EEL3135: Homework #1 Solutions
                  Problem 1:
                       (a) Explain the difference between a continuous-time and discrete-time signal.
                           A continuous-time signal xt() is a continuous functions of time t , a real-valued variable, while a discrete-
                           time signal xn[] is a function of time index n , an integer-valued variable; that is, a discrete-time signal is 
                           only deÞned for a discrete number of points. Oftentimes, discrete-time signals are sampled versions of 
                           continuous-time signals.
                       (b) Give at least one reason why the study of each type of signal (continuous-time and discrete-time) is 
                           important.
                           Continuous-time signals exist in the real-world, and any discrete-time processing system (e.g. computer) 
                           must interface with continuous-time signals both at the input and the output. Discrete-time signals have 
                           grown in importance with the increased use of computers as digital signal processing (DSP) systems. As 
                           such, both types of signals are important in understanding a complete signal processing system.
                  Problem 2:
                       (a) Sketch the magnitude spectrum (frequency-domain representation) for the following continuous-time signal:
                                xt()= 14+ cos()30t + 36cos()0πt                                                                     (1)
                           Be sure to label your plot. [Note: It is not important that you indicate absolute magnitude on your plot, only 
                           the relative magnitude of different frequency components.]
                           The frequencies in xt() occur at ±30 ⁄ 2π = ±15 ⁄ πHz and ±π60     ⁄ 2π = ±30Hz (cycles/second).
                                                                                    Xf()                       problem 2(a) solution
                                                                             2          2
                                               32⁄                                                                    32⁄
                                                                               1
                                          Ð30                          Ð15⁄ π 0 15⁄π                                30          f()Hz
                       (b) Sketch the magnitude spectrum for the following continuous-time signal:
                                xt()= cos()4πt cos()20πt                                                                            (2)
                           Hint: The following trigonometric identity may be useful:
                                                  1               1
                                                  ---             ---
                                cos()α cos()β=      cos()αβ+    + cos()αβÐ                                                          (3)
                                                  2               2
                           Using identity (3), we can rewrite equation (2) as:
                                        1                    1
                                xt()= ---cos(20πt + 4πt) + ---cos(20πt Ð 4πt)
                                        2                    2                                                                    (S-1)
                                        1              1
                                        ---            ---
                                     =    cos()24πt + cos()16πt
                                        2              2
                           The frequencies in xt() occur at ±π24   ⁄ 2π = ±12Hz and ±π16 ⁄ 2π = ±8Hz (cycles/second). The 
                           magnitude spectrum for equation (S-1) is plotted on top of page 2.
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                                                                                                 Xf()                           problem 2(b) solution
                                                                                       14⁄
                                                       Ð12            Ð8                                           8           12                  f()Hz
                     Problem 3:
                               For the continuous-time signal xt()= 12+ cos()π ⋅ 50t + cos()2π ⋅ 80t , explain how the signal would be 
                               changed by an ideal Þlter, whose frequency response is plotted below.
                                                                         filter magnitude frequency response
                                                                   2
                                                                1.5
                                                                   1
                                                                0.5
                                                                   0
                                                                   -100          -50          0           50          100
                                                                                     f  (frequency)
                               After the signal xt() is Þltered by the above ideal Þlter, the output yt() will be given by,
                                     yt()= cos(2π⋅α50t + )                                                                                           (S-2)
                               (Not enough information is given to determine α.) That is the DC component and 80Hz component will be 
                               completely eliminated.
                     Problem 4:
                          (a) Compute the output sequence yn[], n ≥ 0, for the difference equation yn[]= xn[]Ð 3xn[]Ð 1  and the 
                               input sequence xn[] plotted below.
                                                                   6
                                                                      xn[]
                                                                   4
                                                                   2
                                                                   0
                                                                 -2
                                                                 -4
                                                                        -4    -2      0     2      4      6      8     10
                                                                                               n
                                     y[]0  = x[]0 Ð 3x[]Ð1      = Ð3                                                                                 (S-3)
                                     y[]1  = x[]1 Ð 3x[]0      = 13                                                                                  (S-4)
                                                                                      - 2 -
 
                                                                                                                                                                                                                                                                                                                                                                                                                                                                                        
                                     y[]2  = x[]2 Ð 3x[]1      = Ð15                                                                                 (S-5)
                                     y[]3  = x[]3 Ð 3x[]2      = 14                                                                                  (S-6)
                                     y[]4  = x[]4 Ð 3x[]3      = Ð11                                                                                 (S-7)
                                     y[]5  = x[]5 Ð 3x[]4      = Ð15                                                                                 (S-8)
                                     y[]6  = x[]6 Ð 3x[]5      = 9                                                                                   (S-9)
                                     yn[]= 0, n≥7.                                                                                                  (S-10)
                               This output sequence yn[] is plotted on top of the next page.
