Solution manual for signals and systems analysis using transform methods and MATLAB 2nd edition by roberts

Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed ©M J Roberts - 3/16/11 Full file at https://TestbankDirect.eu/ Chapter - Mathematical Description of Continuous-Time Signals Solutions Exercises With Answers in Text Signal Functions () If g t = 7e−2t −3 write out and simplify () g ( − t ) = 7e ( ) = 7e g ( t / 10 + ) = 7e g ( jt ) = 7e g ( jt ) + g ( − jt ) e = 7e g = 7e−9 (a) −2 2−t −3 (b) −t /5−11 (c) − j 2t −3 (d) − j 2t −3 (e) ⎛ g⎜ ⎝ (f) −7 + 2t + e j 2t = 7e−3 cos 2t ( ) ⎛ − jt − ⎞ jt − ⎞ + g⎜ ⎟ ⎠ ⎝ ⎟⎠ e− jt + e jt =7 = cos t 2 () () If g x = x − 4x + write out and simplify (a) (b) (c) (d) () g (u + v ) = (u + v ) g z = z − 4z + ( ( ) ( ) − 4e + = e g ( g ( t )) = g ( t − 4t + ) = ( t g e jt = e jt ) − u + v + = u + v + 2uv − 4u − 4v + jt ( − 4t + ) − ( t j 2t − 4e jt + = e jt − 2 ) ) − 4t + + ( ( )) g g t = t − 8t + 20t − 16t + (e) () g = 4−8+ = What would be the value of g in each of the following MATLAB instructions? (a) t = ; 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’) ; Solutions 2-1 Full file at https://TestbankDirect.eu/ Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed ©M J Roberts - 3/16/11 Full file at https://TestbankDirect.eu/ ⎡ 0.0247 + ⎢ ⎢0.0920 + ⎢ ⎢ 0.0920 − ⎢ ⎢⎣ 0.0247 − j0.155⎤ ⎥ j0.289 ⎥ ⎥ ⎥ j0.289 ⎥ j0.155⎥⎦ Let two functions be defined by ( ( ) ) ⎧⎪1 , sin 20π t ≥ x1 t = ⎨ and ⎩⎪−1 , sin 20π t < () ( ) ( ) ⎧⎪t , sin 2π t ≥ x2 t = ⎨ ⎩⎪−t , sin 2π t < () Graph the product of these two functions versus time over the time range, −2 < t < x(t) -2 t -2 Transformations of Functions () ( ) () ( ) ( ) For each function g t graph g −t , − g t , g t − , and g 2t (a) (b) g(t) g(t) -1 t t -3 g(-t) g(-t) -g(t) -g(t) t -2 -1 t g(t-1) g(t-1) g(2t) 4 t -1 -3 1 t t g(2t) -1 t 1 -3 -3 Find the values of the following signals at the indicated times () ( ) ( ) () ( ) ( ) () ( ( ) (a) x t = rect t / , x −1 = rect −1 / = (b) x t = 5rect t / sgn 2t , x 0.5 = 5rect / sgn = (c) x t = rect t / 10 sgn t − ) (( ( ) )) ( () ) () ( ) ( ) , x = rect / 10 sgn −3 = −9 Solutions 2-2 Full file at https://TestbankDirect.eu/ t -3 t Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed ©M J Roberts - 3/16/11 Full file at https://TestbankDirect.eu/ For each pair of functions in Figure E-7 provide the values of the constants A, (( () ) ) t0 and w in the functional transformation g t = Ag1 t − t0 / w (a) (a) -1 -2 -1 -2 -4 -1 -2 -4 g2(t) g1(t) -1 -2 -4 -4 -2 t -2 t (b) 4 (b) g2(t) g1(t) -1 -2 -4 -2 t -2 t (c) (c) g2(t) g1(t) -1 -2 -4 -2 t -2 t Figure E-7 Answers: A = 2,t0 = 1, w = , (b) (a) (c) A = −2,t0 = 0, w = / , A = −1 / 2,t0 = −1, w = For each pair of functions in Figure E-8 provide the values of the constants A, ( ( () )) t0 and a in the functional transformation g t = Ag1 w t − t0 4 g2(t) (a) g1(t) A = 2, t0 = 2, w = -2 -4 -4 -8 -10 -5 -8 -10 10 -5 10 t t Amplitude comparison yields A = Time scale comparison yields w = −2 ( ( () g 2 = g1 −2 − t0 )) = g (0) ⇒ −4 + 2t = ⇒ t0 = 4 g2(t) (b) g1(t) A = 3, t0 = 2, w = -4 -4 -8 -10 -5 10 -8 -10 -5 10 t t Amplitude comparison yields A = Time scale comparison yields w = () (( g 2 = 3g1 2 − t0 )) = 3g (0) ⇒ − 2t = ⇒ t0 = Solutions 2-3 Full file at https://TestbankDirect.eu/ Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed ©M J Roberts - 3/16/11 4 0 (c) A = -3, t0= -6, w = 1/3 g (t) g1(t) Full file at https://TestbankDirect.eu/ -4 -4 -8 -10 -5 -8 -10 10 -5 10 t t Amplitude comparison yields A = −3 Time scale comparison yields w = / (( ) ( () g = −3g1 / − t0 )) = −3g ( 2) ⇒ −t / = ⇒ t0 = −6 OR Amplitude comparison yields A = −3 Time scale comparison yields w = −1 / g = −3g1 −1 / 3 − t0 = −3g1 ⇒ t0 / − = ⇒ t0 = (( () )) )( () A = -2, t = -2, w = 1/3 8 g2(t) (d) g1(t) -4 -4 -8 -10 -5 -8 -10 10 -5 10 t t Amplitude comparison yields A = −2 Time scale comparison yields w = / (( ) ( () )) = −3g ( 2) ⇒ −t / + / = ⇒ t0 = −2 A = 3, t = -2, w = 1/2 8 4 g2(t) (e) g1(t) g = −2 g1 / − t0 -4 -4 -8 -10 -5 -8 -10 10 -5 10 t t Amplitude comparison yields A = Time scale comparison yields w = / g = 3g1 / − t0 = 3g1 ⇒ −t0 / = ⇒ t0 = −2 (( ) ( () )) () Figure E-8 () In Figure E-9 is plotted a function g1 t which is zero for all time outside the range plotted Let some other functions be defined by () ( g t = 3g1 − t ) , () ( ) g t = −2 g1 t / ⎛ t − 3⎞ g t = g1 ⎜ ⎝ ⎟⎠ () , Find these values () (a) g = −3 (c) 3 ⎡ g t g t ⎤ = × −1 = − ⎣ ⎦t = 2 () () (b) ( ) ( ) g −1 = −3.5 −1 (d) ∫ g (t ) dt −3 () The function g t is linear between the integration limits and the area under it is a triangle The base width is and the height is -2 Therefore the area is -2 −1 ∫ g (t ) dt = −2 −3 Solutions 2-4 Full file at https://TestbankDirect.eu/ Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed ©M J Roberts - 3/16/11 Full file at https://TestbankDirect.eu/ g1(t) -4 -3 -2 -1 -1 -2 -3 -4 t Figure E-9 10 ( ) A function G f is defined by ( ) ( G f = e− j 2π f rect f / ) ( ) ( Graph the magnitude and phase of G f − 10 + G f + 10 −20 < f < 20 ) over the range, ( ) First imagine what G f looks like It consists of a rectangle centered at f = 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− j 2π f = cos −2π f + j sin −2π f = cos 2π f − j sin 2π f ( ) ( ) ) e− j 2π f = cos 2π f + sin 2π f = ( ) The phase (angle) of G f is simply the phase of the complex exponential between f = −1 and f = and undefined outside that range because the phase of the rectangle function is zero between f = −1 and f = 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− j 2π f =  cos 2π f − j sin 2π f )) = tan −1 ( ( ( ( ) ) ( ( ) ) ⎛ sin 2π f ⎞ ⎛ sin 2π f ⎞ −1 ⎜− ⎟ = − tan ⎜ ⎟ ⎝ cos 2π f ⎠ ⎝ cos 2π f ⎠ e− j 2π f = − tan −1 tan 2π 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− j 2π f = −2π f which is simply the coefficient of j in the original complex exponential expression A more general solution would be e− j 2π f = −2π f + 2nπ , n an integer The solution of the original problem is simply this solution except shifted up and down by 10 in f and added ( ) ( ) G f − 10 + G f + 10 = e ( − j π f −10 ) ⎛ f − 10 ⎞ ⎛ f + 10 ⎞ − j π f +10 rect ⎜ + e ( ) rect ⎜ ⎝ ⎟⎠ ⎝ ⎟⎠ Solutions 2-5 Full file at https://TestbankDirect.eu/ Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed ©M J Roberts - 3/16/11 Full file at https://TestbankDirect.eu/ |G( f )| -20 20 