Chap2Solutions - M. J. Roberts - 10/7/06 Chapter 2 -...

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Unformatted text preview: M. J. Roberts - 10/7/06 Chapter 2 - Mathematical Description of Continuous-Time Signals Solutions CT Functions 1. If g t = 7 e (a) (b) (c) (d) (e) () 2t 3 write out and simplify 9 () g ( 2 t ) = 7e ( ) = 7e g ( t / 10 + 4 ) = 7 e g ( jt ) = 7 e g ( jt ) + g ( jt ) e = 7e g 3 = 7e 22 t 3 t /5 11 j 2t 3 3 7 + 2t j 2t (f) 2. 2 jt 3 +g g 2 2 + e j 2t = 7 e 3 cos 2t 2 e jt () () jt 3 2 =7 + e jt = 7 cos t 2 If g x = x 2 (a) (b) (c) (d) () 4 x + 4 write out and simplify 4z + 4 () g (u + v ) = (u + v ) g z = z2 2 4 u + v + 4 = u 2 + v 2 + 2uv 4u 4v + 4 4e jt + 4 = e jt 2 2 ()() g ( g ( t )) = g ( t g e jt = e jt g g t = t4 2 4e jt + 4 = e j 2t 2 2 ( 4t + 4 ) = ( t 4t + 4 ) 4 ( t ( ) 2 ) 2 4t + 4 + 4 ) ( ( )) () 8t 3 + 20t 2 16t + 4 (e) 3. g 2 = 4 8+ 4 = 0 What would be the value of g in each of the following MATLAB instructions? (a) (b) t = 3 ; g = sin(t) ; 0.1411 [-1,1,-1,1,-1] x = 1:5 ; g = cos(pi*x) ; Solutions 2-1 M. J. Roberts - 10/7/06 f = -1:0.5:1 ; w = 2*pi*f ; g = 1./(1+j*w’) ; (c) 0.0247 + 0.0920 + 1 0.0920 0.0247 4. Let two functions be defined by x1 t = j0.155 j0.289 j0.289 j0.155 () 1 , sin 20 t 1, ( )0 sin ( 20 t ) < 0 and x2 t = () t , sin 2 t t, ( )0 sin ( 2 t ) < 0 . Graph the product of these two functions versus time over the time range, 2 < t < 2. x(t) 2 -2 -2 2 t Transformations of CT Functions 5. For each function, g t , graph g (a) g(t) 4 2 () ( t ) , g (t ) , g (t 1) , and g ( 2t ) . (b) g(t) 3 t -1 1 -3 t g(-t) 4 -2 g(-t) 3 -g(t) -g(t) 3 t -1 1 -3 t 4 2 t 1 -1 -3 t Solutions 2-2 M. J. Roberts - 10/7/06 g(t-1) 4 1 3 g(t-1) 3 g(2t) 4 g(2t) 3 1 t -3 1 2 t t -1 2 1 2 -3 t 6. Find the values of the following signals at the indicated times. (a) (b) (c) (d) (e) () () ( ) ()( ) x ( t ) = 2 rect ( t / 4 ) , x ( 1) = 2 rect ( 1 / 4 ) = 2 sin ( 3 / 2 ) 2 x ( t ) = 10sinc ( t ) , x ( 3 / 2 ) = 10 = = 0.21221 3 3 /2 x ( t ) = 5rect ( t / 2 ) sgn ( 2t ) , x (1) = 5rect (1 / 2 ) sgn ( 2 ) = 2.5 x t = 4 tri t , x 1 / 2 = 4 tri 1 / 2 = 4 1 1 / 2 = 2 x t = 2 tri 2 t 1 + 6 rect t / 4 , x 3 / 2 = 2 tri 2 3 / 2 1 + 6 rect 3 / 8 = 2 tri 1 + 6 = 6 () ( () () ( ) () )) ) ( ) ( ) () ( (f) (g) 7. x t = 9 rect t / 10 sgn 3 t ( ) (( )) 2 , x 1 = 9 rect 1 / 10 sgn () ) ( 3) = 9 x t = 10sinc t + 2 / 4 , x -6 = 10sinc (( () ( 1) = 0 For each pair of functions in Figure E-7 provide the values of the constants, A, t0 and a in the functional transformation g 2 t = A g1 t t0 / w . () (( )) Solutions 2-3 M. J. Roberts - 10/7/06 (a) 2 1 0 -1 -2 -4 -2 0 2 4 2 1 0 -1 -2 -4 (a) g1(t) g2(t) -2 0 2 4 t (b) 2 1 0 -1 -2 -4 -2 0 2 4 2 1 0 -1 -2 -4 t (b) g1(t) g2(t) -2 0 2 4 t (c) 2 1 0 -1 -2 -4 -2 0 2 4 2 1 0 -1 -2 -4 t (c) g1(t) g2(t) -2 0 2 4 t t Figure E-7 Answers: (a) A = 2, t0 = 1, w = 1 , (b) (c) 8. A = 2 , t0 = 0, w = 1 / 2 , A = 1 / 2, t0 = 1, w = 2 For each pair of functions in Figure E-8 provide the values of the constants, A, t0 and a in the functional transformation g 2 t = A g1 w t t0 . () (( )) A = 2, t0 = 2, w = -2 8 4 8 4 g1(t) (a) 0 -4 -8 -10 -5 0 5 10 g2(t) 0 -4 -8 -10 -5 0 5 10 t 8 4 t A = 3, t0 = 2, w = 2 8 4 g (t) 1 (b) -4 -8 -10 -5 0 5 10 2 0 g (t) 0 -4 -8 -10 -5 0 5 10 t t Solutions 2-4 M. J. Roberts - 10/7/06 A = 3, t0 = 3, w = 1/3 8 4 8 4 g (t) 1 (c) -4 -8 -10 -5 0 5 10 2 0 g (t) 0 -4 -8 -10 -5 0 5 10 t 8 4 t A = 2, t0 = 2, w = 1/3 8 4 g1(t) (d) 0 -4 -8 -10 -5 0 5 10 g2(t) 0 -4 -8 -10 -5 0 5 10 t 8 4 t A = 3, t0 = 2, w = 1/2 8 4 g1(t) (e) 0 -4 -8 -10 -5 0 5 10 g2(t) 0 -4 -8 -10 -5 0 5 10 t Figure E-8 9. t In Figure E-9 is plotted a CT function, g1 t which is zero for all time outside the range plotted. Let some other functions be defined by g 2 t = 3g1 2 t () () ( ) , g 3 t = 2 g1 t / 4 () () , g 4 t = g1 () t3 2 Find these values. (a) g2 1 = 3 g 4 t g3 t () (b) = 3 2 g3 1 ( 1) = 3.5 (c) () () t=2 () 1= 3 2 (d) 3 g 4 t dt () The function g 4 t is linear between the integration limits and the area under it is a triangle. The base width is 2 and the height is -2. Therefore the area is -2. () Solutions 2-5 M. J. Roberts - 10/7/06 1 g 4 t dt = 2 3 () 4 3 2 1 g1(t) -4 -3 -2 -1 -1 -2 -3 -4 1 2 3 4 t Figure E-9 10. A function, G f , is defined by G f =e () () j2 f rect f / 2 ( ) . Graph the magnitude and phase of G f 20 < f < 20 . ( 10 + G f + 10 ) ( ) over the range, First imagine what G f looks like. It consists of a rectangle centered at f = 0 of width, 2, multiplied by a complex exponential. Therefore for frequencies greater than one in magnitude it is zero. Its magnitude is simply the magnitude of the rectangle function because the magnitude of the complex exponential is one for any f. e j2 f () = cos ( 2 f + j sin j2 f ) ( 2 f = cos 2 f ) ( ) j sin 2 f ( ) e = cos 2 2 f + sin 2 2 f = 1 ( ) ( ) The phase (angle) of G f is simply the phase of the complex exponential between f = 1 and f = 1 and undefined outside that range because the phase of the rectangle function is zero between f = 1 and f = 1 and undefined outside that range and the phase of a product is the sum of the phases. The phase of the complex exponential is e j2 f () = (cos ( 2 f ) e j sin 2 f ( )) = tan 1 1 () cos ( 2 f ) sin 2 f = tan 1 () cos ( 2 f ) sin 2 f j2 f = tan ( tan ( 2 f )) The inverse tangent function is multiple-valued. Therefore there are multiple correct answers for this phase. The simplest of them is found by choosing e j2 f =2f Solutions 2-6 M. J. Roberts - 10/7/06 which is simply the coefficient of j in the original complex exponential expression. e j 2 f = 2 f + 2 n , n an integer . The A more general solution would be solution of the original problem is simply this solution except shifted up and down by 10 in f and