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Unformatted text preview: M. J. Roberts  10/6/06 Chapter 3  Mathematical Description of DiscreteTime Signals
Solutions
Functions 1. In Figure E1, is a circuit in which a voltage x t = A sin 2 f0t () ( ) is connected periodically to a resistor by a switch. The switch rotates at a frequency f s of 500 rpm. The switch is closed at time t = 0 an each time the switch closes it stays closed for 10 ms.
+ x i (t) fs xo(t)
 (a) Figure E1 If A = 5 and f0 = 1 , graph the excitation voltage x i t and the response voltage x o t for 0 < t < 2 . () () () (b) (c) If A = 5 and f0 = 10 , graph the excitation voltage x i t and the response voltage x o t for 0 < t < 1 . This is an approximation of an ideal sampler. If the sampling process were ideal what discretetime signal x n would be produced in parts (a) and (b)? Graph them versus discrete time n. () (a)
x (t)
i 6 2 6 t x (t)
o 6 x[n]
6 2 6 t
6 20 n (b) Solutions 31 M. J. Roberts  10/6/06 xi(t)
6 1 6 t x (t)
o 6 x[n]
6 1 6 t
6 8 n 2. Plot these discretetime functions. (a) x n = 4cos 2 n / 12 ( ) 3sin 2 ( ( n 2 ) / 8) , 24 n < 24 n = 24:23 ; x = 4*cos(2*pi*n/12)  3*sin(2*pi*(n2)/8) ; stem(n,x,’k’,’filled’) ; x[n]
7 24 7 24 n (b) x n = 3ne n /5 , 20 n < 20 n = 20:19 ; x = 3*n.*exp(abs(n/5)) ; stem(n,x,’k’,’filled’) ; x[n]
6 20 6 20 n (c) x n = 21 n / 2 + 14 n3 ( ) 2 ,5
x[n]
2000 n<5 5 2000 5 n 3. Let x1 n = 5cos 2 n / 8 and x 2 n = 8e ( ) . Plot the following combinations of those two signals over the discretetime range, 20 n < 20 . If a signal has some defined and some undefined values, just plot the defined values.
n/6
2 ( ) (a) x n = x1 n x 2 n Solutions 32 M. J. Roberts  10/6/06 Make MATLAB functions for x1 and x 2 and save them in files named x2DT.m. (DT for discrete time.)
function y = x1DT(n), y = 5*cos(2*pi*n/8) ; I = find(round(n) ~= n) ; y(I) = NaN ; function y = x2DT(n) y = 8*exp((n/6).^2) ; I = find(round(n) ~= n) ; y(I) = NaN ; x 1DT.m and Then use the functions in this way in part (a).
n = 20:19 ; x = x1DT(n).*x2DT(n) ; x[n]
40 20 40 20 n (b) x n = 4 x1 n + 2 x 2 n n = 20:19 ; x = 4*x1DT(n) + 2*x2DT(n) ; x[n]
20 20 20 n 40 (c) x n = x1 2 n x 2 3n n = 20:19 ; x = x1DT(2*n).*x2DT(3*n) ; x[ n] 20 20 40 20 n (d) xn= x1 2 n x2 n Solutions 33 M. J. Roberts  10/6/06 x[n]
10000 20 20 n 50000 (e) x n = 2 x1 n / 2 + 4 x 2 n / 3 n = 20:19 ; x = 2*x1DT(n/2) + 4*x2DT(n/3) ;
x[n]
5 20 20 n 40 Scaling and Shifting Functions 4. For each pair of functions in Figure E4 provide the values of the constants in g 2 n = A g1 n n0 / a . ( ) (a)
2 1 0 1 2 10 2 1 0 1 2 1 0 (a)
g2[n] g1[n] 5 n 0 5 10 5 n 0 5 10 (b)
2 1 0 1 2 10 2 1 0 1 2 10 (b)
g2[n] g1[n] 5 n 0 5 10 5 n 0 5 10 Figure E4 (a) A = 1 / 2, n0 = 0, a = 1 / 2 , (b) A = 1, n0 = 1, a = 1 / 2 5. A function, g n is defined by , n< 4 , 4 n <1 . gn= n 4/n , 1 n 2 Solutions 34 M. J. Roberts  10/6/06 Graph g n , g 2 n , g 2 n and g n / 2 .