f Phase of G( f ) / -20 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 ms and height which occur at a uniform rate of 100 pulses per second with the leading edge of the first pulse occurring at time t = ∞ () ( ) ( ) x t = 3∑ ⎡⎣ u t − 0.01n − u t − 0.01n − 0.006 ⎤⎦ n=0 Derivatives and Integrals of Functions () ( ) () Graph the derivative of x t = − e−t u t 12 This function is constant zero for all time before time, t = , therefore its derivative during that time is zero This function is a constant minus a decaying exponential after time, t = , and its derivative in that time is therefore also a positive decaying exponential ⎧e−t , t > x′ t = ⎨ ⎩⎪0 , t < () Strictly speaking, its derivative is not defined at exactly t = 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 = , without changing the effect of this signal on any physical system If we choose 1/2, then we can write the derivative as x ′ t = e−t u t () () x(t) -1 t -1 dx/dt -1 t -1 13 Find the numerical value of each integral (a) ( ) ( ) ( ) ( ) () ∫ ⎡⎣δ t + − 2δ 4t ⎤⎦ dt = ∫ δ t + dt − ∫ δ 4t dt = − × / ∫ δ t dt = −1 / −1 −1 −1 −1 (b) 5/ 5/ ∞ 5/ ∞ ∫ δ (3t ) dt = ∫ ∑ δ (3t − 2n)dt = ∫ ∑ δ (t − 2n / 3)dt = ⎡⎣1 + + 1⎤⎦ = 1/ 1/ n= −∞ 1/ n= −∞ Solutions 2-6 Full file at https://TestbankDirect.eu/ Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed ©M J Roberts - 3/16/11 Full file at https://TestbankDirect.eu/ 14 Graph the integral from negative infinity to time t of the functions in Figure E-14 which are zero for all time t < ∫ g (τ ) dτ t This is the integral −∞ () area under the function g t which, in geometrical terms, is the accumulated from time −∞ to time t For the case of the two back-to-back rectangular pulses, there is no accumulated area until after time t = and then in the time interval < t < the area accumulates linearly with time up to a maximum area of one at time t = In the second time interval < t < the area is linearly declining at half the rate at which it increased in the first time interval < t < down to a value of 1/2 where it stays because there is no accumulation of area for t > 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 second-degree polynomial The rate of accumulation of area is increasing up to time t = and then decreasing (but still positive) until time t = 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) g(t) 1 t 2 3 t Figure E-14 g(t) dt g(t) dt 1 2 t t Even and Odd Functions 15 () An even function g t is described over the time range < t < 10 by ⎧2t , 0 () (a) (b) x(t) x(t) 1 -4 t -1 t t -1 -1 dx/dt dx/dt 6 -4 t -1 -6 -6 Even and Odd Functions 51 Graph the even and odd parts of these signals (a) () ( ) x t = rect t − Solutions 2-28 Full file at https://TestbankDirect.eu/ Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed ©M J Roberts - 3/16/11 Full file at https://TestbankDirect.eu/ () xe t = ( ) ( ) rect t − + rect t + >>>need graph () ( () xo t = , ) ( ) ) () x t = 2sin 4π t − π / rect t (b) () ( ) ( ) () () ( ) ( ) () x o t = 2cos π / sin 4π t rect t x e t = −2sin π / cos 4π t rect t , (c) (d) x (t) x (t) e e -10 t 10 -1 t -4 -2 xo(t) xo(t) -10 t 10 -1 t -4 52 ( rect t − − rect t + -2 Find the even and odd parts