Exam (elaborations) TEST BANK FOR Signals and Systems Analysis of Signals Through Linear Systems By M.J. Roberts (Solution manual)

Exam (elaborations) TEST BANK FOR Signals and Systems Analysis of Signals Through Linear Systems By M.J. Roberts (Solution manual) Mathematical Description of Signals Solutions 1. If g(t) = 7e−2t −3 write out and simplify (a) g(3) = 7e−9 (b) g(2 − t) = 7e−2(2−t)−3 = 7e−7+2t (c) g t e t 10 4 7 5 11 +      = − − (d) g( jt) = 7e− j 2t −3 (e) g g cos jt jt e e e e t ( ) + (− ) j t j t = + − = ( ) − − 2 7 2 3 7 2 2 2 3 (f) g g cos jt jt e e t jt jt −      + − −      = + = ( ) − 3 2 3 2 2 7 2 7 2. If g(x) = x2 − 4x + 4 write out and simplify (a) g(z) = z2 − 4z + 4 (b) g(u + v) = (u + v) − (u + v) + = u + v + uv − u − v + 2 4 4 2 2 2 4 4 4 (c) g(e jt ) = (e jt ) − e jt + = e j t − e jt + = (e jt − ) 2 2 2 4 4 4 4 2 (d) g(g(t)) = g(t2 − 4t + 4) = (t2 − 4t + 4)2 − 4(t2 − 4t + 4) + 4 g(g(t)) = t4 − 8t3 + 20t2 −16t + 4 (e) g(2) = 4 − 8 + 4 = 0 3. What would be the numerical value of “g” in each of the following MATLAB instructions? (a) t = 3 ; g = sin(t) ; 0.1411 (b) x = 1:5 ; g = cos(pi*x) ; [-1,1,-1,1,-1] (c) f = -1:0.5:1 ; w = 2*pi*f ; g = 1./(1+j*w) ; M. J. Roberts - 7/12/03 Solutions 2-2 1 . . . . . . . . + + − −       j j j j 4. Let two functions be defined by x , sin , sin 1 1 20 0 1 20 0 t t t ( ) = ( )≥ − ( )  π π and x , sin , sin 2 2 0 2 0 t t t t t ( ) = ( )≥ − ( )  π π . Graph the product of these two functions versus time over the time range, −2 t 2. t -2 2 x(t) -2 2 5. For each function, g(t), sketch g(−t), −g(t), g(t −1), and g(2t). (a) (b) t g(t) 2 4 t g(t) 1 -1 3 -3 t g(-t) -2 4 t g(-t) 1 -1 3 -3 t -g(t) 2 4 t -g(t) 1 -1 3 -3 t g(t-1) 1 3 4 t g(t-1) 1 2 3 -3 t g(2t) 1 4 t g(2t) 1 3 -3 2 1 2 - 6. A function, G( f ), is defined by M. J. Roberts - 7/12/03 Solutions 2-3 G f e rect f j f ( ) =      − 2 2 π . Graph the magnitude and phase of G( f −10) + G( f +10) over the range, −20 f 20. G f G f e rect rect f e f ( − ) + ( + ) = j f j f −      + +      10 10 − ( − ) − ( + ) 10 2 10 2 2π 10 2π 10 f -20 20 |G( f )| 1 f -20 20 Phase of G( f ) -π π 7. Sketch the derivatives of these functions. (All sketches at end.) (a) g(t) = sinc(t) ′( ) = ( )− ( ) ( ) = ( )− ( ) g cos sin cos sin t t t t t t t t t π π π π π π π π π 2 2 2 (b) g(t) = (1− e−t )u(t) ′( ) = ≥   = ( ) − g − , , t u e t t e t t t 0 0 0 t -4 4 x(t) -1 1 t -4 4 dx/dt -1 1 t -1 4 x(t) -1 1 t -1 4 dx/dt -1 1 (a) (b) 8. Sketch the