added. Gf ( 10 + G f + 10 = e ) ( ) j2 (f 10 ) rect f 2 10 +e j2 ( f +10) rect f + 10 2 |G( f )| 1 -20 20 f Phase of G( f ) π -20 -π 20 f 11. Write an expression consisting of a summation of unit step functions to represent a signal which consists of rectangular pulses of width 6 ms and height 3 which occur at a uniform rate of 100 pulses per second with the leading edge of the first pulse occurring at time t = 0 . x t =3 n=0 () u t 0.01n ( ) u (t 0.01n 0.006 ) 12. Write an expression consisting of a summation of triangle functions to represent a periodic triangular wave whose maximum value is 5 and whose minimum value is 2 with a period of 20 µs and a value of 5 at time t = 0 . x t = 3.5 + 1.5 n= () tri 2 105 t ( ( 2 10 5 n )) tri 2 105 t 10 ( ( 5 ( 2n 1))) Derivatives and Integrals of CT Functions 13. Graph the derivatives of these functions. (All graphes at end.) (a) g t = sinc t () () Solutions 2-7 M. J. Roberts - 10/7/06 gt= (b) () 2 t cos gt = 1 et t = 0 , therefore its derivative during that time is zero. This function is a constant minus a decaying exponential after time, t = 0 , and its derivative in that time is therefore also a positive decaying exponential. e gt= 0 () ( ( t ) sin ( t ) = t cos ( t ) sin ( t ) t ( t) ) u (t ) This function is constant zero for all time before time, 2 2 () t , t>0 , t<0 Strictly speaking, its derivative is not defined at exactly t = 0 . Since the value of a physical signal at a single point has no impact on any physical system (as long as it is finite) we can choose any finite value at time, t = 0 , without changing the effect of this signal on any physical system. If we choose 1/2, then we can write the derivative as g t =e tu t . () () (a) x(t) 1 -4 -1 4 (b) x(t) 1 t -1 -1 4 t dx/dt 1 -4 -1 4 dx/dt 1 t -1 -1 4 t 14. (a) If g t = tri t / 2 what is the value of the first derivative of gt 1, () () ( ) d g t 1 , at t = 2 ? dt 0 , t< 2 1/ 2 , 2 < t < 0 d gt = dt 1/ 2 , 0 < t < 2 0 , t>2 ( ( )) ( ( )) 0 , t< 1 1/ 2 , 1< t <1 d gt 1 = dt 1/ 2 , 1< t < 3 0 , t>3 ( ( )) Solutions 2-8 M. J. Roberts - 10/7/06 At t = 2 , (b) d g t 1 = 1/ 2. dt ( ( )) If g t = sinc 2 t + 1 what is the value of 10 g t / 10 at t = 4 ? () ( ( )) 10 g ( t / 10 ) = 10sinc ( 2 ( t / 10 + 1)) ( ) ( ) At t = 4 , 10 g t / 10 = 10sinc 2 4 / 10 + 1 = 10sinc 2.8 = 10 15. Find the numerical value of each integral. (a) 8 (( )) () sin 2.8 2.8 ( ) = 0.0668 ( t+3 ) 2 () 5/ 2 8 4t dt = 1 ( t + 3 dt ) 8 2 1 5/ 2 () 8 4t dt = 0 2 1 / 4 1 (t ) dt = 1/ 2 1 (b) 5/ 2 2 1/ 2 () 3t dt = 1/ 2 n = ( 3t 2 n dt = ) 1 3 1/ 2 n = (t 2 n / 3 dt = ) 1 1+ 1+ 1 = 1 3 16. Graph the integral from negative infinity to time, t, of the functions in Figure E-16 which are zero for all time t < 0 . This is the integral, t g to time, t. For the case of the two back-tounder the function, g t , from time, back rectangular pulses, there is no accumulated area until after time, t = 0 , and then in the time interval, 0 < t < 1 , the area accumulates linearly with time up to a maximum area of one at time, t = 1 . In the second time interval, 1 < t < 2 , the area is linearly declining at half the rate at which it increased in the first time interval, 0 < t < 1 , down to a value of 1/2 where it stays because there is no accumulation of area for t > 2. In the second case of the triangular-shaped function, the area does not accumulate linearly, but rather non-linearly because the integral of a linear function is a seconddegree polynomial. The rate of accumulation of area is increasing up to time, t = 1 , and then decreasing (but still positive) until time, t = 2 , at which time it stops completely. The final value of the accumulated area must be the total area of the triangle, which, in this case, is one. g(t) 1 1 1 2 () ( )d , which, in geometrical terms, is the accumulated area g(t) 1 2 3 t 1 2 3 t Figure E-16 Solutions 2-9 M. J. Roberts - 10/7/06 ∫ g(t) dt 1 1 2 ∫ g(t) dt 1 1 2 3 t 1 2 3 t Even and Odd CT Functions 17. An even function, g(t), is described over the time range, 0 < t < 10 , by 2t , 0<t<3 g t = 15 3t , 3 < t < 7 . 2 , 7 < t < 10 () (a) What is the value of g t at time, t = 5 ? Since g t is even, g t = g () () () () ( t) g ( 5) = g (5) = 15 3 5= 0. (b) What is the value of the first derivative of g(t) at time, t = 6 ? Since g t is even, d gt = dt () d g dt ( t) d gt dt () = t= 6 d gt dt () = t =6 ( 3) = 3 . 18. Find the even and odd parts of these functions. (a) g t = 2t 2 () 3t + 6 3t + 6 + 2 ge t = () 2t 2 ( t) 2 2 3 t +6 () 4t 2 + 12 = = 2t 2 + 6 2 = 6t = 3t 2 go t = (b) () 2t 2 3t + 6 2 ( t) 2 /4 2 +3 t () 6 g t = 20cos 40 t ge t = () ( ) ) ( 40 t /4 () 20cos 40 t ( / 4 + 20cos 2 ) Using cos z1 + z2 = cos z1 cos z2 ( ) () () sin z1 sin z2 , () () Solutions 2-10 M. J. Roberts - 10/7/06 20 cos 40 t cos ge t = () +20 ) ( / 4) sin ( 40 t ) sin ( / 4) cos ( 40 t ) cos ( / 4 ) sin ( 40 t ) sin ( / 4 ) 2 ( 20 cos 40 t cos ge t = () () +20 ) ( / 4) + sin ( 40 t ) sin ( / 4) cos ( 40 t ) cos ( / 4 ) sin ( 40 t ) sin ( / 4 ) 2 / 4 cos 40 t = 20 / 2 cos 40 t /4 ( g e t = 20cos go t = ( )( ) ( ) ( ) () 20cos 40 t ( ) 20cos 2 ( 40 t /4 ) Using cos z1 + z2 = cos z1 cos z2 20 cos 40 t cos go t = ( ) () () sin z1 sin z2 , () () () 20 ) ( / 4) sin ( 40 t ) sin ( / 4) cos ( 40 t ) cos ( / 4 ) sin ( 40 t ) sin ( / 4 ) 2 ( 20 cos 40 t cos go t = () () 20 ) ( / 4) + sin ( 40 t ) sin ( / 4) cos ( 40 t ) cos ( / 4 ) sin ( 40 t ) sin ( / 4 ) 2 / 4 sin 40 t = 20 / 2 sin 40 t ( g o t = 20sin gt = ( )( ) ( ) ( ) (c) () 2t 2 3t + 6 1+ t 3t + 6 2t 2 + 3t + 6 + 1+ t 1t 2 2 2t 2 ge t = () () ( 2t ge t = 3t + 6 1 t + 2t 2 + 3t + 6 1 + t )( ) ( 2 (1 + t )(1 t ) )( ) Solutions 2-11 M. J. Roberts - 10/7/06 ge t = () 4t 2 + 12 + 6t 2 2 1 t2 2t 2 ( ) = 6 + 5t 2 1 t2 go t = () () 3t + 6 1+ t 2 2 2t 2 + 3t + 6 1t ( 2t go t = go t = 3t + 6 1 t )( ) ( 2t 2 2 (1 + t )(1 t ) =t + 3t + 6 1 + t )( ) () 6t 2 1 t2 ( 4t 3 12t ) 2t 2 + 9 1 t2 (d) () () g (t ) = 0 g t = sinc t o ge t = () sin ( t) / t + sin 2 ( t ) / ( t ) = sin ( t ) t (e) g t = t 2 t 2 1 + 4t 2 () ( )( ) g t = t 2 t 2 1 + 4t 2 odd even even () ( )( ) () () ( ) ) Therefore g t is odd, g e t = 0 and g o t = t 2 t 2 1 + 4t 2 (f) g t = t 2 t 1 + 4t ge t = () )( ) () ( () )( ) ) ( t )( 2 + t )(1 2 4t g e t = 7t 2 t 2 t 1 + 4t + ( )( () go t = 19. () t 2 t 1 + 4t ( )( ) ( t )( 2 + t )(1 2 4t g o t = t 2 4t 2 () ( ) ( t) . Graph the even and odd parts of the functions in Figure E-19. To graph the even part of graphically-defined functions like these, first graph g Then add it (graphically, point by point) to g t and (graphically) divide the sum by Solutions 2-12 () M. J. Roberts - 10/7/06 two. Then, to graph the odd part, subtract g the difference by two. g(t) 1 ( t) from g t (graphically) and divide () g(t) 1 1 t -1 1 2 t Figure E-19 g e(t) 1 g e(t) 1 1 t -1 1 2 t g o(t) 1 g o(t) 1 1 t 1 2 t , (a) 20. (b) -1 Graph the indicated product or quotient g t of the functions in Figure E-20. (a) 1 -1 1 -1 () (b) 1 t g(t) -1 1 -1 t g(t) 1 Multiplication -1 1 1 -1 1 -1 Multiplication t t g(t) 1 -1 1 -1 g(t) 1 t -1 -1 1 t Solutions 2-13 M. J. Roberts - 10/7/06 (c) 1 -1 (d) 1 1 t g(t) t g(t) 1 1 Multiplication t 1 1 Multiplication t g(t) -1 -1 -1 1 g(t) t 1 1 t (e) (f) 1 ... -1 1 1 -1 ... 1 t g(t) t g(t) 1 -1 1 Multiplication -1 1 1 -1 Multiplication t t g(t) 1 g(t) ... -1 -1 1 ... 1 t 1 -1 t (g) 1 -1 (h) 1 -1 -1 1 t g(t) t g(t) 1 1 Division π 1 Division t t g(t) t -1 -1 g(t) 1 1 t Figure E-20 21. Use the properties of integrals of even and odd functions to evaluate these integrals in the quickest way. Solutions 2-14 M. J. Roberts - 10/7/06 1 (a) 1 ( 2 + t dt = ) 1 1 1 2 dt + 1 even t dt = 2 2 dt = 4 1 odd 0 1/ 20 (b) 1/ 20 1/ 20 4 cos 10 t + 8sin 5 t dt = 1/ 20 ( ) () 1/ 20 4cos 10 t dt + even 1/ 20 1/ 20 ( ) 1/ 20 8sin 5 t dt odd () 4 cos 10 t + 8sin 5 t dt = 8 1/ 20 0 1/ 20 ( ) () cos 10 t dt = ( ) 8 10 (c) 1/ 20 4 t cos 10 t dt = 0 odd even odd 1/10 ( ) (d) t sin 10 t dt = 2 1/10 odd odd even 1/10 0 ( ) 1/10 t sin 10 t dt = 2 ( ) t cos 10 t 10 ( ) 1/10 1/10 + 0 0 cos 10 t 10 ( ) dt t sin 10 t dt == 2 1/10 odd odd even ( ) 1 + 100 sin 10 t (10 ) t 1 0 ( ) 1/10 2 0 = 1 50 1 1 1 t (e) 1 e t dt = 2 e dt = 2 e t dt = 2 0 t 0 e =2 1 e 1 even ( 1 ) 1.264 (f) te odd dt = 0 1 odd even Periodic CT Functions 22. Find the fundamental period and fundamental frequency of each of these functions. (a) (b) (c) g t = 10cos 50 t () () () ( ( ) /4 f0 = 25 Hz , T0 = 1 / 25 s g t = 10cos 50 t + ) ) f0 = 25 Hz , T0 = 1 / 25 s g t = cos 50 t + sin 15 t ( ) ( The fundamental period of the sum of two periodic signals is the least common multiple (LCM) of their two individual fundamental periods. The fundamental frequency of the sum of two periodic signals is the greatest common divisor (GCD) of their two individual fundamental frequencies. Solutions 2-15 M. J. Roberts - 10/7/06 f0 = GCD 25,15 / 2 = 2.5 Hz , T0 = 1 / 2.5 = 0.4 s (d) g t = cos 2 t + sin 3 t + cos 5 t 3 / 4 ( ( ) () () ) () ( ) f0 = GCD 1,3 / 2,5 / 2 = 1 / 2 Hz , T0 = 23. 1 =2s 1/ 2 One period of a periodic signal, x t , with period, T0 , is graphed in Figure E-23. Assuming x t has a period T0 , what is the value of x t at time, t = 220ms ? x(t) 4 3 2 1 -1 -2 -3 -4 5ms 10ms 15ms 20ms () () () t T0 Figure E-23 Since the function is periodic with period 15 ms, x 220ms = x 220ms n 15ms where n is any integer. If we choose n = 14 we get x 220ms = x 220ms 14 15ms = x 220ms 210ms = x 10ms = 2 . 24. In Figure E-24 find the fundamental period and fundamental frequency of g t . ( ) ( ) ( ) ( ) ( ) ( ) () g(t) ... ... 1 ... 1 t (a) ... t (b) ... ... 1 + t g(t) ... (c) ... 1 1 ... ... t + t g(t) Figure E-24 (a) (b) (c) f0 = 3 Hz and T0 = 1 / 3 s f0 = GCD 6,4 = 2 Hz and T0 = 1 / 2 s f0 and T0 = 1 s Solutions 2-16 () = GCD ( 6,5) = 1 Hz M. J. Roberts - 10/7/06 Signal Energy and Power of CT Signals 25. Find the signal energy of these signals. (a) (b) Ex = (c) Ex = (d) Ex = x t = 2 rect t xt =Aut () () () 10 2 Ex = 2 rect t () 2 1/ 2 dt = 4 1/ 2 dt = 4 ( ( ) u (t 10 )) 10 2 0 Aut ( ( ) u (t )) dt = A dt = 10 A2 x t =u t ( ) ( ) u (10 t ) ut () ( u 10 t ) 2 0 dt = dt + dt 10 x t = rect t cos 2 t rect t cos 2 t () () ( ) 2 () ( ) 1/ 2 1 dt = cos 2 t dt = 2 1/ 2 2 1/ 2 () 1/ 2 1/ 2 (1 + cos ( 4 t )) dt 1 Ex = 2 1/ 2 dt + 1/ 2 1/ 2 cos 4 t dt = =0 () 1 2 (e) Ex = x t = rect t cos 4 t () () ( ) 2 1/ 2 rect t cos 4 t () ( ) 1/ 2 dt = 1/ 2 cos 2 4 t dt = () 1 2 1/ 2 1/ 2 (1 + cos (8 t )) dt Ex = 1 2 1/ 2 dt + 1/ 2 1/ 2 cos 8 t dt = =0 () 1 2 (f) Ex = x t = rect t sin 2 t () () ( ) 2 rect t sin 2 t () ( ) 1 dt = sin 2 t dt = 2 1/ 2 2 1/ 2 () 1/ 2 1/ 2 (1 cos 4 t dt ( )) Solutions 2-17 M. J. Roberts - 10/7/06 1 Ex = 2 1/ 2 1/ 2 dt 1/ 2 1/ 2 cos 4 t dt = =0 () 1 2 26. A CT signal is described by x t = A rect t + B rect t 0.5 . What is its signal energy? Ex = A rect t + B rect t 0.5 () () ( ) () ( ) 2 dt Since these are purely real functions, Ex = Ex = 2 2 ( A rect (t ) + B rect (t 2 2 0.5 )) 2 dt ( A rect (t ) + B rect (t 1/ 2 1 0.5 + 2 AB rect t rect t 0.5 dt 1/ 2 ) () ( )) Ex = A 27. 