                                                                                                                    yn[]
                                                                10
                                                                 5
                                                                 0
                                                                -5
                                                              -10
                                                              -15
                                                                    0         2          4     n    6          8         10
                          (b) Repeat part (a) for the following difference equation:
                                     yn[]= Ðyn[]Ð1 +2xn[]Ðxn[]Ð1                                                                                        (4)
                               and 0 ≤≤n     10. Assume y[]Ð1       = 0.
                                     y[]0  = Ðy[]Ð1 + 2x[]0 Ð x[]Ð1         = Ð6                                                                    (S-11)
                                     y[]1  = Ðy[]0 + 2x[]1 Ð x[]0        = 17                                                                       (S-12)
                                     y[]2  = Ðy[]1 + 2x[]2 Ð x[]1        = Ð27                                                                      (S-13)
                                     y[]3  = Ðy[]2 + 2x[]3 Ð x[]2        = 40                                                                       (S-14)
                                     y[]4  = Ðy[]3 + 2x[]4 Ð x[]3        = Ð37                                                                      (S-15)
                                     y[]5  = Ðy[]4 + 2x[]5 Ð x[]4        = 27                                                                       (S-16)
                                     y[]6  = Ðy[]5 + 2x[]6 Ð x[]5        = Ð24 (note xn[]= 0, n > 5)                                                (S-17)
                                     y[]7  = Ðy[]6 + 2x[]7 Ð x[]6        = 24                                                                       (S-18)
                                     y[]8  = Ðy[]7 + 2x[]8 Ð x[]7        = Ð24                                                                      (S-19)
                                     y[]9  = Ðy[]8 + x[]9 Ð x[]8       = 24                                                                         (S-20)
                                     y[]10   = Ðy[]9 + 2x[]10 Ð x[]9        = Ð24                                                                   (S-21)
                               This output sequence yn[] is plotted on the top of the next page.
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                                             40                                   yn[]
                                             20
                                              0
                                            -20
                                                0      2       4   n   6      8      10
               Problem 5:
                      Assume you want to sample and Þlter the continuous-time signal xt()= sin()2πt ; further assume that the 
                      discrete-time Þlter you want to apply to the sampled signal xn[] is given by the difference equation below:
                                 1      1         1         1         1
                                 ---    ---       ---       ---       ---
                          yn[]= xn[]+ xn[]Ð1 + xn[]Ð2 + xn[]Ð3 + xn[]Ð4                                    (5)
                                 5      5         5         5         5
                      For which of the following sampling frequencies (samples/second) —1Hz, 3Hz, 5Hz and 7Hz —will the 
                      output yn[] be zero for all n ?
                      The plot below shows the sequences that would be generated at the four different sampling frequencies. Note 
                      that the Þlter in equation (5) will output zero for all n , if Þve consecutive samples of the input sequence 
                      average to zero. This is the case for the 1Hz and 5Hz sampling frequencies, and is not true for the 3Hz and 
                                                                                                   1
                      7Hz sampling frequencies. Therefore, yn[] will be zero for sampling frequencies 1Hz and 5Hz.
                           1                               1Hz        1                            3Hz
                         0.5                                        0.5
                           0                                          0
                         -0.5                                      -0.5
                          -1                                         -1
                             0       5       10n     15       20       0       5       10 n     15      20
                           1                               5Hz        1                              7Hz
                         0.5                                        0.5
                           0                                          0
                         -0.5                                      -0.5
                          -1                                         -1
                             0       5       10 n    15       20       0       5       10 n     15      20
                     1. Note that in general, the 1Hz sampling frequency will not result in zero output, since we could easily 
                        offset the sampling so that we would get a constant sequence of samples not equal to zero.
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