integral from negative infinity to time, t, of these functions which are zero for all time before time, t = 0. M. J. Roberts - 7/12/03 Solutions 2-4 g(t) t 1 1 2 3 12 g(t) t 1 1 2 3 ∫ g(t) dt ∫ g(t) dt t 1 1 2 3 12 t 1 1 2 3 9. Find the even and odd parts of these functions. (a) g(t) = 2t2 − 3t + 6 ge t t t t t t ( ) = t − + + −( ) − −( )+ = + = + 2 3 6 2 3 6 2 4 12 2 2 6 2 2 2 2 go t t t t t t ( ) = t − + − −( ) + −( )− = − = − 2 3 6 2 3 6 2 6 2 3 2 2 (b) g cos t t ( ) = −      20 40 4 π π g cos cos e t t t ( ) = −      + − −     20 40 4 20 40 4 2 π π π π Using cos z z cos z cos z sin z sin z 1 2 1 2 1 2 ( + ) = ( ) ( ) − ( ) ( ) g cos cos sin sin cos cos sin sin e t t t t t ( ) = ( ) −      − ( ) −            + − ( ) −      − − ( ) −                      20 40 4 40 4 20 40 4 40 4 2 π π π π π π π π g cos cos sin sin cos cos sin sin e t t t t t ( ) = ( )      + ( )            + ( )      − ( )                      20 40 4 40 4 20 40 4 40 4 2 π π π π π π π π M. J. Roberts - 7/12/03 Solutions 2-5 g cos cos cos e t t t ( ) =      20 ( ) = ( ) 4 40 20 2 40 π π π g cos cos o t t t ( ) = −      − − −      20 40 4 20 40 4 2 π π π π Using cos z z cos z cos z sin z sin z 1 2 1 2 1 2 ( + ) = ( ) ( ) − ( ) ( ) g cos cos sin sin cos cos sin sin o t t t t t ( ) = ( ) −      − ( ) −            − − ( ) −      − − ( ) −                      20 40 4 40 4 20 40 4 40 4 2 π π π π π π π π g cos cos sin sin cos cos sin sin o t t t t t ( ) = ( )      + ( )            − ( )      − ( )                      20 40 4 40 4 20 40 4 40 4 2 π π π π π π π π g sin sin sin ot t t ( ) =      20 ( ) = ( ) 4 40 20 2 40 π π π (c) g t t t t ( ) = − + + 2 3 6 1 2 ge t t t t t t ( ) = t − + + + + + − 2 3 6 1 2 3 6 1 2 2 2 ge t t t t t t t ( ) = t t ( − + )( − ) + ( + + )( + ) ( + )( − ) 2 3 6 1 2 3 6 1 1 1 2 2 2 ge t t t t t t ( ) = + + ( − ) = + − 4 12 6 2 1 6 5 1 2 2 2 2 2 go t t t t t t ( ) = t − + + − + + − 2 3 6 1 2 3 6 1 2 2 2 M. J. Roberts - 7/12/03 Solutions 2-6 go t t t t t t t ( ) = t t ( − + )( − ) − ( + + )( + ) ( + )( − ) 2 3 6 1 2 3 6 1 1 1 2 2 2 go t t t t t t t t ( ) = − − − ( − ) = − + − 6 4 12 2 1 2 9 1 3 2 2 2 (d) g(t) = sinc(t) g sin sin sin e t t t t t t t ( ) = ( ) + (− ) − = ( ) π π π π π 2 π go (t) = 0 (e) g(t) = t(2 − t2)(1+ 4t2) g(t) = t ( − t )( + t ) odd even even { 2 2 1 4 2 Therefore g(t) is odd, g g e o (t) = 0 and (t) = t(2 − t2)(1+ 4t2) (f) g(t) = t(2 − t)(1+ 4t) ge t t t t t t t ( ) = (2 − )(1+ 4 ) + (− )(2 + )(1− 4 ) 2 