2 1/ 2 dt + B 2 0 dt + 2 AB dt = A2 + B 2 + AB 0 Find the average signal power of the periodic signal x(t) in Figure E-27. x(t) 3 2 1 -4 -3 -2 -1 -1 -2 -3 1 2 3 4 t Figure E-27 1 P= T0 28. t0 + T0 xt t0 () 2 1 dt = 3 2 xt 1 () 2 2 1 42 4 t3 dt = 2t dt = t dt = 31 31 33 1 1 1 = 1 8 9 Find the average signal power of these signals. (a) xt =A () 1 Px = lim T T T /2 T /2 A2 A dt = lim T T 2 T /2 dt = lim T /2 T A2 T = A2 T Solutions 2-18 M. J. Roberts - 10/7/06 (b) x t =u t () () () ( Px = lim T 1 T T /2 ut T /2 () 2 dt = lim T 1 T T /2 dt = lim 0 T 1T 1 = T2 2 (c) 1 Px = T0 x t = A cos 2 f0t + T0 / 2 ) 2 A cos 2 f0t + T0 / 2 ( ) A2 dt = T0 T0 / 2 cos 2 2 f0t + T0 / 2 ( ) dt A2 Px = 2T0 T0 / 2 T0 / 2 (1 + cos ( 4 ( f 0t + 2 )) sin 4 f0t + 2 A2 dt = t+ 2T0 4 f0 ( ) T0 / 2 T0 / 2 Px = sin 4 f0T0 / 2 + 2 A2 T0 + 2T0 4 f0 ) sin =0 ( 4 f0T0 / 2 + 2 4 f0 ) = A2 2 The average signal power of a periodic power signal is unaffected if it is shifted in time. Therefore we could have found the average signal power of A cos 2 f0t instead, which is somewhat easier algebraically. ( ) CT Functions 29. Given the function definitions on the left, find the function values on the right. (a) g t = 100sin 200 t + g 0.001 = 100sin 200 (b) (c) g t = 13 4t + 6t 2 g t = 5e 2t e () () () ( /4 ) / 4 = 100sin ( ) ( 0.001 + ) ( / 5+ 2 / 4 = 98.77 ) g 2 = 13 4 2 + 6 2 = 29 () () () = j2 t g 1 / 4 = 5e 30. ( ) 2/ 4 e j2 /4 = 5e 1/ 2 e j /2 j3.03 Let the CT unit impulse function be represented by the limit, ( x ) = lim (1 / a ) tri ( x/ a ) , a > 0 . The function, (1 / a ) tri ( x / a ) has an area of one regardless of the value of a. a 0 Solutions 2-19 M. J. Roberts - 10/7/06 (a) ( 4 x ) = lim (1 / a ) tri ( 4 x / a ) ? This is a triangle with the same height as (1 / a ) tri ( x/ a ) but 1/4 times the base What is the area of the function, a 0 width. Therefore its area is 1/4 times as great or 1/4. ( )( )? This is a triangle with the same height as (1 / a ) tri ( x/ a ) but 1/6 times the base (b) What is the area of the function, 6 x = lim 1 / a tri 6 x / a a 0 ( ) width. (The fact that the factor is “-6” instead of “6” just means that the triangle is reversed in time which does not change its shape or area.) Therefore its area is 1/6 times as great or 1/6. (c) What is the area of the function, and for b negative ? It is simply 1 / b . 31. Using a change of variable and the definition of the unit impulse, prove that ( bx ) = lim (1 / a ) tri ( bx / a ) for b positive a 0 ( a (t t )) = (1 / a ) (t t ) 0 0 . (x) = 0 a t t0 ,x 0, (x) d x = 1 ( ) 0 or t dt t0 ( ) = 0 , where a t t0 a t t0 and = 1 a = Strength = Let Then, for a > 0, Strength = and for a < 0, Strength = a t t0 = ( ) ( ) adt = d ( ) da = 1 a ( )d 1 a = 1 a = 11 = aa = 11 = aa ( ) da ( )d ( )d Therefore for a > 0 and a < 0, Strength = 32. 1 and a a t t0 ( ) (t t0 ) . Using the results of Exercise 31, show that (a) Solutions 2-20 M. J. Roberts - 10/7/06 1 ( ax ) = (1 / a ) n= (x 1 n/a ) ( ax n) . ax From the definition of the periodic impulse, Then, using the property from Exercise 31, (b) Show that the average value of 1 ( ax ) = 1 ( ax ) = ( n/a ) = 1 a (x n/a . ) ( ax ) is one, independent of the value of a 1/ 2 a 1 1/ 2 a The period is 1 / a . Therefore 1 ( ax ) 1 = 1/ a t0 + 1/ a 1 t0 ( ax ) dx = a 1/ 2 ( ax ) dx = a =1 1/ 2 a ( ax ) dx 1/ 2 a Letting = ax 1 ( ax ) = 1/ 2 ( )d (d) Even though ( at ) = (1 / a ) (t ) , ( ax ) (1 / a ) ( x ) 1 1 1 ( ax ) = n= ( ax n) (1 / a ) 1 n= ( x n) = (1 / a ) ( x ) 1 ( ax ) (1 / a ) ( x ) 1 QED Scaling and Shifting Functions 33. Graph these CT singularity and related functions. (a) (c) (e) (g) g t = 2u 4 t g t = 5s gn t () () () () ( ) 4 (b) g t = u 2t () ( ) () () () ( ) () ) ( ) ) (d) (f) (h) g t = 1 + s gn 4 t g t = 3 ramp 2t g t = 6 3t + 9 g t = 5ramp t + 1 g t = 2 t+3 ( ( ) ( Solutions 2-21 M. J. Roberts - 10/7/06 (a) g(t) 2 4 t (b) g(t) 1 t 5 (c) g(t) (d) g(t) 2 4 t 4 -5 t (e) g(t) 10 -6 (f) g(t) 1 t (g) g(t) 2 (h) g(t) 2 -1 1 t -3 t -3 t (i) (k) (m) (o) gt = 4 g t =8 () ( 2 (t 1)) ( 4t ) () (j) (l) (n) (p) g t =2 () () () () 1 (t 2 1/ 2 ) () () () 1 gt = 6 (t + 1) g t = 2 rect t / 3 g t = tri 4t g t = 4 rect t + 1 / 2 g t = 6 tri t 1 / 2 (( ) ) () (j) g(t) (( ) ) (l) ... g(t) -1 1 3 (i) g(t) 1 t 2 (k) g(t) 2 ... t -2 ... 1 ... (n) g(t) 4 ... t 1 4 ... t -6 (m) g(t) 2 (o) g(t) 1 (p) g(t) -1 1 3 t -3 2 3 2 t -2 t -1 4 1 4 t -6 (q) (s) g t = 5sinc t / 2 () () () () (r) (t) g t = sinc 2 t + 1 g t = 5drcl t / 4,7 () ( ( )) ) g t = 10drcl t ,4 () ( Solutions 2-22 M. J. Roberts - 10/7/06 (q) g(t) 5 (r) g(t) -1 t t 2 -1 (s) g(t) 10 (t) g(t) 5 4 -10 t -1 8 t -3rect(t-2) 3 2 5 2 0.1rect t-3 4 0.1 () 1 35 (u) t (v) -3 t -4tri 3+t 2 (w) () t -4 (x) 4sinc[5(t-3)] 4 -5 3 -1 -1 123456 t 4sinc(5t-3) 4 (y) 34. -1 123456 t Graph these CT functions. (a) (c) g t =u t ( ) ( ) u (t 1) () () ( 2 (b) g t = rect t 1 / 2 () () ( ) g t = 4 ramp t u t ) (d) g t = s gn t sin 2 t () ( ) Solutions 2-23 M. J. Roberts - 10/7/06 (e) (g) g t = 5e () () t/4 ut () (f) (h) g t = rect t cos 2 t g t = rect t tri t () () () ( ) () () (d) g(t) g t = 6 rect t cos 3 t () ( ) (b) g(t) 1 (a) g(t) 1 (c) g(t) 2 t -8 -16 -1 4 t 1 t 1 1 t 1 (e) g(t) 5 t -1 2 4 (f) g(t) 1 1 2 -1 (g) g(t) 6 t (h) g(t) 1 1 2 1 2 -1 2 1 2 t t -1 2 -6 (i) (k) g t = rect t tri t + 1 / 2 g t = tri2 t () () () () ( () ) (j) (l) g t = u t + 1 / 2 ramp 1 / 2 t g t = sinc 2 t gt = () ( () () ) ( ) () () (m) (o) (p) g t = sinc t (n) d tri t dt ( ( )) g t = rect t + 1 / 2 gt = () () ( ) 2 rect t 1 / 2 ( ) 1d t ( +1 ) ( )+ ( ) (i) g(t) 1 (j) g(t) 1 (k) g(t) 1 t (l) g(t) 1 -1 2 1 2 t -1 2 1 2 t -1 1 -1 1 t (m) g(t) 1 -1 (n) g(t) 1 1 -1 1 t t -1 (o) g(t) 1 1 -1 t -1 (p) g(t) 1 1 -1 t Solutions 2-24 M. J. Roberts - 10/7/06 6 (q) g t = 3tri 2t / 3 + 3rect t / 3 () ( ) () -3 2 3 , t 3 2 6 (r) g t = 6 tri t / 3 rect t / 3 () () () -3 2 3 t 3 2 4 (s) 4 sinc 2t sgn ( ) ( t) t -1 2 1 2 , -4 (t) g t = 2 ramp t rect t 1 / 2 () () () (( ) ) 4 4 t 2 (u) g t = 4 tri t (( 2 /2 u 2 t ) )( ) 2 3 t , (v) g t = 3rect t / 4 () () () ( 6 rect t / 2 () t -2 -1 -3 12 (w) g t = 10drcl t / 4,5 rect t / 8 ) () (g) g(t) 10 -8 -2 8 t 35. Graph the following CT functions. Solutions 2-25 M. J. Roberts - 10/7/06 (a) g t = 3 3t + 6 () () ( 4 ( t 2 )) ( ) (t ) + (3 / 2) (t 2) g(t) 3 2 1 t 2 Using the impulse scaling property, g t = (b) g t =2 () 1 ( ( t /5 ) g(t) g t =2 n= () t / 5 n = 10 n= ) (t + 5n) , 10 ... -10 -5 ... 