Solution manual for signals and systems analysis using transform methods and matlab 3rd edition robe by mary.li526

Solutions 2-1

for Signals and Systems Analysis Using Transform Methods and MATLAB 3rd Edition Roberts 0078028124

2 - Mathematical Description of Continuous-Time Signals Solutions Exercises With Answers in Text Signal Functions 1. If g(t) = 7e 2t 3 write out and simplify (a) g(3) = 7e 9 = 8 6387  10 4 (b) g(2 t) = 7e 2(2 t) 3 = 7e 7+2t (c) g(t /10 + 4) = 7e t 5 11 (d) g( jt) = 7e j2t 3 (e) g( jt) + g( jt) 2 = 7e 3 e j 2t + e j2t 2 = 7e 3 cos(2t) = 0 3485cos(2t) g( jt 3 ( jt 3   + g  jt jt (f)  2 )  2 ) = 7 e + e = 7cos(t) 2 2 2. If g(x) = x 2 4x + 4 write out and simplify (a) g(z) = z 2 4z + 4 (b) g(u + v) = (u + v)2 4(u + v) + 4 = u 2 + v 2 + 2uv 4u 4v + 4

Solution Manual

9780078028120 Chapter

3. Find the magnitudes and phases of these complex quantities.

Solutions 2-2

prior written consent of McGraw-Hill Education.

without

or distribution

(c) g(e jt ) = (e jt )2 4e jt + 4 = e j2t 4e jt + 4 = (e jt 2)2 (d) g(g(t)) = g(t2 4t + 4) = (t2 4t + 4)2 4(t2 4t + 4) + 4 g(g(t)) = t 4 8t 3 + 20t2 16t + 4 (e) g(2) = 4 8 + 4 = 0

the

(a) What value does the magnitude of this function approach as f approaches positive infinity?

Solutions 2-3

(a) e (3+ j2 3) e (3+ j2 3) = e 3 e j2.3 = e 3 cos(2.3) jsin(2.3) e (3+ j2 3) = e 3 cos 2 (2.3) + sin2(2.3) = e 3 = 0.0498 (b) e 2 j6 e 2 j6 = e 2 e j6 = e 2 = 7 3891 (c) 100 8 + j13 100 = 100 = 100 = 6.5512 8 + j13 8 + j13 82 +132 4. Let G( f ) = j4 f 2 + j7 f /11

lim j4 f = j4 f = 4 = 44  6 285 f  2 + j7 f /11 j7 f /11 7 /11 7

(b) What value (in radians) does the phase of this function approach as f approaches zero from the positive side?

5. Let X( f ) = jf

(a) Find the magnitude X(4) and the angle in radians

(b) What value (in radians) does approach as f approaches zero from the positive side?

6. For each function g(t) graph

Solutions 2-4

f 0+ j 2

lim f 0+ [ j4 f [ (2 + j7 f /11) = lim [ j4 f [ 2 =  /2 0 =  /2

jf +

X( f ) = jf  X(4) = j4 = 4e = 0.3714e j1 19 jf +

j4 + 10 10.77e j 0 3805 X(4) = 0.3714

Shifting and Scaling

t), g(t 1),

)

(a) (b) g(t) g(t) 4 3 2 t 1 t -3 g(-t) g(-t) -g(t) -g(t) 4 3 -2 t 1 t -3 3 2 t -1 t 4 -3

(

and

(2t

7. Find the values of the following signals at the indicated times.

8. For each pair of functions in Figure E-8 provide the values of the constants A,

9. For each pair of functions in Figure E-9 provide the values of the constants A,

Solutions 2-5

(a) (a) 2 2 1 1 0 0 -1 -1 -2 -2 -4 -2 t 0 2 4 -4 -2 t 0 2 4 (b) (b) 2 2 1 1 0 0 -1 -1 -2 -2 -4 -2 t 0 2 4 -4 -2 t 0 2 4 (c) (c) 2 2 1 1 0 0 -1 -1 -2 -2 -4 -2 0 2 4 -4 -2 0 2 4 t t g ( t ) g1( t ) g 1( t ) 1 g ( t ) g ( t ) g ( t ) 2 2 2 g(t-1) g(t-1) g(2t) g(2t) 4 1 3 t 3 1 2 t -3 4 31 1 t 1 t 2 -3

(a) x(t) = 2rect(t /4) , x( 1) = 2rect( 1/4) = 2 (b) x(t) = 5rect(t /2)sgn(2t) , x(0.5) = 5rect(1/4)sgn(1) = 5 (c) x(t) = 9rect(t /10)sgn(3(t 2)) , x(1) = 9rect(1/10)sgn( 3) = 9

t0 and w in

functional transformation g2 (t) = Ag1 ((t t0 ) / w). Figure E-8

(a) A = 2,t0 = 1,w = 1 , (b) A = 2,t0 = 0,w = 1/2 , (c) A = 1/2,t0 = 1,w = 2

the

Answers:

Solutions 2-6

0 0 g ( t ) g ( t ) g ( t ) 1 1 1 g ( t ) g ( t ) g ( t ) 2 2 2 t0 and a in the functional transformation g2 (t) = Ag1 (w(t t0 )). A = 2, t0 = 2, w = -2 8 8 4 4 (a) 0 0 -4 -4 -8-10 -5 0 5 10 -8-10 -5 0 5 10 t t Amplitude comparison yields A = 2 Time scale comparison yields w = 2 g2 (2) = 2g1 ( 2(2 t0 )) = 2g1 (0)  4 + 2t0 = 0  t0 = 2 A = 3, t = 2, w = 2 8 8 4 4 0 0 (b) -4 -4 -8 -10 -5 0 5 10 -8 -10 -5 0 5 10 t t Amplitude comparison yields A = 3. Time scale comparison yields w = 2. g2 (2) = 3g1 (2(2 t0 )) = 3g1 (0)  4 2t0 = 0  t0 = 2 A = -3, t = -6, w = 1/3 8 8 4 4 (c) 0 0 -4 -4 -8-10 -5 0 5 10 -8-10 -5 0 5 10 t t Amplitude comparison yields A = 3. Time scale comparison yields w = 1/3. g2 (0) = 3g1 ((1/3)(0 t0 )) = 3g1 (2)  t0 /3 = 2  t0 = 6 OR Amplitude comparison yields A = 3. Time scale comparison yields w = 1/3. g2 (3) = 3g1 (( 1/3)(3 t0 )) = 3g1 (0)  t0 /3 1 = 0  t0 = 3

10. In Figure E-10 is plotted a function

which is zero for all time outside the range plotted. Let some other functions be defined by

Find these values.

The function g4 i) 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 2-7

g 1 ( t ) g 1 ( t ) g 2 ( t ) g 2 ( t ) A = -2, i0 = -2, w = 1/3 8 8 4 4 (d) 0 0 -4 -4 -8 -10 -5 0 5 10 -8 -10 -5 0 5 10 i i Amplitude comparison yields A = 2. Time scale comparison yields w = 1/3. g2 4) = 2g1 1/3)4 i0 )) = 3g1 2)  i0 /3 + 4/3 = 2  i0 = 2 A = 3, i0 = -2, w = 1/2 8 8 4 4 (e) 0 0 -4 -4 -8 -10 -5 0 5 10 -8 -10 -5 0 5 10 i i Amplitude comparison yields A = 3 Time scale comparison yields w = 1/2 g2 0) = 3g1 1/2)0 i0 )) = 3g1 1)  i0 /2 = 1  i0 = 2 Figure E-9

g1 i)

( i 3 g2 i) = 3g1 2 i) , g3 i) = 2g1 i /4) , g4 i) = g1  ( 2 

(a) g2 1) = 3 (b) g3 1) = 3.5 3 3 1 (c)  g4 i)g3 i)i=2 = 2   1) = 2 (d)  g4 i)di 3

Solutions 2-8

1  g4 i)di = 2 3

Figure E-10

over the range, 20

20.

e j2 f = cos( 2 f ) + j sin( 2 f ) = cos(2 f ) j sin(2 f ) e j2 f = cos 2(2 f ) + sin2(2 f ) = 1

D e j2 f = D (cos(2 f ) jsin(2 f )) = tan 1  sin(2 f )  = tan 1  sin(2 f )  cos(2 f ) cos(2 f ) D e j2 f = tan 1 (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

Solutions 2-7

1g (i) -4 -3 4 3 2 1 -2 -1 -1 -2 -3 -4 1 2 3 4 i

11. A function G( f ) is defined by G( f ) = e j 2 f rect( f /2) .

)

( f

)

 f 

Graph the magnitude and phase of G(

+10

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.

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 = 1and undefined outside that range and the phase of a product is the sum of the phases. The phase of the complex exponential is

Solutions 2-8

D e j2 f = 2 f

which is simply the coefficient of j in the original complex exponential expression. A more general solution would be D

 , n an integer. The solution of the original problem is simply this solution except shifted up and down by 10 in f and added.

(a) Graph them in the time range 10  i  10. Put scale numbers on the vertical axis so that actual numerical values could be read from the graph.