nonall to function y = gDT(n) I = find(round(n) ~= n) ; % Find all % integer “n’s” n(I) = NaN ; % Set them % “NaN” y1 = 2 ; y2 = n ; num3 = 4*ones(length(n),1) ; den3 = n ; I = find(den3 == 0) ; num3(I) = 1 ; den3(I) = 1 y3 = num3./den3 ; y = y1.*(n<4) + y2.*(n>=4 & n<1) + y3.*(n>=1)
g[n]
4 10 10 4 ; ; n g[ n]
4 10 10 4 g[2 n]
4 n 10 10 4 n g[2n]
4 10 10 4 g[n/2]
4 n 10 10 4 n Differencing and Accumulation 6. Graph the backward differences of the discretetime functions in Figure E6. (a)
g[n]
1 4 1 20 (b)
g[n]
1 n 4 1 20 n g[n]  g[n1]
1 4 1 20 g[n]  g[n1]
1 n 4 1 20 n (c)
g[n] = (n/10)
4 2 g[n]  g[n1]
0.5 4 0.25 4 20 n 20 n Figure E6 Solutions 35 M. J. Roberts  10/6/06 7. The signal x[n] is defined in Figure E7. Let y[n] be the first backward difference of x[n] and let z[n] be the accumulation of x[n]. (Assume that x[n] is zero for all n < 0).
x[n]
4 20 n 6 Figure E7 (a) What is the value of y[4]? y 4 = x 4 (b) What is the value of z[6]? z6= 8. Let g n = u n + 3 (a)
6 m= x 3 = 1 2= 3 x m = 1 3+ 1+ 2 1 5 1 = 8 un 5. What is the sum of all the values of g n ? g n consists of a sequence of unit discretetime impulses from n = 3 through n = +4 , for a total of eight unit discretetime impulses. Therefore the sum is 8. (b) If h n = g 3n , what is the sum of all the values of h n ? The transformed function is h n = g 3n = u 3n + 3 u 3n 5 . The first unit sequence begins at n = 1 and the second one begins at n = 2 . So there are unit discretetime impulses in h n at 1, 0 and 1 for a total of three. The numerical value is, therefore, 3. 9. Graph the accumulation, g n of each of these discretetime functions h n which are zero for all times n < 16 . (a) (c) (e) hn= n (b) h n =u n h n = cos 2 n / 8 u n h n = cos 2 n / 16 u n h n = cos 2 n / 16 u n + 8 ( ( ) ) (d) ( ) Solutions 36 M. J. Roberts  10/6/06 h[n]
1 h[n]
1 16 (a) 16 n g[n]
1 (b) 16 g[n]
16 16 n 16 16 n 16 16 n h[n]
1 16 1 16 h[n] n
1 16 1 16 n (c)
16 g[n]
3 3 16 (d)
n g[n]
3 16 3 16 n h[n]
1 16 1 16 n (e) g[n]
3 16 3 16 n Even and Odd Functions 10. Find and graph the even and odd parts of these functions. (a) g n =u n ge n = un 4 un u n 4 +u 2 n u n4 ge n = 1 u n +u 2
1+ n n un 4 u n+4 (b) (c) g n =e n/ 4 un g n = cos 2 n / 4 ( ) (d) g n = sin 2 n / 4 u n ( ) Solutions 37 M. J. Roberts  10/6/06 g[n]
1 10 1 10 g[n]
1 n 10 1 10 n ge[n]
1 10 1 10 ge[n]
1 n 10 1 10 n g [n]
o 1 10 1 10 g [n]
o 1 n 10 1 10 n g[n]
10 1 1 10 g[n]
1 n 10 1 10 n g [n]
e 10 1 1 10 g [n]
e 1 n 10 1 10 n go[n]
1 10 1 10 go[n] n
10 1 1 10 n 11. Graph g n for the signals in Figure E11. (a)
g1[n]
1 10 1 10 (b)
g1[n]
1 n g[n] 4 20 1 n g[n] g2[n]
1 10 10 1 g2[n]
1 4 1 20 n Multiplication n Multiplication (c)
g1[n]
1 4 1 20 (d)
g1[n]
1 n g[n] 10 1 10 n g[n] g2[n]
1 4 1 20 g[n]
1 10 1 10 n Multiplication n Multiplication Figure E11 Solutions 38 M. J. Roberts  10/6/06 (a)
g[n]
1 10 10 1 (b)
g[n]
1 n 4 1 20 n (c)
g[n]
1 (d)
g[n]
1 4 1 20 n 10 1 10 n Periodic Functions 12. Find the fundamental discretetime period and fundamental discretetime frequency of these functions. (a) (b) (c) (d) (e) g n = cos 2 n / 10 g n = cos ( ( ( ) ( ) N 0 = 10 , F0 = 1 / 10 N 0 = 20 , F0 = 1 / 20 N 0 = 35 , F0 = 1 / 35 N 0 = 20 , F0 = 1 / 20 N 0 = 12 , F0 = 1 / 12 n / 10 ) ) g n = cos 2 n / 5 + cos 2 n / 7 g n = e j2 g n =e