5 10 15 20 t (c) gt = () ( 1 (t ) rect (t / 11) g(t) g t = rect t / 11 () ) n= (t n) = 5 n= 5 (t n) , 1 t 12345 -5 -4 -3 -2 -1 (d) g t = 5sinc t / 4 () ( ) (t ) 2 g t = 5sinc t / 4 () () n= (t 5 2n = 5 n= ) sinc n / 2 ( ) (t 2n ) g(t) t 2 6 10 Solutions 2-26 M. J. Roberts - 10/7/06 (e) gt = () t 2 () 2 ( 1d g(t) 1 ) -2 -1 1 2 3 t 36. A function, g t , has the following description. It is zero for t < 5 . It has a slope of –2 in the range 5 < t < 2 . It has the shape of a sine wave of unit amplitude and with a frequency of 1 / 4 Hz plus a constant in the range, 2 < t < 2 . For t > 2 it decays exponentially toward zero with a time constant of 2 seconds. It is continuous everywhere. (a) Write an exact mathematical description of this function. 0 gt = , t< 5 10 2t , 5 < t < 2 () () sin ( t/2 t/2 ) , 2<t<2 6e (b) (c) (d) (e) , t>2 Graph g t in the range, 10 < t < 10 . Graph g 2t in the range, 10 < t < 10 . Graph 2 g 3 t in the range, 10 < t < 10 . Graph 2 g t + 1 / 2 in the range, 10 < t < 10 . () ( () ) (( ) ) Solutions 2-27 M. J. Roberts - 10/7/06 g(t) -10 10 g(2t) t -10 10 t -8 -8 2g(3- t) -10 10 -2g(( t+1)/2) t 16 -16 -10 10 t 37. Using MATLAB, for each function below plot the original function and the transformed function. % Plotting functions and transformations of those functions close all ; % (a) part tmin = -2 ; tmax = 2 ; N = 400 ; dt = (tmax - tmin)/N ; t = tmin + dt*[0:N]’ ; g0 = g322a(t) ; g1 = 5*g322a(2*t) ; subplot(2,1,1) ; p = plot(t,g0,’k’) ; set(p,’LineWidth’,2) ; grid on ; ylabel(‘g(t)’) ; subplot(2,1,2) ; p = plot(t,g1,’k’) ; set(p,’LineWidth’,2) ; grid on ; xlabel(‘t’) ; ylabel(‘5g(2t)’) ; % (b) part figure ; tmin = -3 ; tmax = 8 ; N = 100 ; dt = (tmax - tmin)/N ; t = tmin + dt*[0:N]’ ; g0 = g322b(t) ; g1 = -3*g322b(4-t) ; subplot(2,1,1) ; p = plot(t,g0,’k’) ; set(p,’LineWidth’,2) ; grid on ; ylabel(‘g(t)’) ; subplot(2,1,2) ; p = plot(t,g1,’k’) ; set(p,’LineWidth’,2) ; grid on ; xlabel(‘t’) ; ylabel(‘-3g(4-t)’) ; % (c) part figure ; tmin = 0 ; tmax = 96 ; N = 400 ; dt = (tmax - tmin)/N ; t = tmin + dt*[0:N]’ ; g0 = g322c(t) ; g1 = g322c(t/4) ; subplot(2,1,1) ; p = plot(t,g0,’k’) ; set(p,’LineWidth’,2) ; grid on ; ylabel(‘g(t)’) ; subplot(2,1,2) ; p = plot(t,g1,’k’) ; set(p,’LineWidth’,2) ; grid on ; xlabel(‘t’) ; ylabel(‘g(t/4)’) ; % (d) part Solutions 2-28 M. J. Roberts - 10/7/06 figure ; fmin = -20 ; fmax = 20 ; N = 200 ; df = (fmax - fmin)/N ; f = fmin + df*[0:N]’ ; G0 = G322d(f) ; G1 = abs(G322d(10*(f-10)) + G322d(10*(f+10))) ; subplot(2,1,1) ; p = plot(f,G0,’k’) ; set(p,’LineWidth’,2) ; grid on ; ylabel(‘G(f)’) ; subplot(2,1,2) ; p = plot(f,G1,’k’) ; set(p,’LineWidth’,2) ; grid on ; xlabel(‘f’) ; ylabel(‘|G(10(f-10)) + G(10*(f+10))|’) ; function y = g322a(t) g = (1-abs(t)).*(-1 < t & t < 1) ; y = 10*cos(20*pi*t).*g ; function y = g322b(t) y = -2*(t <= -1) + 2*t.*(-1 < t & t <= 1) + ... (3-t.^2).*(1 < t & t <= 3) - 6*(t > 3) ; function y = g322c(t) y = real(exp(j*pi*t) + exp(j*1.1*pi*t)) ; function y = G322d(f) y = abs(5./(f.^2 - j*2 + 3)) ; (a) g t = 10cos 20 t tri t () ( ) () 10 5 g 2t () vs. t Original g(t) -2 -10 2 t Transformed g(t) 50 -2 -50 2 t (b) gt = () 2 , t< 1 2t , 1 < t < 1 3 t2 , 1 < t < 3 6 , t>3 3g 4 t ( ) vs. t Solutions 2-29 M. J. Roberts - 10/7/06 Original g(t) 2 -4 8 t -6 Transformed g(t) 20 -4 -10 8 t (c) g t = Re e j t + e j1.1 Original g(t) 2 () ( t ) g t/4 () t vs. t 100 -2 Transformed g(t) 2 100 -2 t (d) Gf= () f 2 5 j2 + 3 G 10 f Original g(t) 1.5 (( 10 + G 10 f + 10 )) (( )) vs. f -20 20 t Transformed g(t) 1.5 -20 20 t Solutions 2-30 M. J. Roberts - 10/7/06 38. The voltage illustrated in Figure E38 occurs in an analog-to-digital converter. Write a mathematical description of it. Signal in A/D Converter x(t) 5 -0.1 0.3 t (ms) Figure E38 Signal occurring in an A/D converter t 5 10 x t = tri 5 10 5 3 9. () 5 t 1.5 10 + tri 5 10 5 4 A signal occurring in a television set is illustrated in Figure E39. mathematical description of it. Signal in Television x(t) 5 -10 60 Write a t (µs) -10 Figure E39 Signal occurring in a television set x t = 10 rect 40. () t 2.5 10 5 10 6 6 The signal illustrated in Figure E40 is part of a binary-phase-shift-keyed (BPSK) binary data transmission. Write a mathematical description of it. BPSK Signal x(t) 1 4 -1 t (ms) Figure E40 BPSK signal Solutions 2-31 M. J. Roberts - 10/7/06 xt = () sin 8000 t rect + sin 8000 t rect ( ) t 0.5 10 10 3 t 3 sin 8000 t rect 3 ( ) t 1.5 10 10 3 3 ( ) 2.5 10 10 3 sin 8000 t rect ( ) t 3.5 10 10 3 3 4 1. The signal illustrated in Figure E41 is the response of an RC lowpass filter to a sudden change in excitation. Write a mathematical description of it. On a decaying exponential, a tangent line at any point intersects the final value one time constant later. Theconstant value before the decaying exponential is -4 V and the slope of the tangent line at 4 ns is -2.67V/4 ns or -2/3 V/ns. RC Filter Signal x(t) -1.3333 -4 -6 4 20 t (ns) (8000 t ) rect (t 42. 0.5 10 3 ) Figure E41 Transient response of an RC filter sin 8000 t rect t 1.5 10 3 + sin 8000 t rect t ( t 4 /3 xt = 4 21 e( ) () ) ( ( ( ) u (t 4) ) ) ( 2.5 10 3 ) sin 8000 t rect t 3.5 ( ) ( (times in ns) Describe the signal in Figure E42 two ways. x(t) 15 ... 4 Figure E42 (a) As a ramp function minus a summation of step functions. x t = 3.75ramp t (b) t () () 15 n=0 ut ( 4n ) As a summation of products of triangle functions and rectangle functions. x t = 15 n=0 () rect t t 2 4n tri 4 4 4n 4 Solutions 2-32 M. J. Roberts - 10/7/06 43. Mathematically describe the signal in Figure E-43 . x(t) Semicircle ... 