ge(t) = 7t2 go t t t t t t t ( ) = (2 − )(1+ 4 ) − (− )(2 + )(1− 4 ) 2 go (t) = t(2 − 4t2) 10. Sketch the even and odd parts of these functions. M. J. Roberts - 7/12/03 Solutions 2-7 t g(t) 1 1 t g(t) 1 2 1 -1 t g (t) 1 1 t g (t) 1 2 1 -1 t g (t) 1 1 t g (t) 1 2 1 -1 e e o o (a) (b) 11. Sketch the indicated product or quotient, g(t), of these functions. t 1 -1 1 -1 t -1 1 1 g(t) Multiplication t 1 -1 1 -1 t 1 -1 -1 1 g(t) g(t) g(t) Multiplication (a) (b) t 1 -1 1 -1 t -1 1 1 -1 M. J. Roberts - 7/12/03 Solutions 2-8 t 1 1 g(t) Multiplication (c) t -1 1 t 1 1 g(t) g(t) Multiplication (d) t 1 1 t -1 -1 1 g(t) t -1 1 1 (e) (f) -1 1 t 1 -1 t -1 1 1 g(t) Multiplication ... ... t 1 1 -1 t 1 -1 1 g(t) Multiplication g(t) t -1 1 1 -1 ... ... g(t) t 1 1 -1 t 1 1 g(t) Division Division (g) t -1 -1 -1 1 1 1 g(t) g(t) (h) t 1 t π -1 1 1 t g(t) t -1 12. Use the properties of integrals of even and odd functions to evaluate these integrals in the quickest way. M. J. Roberts - 7/12/03 Solutions 2-9 (a) 2 2 2 2 4 1 1 1 1 1 1 0 1 ( + ) = + = = − − − ∫ t dt ∫ dt ∫ t dt ∫ dt even odd { { (b) 4 10 8 5 4 10 8 5 8 10 8 10 1 20 1 20 1 20 1 20 1 20 1 20 0 1 20 cos π sin π cos π sin π cos π π [ ( t) + ( t)]dt = ( t)dt + ( t)dt = ( t)dt = − − − ∫ ∫ ∫ ∫ even odd (c) 4 10 0 1 20 1 20 t tdt odd even odd { 14243 1 2 4 3 4 cos( π ) = − ∫ (d) t t dt t t dt t t t dt odd odd even { 14243 1 2 4 3 4 sin sin cos cos 10 2 10 2 10 10 10 10 1 10 1 10 0 1 10 0 1 10 0 1 10 π π π π π π ( ) = ( ) = − ( ) + ( )       − ∫ ∫ ∫ t tdt t odd odd even { 14243 1 2 4 3 4 sin sin 10 2 1 100 10 10 1 50 1 10 1 10 2 0 1 10 π π π π π ( ) == + ( ) ( )       = − ∫ (e) e− t dt e t dt e tdt e t e − − − − − ∫ = ∫ = ∫ = [− ] = ( − ) ≈ even { 1 1 0 1 0 1 0 2 2 2 1 2 1 1 1.264 (f) t e t dt odd even odd { { 123 − − ∫ = 1 1 0 13. Find the fundamental period and fundamental frequency of each of these functions. (a) g(t) = 10cos(50πt) f T 0 0 25 1 25 = Hz , = s (b) g cos t t ( ) = +      10 50 4 π π f T 0 0 25 1 25 = Hz , = s (c) g(t) = cos(50πt) + sin(15πt) f T 0 0 25 15 2 2 5 1 2 5 0 4 =      GCD , = . , = = . Hz . s (d) g cos sin cos t t t t ( ) = ( ) + ( ) + −      2 3 5 3 4 π π π π M. J. Roberts - 7/12/03 Solutions 2-10 f T 0 0 1 3 2 5 2 1 2 1 1 2 2 =      GCD , , = Hz , = = s 14. Find the fundamental period and fundamental frequency of g(t). g(t) t ... ... 1 t ... ... 1 g(t) t ... ... 1 (a) (b) + t ... ... 1 g(t) t ... ... 