n / 20 +e +e j 2 n / 20 j2 n/3 j2 n/ 4 13. Graph the following functions and determine from the graphs the fundamental period of each one (if it is periodic). (a) (b) (c) g n = 5sin 2 n / 4 + 8cos 2 n / 6 ( ( ) ( ) )
(d) g n = Re e jn + e g n = 5sin 7 n / 12 + 8cos 14 n / 8 g n = Re e j n + e ) ( ( jn / 3 ) ( jn / 3 ) Solutions 39 M. J. Roberts  10/6/06 (a)
g[n]
12 (b)
g[n]
12 24 12 24 n 24 12 24 n N0 = 12 N0 = 24 (c)
g[n]
2 (d)
g[n]
2 24 2 24 n 24 2 24 n N0 = 6 Not Periodic 14. (a) What is the maximum value, over all discrete time, of the function, g n = tri n / 2 sin 2 n / 8 ? ( )( ) There are only three nonzero values of this function, at n = 1,0,1 . The only positive value occurs at n = 1 . The value of g n there is g 1 = tri 1 / 2 sin (b) If g n = 15cos ( )( / 4 = 1/ 2 )( ) ( 2 / 2 = 2 / 4 = 0.3535 ) ( 2 n / 12 ) a nd h n = 15cos 2 Kn ( ) what are the two smallest positive values of K for which g n = h n for all n? g n = 15cos 2 n / 12 because it is an even function. Therefore K = 1 / 12 works. Since this is a discretetime function adding 1 to or subtracting 1 from the fundamental discretetime frequency does not change the function values. The fundamental frequency is 1 / 12 . Therefore changing the frequency to 13/12 would yield the same function values. But changing it to 11 / 12 would also yield the same values and, because the function is even, 11 / 12 also works. 1 / 12and 11 / 12 or 0.08333 and 0.91667 Signal Energy and Power 15. Find the signal energy of these signals. (a) x n = A rect N n
0 ( ) Solutions 310 M. J. Roberts  10/6/06 Just looking at this simple function, it obviously has 2 N 0 + 1 discretetime impulses, each of strength, A. Squaring each one and then adding the squares the energy must be 2 N 0 + 1 A2 . ( ) Ex = 2 A rect N n
0 =A N0 2 N0 (1) = ( 2 N 0 + 1 A2
0 ) (b) x n =A n Ex = A
2 n 2 = A2 0 (1) = A 2 (c) xn= N0 n Ex = N0 n =
n = mN 0 (1)
=
0 (d) x n = ramp n Ex = ramp n 2 n2 This function takes off at n = 0 and rises linearly for all positive time. Therefore its energy must be infinite. (e) x n = ramp n 2 ramp n 4 + ramp n 8 Even though each of these three individual functions has infinite energy, the sum of the three functions does not. This can be seen by drawing a graph of the function. This is a dramatic demonstration that the energy of a sum of functions is not necessarily the sum of the energies of the functions. Ex = ramp n 2 ramp n 4 + ramp n 8
2 Ex = 02 + 12 + 22 + 32 + 42 + 32 + 22 + 12 + 02 = 1 + 4 + 9 + 16 + 9 + 4 + 1 = 44 16. A d iscretetime signal consists of the periodic alternating sequence, 4, 2,4, 2,4, 2, . What is the average signal power of this signal? The fundamental period is 2. The signal energy in one period is 42 + 2 = 20 . The average signal power is the signal energy in one period divided by the period. The average signal power is 10. 17. A discretetime signal x n is periodic with period N 0 = 6 . Some selected values of x n a re x 0 = 3 , x 1 = 1, x 4 = 2, x 8 = 2 , x 3 = 5, x 7 = 1 , x 10 = 2 amd x 3 = 5 . What is its average signal power? ( ) () 2 Solutions 311 M. J. Roberts  10/6/06 We need the values in one period. x 0 = 3 , x 1 = x 1 + 6 = x 7 = 1 , x 2 =x 2 6 =x x 5 =x 5 6 =x Px = 1 / N 0 18. 4 = 2, 1 = 1.