9 9 Figure E-43 ... t The semicircle centered at t = 0 is the top half of a circle defined by x 2 t + t 2 = 81 Therefore x t = 81 t 2 , () () 9<t<9 . This one period of this periodic function. The other periods are just shifted versions. xt = n= () rect t 18n 18 81 (t 18n ) 2 (The rectangle function avoids the problem of imaginary values for the square roots of negative numbers.) 44. Let two signals be defined by x1 t = () 1 , cos 2 t 0, ( )1 cos ( 2 t ) < 1 and x 2 t = sin 2 t / 10 () ( ) . Plot these products over the time range, 5 < t < 5 . (a) (c) x1 2t x 2 ( ) ( t) ( )(( )) (d) (b) x1 t x1 t / 5 x 2 20t 2 / 5 x 2 20t ( )() x1 t / 5 x 2 20 t + 1 (( ))() Solutions 2-33 M. J. Roberts - 10/7/06 (a) x1(t)x2(t) 1 (b) x1(t)x2(t) 1 -5 -1 5 t -5 -1 5 t (c) x1(t)x2(t) 1 (d) x1(t)x2(t) 1 -5 -1 5 t -5 -1 5 t 45. Given the graphical definition of a function in Figure E-45, graph the indicated transformation(s). (a) g(t) 2 1 -2 -2 g t = 0 , t > 6 or t < 2 (b) 1 23456 t () g (t ) gt g 2t () 3g ( t ) () g(t) 2 1 -2 123456 -2 t gt t () t+4 2g t 1 / 2 (( ) ) g t is periodic with fundamental period, 4 Figure E-45 (a) () Solutions 2-34 M. J. Roberts - 10/7/06 () The transformation g ( t ) The transformation, g t g 2t , simply compresses the time scale by a factor of 2. () 3 g ( t ) time inverts the signal, amplitude inverts the g(2t) 2 1 signal and then multiplies the amplitude by 3. -2 -2 2 4 6 t -3g(-t) 6 3 -4 (b) -2 -6 2 4 6 t g(t + 4) 2 1 -2 -2 1234567 t -2g( t -1 ) 2 4 2 -2 -4 46. For each pair of functions graphed in Figure E-46 determine what transformation has been done and write a correct functional expression for the transformed function. (a) 12345678 t Solutions 2-35 M. J. Roberts - 10/7/06 g(t) 2 -2 -1 (b) 2 123456 t -4 -3 -2 -1 -1 1234 t g(t) 2 -2 123456 t -2 -1 123456 t In (b), assuming g t is periodic with fundamental period, 2, find two different transformations which yield the same result Figure E-46 (a) It should be visually obvious that the transformed signal has been time inverted and time shifted. By identifying a few corresponding points on both curves we see that after the time inversion the shift is to the right by 2. This corresponds to two successive transformations, t t , followed by t t 2 . The overall effect of the t 2 =2 t. Therefore the two successive transformations is then t transformation is gt g2 t () ( ) () ( ) 47. (1 / 2) g (t + 1) or g (t ) (1 / 2) g (t 1) Let a function be defined by g ( t ) = tri ( t ) . Below are four other functions based on (b) gt this function. All of them are zero for large negative values of t. () () (( ) ) g ( t ) = g ( t + 2 ) 4 g (( t + 4 ) / 3) g1 t = 5 g 2 t / 6 3 g 2 t = 7 g 3t g4 ( ) ( ) 4 g (t 4) (t ) = 5 g (t ) g (t 1 / 2) () (a) Which of these transformed functions is the first to become non-zero g3 t (becomes non-zero at the earliest time)? Which of these transformed functions is the last to go back to zero and stay there? g1 t (b) () Solutions 2-36 M. J. Roberts - 10/7/06 (c) Which of these transformed functions has a maximum value that is greater than all the other maximum values of all the other transformed functions? g2 t () (d) Which of these transformed functions has a minimum value that is less than all the other minimum values of all the other transformed functions? g1 t () 48. Name a function of continuous time t for which the two successive transformations t t and t t 1 leave the function unchanged. cos 2 t , 1 t , etc... (Any even periodic function with a period of one.) () () 49. Graph the magnitude and phase of each function versus f. (a) G f = sinc f e () () j f /8 This function is a ratio of two functions, jf and 1 + jf/10. The magnitude of the ratio is the ratio of the magnitudes. At very low values of f , the ratio approaches 0 because the numerator approaches 0 and the denominator approaches 1. At very high values of f the denominator is approximately jf/10 and the magnitude of the ratio approaches 10. All these statements are equally true for positive and negative f. Therefore the magnitude is an even function of f. The phase of the ratio is the phase of the numerator minus the phase of the denominator. For any positive f , the phase of the numerator is the phase of j times a positive constant. That is some number on the positive imaginary axis in the complex plane. So the phase is / 2 radians or 90°. For very small positive f, the denominator is approximately just the real number, 1, whose phase is 0. Therefore for very small positive f approaching 0 the phase approaches / 2 . For very large positive f , the phase of the denominator approaches / 2 also and the difference between the numerator and denominator phases approaches 0. The behavior for / 2 . So the negative f is similar except that the phase of the numerator is now phase for negative f is exactly the negative of the phase for the corresponding positive f. That is, the phase is an odd function of f. (b) Gf= () jf 1 + jf / 10 Solutions 2-37 M. J. Roberts - 10/7/06 (a) |G( f )| 1 (b) |G( f )| 10 -16 16 f -100 100 f Phase of G( f ) π Phase of G( f ) π -16 -π 16 f -100 -π 100 f (c) (d) G f = rect Gf= () () f f + 1000 1000 + rect 100 100 e j f /500 250 1 f 2 + j3 f (c) |G( f )| 1 0.02 (d) |G( f )| -1100 1100 f -50 50 f Phase of G( f ) π Phase of G( f ) π -1100 -π 1100 f -50 -π j f /50 50 f (e) Gf= () 0.01 ( f ) sinc ( 25 f ) e 100 Solutions 2-38 M. J. Roberts - 10/7/06 (e) |G( f )| 0.01 -0.2 0.2 f Phase of G( f ) π -0.2 -π 0.2 f 50. Graph versus f , in the range, 4 < f < 4 , the magnitude and phase of (a) X f = sinc f () () The phase in this plot is the phase of a purely real function. If we only plotted purely real functions we would not need to graph magnitude and phase separately. A