1 (c) + (a) f T 0 0 3 1 3 = Hz and = s (b) f T 0 0 6 4 2 1 2 = GCD( , ) = Hz and = s (c) f T 0 0 = GCD(6,5) = 1Hz and = 1 s 15. Plot these DT functions. (a) x n cos sin n n [ ]=      −  ( − )     4 2 12 3 2 2 8 π π , −24 ≤ n 24 n -24 24 x[n] -7 7 (b) xn ne n [ ]= − 3 5 , −20 ≤ n 20 n -20 20 x[n] -6 6 (c) x n n n [ ] =      21 + 2 14 2 3 , −5 ≤ n 5 M. J. Roberts - 7/12/03 Solutions 2-11 n -5 5 x[n] -2000 2000 16. Let x cos 1 5 2 8 n n [ ]=      π and x2 8 6 2 n e n [ ]= − −     . Plot the following combinations of those two signals over the DT range, −20 ≤ n 20. If a signal has some defined and some undefined values, just plot the defined values. (a) x[n] = x [n]x [n] 1 2 n -20 20 x[n] -40 40 (b) x[n] = 4 x [n] + 2 x [n] 1 2 n -20 20 x[n] -40 20 (c) x[n] = x [ n]x [ n] 1 2 2 3 n -20 20 x[n] -40 20 (d) x x x n n n [ ]= [ ] [− ] 1 2 2 n -20 20 x[n] -50000 10000 (e) x n x x n n [ ]=       +       2 2 4 3 1 2 n -20 20 x[n] -40 5 M. J. Roberts - 7/12/03 Solutions 2-12 17.A function, g[n] is defined by g , , , n n n n n n [ ]= − − − ≤ ≤      2 4 4 1 4 1 . Sketch g[−n], g[2 − n], g[2n] and g n 2       . n -10 10 g[n] -4 4 n -10 10 g[- n] -4 4 n -10 10 g[2- n] -4 4 n -10 10 g[2n] -4 4 n -10 10 g[n/2] -4 4 18. Sketch the backward differences of these DT functions. n -4 20 g[n] -1 1 n -4 20 Δg[n-1] -1 1 n -4 20 g[n] -1 1 n -4 20 Δg[n-1] -1 1 n -4 20 g[n] = (n/10)2 4 n -4 20 Δg[n-1] -0.25 0.5 (a) (b) (c) 19. Sketch the accumulation, g[n], from negative infinity to n of each of these DT functions. M. J. Roberts - 7/12/03 Solutions 2-13 (a) h[n] =δ[n] (b) h[n] = u[n] (c) h n cos u n n [ ] =      [ ] 2 16 π (d) h n cos u n n [ ] =      [ ] 2 8 π (e) h n cos u n n [ ] =      [ + ] 2 16 8 π (a) n -16 16 h[n] 1 n -16 16 g[n] 1 (b) n -16 16 h[n] 1 n -16 16 g[n] 16 (c) n -16 16 h[n] -1 1 n -16 16 g[n] -3 3 (d) n -16 16 h[n] -1 1 n -16 16 g[n] -3 3 (e) n -16 16 h[n] -1 1 n -16 16 g[n] -3 3 20. Find and sketch the even and odd parts of these functions. (a) g[n] = u[n]− u[n − 4] (b) g n e u n n [ ]= [ ] − 4 (c) g n cos n [ ]=      2 4 π (d) g n sin u n n [ ] =      [ ] 2 4 π M. J. Roberts - 7/12/03 Solutions 2-14 n -10 10 g[n] -1 1 n -10 10 g e [n] -1 1 n -10 10 g o [n] -1 1 n -10 10 g[n] -1 1 n -10 10 g e [n] -1 1 n -10 10 g o [n] -1 1 n -10 10 g[n] -1 1 n -10 10 g e [n] -1 1 n -10 10 g o [n] -1 1 n -10 10 g[n] -1 1 n -10 10 g e [n] -1 1 n -10 10 g o [n] -1 1 21. Sketch g[n]. Multiplication Multiplication Multiplication Multiplication g [n] 1 g [n] 1 g[n] g[n] g[n] g[n] g [n] 2 g [n] 2 g [n] 1 