2 x 3 = 5, x 4 =x 4 6 6 =x 8 = 2, ( )
N0 xn = 1/ 6 ()
2n 5 xn
n=0 2 = 1 / 6 9 + 1 + 4 + 25 + 4 + 1 = ( )( ) 44 = 7.333… 6 Find the signal power of these signals. (a) xt =A
n= () rect t ( ) It will help to visualize this signal before beginning the analysis. It is a “square” wave with fundamental period, T0 = 2 , alternating between 0 and A, spending half its time at each level. So the square of the signal alternates between 0 and A2 , spending half its time at each level. Therefore, without any math, its average signal power is obviously A2 / 2 . 1 Px = T0 (b)
T0 / 2 A
T0 / 2 n= rect t ( 2n ) 2 A2 dt = 2 1 rect t
1 () 2 A2 dt = 2 1/ 2 A2 dt = 2 1/ 2 x t = 2A () 1 + rect t 2 n= 1 + rect t 2 n= ( 2n )
2 1 Px = T0 Px = 2 A T0 / 2 2A
T0 / 2 ( 2n ) 4 A2 dt = 2 1 1 1 + rect t 2 () 2 dt 1 2 1 1 + rect t 2 1 2
2 () 2 1 dt = 4 A
2 2 0 1 + rect t 2 () 2 dt 1/ 2 1 Px = 4 A 2 0 dt +
1/ 2 1 2 dt = A2 (c) x n =A Px = lim
N 1 2N N1 n= N A = lim
N 2 A2 2N N1 n= N () 1 = lim
N A2 2 N = A2 2N () Solutions 312 M. J. Roberts  10/6/06 (d) x n =u n Px = lim
N 1 2N N1 un
n= N 2 = lim
N 1 2N N1 n=0 (1) = lim 2N = 1 N2
N (e) x n =A
m= rect 2 n 8m Graph the function and the square of the function first. Then you can find the average signal power without much work. 1 Px = 2 N0 Px = A2 28
N0 1 2 A
n = N0 m= rect 2 n 8m
2 7 = A2 7 8 m= 2 2 8 n= rect 2 n 8m 6 n= 8 (1) + (1) + ( )
n= 2 n=6 10 A2 5 A2 1= = 28 8 (f) xn= N0 n Px = 1 N0 2 n = N0 N0 n = 1 N0
2 (g) Functions 19. x n = ramp n Px = lim
N 1 2N N1 ramp n
n= N = lim
N 1 2N N1 n2
n=0 Graph these discretetime exponential and trigonometric functions. (a) (c) (e) g n = 4 cos 2 n / 10 g n = 4 cos 1.8 n g n = 3/ 4 ( ( ) (b) (d) g n = 4 cos 2.2 n g n = 2cos 2 n / 6 (f) ( )
3sin 2 n / 6 ) ( )
n ( ) ( ) n g n = 2 0.9 sin 2 n / 4 () ( ) Solutions 313 M. J. Roberts  10/6/06 (a)
g[n]
4 (b)
g[n]
4 (c)
g[n]
4 5 4 20 n 5 4 20 n 5 4 20 n (d)
g[n]
4 (e)
g[n]
4 (f)
g[n]
4 5 4 20 n 5 4 20 n 5 4 20 n 20. Given the function definitions on the left, find the function values on the right. (a) gn= 3n + 6 e 10
2n g3=
n 3 3 +6 10 () 26 e ( ) = 0.0000092 (b) g n = Re 1+ j 2 A complex number of the form, x + jy , raised to the nth power can be expressed as re j and () n = r n e jn where r is the magnitude of the number,
1 x2 + y2 , is the angle of the number, tan g 5 = Re 1+ j 2 ( y / x) .
