k            1000         1000            k            
            below.
                                                             X()
                         2                                                  2         
                                                   2            2
                                              10002      1000                        k2
            d) X  10003rect  k   3000rect shown below
                                  k            1000         1000            k           2
                                                                  X()
                            2                                                       2           
                                          300023000       3000              300023000      3000
            e) X  3000rect  k   rect  k  
                               k              1000            1000                  1000            1000
                                         3                        3
                     3000rect 33krect 33k
                             k          2                        2
                                          k2
                       6000rect
            shown below.      k          23
                                                                     X()
                                                      6000
                          2                           2        2                        2            
                                                              6       6