of each of these functions () () ge t = ( g t = 10sin 20π t (a) ( ) ) ( 10sin 20π t + 10sin −20π t () () ge t = ( ) ( 10sin 20π t − 10sin −20π t ) = 10sin 20π t ( ) ( ) 20t + 20 −t () () = , go t = ( ) 20t − 20 −t = 20t x t = + 7t (c) ( ) + 7t + + −t () x (t ) = + t xe t = (d) () xe t = () x t = 6t (f) () g ( t ) = 4t cos (10π t ) ge t = () ge t = ( = + 7t ( )=0 , go t = ) ( ) ( 4t cos 10π t + −t cos −10π t ( ) ( ) ( 4t cos 10π t − −t cos −10π t () , xo t = , xo t = 6t + −t 2 ( ) =1 + t + + −t (e) () () , go t = g t = 20t (b) go t = )=0 ( ) + 7t − − −t + t − − −t () 6t − −t =0 ( )=t () 2 ( ) = 6t ) = 4t cos (10π t ) + ( −t ) cos (10π t ) = ) = 4t cos (10π t ) − ( −t ) cos (10π t ) = 4t cos 10π t ( ) Solutions 2-29 Full file at https://TestbankDirect.eu/ Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed ©M J Roberts - 3/16/11 Full file at https://TestbankDirect.eu/ ( ) cos π t () g t = (g) πt cos π t ( ) + cos ( −π t ) () πt ge t = −π t ( ) − cos ( −π t ) cos π t () πt go t = () g t = 12 + (h) () ge t = 12 + (i) πt = −π t ( ) + cos (π t ) cos π t πt = πt = =0 ( ) cos π t πt ( ) sin 4π t 4π t ( ) + 12 + sin ( −4π t ) sin 4π t 4π t −4π t = ( ) + 12 + sin ( 4π t ) sin 4π t 12 + 4π t 4π t = 12 + ( ) sin 4π t 4π t 2 sin 4π t sin −4π t sin 4π t sin 4π t 12 + − 12 − − 4π t −4π t = 4π t 4π t = go t = 2 g t = + 7t cos 32π t ( ) () () ( ) ( ( ( ) () (8 + 7t ) cos (32π t ) + (8 − 7t ) cos ( −32π t ) = 8cos 32π t ( ) () (8 + 7t ) cos (32π t ) − (8 − 7t ) cos ( −32π t ) = 7t cos 32π t ( ) () ( ) ( () go t = ) (8 + 7t ) sin (32π t ) + (8 + ( −t ) ) sin ( −32π t ) 2 ge t = () ( ) ) g t = + 7t sin 32π t (j) ) ge t = go t = 53 −π t ( ) + cos (π t ) cos π t (8 + 7t ) sin (32π t ) − (8 + ( −t ) ) sin ( −32π t ) =0 2 ( ) ( = + 7t sin 32π t ) Is there a function that is both even and odd simultaneously? Discuss The only function that can be both odd and even simultaneously is the trivial signal, x t = Applying the definitions of even and odd functions, () () xe t = 0+0 0−0 = = x t and x o t = =0=x t 2 () () () proving that the signal is equal to both its even and odd parts and is therefore both even and odd 54 () Find and graph the even and odd parts of the function x t in Figure E-54 Solutions 2-30 Full file at https://TestbankDirect.eu/ Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed ©M J Roberts - 3/16/11 Full file at https://TestbankDirect.eu/ x(t) 1 t -5 -4 -3 -2 -1 -1 Figure E-54 xe(t) xo(t) 2 1 t -5 -4 -3 -2 -1 -1 -5 -4 -3 -2 -1 -1 t Periodic Functions 55 For each of the following signals decide whether it is periodic and, if it is, find the fundamental period (b) () ( ) Periodic Fundamental frequency = 200 Hz, Period = ms g ( t ) = 14 + 40cos ( 60π t ) Periodic Fundamental frequency = 30 Hz (c) Period = 33.33 ms g t = 5t − 2cos 5000π t (a) g t = 28sin 400π t (d) () ( ) Not periodic g ( t ) = 28sin ( 400π t ) + 12cos (500π t ) Periodic (e) periods of ms and ms Least common multiple is 20 ms Period of the overall signal is 20 ms g t = 10sin 5t − 4cos 7t Periodic The Periods of the two sinusoids are 2π / s () ( ) ( ) () ( ) ( 3t ) Two sinusoidal components with and 2π / s Least common multiple is 2π Period of the overall signal is 2π s (f) 56 g t = 4sin 3t + 3sin Not periodic because least common multiple is infinite Is a constant a periodic signal? Explain why it is or is not periodic and, if it is periodic what is its fundamental period? A constant is periodic because it repeats for all time The fundamental period of a periodic signal is defined as the minimum positive time in which it repeats A constant repeats in any time, no matter how small Therefore since there is no minimum positive time in which it repeats it does not have a fundamental period Signal Energy and Power of Signals 57 Find the signal energy of each of these signals (a) (b) ( ) rect −t ( ) rect 8t ∞ , E= ∞ , E= ∫ −∞ ∫ −∞ ( ) ⎡ rect 8t ⎤ dt = ⎣ ⎦ 1/ ∫ 1/16 ∫ −1/16 dt = −1/ dt = 2 ⎡ ⎛ t ⎞⎤ ⎛t⎞ 3rect ⎜ ⎟ , E = ∫ ⎢3rect ⎜ ⎟ ⎥ dt = ∫ dt = 36 ⎝ 4⎠ ⎦ ⎝ 4⎠ −∞ ⎣ −2 ∞ (c) ( ) ⎡ rect −t ⎤ dt = ⎣ ⎦ Solutions 2-31 Full file at https://TestbankDirect.eu/ Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed ©M J Roberts - 3/16/11 Full file at https://TestbankDirect.eu/ (d) ( 2sin 200π t ∞ E= ) ∞ ( ) ( ) ( ) ⎤⎥ ⎡ cos 400π t E = ⎢t + 400π ⎢⎣ (e) ∞ ⎡1 ∫ ⎡⎣ 2sin 200π t ⎤⎦ dt = ∫ sin 200π t dt = ∫ ⎢⎣ − sin 400π t −∞ −∞ −∞ () δ t ( )⎤⎥ dt ⎦ ∞ ⎥⎦ −∞ →∞ (Hint: First find the signal energy of a signal which approaches an impulse some limit, then take the limit.) () ( ) ( δ t = lim / a rect t / a (f) ( ( )) d rect t dt () x t = ( ( )) ( d rect t = δ t + / − δ t − / dt ) ( ∞ Ex = ) a/2 ⎡ ⎛ t ⎞⎤ ⎛t⎞ 1 a lim rect dt = lim rect ⎜ ⎟ dt = lim → ∞ ∫ ⎢⎣ a→0 a ⎜⎝ a ⎟⎠ ⎥⎦ a→0 a ∫ a→0 a a ⎝ ⎠ −∞ −a/2 ∞ E= a→0 ∫ ⎡⎣δ (t + / 2) − δ (t − / 2)⎤⎦ ) dt → ∞ −∞ t (g) ( ) ∫ rect ( λ ) d λ = ramp (t + / 2) − ramp (t − / 2) x t = −∞ 1/ Ex = ∫ (t + / 2) ∞ dt + ∫ dt → ∞ −1/ 2  1/ infinite finite (h) () x t = e( ) −1− j8π t ∞ Ex = ∫ x (t ) () u t ∞ dt = −∞ ∫ e( ) −1− j8π t −∞ () u t ∞ dt = ∫ e( ) ( −1+ j8π )t −1− j8π t e dt ∞ ⎡ e−2t ⎤ Ex = ∫ e dt = ⎢ ⎥ = ⎣ −2 ⎦0 ∞ 58 −2t Find the average signal power of each of these signals: (a) () ( x t = 2sin 200π t Px = ) This is a periodic function Therefore T /2 T /2 ⎡1 ⎤ ⎡ 2sin 200π t ⎤ dt = ⎢ − cos 400π t ⎥ dt ∫ ⎦ T −T / ⎣ T −T∫/ ⎣ 2 ⎦ ( ( ) ) T /2 ( ( ) ) ( ) ⎡ sin 400π t ⎤ ⎡ T sin 200π T T sin −200π T ⎤ Px = ⎢t − = ⎢ − + + ⎥ ⎥=2 T ⎢⎣ 400π ⎥⎦ T ⎣⎢ 400π 400π ⎥⎦ −T / For any sinusoid, the average signal power is half the square of the amplitude Solutions 2-32 Full file at https://TestbankDirect.eu/ Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed ©M J Roberts - 3/16/11 Full file at https://TestbankDirect.eu/ () () x t = δ1 t (b) This is a periodic signal whose period, T, is Between −T / and +T / , there is one impulse whose energy is infinite Therefore the average power is the energy in one period, divided by the period, or infinite () x t = e j100π t This is a periodic function Therefore (c) Px = T0 ∫ () T0 x t T /2 dt = 1/100 e j100π t dt = 50 ∫ e j100π t e− j100π t dt ∫ T0 −T / −1/100 1/100 Px = 50 ∫ dt = −1/100 59 A signal x is periodic with fundamental period T0 = This signal is described over the time period < t < by (( ) ) (( ) ) rect t − / − rect t − / What is the signal power of this signal? The signal x can be described in the time period < t < by ⎧0 ⎪ ⎪⎪1 x t = ⎨−3 ⎪−4 ⎪ ⎪⎩0 () , < t