5 = 1 2 (c) g n = j 2 n + j10 n 4 ( ) 2 g 4 = j2 ( ( 4 )) 2 + j10 ( 4) 4 = 635.7 + j125.7 Shifting and Scaling Functions 21. Graph these discretetime singularity functions. (a) (c) g n = 2u n + 2 g n = 2 ramp n (b) (d) g n = u 5n g n = 10 ramp n / 2 Solutions 314 M. J. Roberts  10/6/06 (e) g n =7 n1 (a) (f) (b)
g[n]
1 g n =7 (c)
g[n]
5 2n 1 ( ) g[n]
2 20 n 5 20 n 5 20 n 10 (d)
g[n]
100 (e)
g[n]
7 (f)
g[n]
7 5 20 n 5 20 n 5 20 n (g) (i) (k) gn= 4 g n =8
4 2n / 3 n (h) (j) (l) (h)
g[n] gn= 4 g n =8
4 2n / 3 1 2n g n = rect 4 n (g)
g[n]
5 20 g n = 2 rect 5 n / 3 (i)
g[n]
20 n 5 n 8 4 4 5 20 n (j)
g[n]
8 (k)
g[n]
1 (l)
g[n]
2 5 20 n 5 20 n 20 20 n (m) (o) g n = tri n / 5 ( )
)) (n) g n = sinc n / 4 ( ) g n = sinc n + 1 / 4 (( Solutions 315 M. J. Roberts  10/6/06 (m)
g[n]
1 (n)
g[n]
1 (o)
g[n]
1 20 20 1 n 20 1 20 n 10 10 n (p) g n = drcl n / 10,9 ( )
(p)
g[n]
1 20 1 20 n 22. Graph these combinations of discretetime functions. (a) (c) (d) g n =u n +u n (b) n n/2 (b)
g[n]
1 10 10 g n =u n u n g n = cos 2 n / 12 g n = cos 2 n / 12 (a)
g[n]
2 ( ( ) ) 3 3 (c)
g[n]
1 n 5 1 20 n 10 10 n 1 (d)
g[n]
1 5 1 20 n (e) g n = cos gn=
n 2 ( n + 1)
12 u n +1 cos 2n un 12 (f) cos
m= 0 2m um 12 Solutions 316 M. J. Roberts  10/6/06 (g) (h) gn= gn= n m= 0 n m= ( ( 4 m m+ 4 m2 ) 4 3 m rect 4 m
n +1 n ) (i) gn= 2 n +1 (e) 2 n (j) (f)
g[n]
2 20 gn= m
m= m= m (g)
g[n]
1 g[n]
1 5 1 20 n 5 2 n n 5 20 (h)
g[n]
6 1 (i)
g[n] (j)
g[n]
1 5 20 1 n n 10 10 n 5 10 23. Graph the magnitude and phase of each function versus k. (a) (b) (c) G k = 20sin 2 k / 8 e ( ) j k /4 G k = 20cos 2 k / 8 sinc k / 40 Gk= ( ) ( )
k 2 k 4+ k 8 ej ( k +8 2 k+4 + ) k /8 Solutions 317 M. J. Roberts  10/6/06 (a)
G[k]
20 (b)
G[k]
20 (c)
G[k]
2 16 16 k 16 16 k 16 16 k Phase of G[k]
π 16 16 Phase of G[k]
π Phase of G[k]
π k 16 π 16 k 16 π 16 k π 24. Using MATLAB, for each function below plot the original function and the shifted and/or scaled function. (a) gn= 5,n 0 5 3n , 0 < n
2 4 23 + n , 4 < n 8 41 , n > 8 (a)
g[n]
50 g 3n vs. n 10 10 20 n g[3n]
50 10 10 20 n (b) g n = 10cos 2 n / 20 cos 2 n / 4 ( )( ) 4g 2 n + 1 ( ) vs. n Solutions 318 M. J. Roberts  10/6/06 (b)
g[n]
10 40 10 n 4g[2(n+1)]
40 40 40 n (c) g n = 8e j 2
n /16 un (c)
g[n] g n / 2 vs. n 10 10 30 n g[n/2]
10 10 30 n 25. Given the graphical definition of a function, g[n], graph the indicated function(s), h[n]. (a)
g[n]
6 4 2 8 6 4 2 2 4 6 2 4 6 8 h n = g 2n 4
n g n =0 , n >8 (b) Solutions 319 M. J. Roberts  10/6/06 g[n]
6 4 2 8 6 4 2 2 4 6 2 4 6 8 h n =g n/2
n g n =0 , n >8 (c)
g[n]
6 4 2 8 6 4 2 2 4 6 2 4 6 8 h n =g n/2
n g n is periodic Figure E25 (a)
g[2n  4]
6 4 2 8 6 4 2 2 4 6 2 4 6 8 n (b)
g[
6 4 2 8 6 4 2 2 4 6 2 4 6 8 n 2 n (c)
g[ 2 ]
6 4 2 8 6 4 2 2 4 6 2 4 6 8 n n Solutions 320 M. J. Roberts  10/6/06 26. Graph the following discretetime functions.
