Chap3Solutions - M. J. Roberts - 10/6/06 Chapter 3 -...

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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 E-1, 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 E-1 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 discrete-time 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 3-1 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 discrete-time 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*(n-2)/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 discrete-time 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 3-2 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 3-3 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 E-4 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 E-4 (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 3-4 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 discrete-time functions in Figure E-6. (a) g[n] 1 -4 -1 20 (b) g[n] 1 n -4 -1 20 n g[n] - g[n-1] 1 -4 -1 20 g[n] - g[n-1] 1 n -4 -1 20 n (c) g[n] = (n/10) 4 2 g[n] - g[n-1] 0.5 -4 -0.25 -4 20 n 20 n Figure E-6 Solutions 3-5 M. J. Roberts - 10/6/06 7. The signal x[n] is defined in Figure E-7. 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 E-7 (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 discrete-time impulses from n = 3 through n = +4 , for a total of eight unit discrete-time 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 discrete-time 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 discrete-time 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 3-6 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 3-7 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 E-11. (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 E-11 Solutions 3-8 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 discrete-time period and fundamental discrete-time 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 3-9 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 non-zero 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 discrete-time function adding 1 to or subtracting 1 from the fundamental discrete-time 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 3-10 M. J. Roberts - 10/6/06 Just looking at this simple function, it obviously has 2 N 0 + 1 discrete-time 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 iscrete-time 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 discrete-time 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 3-11 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 3-12 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 discrete-time 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 3-13 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 discrete-time 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 3-14 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 3-15 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 discrete-time 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 3-16 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 3-17 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 3-18 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 3-19 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 E-25 (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 3-20 M. J. Roberts - 10/6/06 26. Graph the following discrete-time 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 3-21 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 3-22 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 3-23 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 3-24 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 discrete-time 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 3-25 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 “discrete-k” 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 3-26 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 discrete-time function in Figure E-32. g[n] 6 4 2 -8 -6 -4 -2 -2 -4 -6 2 4 6 8 n Figure E-32 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 3-27 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 3-28 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 discrete-time 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 3-29 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 3-30 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 3-31 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 3-32 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 discrete-time 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 3-33 ...
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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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