simple real plot of the function would be sufficient and clearer. But most transforms that we will later graph are complex functions and magnitude and phase plots are good ways of representing them. Since this function is purely real its value always lies on the real axis of the complex plane. When it is positive the simplest phase answer is 0. When it is negative the simplest phase answer is either positive or negative radians. Later, in the study of transform methods applied to systems, we will find that we always have an even magnitude and an odd phase. For that reason, it is consistent and logical to choose phase values so as to make the plot an odd function. Here that for positive f and is done by making the phase for negative values of sinc f be for negative f. () |X( f )| 1 -4 -3 -2-1 1234 f X( f ) π -4 -3 -2-1 1234 -π f Solutions 2-39 M. J. Roberts - 10/7/06 |X( f )| 2 -4 -3 -2-1 1234 f (b) X f = 2sinc f e () () j4 f X( f ) -4 -3 -2-1 4π -4π 1 2 3 4 π f |X( f )| 5 (c) X f = 5rect 2 f e+ j 2 () () 1 4 1 4 f f X( f ) π 2 π 2 f |X( f )| 10 (d) X f = 10sinc 2 f / 4 () ( ) -4 -3 -2-1 1234 f X( f ) -4 -3 -2-1 1234 f |X( f )| 5 (e) X f = j5 () ( f + 2) j5 (f 2 ) -4 -3 -2-1 1234 f X( f ) π 2 -4 -3 -2-1 (f) X f = 1/ 2 ()( ) ( f )e 1/ 4 π 2 1234 f jf Solutions 2-40 M. J. Roberts - 10/7/06 |X( f )| 1 2 ... 311 -1 4 2 4 1131 424 ... f Xf= () e jf 2 f n= n 4 X( f ) 4π 3π 2π π -4 -3 -2 -1π -2π -3π -4π 1234 f Generalized Derivative 51. Graph the generalized derivative of g t = 3sin () ( t / 2 rect t . ) () Except at the discontinuities at t = ±1 / 2 , the derivative is either zero, for t > 1 / 2 , or it is the derivative of 3sin t / 2 , 3 / 2 cos t / 2 , for t < 1 / 2 . At the discontinuities the generalized derivative is an impulse whose strength is the difference between the limit approached from above and the limit approached from below. In both cases that strength is 3 / 2 . ( )( )( ) d (g(t)) dt 3π 2 t 3 2 Alternate solution: g t = 3sin () ( t / 2 u t +1/ 2 )( ) u (t 1/ 2 ) )( ) u (t ) 1/ 2 d g t = 3sin dt ( ( )) ( t/2 ) (t + 1 / 2) (t /4 1 / 2 + 3 / 2 cos )( )( t / 2 u t +1/ 2 ) d g t = 3sin dt ( ( )) ( ) (t + 1 / 2) 3sin ( /4 ) (t 1 / 2 + 3 / 2 cos )( )( ) t / 2 rect t () d g t = 3 2/2 dt ( ( )) (t + 1 / 2) + (t 1 / 2 + 3 / 2 cos )( )( t / 2 rect t () Solutions 2-41 M. J. Roberts - 10/7/06 Derivatives and Integrals of CT Functions 52. What is the numerical value of each of the following integrals? (a) 20 (t ) cos ( 48 t ) dt = cos (0) = 1 , (b) (t 5) cos ( t ) dt = cos (5 ) = 1 (c) 0 20 3 (t 8) tri (t / 32) dt = tri (8 / 32) = 4 (d) 0 (t 8) rect (t / 16) dt = rect (8 / 16) = 1 / 2 (t 1.5 sinc t dt = sinc 1.5 = 2 (e) 2 2 ) () () sin 3 / 2 3 /2 ( )= 2 3 (f) 2 (t 1.5 sinc 4t dt = sinc 4 1.5 = ) () ( ) sin 6 =0 6 53. What is the numerical value of each of the following integrals? (a) 1 (t ) cos ( 48 t ) dt = 1 n= (t n) cos ( 48 t ) dt = cos 48n n= n= (t n) cos ( 48 t ) dt 1 (t ) cos ( 48 t ) dt = ( )= n= x(t) 1 -0.1 -1 0.1 t (b) 1 (t ) sin ( 2 t ) dt 1 (t ) sin ( 2 t ) dt = n= (t n) sin ( 2 t ) dt = sin 2 n n= ( )=0 Solutions 2-42 M. J. Roberts - 10/7/06 x(t) 1 -2 -1 2 t 20 (c) 20 4 0 4 (t 2) rect (t ) dt ) () 20 n= 4 0 4 ( t 2 rect t dt = 4 0 (t 2 4 n rect t dt = 4 n= ) () rect 2 + 4 n = 0 ( ) x(t) 4 -8 2 2 8 t (d) 2 1 (t ) sinc (t ) dt = n= (t n) sinc (t ) dt = x(t) 4 sinc n = 1 n= () 2 -8 8 t 54. Graph the derivatives of these functions. (a) (b) g t = sin 2 t s gn t g t = 2 tri t / 2 () () () () () g t =2 () () () cos ( 2 t ) cos 2 t , t < 0 ,t 0 1 1 , 2<t<0 g t = 1 , 0<t<2 0 , otherwise 1+ t , 2 < t < 0 g t = 1 t , 0<t<2 0 , otherwise () (c) g t = cos 2 t () () g t =2 () sin 2 t ( ) , cos ( 2 t ) < 0 sin ( 2 t ) , cos ( 2 t ) > 0 Solutions 2-43 M. J. Roberts - 10/7/06 (a) x(t) 1 -4 -1 4 (b) x(t) 1 1 (c) x(t) 4 t -4 -1 t -1 -1 1 t dx/dt 6 -4 -6 4 dx/dt 1 dx/dt 6 t -4 -1 4 t -1 -6 1 t Even and Odd CT Functions 55. Graph the even and odd parts of these CT signals. (a) xe t = x t = rect t 1 () ( ) ( ) , xo t = () rect t 1 + rect t + 1 2 ( ) () rect t 1 ( ) 2 rect t + 1 ( ) (b) x t = tri t 3 / 4 + tri t + 3 / 4 () ( ( ) ( ) , (b) xe(t) 1 x e t = tri t 3 / 4 + tri t + 3 / 4 (a) xe(t) 1 () ) ( ) x0 t = 0 () -3 -1 3 t -3 -1 3 t xo(t) 1 xo(t) 1 -3 -1 3 t -3 -1 3 t (c) x t = 4sinc t 1 / 2 () (( ) ) (( t 1 /2 xe t = () 4sinc t 1 / 2 + 4sinc 2 (( ) ) ) )=2 sin t 1 (t 1) (( ) /2 /2 ) + sin (( ( t1 t1 ) ) /2 /2 ) Solutions 2-44 M. J. Roberts - 10/7/06 xe t = 2 Using () sin (t 1) t 2 2 /2 sin + ( ) t 2 t1 ) 2 /2 sin =4 (t 1) t 2 2 sin (t + 1) t 2 2 sin x + y = sin x cos y + cos x sin y ( () () (t 1) ) () () )( =1 sin xe t = 4 ( t / 2 cos )( =0 / 2 + cos =0 ( t / 2 sin /2 ) /2 () sin ( t / 2 cos )( / 2 + cos (t + 1) ) ( t 2 t / 2 sin )( =1 ) xe t = 4 () cos (t + 1) () ( t/2 ) cos (t 1) t 2 ( t/2 ) 2 = 4cos (t 1) (t + 1) (t + 1) (t 1) x e t = 4cos (t 1) 2 o =8 cos (t 1) 2 ( t/2 ) ) xe t = 8 () cos (t 1) 2 ( t/2 ) Similarly, x (t ) = 8t cos (t 1) 2 ( t/2 (d) x t = 2sin 4 t () ( / 4 rect t ) () x o t = 2cos x e t = 2 sin () ( / 4 cos 4 t rect t , )() () () ( / 4 sin 4 t rect t )() () Solutions 2-45 M. J. Roberts - 10/7/06 (c) xe(t) 4 (d) xe(t) 2 -10 -4 10 t -1 -2 1 t xo(t) 4 xo(t) 2 -10 -4 10 t -1 -2 1 t 56. Find the even and odd parts of each of these CT functions. (a) g t = 10sin 20 t () ( ) )=0 , xo t = ge t = () 10sin 20 t + 10sin 2 g t = 20t 3 ge t = ( ) ( 20 t () 10sin 20 t ( ) 10sin 2 ( 20 t ) = 10sin 20 t () (b) () () 20t 3 + 20 2 2 ( t) 2 3 = 0 , go t = () 20t 3 20 2 ( t) 3 = 20t 3 (c) x t = 8 + 7t xe t = () (d) () x (t ) = 1 + t x t = 6t 8 + 7t 2 + 8 + 7 2 ( t) = 8 + 7t , x o t = 2 () 8 + 7t 2 87 2 ( t) 2 =0 xe t = (e) () 1+ t + 1+ 2 6t + 6 2 ( t) = 1 , xo t = () 1+ t 1 2 6t 6 2 ( t) = t () (f) () g ( t ) = 4t cos (10 t ) ge t = ge