g[n] (a) g n =5 n 2 +3 n +1 5 3 n 1 g[n] 2 (b) g n =5 2n + 3 4n 2 ( )
g[n] 5 3 n 2 (c) g n =5 u n 1 ( u4 n ) 5 n 5
g[n] 2 4 6 (d) g n = 8rect 4 n + 1
6 4 2 8 n 2 4 6 g[n]
10 (e) g n = 8cos 2 n / 7 ( ) 5 n
5 5 10 15 20 10 g[n] n
2 20 2 4 6 8 (f) g n = 10en / 4 u n 40 60 80 g[n] (g) g n = 10 1.284 u n ( ) n n
2 20 2 4 6 8 40 60 80 Solutions 321 M. J. Roberts  10/6/06 g[n]
1 (h) g n = j/4 u n
4 2 ( ) 0.8 n 0.6 0.4 0.2 2 4 6 8 n (i) g n = ramp n + 2
2 2 ramp n + ramp n 2
g[n] 1.5 1 0.5 4 2 2 4 6 8 n (j) g n = rect 2 n
g[n]
1 0.8 0.6 0.4 0.2 2 n 8 6 4 2 2 4 6 8 n (k) g n = rect 2 n 2 n +1
g[n]
1 0.8 0.6 0.4 0.2 8 6 (l) g n = 3sin 2 n / 3 rect 4 n
g[n]
3 2 1 ( 4 ) 2 2 4 6 8 n n
15 10 5 1 2 3 5 10 15 Solutions 322 M. J. Roberts  10/6/06 (m)
g[n]
5 g n = 5cos 2 n / 8 u n / 2 ( ) n
5 5 10 15 20 25 30 5 27. Graph versus k , in the range, 10 < k < 10 , the magnitude and phase of
X[k]
1 (a) X k = sinc k / 2 ( ) 20 20 k Phase of X[k]
π 20 π 20 k X[k]
1 (b) X k = sinc k / 2 e ( ) j2 k /4 20 20 k Phase of X[k]
π 20 π 20 k Solutions 323 M. J. Roberts  10/6/06 X[k]
1 (c) X k = rect 3 k e j2 k /3 15 15 k Phase of X[k]
π 15 π 15 k X[k]
1 (d) Xk= 1 1 + jk / 2 15 15 k Phase of X[k]
π 15 π 15 k X[k]
2 (e) Xk= jk 1 + jk / 2 15 15 k Phase of X[k]
π 15 π 15 k Solutions 324 M. J. Roberts  10/6/06 X[k]
1 (f) Xk= 2 ke j2 k /4 15 15 k Phase of X[k]
π 15 π 15 k Differencing and Accumulation 28. Graph the accumulation of each of these discretetime functions. (a) g n = cos 2 n u n ( ) (b) g n = cos 4 n u n ( ) (a)
Accumulation of g[n]
20 5 20 n (b)
Accumulation of g[n]
20 5
n 20 n 29. In the equation,
m= u m = g n n0 / N w , ( ) (a) (b) What is name of the function, g? Ramp Find the values of n0 and N w . n0 = 1 , Nw = 1 30. What is the numerical value of each of the following accumulations? Solutions 325 M. J. Roberts  10/6/06 10 (a)
n=0 6 ramp n = 0 + 1 + 2 + + 10 = 55 (b) 1 = 1+ 1/ 2 + n n=0 2
N1 + 1 / 26 . =1
N N,
n Using
n=0 1 1 / 128 127 1 1 1/ 2 = = = n 1/ 2 256 1 1/ 2 n=0 2
6 ( =1 1 , otherwise ) 7 (c)
n= u n / 2n Using
n=0 n = 1 1 , <1 u n / 2n =
n=0 10 1 =2 1 1/ 2 (d)
n = 10 10 3 n =7 2n = 7 (e)
n = 10 3 (f)
n= sinc n = 1 () Even and Odd Functions 31. Find and graph the magnitude and phase of the even and odd parts of each of this “discretek” function. 10 Gk= 1 j 4k 10 10 + 1 j 4k 1 + j 4k k= = 2 1 Ge ( 10 10 = 1 + 16 k 2 j 4k 1 + j 4k )( ) Go k = 10 1 j 4k 2 10 1 + j 4k = (1 j 40 k j 40 k = 1 + 16 k 2 j 4k 1 + j 4k )( ) Solutions 326 M. J. Roberts  10/6/06 G [k]
e 10 G [k]
o 10 10 10 k 10 10 k Phase of G [k]
e π Phase of G [k]
o π 10 π 10 k 10 π 10 k 32. Find and graph the even and odd parts of the discretetime function in Figure E32.