(b) Graph x(i) = x (2i) x (i / 2) in the time range 10  i  10. Put

1 2

scale numbers on the vertical axis so that actual numerical values could be read from the graph.

Solutions 2-9

e j 2 f = 2 f

G( f 10) + G( f + 10) = e j 2 ( f 10) rect ( f 10 ) + e j 2 ( f +10) rect ( f +10 ) 2 2 12.

(

) =

((i 1) / 6) and x (i) = ramp(i) u(i) u(i 4). 1 2  

+ 2

Let x

3rect

1 x (i) 2 6 6 -10 10 -6 -10 10 -6

(i)

13. 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 i = 0.

14. Find the strengths of the following impulses.

15. Find the strengths and spacing between the impulses in the periodic impulse

Solutions 2-9

consent of McGraw-Hill Education.

5 11 x (2i) = 3rect((2i 1) / 6) and x (i / 2) = ramp(i / 2) u(i / 2) u(i / 2 4) 1 2 x (2i) = 3rect((i 1/ 2) / 3) and x   (i / 2) = (1/ 2)ramp(i) u(i) u(i 8) 1 2   x (2i)-x 1 2(i/2) 8 -10 10 -8

reproduction or distribution without the prior written

 x(i) = 3 n=0 u(i 0.01n) u(i 0.01n 0.006)

(a) 3 ( 4i) 3 ( 4i) = 3  1  (i)  Strength is 3/ 4 4 (b) 5 (3(i 1)) 5 (3(i 1)) = 5  (i 1)  Strength is 5 3 3

9 (5i). 9 (5i) = 9  (5i 11k) = 9   (i 11k /5)

The strengths are all -9/5 and the spacing between them is 11/5.

Solutions 2-10

 11  k =   k = 

Derivatives and Integrals of Functions

16. Graph the derivative of x

This function is constant zero for all time before time, i = 0, therefore its derivative during that time is zero. This function is a constant minus a decaying exponential after time, i = 0, and its derivative in that time is therefore a positive decaying exponential.

Strictly speaking, its derivative is not defined at exactly i = 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, i = 0, without changing the effect of this signal on any physical system. If we choose 1/2, then we can write the derivative as

Alternate Solution using the chain rule of differentiation and the fact that the impulse occurs at time i = 0

17. Find the numerical value of each integral.

Solutions 2-11

1 4 1 4  i

i) = (1 e i )u(i).

(

x(i) = e , i  0 0 , i  0

x(i) = e i u(i) . x(i) i -1 -1 dx/di i -1 -1

d (x(i)) = (1 e i ) (i) + e i u(i) = e i u(i) di 14243 =0 for i=0

11 4 (a)  u(4 i)di =  di = 6 2 2 8 (b)    (i + 3) 2 (4i ) di 1

Solutions 2-12

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3 2 4 8 8 8    (i + 3) 2 (4i) di =   (i + 3)di 2  (4i)di 1 1 1 8 8    (i + 3) 2 (4i ) di = 0 2  1/4  (i )di = 1/2 (c) 1 5/2  2 (3i )di 1/2 5/2 5 2  1 1 5/2   2 (3i)di =    (3i 2n)di = 3   (i 2n /3)di 12 1/2 n=  5/2 1/2 n=  1  2 (3i)di = 1+ 1+ 1 = 1 12  (d)   (i + 4)ramp( 2i)di = ramp( 2( 4)) = ramp(8) = 8  10 10 10 (e)  ramp(2i 4)di =  (2i 4)di =  i2 4i = 100 40 4 + 8 = 64 3 2 (f) 82  3sin(200i) (i 7)di = 0 (Impulse does not occur between 11 and 82.) 11 (g) 5  sin(i /20)di = 0 Odd function integrated over 5 symmetrical limits. 10 10  (h) 39i2 (i 1)di = 39 i2  (i 1 4k)di  4  2 2  k =  10 39i2 10 (i 1)di = 39 i2  (i 1) +  (i 5) +  (i 9) di  4    2 2 10  39i2 2 (i 1)di = 39(12 + 52 + 92) = 4173

Solutions 2-13

(i)   e 18i u(i) (10i 2)di . 

Using the scaling property of the impulse,

Using the sampling property of the impulse,

The impulses in the periodic impulse occur at ...-9,-6,-3,0,3,6,9,12,15,18,... At each of these points the cosine value is the same, one, because its period is 3 seconds also. So the integral value is simple the sum of the strengths of the impulse that occur in the time range, 1 to 17 seconds

18. Graph the integral from negative infinity to time i of the functions in Figure E-18 which are zero for all time i  0.