g [n] 1 g [n] 2 n -10 10 -1 1 n -10 10 -1 1 n -4 20 -1 1 n -4 20 -1 1 n -4 20 -1 1 n -4 20 -1 1 n -10 10 -1 1 n -10 10 g[n] -1 1 (a) (b) (c) (d) M. J. Roberts - 7/12/03 Solutions 2-15 n -10 10 g[n] -1 1 (a) n -4 20 g[n] -1 1 (b) n -4 20 g[n] -1 1 (c) n -10 10 g[n] -1 1 (d) 22. Find the fundamental DT period and fundamental DT frequency of these functions. (a) g n cos n [ ]=      2 10 π N F 0 0 10 1 10 = , = (b) g n cos n [ ]=      π 10 N F 0 0 20 1 20 = , = (c) g n cos cos n n [ ]=      +      2 5 2 7 π π N F 0 0 35 1 35 = , = (d) gn e e j n j n [ ]= +− 2 20 2 20 π π N F 0 0 20 1 20 = , = (e) gn e e j n j n [ ]= + − − 2 3 2 4 π π N F 0 0 12 1 12 = , = 23. Graph the following functions and determine from the graphs the fundamental period of each one (if it is periodic). (a) g n sin cos n n [ ]=      +      5 2 4 8 2 6 π π (b) g n sin cos n n [ ]=      +      5 7 12 8 14 8 π π (c) g n Re e j n e j n [ ]= +       π − π 3 (d) g n Re e jn e j n [ ]= +       − 3 M. J. Roberts - 7/12/03 Solutions 2-16 n -24 24 g[n] -12 12 (a) n -24 24 g[n] -12 12 (b) n -24 24 g[n] -2 2 (c) n -24 24 g[n] -2 2 (d) N = 12 0 N = 6 0 N = 24 0 Not Periodic 24. Find the signal energy of these signals. (a) x(t) = 2rect(t) E t dt dt x = ( ) = = −∞ ∞ − ∫ 2 4 ∫ 4 2 1 2 1 2 rect (b) x(t) = A(u(t) − u(t −10)) E A t t dt A dt A x = ( ( ) − ( − )) = = −∞ ∞ ∫ u u 10 ∫ 10 2 2 0 10 2 (c) x(t) = u(t) − u(10 − t) E t t dt dt dt x = ( ) − ( − ) = + →∞ −∞ ∞ −∞ ∞ ∫ u u 10 ∫ ∫ 2 0 10 (d) x(t) = rect(t)cos(2πt) E t t dt t dt t dt x = ( ) ( ) = ( ) = ( + ( )) −∞ ∞ − − ∫ rect cos 2 ∫ cos 2 ∫ cos 1 2 1 4 2 2 1 2 1 2 1 2 1 2 π π π E dt t dt x= + ( )       = − − = ∫ ∫ 1 2 4 1 2 1 2 1 2 1 2 1 2 0 cos π (e) x(t) = rect(t)cos(4πt) M. J. Roberts - 7/12/03 Solutions 2-17 Ex = (t) ( t) dt = ( t)dt = ( + ( t))dt −∞ ∞ − − ∫ rect cos 4 ∫ cos 4 ∫ cos 1 2 1 8 2 2 1 2 1 2 1 2 1 2 π π π E dt t dt x= + ( )       = − − = ∫ ∫ 1 2 8 1 2 1 2 1 2 1 2 1 2 0 cos π (f) x(t) = rect(t)sin(2πt) E t t dt t dt t dt x = ( ) ( ) = ( ) = ( − ( )) −∞ ∞ − − ∫ rect sin 2 ∫ sin 2 ∫ cos 1 2 1 4 2 2 1 2 1 2 1 2 1 2 π π π E dt t dt x= − ( )       = − − = ∫ ∫ 1 2 4 1 2 1 2 1 2 1 2 1 2 0 cos π (g) x n Arect n N [ ]= [ ] 0 E A n A N A x N N N = [ ] = ( ) = ( + ) −∞ ∞ − Σ rect Σ 0 0 2 0 2 0 1 2 1 2 (h) x[n] = Aδ[n] xn A n E A n A A x [ ]= [ ] = [ ] = ( ) = −∞ ∞Σ Σ δ δ 2 2 0 0 1 2 (i) x n comb n N [ ]= [ ] 0 E n x N n mN = [ ] = ( )→∞ −∞ ∞ −∞ = ∞ Σcomb Σ 0 0 2 1 (j) x[n] = ramp[n] E n n x = [ ]= →∞ −∞ ∞ ∞ Σramp