g[n]
6 4 2 8 6 4 2 2 4 6 2 4 6 8 n Figure E32
ge[n]
6 4 2 8 6 4 2 2 4 6 2 4 6 8 go[n]
6 4 2 n 8 6 4 2 2 4 6 2 4 6 8 n 33. Graph the even and odd parts of these signals. All plots below. (a) x n = rect 5 n + 2 Solutions 327 M. J. Roberts  10/6/06 xe[n]
1 10 1 10 n xo[n]
1 10 1 10 n (b) xn=
3 3 n1 n1 = 1 2
1 xe n = n 1+ 2 3 ( 3 n 1+ 3 n +1 )
n xe[n] 10 1 10 xo[n]
1 10 1 10 n (c) x n = 15cos 2 n / 9 + ( /4 )
)(
/4 x n = 15 cos 2 n / 9 cos ( )( /4 ) sin 2 n / 9 sin ( ) = 15 2 (cos ( 2 n/9 ) sin 2 n / 9 ( )) = 1/ 2 = 1/ 2 x e n = 15 / 2 cos 2 n / 9 ( ) ( ) , xo n = (15 / 2 )sin (2 n / 9) Solutions 328 M. J. Roberts  10/6/06 xe[n]
20 10 20 10 n xo[n]
20 10 10 n (d) x n = sin 2 n / 4 rect 5 n 1
xe[n]
1 ( ) 20 10 1 10 n xo[n]
1 10 1 10 n Periodic Functions 34. Using MATLAB, plot each of these discretetime functions. If a function is periodic, find the period analytically and verify the period from the plot. (a) (b) g n = sin 3 n / 2 = sin 6 n / 4 ( (
( ) )
) ( ) )
) Period is 4 g n = sin 2 n / 3 + cos 10 n / 3 (
( g n = sin 2 n / 3 + cos 6 n / 3 + 4 n / 3 = sin 2 n / 3 + cos 4 n / 3
Period = 3 Period = 3 ( ) ( ) Period is 3 (c) g n = 5cos 2 n / 8 + 3sin 2 n / 5
Period of 8 Period of 5 ( ) ( ) Solutions 329 M. J. Roberts  10/6/06 LCM of the fundamental periods is 40, therefore the fundamental period is 40.
g[n]
1 60 1 (a)
n g[n]
2 (b)
n 60 2 g[n]
8 (c)
60 n 8 (d) (e) g n = 10cos n / 4 ( ) Not periodic. g n = 3cos 2 n / 7 sin 2 n / 6 ( )( )
( ) (A trigonometric identity will be useful here.) gn= gn= = 3 sin 2 n / 6 2 n / 7 + sin 2 n / 6 + 2 n / 7 2 3 sin 12 n / 42 + sin 40 n / 42 2 3 sin 40 n / 42 sin 12 n / 42 2 ( ) ( ) ( ) ( ) ( ) The period is 42.