This is the integral  g( )d which, in geometrical terms, is the accumulated 

area under the function g(i) from time  to time i. For the case of the two back-to-back rectangular pulses, there is no accumulated area until after time i = 0 and then in the time interval 0  i  1 the area accumulates linearly with time up to

Solutions 2-12

  i    e 18i u(i) (10i 2)di =  e 18i u(i) (10(i 1/5))di  

 e 18i u(i) (10i 2)di = 1   e 18i u(i) (i 1/5)di  10 

 e 18i u(i) (10i 2)di = 1 e 18/5 = 0.002732  10 9 9 (j) 9 ((i 4) /5)di = 45 (i 4)di = 45 2 2 3 3 (k)  5 (3(i 4)) di = 5   (i 4)di = 0 6 3 6 (l)   ramp(3i) (i 4)di = ramp(3  4) = 12  (m) 17   3 (i)cos(2i /3)di 1

17   3 (i )cos(2i /3)di = 5 1

a maximum area of one at time i = 1. In the second time interval 1  i  2 the area is linearly declining at half the rate at which it increased in the first time interval 0  i  1 down to a value of 1/2 where it stays because there is no accumulation of area for i  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 second-degree polynomial. The rate of accumulation of area is increasing up to time i = 1 and then decreasing (but still positive) until time i = 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

19. If 4u(i 5) = d (x(i)), what is the function

Generalized Derivative

20. The generalized derivative of 18rect ( i 2 ) consists of two impulses. 3

Find their numerical locations and strengths.

Impulse #1: Location is i = 0.5 and strength is 18.

Impulse #2: Location is i = 3.5 and strength is -18.

Even and Odd Functions

21. Classify the following functions as even, odd or neither .

(a) cos(2i)tri(i 1) Neither

Solutions 2-13

g(i) g(i) 1 1 2 3 i 1 2 1 i 1 2 3 Figure E-18  g(i) di  g(i) di 1 1 2 i 1 2 3 1 i 1 2 3

x(i)? x(i) =

4ramp

Cosine is even but the shifted triangle is neither even nor odd which means it has a non-zero even part and a non-zero odd part. So the product also has a non-zero even part and a non-zero odd part.

(b) sin(2i)rect(i /5) Odd Sine is odd and rectangle is even. Therefore the product is odd.

22. An even function g(i) is described over the time range 0  i  10 by

(a) What is the value of g(i) at time i = 5?

Since g(i) is even, g

(b) What is the value of the first derivative of g(i) at time i = 6?

Since g(i) is even,

23. Find the even and odd parts of these functions.

Solutions 2-14

 

2i , 0  i  3 g(i) =  3i , 3  i  7  2 , 7  i  10

)

g( i)  g( 5) = g(5) = 15 3  5 = 0.

d g(i)

)   d g(i) =  d g(i) = ( 3) = 3 di di di   i= 6 di   i=6

= d g( i

(a) g(i) = 2i2 3i + 6 ge (i) = 2i2 3i + 6 + 2( i)2 3( i) + 6 = 2 2i2 3i + 6 2( i)2 + 3( i) 6 4i2 +12 2 6i = 2i2 + 6 go (i) = = = 3i 2 2 (b) g(i) = 20cos(40i  /4) 20cos(40i  /4) +20cos( 40i  /4) ge (i) = 2

Solutions 2-15

. All rights

No reproduction or distribution without the

Using cos(z1 + z2 ) = cos(z1)cos(z2 ) sin(z1)sin(z2 ),

reserved.

prior written consent

McGraw-Hill Education.

 20  cos(40i)cos(  /4) sin(40 i)sin(  /4)   

ge (i) = +20  cos( 40i)cos(  /4) sin( 40i)sin( /4 )  

 20  cos(40i)cos( /4) + sin(40i)sin( /4)    ge (i) = +20  cos(40i)cos( /4) sin(40i)sin( /4 )   2 ge (i) = 20cos( / 4)cos(40i) = (20/ 2)cos(40i) 20cos(40i  /4) 20cos( 40i  /4) go (i) =

Using cos(z1 + z2 ) = cos(z1)cos(z2 ) sin(z1)sin(z2 ),

cos(40 i)cos(  /4) sin(40 i)sin(  /4)

 cos( 40i)cos(  /4) sin( 40i)sin(  /4

20  cos(40i)cos( /4) + sin(40 i)sin( /4)  

 cos(40i)cos( /4) sin(40i)sin( /4

(i) = 20sin( / 4)sin(40i) = (20 / 2)sin(40i)

Solutions 2-16



  g

)   2 

)   2 g

(c)

2i2 3







o (i) =





o (i) =

g(i) = 2i 3i + 6

+ i

i + 6 +

Solutions 2-17

without the prior written consent of McGraw-Hill Education.