Σ 2 2 0 (k) x[n] = ramp[n]− 2ramp[n − 4] + ramp[n − 8] M. J. Roberts - 7/12/03 Solutions 2-18 Ex = [n]− [n − ] + [n − ] = ( + + + + + + + + ) −∞ ∞Σ ramp 2ramp 4 ramp 8 0 1 2 3 4 3 2 1 0 2 2 2 2 2 2 2 2 2 2 Ex= 1+ 4 + 9 +16 + 9 + 4 +1= 44 25. Find the signal power of these signals. (a) x(t) = A P T A dt A T dt A T T A x T T T T T T T = = = = →∞ − →∞ − →∞ lim ∫ lim ∫ lim 1 2 2 2 2 2 2 2 2 (b) x(t) = u(t) P T t dt T dt T T x T T T T T T = ( ) = = = →∞ − →∞ →∞ lim ∫ u lim ∫ lim 1 1 1 2 1 2 2 2 2 0 2 (c) x(t) = Acos(2 f t + ) 0 π θ P T A ft dt A T f t dt x T T T T = ( + ) = ( + ) − − ∫ ∫ 1 2 2 0 0 2 2 2 2 0 2 0 2 2 0 0 0 0 cos π θ cos π θ P A T f t dt A T t f t f x T T T T = ( + ( + )) = +  ( + )      − − ∫ 2 0 0 2 2 2 0 0 0 2 2 2 1 4 2 2 4 2 4 0 0 0 0 cos sin π θ π θ π P A T T f T f f T f A x= + +      − − +            = = 2 0 0 0 0 0 0 0 0 0 2 2 4 2 2 4 4 2 2 4 2 sin π θ sin π π θ π (d) x t A rect t n n ( ) = ( − ) =−∞ ∞Σ 2 P T A t n dt A t dt A dt A x T n T = ( − ) = ( ) = = =−∞ ∞ − − − ∫ Σ ∫ ∫ 1 2 2 2 2 0 2 2 2 2 2 1 1 2 1 2 1 2 2 0 0 rect rect (e) x t A rect t n n ( ) = − + ( − )      =−∞ ∞Σ 2 1 2 2 M. J. Roberts - 7/12/03 Solutions 2-19 P T A t n dt A t dt x T n T = − + ( − )       = − + ( ) =−∞ ∞ − − ∫ Σ ∫ 1 2 1 2 2 4 2 1 2 0 2 2 2 2 2 1 1 0 0 rect rect P A t dt A t dt x= − + ( ) = − + ( ) − 2 ∫ ∫ 1 2 4 1 2 2 2 1 1 2 2 0 1 rect rect P A dt dt A x =      + −            4 ∫ ∫ = 1 2 1 2 2 2 0 1 2 2 1 2 1 2 (f) x[n] = A P N A A N A N N A x N n N N N n N N N = = ( ) = ( )= →∞ =− − →∞ =− − →∞ lim Σ lim Σ lim 1 2 2 1 2 2 2 1 2 1 2 2 (g) x[n] = u[n] P N n N N N x N n N N N n N N = [ ] = ( )= = →∞ =− − →∞ = − →∞ lim Σ u lim Σ lim 1 2 1 2 1 2 1 2 2 1 0 1 (h) x n A rect n m m [ ] = [ − ] =−∞ ∞Σ 2 8 P N A n m A n m x n N m N n m = [ − ] = × [ − ] =−∞ ∞ =− − =−∞ ∞ =− Σ Σ Σ Σ 1 2 8 2 8 8 0 2 1 2 2 2 2 8 7 0 0 rect rect P A A A x n n n = ×  ( ) + ( ) + ( )      = × = =− − =− = Σ Σ Σ 2 8 6 2 2 6 7 2 2 2 8 1 1 1 10 2 8 5 8 (i) x n comb n N [ ]= [ ] 0 P N n N x N n N = [ ] = = Σ 1 1 0 2 0 0 0 comb (j) x[n] = ramp[n] P N n N n x N n N N N n N = [ ]= →∞ →∞ =− − →∞ = − lim Σ ramp lim Σ 1 2 1 2 2 1 2 0 1 26.Using MATLAB, plot the CT signal, x(t) = sin(2πt), over the time range, 0 t 10, with the following choices of the time resolution, Δt , of the plot. Explain why the plots look the way they do. (a) Δt = 1 24 (b) Δt = 1 12 (c) Δt = 1 4 (d) Δt = 1 2 (e) Δt = 2 3 (f) Δt = 5 6 M. J. Roberts - 7/12/03 Solutions 2-20 (g) Δt = 1 t 10 x(t)