t = ( t) = 0 , go t = () ( t ) = 6t () 4t cos 10 t + 4 ( ) ( t ) cos ( 2 10 t ) = 4t cos (10 t ) + 4 ( t ) cos (10 t ) = 0 2 Solutions 2-46 M. J. Roberts - 10/7/06 go t = () 4t cos 10 t ( ) 4 ( t ) cos ( 2 10 t ) = 4t cos (10 t ) 4 ( t ) cos (10 t ) = 4t cos 10 t () 2 (g) gt = () cos ( t) t cos ge t = cos go t = () ( t ) + cos ( t ) t t 2 cos 2 cos = cos ( t ) + cos ( t ) t t 2 =0 () ( t) t ( t) t = ( t ) + cos ( t ) t t 2 = cos ( t) t (h) g t = 12 + sin 4 t 4t () sin 4 t 4t () 4t ge t = () 12 + ( ) + 12 + sin ( () ) = 4t 12 + sin 4 t 4t ( ) + 12 + sin ( 4 t ) 4t 2 sin 4 t = 4t 2 2 sin 4 t 12 + 4t go t = = 12 + sin 4 t 4t () (i) () 2 g ( t ) = (8 + 7t ) cos ( 32 t ) ge t = 12 sin ( 4t ) () sin 4 t 4t () =0 4t () (8 + 7t ) cos (32 t ) + (8 2 7t cos )( 32 t ) = 8cos 32 t () go t = (j) () (8 + 7t ) cos (32 t ) (8 2 7t cos )( 32 t ) = 7t cos 32 t () g t = 8 + 7t 2 sin 32 t 2 () ( )( ) 2 ge t = 2 () (8 + 7t ) sin (32 t ) + (8 + 7 ( t ) ) sin ( 2 2 32 t ) =0 go t = 57. () (8 + 7t ) sin (32 t ) (8 + 7 ( t ) )sin ( 32 t ) = (8 + 7t )sin (32 t ) 2 2 Is there a function that is both even and odd simultaneously? Discuss. Solutions 2-47 M. J. Roberts - 10/7/06 The only function that can be both odd and even simultaneously is the trivial signal, x t = 0 . Applying the definitions of even and odd functions, () xe t = () 0+0 00 = 0 = x t and x o t = =0=x t 2 2 () () () proving that the signal is equal to both its even and odd parts and is therefore both even and odd. 58. Find and graph the even and odd parts of the CT function x t in Figure E-58 () x(t) 2 1 -5 -4 -3 -2 -1 -1 Figure E-58 12345 t xo(t) xe(t) 2 1 12345 t 2 1 -5 -4 -3 -2 -1 -1 12345 t -5 -4 -3 -2 -1 -1 Periodic CT Functions 59. For each of the following signals decide whether it is periodic and, if it is, find the period. (a) (b) (c) (d) () ( ) Periodic. Fundamental frequency = 200 Hz, Period = 5 ms. g ( t ) = 14 + 40cos ( 60 t ) Periodic. Fundamental frequency = 30 Hz Period = g t = 28sin 400 t 33.33...ms. g t = 5t 2 cos 5000 t () ( ) Not periodic. g ( t ) = 28sin ( 400 t ) + 12cos (500 t ) Periodic. Two sinusoidal components with periods of 5 ms and 4 ms. Least common multiple is 20 ms. Period of the overall signal is 20 ms. Solutions 2-48 M. J. Roberts - 10/7/06 (e) (f) g t = 10sin 5t () () () 4 cos 7t () Periodic. The Periods of the two sinusoids are 2 / 5 s Not periodic because least common multiple is infinite. and 2 / 7 s. Least common multiple is 2 . Period of the overall signal is 2 s. g t = 4sin 3t + 3sin () ( 3t ) Signal Energy and Power of CT Signals 60. Find the signal energy of each of these signals. (a) (b) 2 rect ( t) , E= 2 rect rect 8t ( t) 2 2 1/ 2 dt = 4 1/ 2 1/16 dt = 4 1 8 rect 8t () t 4 , , E= , E= () dt = 1/16 2 dt = 2 (c) (d) 3 rect tri 2t 3rect t 4 dt = 9 dt = 36 2 () Keep in mind that the square of a rectangle function is another rectangle function but the square of a triangle function is not another triangle function. Use the definition of the triangle function and the fact that a triangle function is even to reduce the work. E= 0 tri 2t 2 () 1/ 2 0 2 1/ 2 dt = 1/ 2 (1 2 2t ) dt = 2 1/ 2 1/ 2 2 (1 2 2 2t + 2t 0 2 ) dt 1/ 2 2 E= 1/ 2 (1 + 4t + 4t ) dt + ( t 3 tri 4 0 t3 1 4t + 4t dt = t + 2t + 4 3 t 3tri 4 2 4 ) +t 1/ 2 t3 2t + 4 3 2 = 0 1 3 (e) , E= 4 dt = 9 4 1 t 4 4 dt = 9 4 0 t t 12 + 44 t2 t3 + 4 48 dt 4 E=9 4 t t2 1+ + dt + 2 16 2 sin 200 t 0 t t2 1 + dt = 9 2 16 t2 t3 t+ + 4 48 +t 4 = 24 0 (f) ( ) E= 2sin 200 t ( ) 2 dt = 4 sin 2 200 t dt = 4 cos 400 t 400 Solutions 2-49 ( ) 1 2 1 sin 400 t dt 2 ( ) E=2 t+ ( ) M. J. Roberts - 10/7/06 (g) (Hint: First find the signal energy of a signal which approaches an impulse some limit, then take the limit.) (t ) (t ) = lim (1 / a ) rect (t / a ) a 0 E= t 1 lim rect a 0a a d rect t dt 2 1 dt = lim 2 a 0a a/2 rect a/2 t a dt = lim 2 a 0a a (h) xt = () ( ( )) 1/ 2 d rect t = dt Ex = ( ( )) ( t + 1 / 2 ) ( t (t + 1 / 2) (t t ) 1/ 2 ) 2 dt (i) xt = 1/ 2 () rect ( )d 2 = ramp t + 1 / 2 ( ) ramp t 1 / 2 ( ) Ex = 1/ 2 (t + 1 / 2) finite 1 j8 dt + 1/ 2 dt infinite (j) x t = e( Ex = () )t ut () e( 1 j8 xt () 2 dt = )t ut () 2 dt = e( 0 1 j8 )t ( e 1+ j 8 )t dt Ex = e 2t dt = 0 e 2t 2 = 0 1 2 61. Find the average signal power of each of these signals: (a) x t = 2sin 200 t 1 Px = T T /2 () ( ) This is a periodic function. Therefore 2sin 200 t T /2 ( ) 2 4 dt = T T /2 T /2 1 2 1 cos 400 t dt 2 ( ) Solutions 2-50 M. J. Roberts - 10/7/06 2 t Px = T sin 400 t 400 ( ) T /2 = T /2 2T T2 sin 200 T 400 ( ) + T + sin ( 2 200 T 400 ) =2 For any sinusoid, the average signal power is half the square of the amplitude. (b) xt = and +T / 2 , there is one impulse whose energy is infinite. Therefore the average power is the energy in one period, divided by the period, or infinite. (c) 1 Px = T0 Px = 50 1/100 () () 1 (t ) This is a periodic signal whose period, T, is 1. Between T /2 x t = e j100 t This is a periodic function. Therefore T0 / 2 2 1/100 T0 xt () 2 1 dt = T0 e T0 / 2 j100 t dt = 50 1/100 e j100 t e j100 t dt 1/100 dt = 1 62. A CT signal, x, is periodic with fundamental period, T0 = 6 . This signal is described over the time period, 0 < t < 6 , by rect t (( 2 /3 )) 4 rect t (( 4 /2 . )) What is the signal power of this signal? The signal, x, can be described in the time period, 0 < t < 6 , by 0 , 0 < t <1/ 2 1 , 1/ 2 < t < 3 x t = 3 , 3< t < 7 / 2 4 , 7/2<t <5 0 , 5<t <6 () The signal power is the signal energy in one fundamental period divided by the fundamental period. P= 12 0 6 12 +1 2 5 + 2 ( 3) 2 1 + 2 ( 4) 2 2.5 + 4.5 + 24 3 + 02 1 = = 5.167 6 2 Solutions 2-51 ...
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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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