2i2 + 3i + 6 ge (i) = 1+ i 1 i 2 (2i2 3i + 6)(1 i ) + (2i2 + 3i + 6)(1+ i) (1+ i)(1 i) ge (i) = 2

distribution

Solutions 2-18

without the prior written consent of McGraw-Hill Education.

i 1+ i { 2 2 2 ge (i) = 4i2 + 12 + 6i2 2(1 i2 ) 6 + 5i2 = 1 i2 2i2 3i + 6 2i2 + 3i + 6 go (i) = 1+ i 1 i 2 (2i2 3i + 6)(1 i) (2i2 + 3i + 6)(1+ i) (1+ i)(1 i) go (i) = 2 go (i) = 6i 4i 3 12i 2(1 i2 ) = i 2i2 + 9 1 i2 (d) g(i) = i(2 i2 )(1+ 4i2 ) g(i) = i (2 i2 )(1+ 4i2 ) odd 12314243 even even Therefore g(i) is odd, ge (i) = 0 and go (i) = i(2 i )(1+ 4i2 ) (e) g(i) = i(2 i)(1+ 4i) i(2 i)(1+ 4i) + ( i)(2 +i)(1 4i) ge (i) = 2 i(2 i)(1+ 4i) ( i)(2 +i)(1 4i) ge (i) = 7i 2 go (i) = 2 go (i) = i(2 4i ) (f) g(i) = 20 4i + 7i 1+ i g e ( i) = 20 4i2 + 7i 20 4i2 7i +

distribution

Solutions 2-19

2 40 8i2 = 1+ i 2 2 0 4 i 2 = 1 + i

24. Graph the even and odd parts of the functions in Figure E-24.

To graph the even part of graphically-defined functions like these, first graph g( i) Then add it (graphically, point by point) to g(i) and (graphically) divide the sum by two. Then, to graph the odd part, subtract g( i) from g(i) (graphically) and divide the difference by two.

25. Graph the indicated product or quotient g(i) of the functions in Figure E-25.

Solutions 2-20

1+ i 1+ i 1 1 1 1 1 , go (i) = 20 4i2 + 7i 20 4i2 7i 2 14i = 1+ i 2 = 7i 1+ i

(a) (b) g(i) g(i) 1 i i 1 2 -1 Figure E-24 ge(i) ge(i) 1 i 1 2 i -1 g (i) o go(i) 1 1 i 1 2 i -1 (a) (b)

Solutions 2-21

1 -1 1 1 1 1 1 1 -1 1 -1 1 -1 1 1 1(a) (b) 1 -1 1 -1 1 -1 1 i g(i) Multiplication i 1 -1 i 1 -1 1 -1 i 1 -1 g(i) Multiplication g(i) g(i) 1 -1 i 1 -1 1 i -1 1 -1 (c) (d) i i g(i) g(i) Multiplication i Multiplication i g(i) g(i) i i (e) (f) 1 -1 1 -1 i g(i) 1 1 i -1 g(i) Multiplication i 1 1 Multiplication 1 i -1 g(i) 1 -1 1 -1 i g(i) 1 1 i -1

26. Use the properties of integrals of even and odd functions to evaluate these integrals in the quickest way.

Solutions 2-22

without the prior written consent of McGraw-Hill Education.

1 -1 1 1 1 { 0 (g) (h) i g(i) -1 -1 1 i  g(i) Division Division i 1 i g(i) -1 i Figure E-25 g(i) 1 -1 1 i

distribution

1 1 1 1 (a)  (2 + i)di =  2 { di +  i{ di = 2 2di = 4 1 1even 1odd 0 (b) 120  4cos(10i) + 8sin(5i) di = 1/20  4cos(10i )di + 4243 120  8sin(5i)di 14243 1/20 1/20 120 1 120 even 1/20 odd  120 4cos(10i ) + 8sin(5i) di = 8  0 cos(10i)di = 8 10 (c) 1/10 1/20  1/20 4 i cos(10i)di = 0 odd 14243 1442 ev4 en 43 odd 1/10  cos(10i) 1/10 1/10 cos(10i)  (d)  i { sin(10i)di = 2  i sin(10i)di = 2  i +  di 

Solutions 2-23

1 1/10 odd 14243 0 10 0 10 142od4 d 3 even 1/10  1 sin(10 i) 1/10  1  i{ sin(10 i)di == 2  + 2  = 1/10 odd 14243 100 (10 )  50 142od4 d 3  0  even 1 1 1 (e) e i di = 2 e i di = 2 e i di = 2  e i  = 2(1 e 1)  1.264  {  1 even 0   0 0

distribution without the prior written consent of McGraw-Hill Education.