g[n]
10 60 10 (d)
n g[n]
4 (e)
n 60 4 Signal Energy and Power 35. Find the signal energy of each of these signals: Solutions 330 M. J. Roberts  10/6/06 (a) x n = 5rect 4 n n +5 n3 n3
2 Ex = xn
n= 2 = 25
n= rect 4 n 2 = 25 4 n= 4 (1) = 225 (b) x n =2 Ex = 2
n= n +5 =
n= 4 n 2 + 25 n3 2 = 29 (c) x n =u n /n xn= The energy is infinite. (d) ( 1 / 3 u n Ex = ) n n= ( 1/ 3 u n ) n 2 =
n=0 (1 / 9) ( n = 1 9 = 1 1/ 9 8 (e) x n = cos Ex = ( n/3 u n cos )( un 6 ) n= ( n/3 u n )( un 6
2 2 ) 2 = 5 cos 2
n=0 2 n/3 ) E x = 12 + 1 / 2 + 36. ( )(
2 1/ 2 + ) ( 1) + ( 1/ 2 + 1/ 2 = 3 )( ) 2 Find the average signal power of each of these signals. (a) x n = u n Px = lim
N 1 2N N1 xn
n= N 2 = lim
N 1 2N N1 un
n= N 2 2 = lim
N 1 2N N1 n=0 (1) = 1 2 (b) x n = ( 1) n Px = lim
N 1 2N N1 n= N ( 1) n = lim
N 1 2N N1 n= N (1) = 1 (c) x n = A cos 2 F0 n + Px = lim
N ( )
A cos 2 F0 n + 1 2N N1 n= N ( ) 2 = lim
N A2 2N N1 n= N cos 2 2 F0 n + ( ) Using the trigonometric identity, cos x cos y = () () 1 cos x y + cos x + y 2 ( ) ( ) , Solutions 331 M. J. Roberts  10/6/06 Px = lim
N A2 2N N1 n= 1 1 + cos 4 F0 n + 2 N2 ( ) = A2 A2 + lim 2 N 4N N1 n= N cos 4 F0 n + 2 ( ) So the average power is a constant, A2 / 2 , plus another term, lim A2 4N
N1 n= N N cos 4 F0 n + 2 . ( ), whose value depends on the parameters, F0 and Case 1. 2 F0 , an integer. A2 A2 Px = + lim 2 N 4N Case 2. 2 F0 , not an integer.
N1 n= N cos 2 () A2 = 1 + cos 2 2 () Subcase 1. The signal is periodic. If it is periodic with period, N 0 = 1 / F0 . The square of the function is periodic with period, N 0 = 1 / 2 F0 . For a periodic signal,
2 1 k+N 1 x n where N is any integer number of periods of the signal and N n= k where k is any integer. Then the average power is Px = Px = A2 A2 + 2 4N k+N 1 n= k cos 2 () A2 k + N 1 cos 2 is zero because the samples are taken at equal 4 N n= k angular intervals over exactly an integer number of periods. Therefore the average power is Px = A2 / 2 . The summation, () Subcase 2. The signal is not periodic. This is the hardest case to examine. We cannot use the periodic formula so we must use Px = A2 A2 + lim 2 N 4N
N1 n= N cos 4 F0 n + 2 ( ) The summation can be geometrically visualized as the real part of a sum of vectors in the complex plane, all with unit length and separated from their nearest neighbors by the angle, 4 F0 . Since this angle is not any integer multiple of radians (that case has already been considered), the summation is of an infinity of vectors at angles which uniformly fill up the full 2 Solutions 332 M. J. Roberts  10/6/06 radians of a full circle. We don’t know what that sum is in the limit, but we do know it cannot grow to infinity because of the cancellation of the vectors arrayed in the circle and the sum is being divided by N which is going to infinity. Therefore, in the limit, the average power is Px = A2 / 2 , just as in the periodic case. (d) x n = A , n= 0 , n= , 0,1,2,3,8,9,10,11,16,17,18,19, , 4,5,6,7,12,13,14,15,20,21,22,23, This is a discretetime rectangular wave with 50% duty cycle. Therefore the average power is the average of the square of the signal’s values or Px = A2 / 2 . (e) x n =e
j n/ 2 This is a periodic function. Px = 1 N0 xn
n = N0 2 = 13 xn 4 n=0 2 = 13 e 4 n=0 j n/ 2 2 = 13 1=1 4 n=0 Solutions 333 ...
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This note was uploaded on 12/10/2010 for the course DCIS 32 taught by Professor Go during the Spring '10 term at College of E&ME, NUST.
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