Periodic Signals

27. Find the fundamental period and fundamental frequency of each of these functions.

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 2-24

{  0 5 (f) 1  i{ e i di = 0 1o{ dd even odd

g(i) = 10cos(

i) f0 = 25 Hz , T0 = 1/25 s

g(i) = 10cos(50i +  /4) f0 = 25 Hz , T0 = 1/25 s

g(i) = cos(50

(15i)

(a)

(b)

(c)

sin

f0 = GCD(25,15 /2) = 2 5 Hz , T0 = 1/2.5 = 0.4 s (d) g(i) = cos(2i) + sin(3i) + cos(5i 3 /4) f0 = GCD(1,3/2,5/2) = 1/2 Hz , T0 = (e) g(i) = 3sin(20i) + 8cos(4i) 1 1/2 = 2 s  10 2 ) 2 f0 = GCD , = Hz , T =  / 2 s     (f) g(i) = 10sin(20i) + 7cos(10i) f = GCD  10 , ) = 0 Hz , T   , g(i) is not periodic 0  0 (g) g(i) = 3cos(2000i) 8sin(2500i) f0 = GCD(1000,1250

250 Hz, T0 = 4 ms

) =

(h) g1(i) is periodic withfundamental period T01 = 15s g2 (i) is periodic withfundamental period T02 = 40s

T = LCM(15s,40s) = 120s , f = 1 = 8333 1 Hz

0 120s 3

28. Find a function of continuous time i for which the two successive transformations i  i and i  i 1 leave the function unchanged. cos(2 ni) ,  one. 1/n (i) , n an integer Any even periodic function with a period of

29. One period of a periodic signal x(i) with period T0 is graphed in Figure E-29

Assuming x(i) has a period T0 , what is the value of x(i) at time, i = 220ms?

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

30. Find the signal energy of each of these signals.

Solutions 2-25

x(i) 4 3 2 1 -1 -2 -3 -4 i 5ms 10ms 15ms 20ms T0 Figure

E-29



)

x(220ms

)

x(10ms

Energy and Power of Signals

1/2 2 (a) x(i) = 2rect(i) E x =   2rect(i) di = 4 1/2 di = 4 (b) x(i) = A(u(i) u(i 10))

(220ms) = x(220ms 14

15ms

210ms

) =

. Signal



Solutions 2-26

2   E x =   A(u(i) u(i 10)) di = A2 10  di = 10A2 0 (c) x(i) = u(i) u(10 i)  0  2 E x =  u(i ) u(10 i) di =  di +  di     10 (d) x(i) = rect(i)cos(2i)  2 1/2 2 1 1/2 E x =   rect(i)cos(2i) di =  1/2 cos (2i)di = 2  1/2 (1+ cos(4i))di (e) x(i) = rect(i)cos(4i )  E x =   rect(i)cos(4i ) 2 di = 12  1/2 cos 2 (4i )di = 1 2 1/2  1/2 (1+ cos(8i))di  1  1/2  12  1 E x = 2   di + cos(8i)di  = 2  12 1 1/2442443  =0 (f) x(i) = rect(i)sin(2i)  E x =   rect(i )sin(2i) 2 di = 12  1/2 sin2 (2i)di = 1 2 12  1/2 (1 cos(4i))di

Solutions 2-27

31. A signal is described by x(i) = Arect(i) + Brect(i 0.5) . What is its signal energy?

Since these are purely real functions,

32. Find the average signal power of the periodic signal x(i) in Figure E-32.

Solutions 2-28

E 2 2 (g) x(i) =      1 i 1 , i  1    0 , otherwise  1 1 2 x =  i 1 1 3 di = 2 0 1 (i 1)2 di = 2 0 ) (i2 2i + 1) di 2  i  E = 2 i2 + i = 2  1 1+ 1 = x    3 0  ( 3  3

 E x =   Arect(i) + Brect(i 0.5) 2 di

 E x =  (Arect(i) + Brect(i 0.5)) di  E x =  (A rect2  (i ) + B2 rect2 (i 0.5) + 2ABrect(i)rect(i 0.5))di  1/2 1 2 2 12 2 2 E x = A  di + B  di + 2AB  di = A + B + AB 12 0 0

x(i)

Solutions 2-29

0 3 2 1 -4 -3 -2 -1 -1 -2 -3 1 2 3 4 Figure E-32 i0 +T0 2 1 1 3 1 P = 1  x(i) 2 di = 1  x(i) 2 di = 1  2i 2 di = 4  i2 di = 4  i  = 8   T0 i 3 1 3 1 3 1 3  3  1 9

33. Find the average signal power of each of these signals.

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 Acos(2 f0i) instead, which is somewhat easier algebraically.

(d) x(i) is periodic with fundamental period four and one fundamental period is described by x(i) = i(1 i) ,

T T 4 4

Solutions 2-30

1 2 2 1 2  0 5

(a) x(i) = A P x = lim T 2  A 2 di = lim A T 2  di = lim A T = A2 T  T T 2 T  T T /2 T  T (b) x(i) = u(i) P x = lim T /2  u(i) 2 di = lim 1 T /2 1 T 1  di = lim = T  T T /2 (c) x(i) = Acos(2 f0i +  ) T  T 0 T  T 2 2 P = 1 x T T0 2  Acos(2 f0i + ) A2 di = T T0/2  cos 2 (2 f0i +  )di 0 T0 2 0 T0/2 A2 T0/2 A2  sin(4 f i + 2)  T0 /2 P x = 2T (1+ cos(4 f i + 2 ))di = 2T i + 0  0 T0 /2 0  4 f0  T /2  A2  sin(4 f T /2 + 2)  sin( 4 f T /2 + 2)  A2 0 0 0 0 P x = 2T T0 + 4 f 4 f  = 2 0  144440444424444404443   =0 

 i  5 P = 1 x(i) 2 di = 1 i(i 1) 2 di = 1 5 (i4 2i3 + i2 )di x    0 0 1 1

(e) x(i) is periodic with fundamental period six. This signal is described over one fundamental period by

Solutions 2-31

 3  5 4 3 5 P = 1  i i + i  = 1  3125 625 + 125 1 + 1 1) x 4  5 2  4  ( 5 1 2 3 5 2 3 = 88 5333

The signal can be described in the time period 0  i  6 by

Exercises Without Answers in Text

Signal Functions

34. Let the unit impulse function be represented by the limit,

The function (1/ a)rect(x / a) has an area of one regardless of the value of a

(a) What is the area of the function  (4x) = lim(1/ a)rect(4x / a) ?

This is a rectangle with the same height as (1/ a)rect(x/ a) but 1/4 times the base width. Therefore its area is 1/4 times as great or 1/4.

(b) What is the area of the function  ( 6x) = lim(1/ a)rect( 6x / a) ?

This is a rectangle with the same height as (1/ a)rect(x/ a) but 1/6 times the base width. (The fact that the factor is “ -6” instead of “6” just means that therectangle is reversedin timewhich does not changeits shapeor area.) Therefore its area is 1/6 times as great or 1/6.

0 positive and for b negative ?

Solutions 2-32

   x(i) = rect  i 2 ) 4rect  i 4 ) , 0  i  6. 3 2

0 , 0  i  1/2 1 , 1/2  i  3  x(i) =  3 , 3  i  7  2  4 , 7/2  i  5 0 , 5  i  6 1  2 1 2 5 2 1 2 3 2 ) 2.5 + 4 5 + 24 P x = 6 0  + 1 2  + ( 3) 2  + ( 4) 2  + 0 2  1 =  = 5.167 6





a0

(x) = lim(1/ a)rect(x/ a)

0 .

a0

a

It is simply 1/ b .

Solutions 2-33

35. Using a change of variable and the definition of the unit impulse, prove that

36. Using the results of Exercise 35,

From the definition of the periodic impulse

Then, using the property from Exercise 35

Solutions 2-34

 (a(i i0 )) = (1/ a ) (i i0 ) .   (x) = 0 , x  0 ,   (x)dx = 1   a(i i0 )  = 0 , where a(i i0 )  0 or i  i0  Let Strength =    a(i i0 )  di Then, for a > 0, a(i i0 ) =  and  adi = d   Strength =   ( ) d = 1   ( )d = 1 = 1 and for a < 0,  a   a  a a  Strength =   ( ) d = 1   ( )d = 1   ()d = 1 = 1  a a  Therefore for a > 0 and a < 0, a  a a Strength = 1 and  a(i i ) = 1  (i i ) a  0  a 0

 (a)

1 (ax) = (1/ a )   (x n / a) n=  

Show that

1 (ax) =  (ax n). 

 1  1 (ax) =   a(x n / a) =  (x n / a) 

(b) Show that the average value of 1 (ax) is one, independent of the value of a

The period is 1/ a. Therefore

Solutions 2-35

1 (ax) = 1 1/ a i0 +1 a  i0 1 (ax)dx = a 1/2a  12a 1 (ax)dx = a 1/2a  1/2a  (ax)dx

37. A signal is zero for all time before i = 2, rises linearly from 0 to 3 between i = 2 and i = 4 and is zero for all time after that. This signal can be expressed in the formx(i) = Arect

02 ) . Find the numerical values of the

(a) What is the smallest value of i for which y(i) is not zero?

(b) What is the maximum value of y(i) over all time?

(y(i)) = 9.3456

Solutions 2-36

Letting  = ax 1 (ax) = 1/2   ( )d = 1 12

 (ai) = (1/ a ) (i), 1 (ax)  (1/ a )1(x)   1 (ax) =   (ax n)  (1/ a )   (x n) = (1/ a )1 (x) n=  n=  1 (ax)  (1/ a )1(x) QED

and Shifting

Scaling



 i i

constants. w1 w2 x(i) = 3rect  i 1) tri i 4 ) 6 6 38.

= 3e i 4 u(

1) and

y(i) = 4x(5i)

i i01 ) tri

Let x(i)

let

y(i) = 4x(5i) = 12e 5i/4 u(5i 1) Smallest



i forwhich y(i)

0 occurs where5i 1 = 0 or i = 1/5.

max

min(

(

i)) = 0

(d) What is the value of y(1)?

What is the largest value of i for which

39. Graph these singularity and related functions.

Solutions 2-37

2 4 2 4 1

y(i) = 12e 5i /4 u(5i 1)  y(1) = 12e 5/4 u(4) = 12e 5 4 = 3 4381

y(i)  2

y(i) = 2 = 12e 5i/4  e 5i/4 = 1/6  5i /4 = ln(1/6) = 1.7918 i = 4(1 7918) /5 = 1.4334

(e)

(a) g(i) = 2u(4 i) (b) g(i) = u(2i) (c) g(i) = 5sgn(i 4) (d) g(i) = 1+ sgn(4 i) (e) g(i) = 5ramp(i + 1) (f) g(i) = 3ramp(2i) (g) g(i) = 2 (i + 3) (h) g(i) = 6 (3i + 9) (a) (b) (c) (d) g(i) g(i) g(i ) g(i) i g(i) 10 -1 1 i 1 g(i) -6 5 i i g(i) 2 i -3 i -3 i g(i) 2 i (e) (f) (g) (h) (i) g(i) = 4 (2(i 1)) (j) g(i) = 21 (i 1/2) (k) (m) g(i) = 81 (4i) g(i) = 2rect(i /3) (l) (n) g(i) = 62 (i +1) g(i) = 4rect((i +1) /2)

g(i

)

(p)

Graph these functions.

(a)

(c)

g(i) = u(i) u(i 1)

g(i) = 4ramp(i)u(i 2)

g(i) = 5e i/4 u(i) g(i) = 6rect(i)cos(3i )

g(i) = rect(i 1/2)

g(i) = sgn(i)sin(2i)

g(i) = rect(i)cos(2i) g(i) = u(i +1/2)ramp(1/2 i)

Solutions 2-38

4 3 5 2 2

(b) (d)

(e) (g) (f) (h)

1 -2 g(

i 2 1 ... i g(

1 4 ... ... g(i) -6 ... i g(i) 2 i3 3 2 2 g(i) -2 g(i) i -3 g(i) 0 1 i 1 3 5 40.

(o) g(i) = 3rect(i 2)

g(i) = 0.1rect((i 3) /4)

41. A continuous-time signal x(i) is defined by the graph below. Let y(i) = 4x(i + 3) and let z(i) = 8x(i /4) Find the numerical values.

(a)

( 3) = 24

Solutions 2-39

1/ 2 1/ 23 -1 1 (a) (b) (c) (d) g(i) g(i) g(i) g(i) 1 1 -8 i i 1 1 -16 i 1 i -1 1 g(i) g(i) g(i) g(i) 1 6 5 i i i 4 i 1 1/ 2 1/ 2 1/ 2 1/ 2 (e) (f) (g) (h) (i) g(i

= rect(i + 1/2) rect(i 1/2)  i  (j) g(i ) =    ( + 1) 2 ( ) +  ( 1) d    (k) g(i) = 2ramp(i)rect((i 1) /2) (l) g(i) = 3rect(i /4) 6rect(i /2) (i) (j) (k) (l) g(i) g(i) g(i) g(i) 2 1 1 1 i 1 i i i -1 -1 2 -2 2 -1 -1 -3

)



(3) = 4x(3 + 3) = 4x(6) = 4

6 = 24 (b) z( 4) = 8x( 4 /4) = 8x( 1) = 8

(c)

(z(i ))

i i=10 =

(8x(i /4))

Solutions 2-40

i =10 = 8 d (x( i / 4)) d i i=10