Chapter 2 - Mathematical Description of Continuous-Time ...€¦ · !M. J. Roberts - 3/16/11 Solutions 2-1 Chapter 2 - Mathematical Description of Continuous-Time Signals Solutions

©M. J. Roberts - 3/16/11

Solutions 2-1

Chapter 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

(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− j2t + e j2t

2= 7e−3 cos 2t( )

(f)

g jt − 32

⎛⎝⎜

⎞⎠⎟+ g − jt − 3

2⎛⎝⎜

⎞⎠⎟

2= 7

e− jt + e jt

2= 7cos t( )

2. If

g x( ) = x2 − 4x + 4 write out and simplify

(a)

g z( ) = z2 − 4z + 4

(b) g u + v( ) = u + v( )2

− 4 u + v( ) + 4 = u2 + v2 + 2uv − 4u − 4v + 4

(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( )( ) = t4 − 8t3 + 20t2 −16t + 4

(e)

g 2( ) = 4 − 8 + 4 = 0

3. What would be the 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’) ;

Full file at https://testbanku.eu/Solution-Manual-for-Signals-and-Systems-Analysis-Using-Transform-Methods-and-MATLAB-2nd-Edition-by-Roberts

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©M. J. Roberts - 3/16/11

Solutions 2-2

0.0247 + j0.1550.0920 + j0.289

10.0920 − j0.2890.0247 − j0.155

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

4. Let two functions be defined by

x1 t( ) = 1 , sin 20πt( ) ≥ 0

−1 , sin 20πt( ) < 0

⎧⎨⎪

⎩⎪ and

x2 t( ) = t , sin 2πt( ) ≥ 0

−t , sin 2πt( ) < 0

⎧⎨⎪

⎩⎪ .

Graph the product of these two functions versus time over the time range, −2 < t < 2 .

t-2 2

x(t)

-2

2

Transformations of Functions 5. For each function

g t( ) graph

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)

31

4

t

g(t-1)

1 2

3

-3

t

g(2t)

1

4

t

g(2t)

1

3

-32

12-

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

(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 2( ) = 5

(c)

x t( ) = 9rect t / 10( )sgn 3 t − 2( )( ) , x 1( ) = 9rect 1 / 10( )sgn −3( ) = −9




©M. J. Roberts - 3/16/11

Solutions 2-3

7. For each pair of functions in Figure E-7 provide the values of the constants A,

t0 and w in the functional transformation g2 t( ) = Ag1 t − t0( ) / w( ) .

-4 -2 0 2 4-2-1012

tg 1 (t

)

(a)

-4 -2 0 2 4-2-1012

t

g 2 (t)

(a)

-4 -2 0 2 4-2-1012

t

g 1 (t)

(b)

-4 -2 0 2 4-2-1012

t

g 2 (t)

(b)

-4 -2 0 2 4-2-1012

t

g 1 (t)

(c)

-4 -2 0 2 4-2-1012

t

g 2 (t)

(c)

Figure E-7

Answers: (a) A = 2,t0 = 1,w = 1 , (b) A = −2,t0 = 0,w = 1 / 2 ,

(c) A = −1 / 2,t0 = −1,w = 2

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

t0 and a in the functional transformation g2 t( ) = Ag1 w t − t0( )( ) .

(a)

-10 -5 0 5 10-8

-4

0

4

8

t

g 1(t)

-10 -5 0 5 10-8

-4

0

4

8

A = 2, t0 = 2, w = -2

t

g 2(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

(b) g 1(t)

A = 3, t0 = 2, w = 2

g 2(t)

-10 -5 0 5 10-8

-4

0

4

8

t-10 -5 0 5 10-8

-4

0

4

8

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




©M. J. Roberts - 3/16/11

Solutions 2-4

(c) g 1(t)

A = -3, t = -6, w = 1/30

g 2(t)

-10 -5 0 5 10-8

-4

0

4

8

t-10 -5 0 5 10-8

-4

0

4

8

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

(d) g 1(t)

0

g 2(t)

-10 -5 0 5 10-8

-4

0

4

8

t-10 -5 0 5 10-8

-4

0

4

8

t

A = -2, t = -2, w = 1/3

Amplitude comparison yields A = −2 . Time scale comparison yields w = 1 / 3 .

g2 4( ) = −2g1 1 / 3( ) 4 − t0( )( ) = −3g1 2( )⇒−t0 / 3+ 4 / 3 = 2 ⇒ t0 = −2

(e) g 1(t)

g 2(t)

-10 -5 0 5 10-8

-4

0

4

8

t-10 -5 0 5 10-8

-4

0

4

8

t

0A = 3, t = -2, w = 1/2

Amplitude comparison yields A = 3. Time scale comparison yields w = 1 / 2 .

g2 0( ) = 3g1 1 / 2( ) 0 − t0( )( ) = 3g1 1( )⇒−t0 / 2 = 1⇒ t0 = −2

Figure E-8

9. 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

g2 t( ) = 3g1 2 − t( ) ,

g3 t( ) = −2g1 t / 4( ) ,

g4 t( ) = g1

t − 32

⎛⎝⎜

⎞⎠⎟

Find these values. (a)

g2 1( ) = −3 (b)

g3 −1( ) = −3.5

(c)

g4 t( )g3 t( )⎡⎣ ⎤⎦t=2=

32× −1( ) = −

32

(d)

g4 t( )dt−3

−1

∫

The function

g4 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.

g4 t( )dt

−3

−1

∫ = −2




©M. J. Roberts - 3/16/11

Solutions 2-5

t

g (t)

1-1-2-3-4 2 3 4

1234

-4-3-2-1

1

Figure E-9

10. A function

G f( ) is defined by

G f( ) = e− j2π f rect f / 2( ) .

Graph the magnitude and phase of

G f −10( ) + G f +10( ) over the range,

−20 < f < 20 . 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− j2π f = cos −2π f( ) + j sin −2π f( ) = cos 2π f( ) − j sin 2π f( )

e− j2π f = cos2 2π f( ) + sin2 2π 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 = 1and undefined outside that range and the phase of a product is the sum of the phases. The phase of the complex exponential is

e− j2π f = cos 2π f( ) − j sin 2π f( )( ) = tan−1 −

sin 2π f( )cos 2π f( )

⎛

⎝⎜

⎞

⎠⎟ = − tan−1

sin 2π f( )cos 2π f( )

⎛

⎝⎜

⎞

⎠⎟

e− j2π 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− j2π f = −2π f

which is simply the coefficient of j in the original complex exponential expression. A more general solution would be e− j2π 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− j2π f −10( ) rect

f −102

⎛⎝⎜

⎞⎠⎟+ e− j2π f +10( ) rect

f +102

⎛⎝⎜

⎞⎠⎟




©M. J. Roberts - 3/16/11

Solutions 2-6

f -20 20

|G( f )|1

f -20 20

Phase of G( 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 u t − 0.01n( ) − u t − 0.01n − 0.006( )⎡⎣ ⎤⎦

n=0

∞

∑

Derivatives and Integrals of Functions 12. Graph the derivative of

x t( ) = 1− e− t( )u t( ) .

This function is constant zero for all time before time, 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.

′x t( ) = e− t , t > 0

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

′x t( ) = e− t u t( ) .

t-1 4

x(t)

-1

1

t-1 4

dx/dt

-1

1

13. Find the numerical value of each integral.

(a)

δ t + 3( ) − 2δ 4t( )⎡⎣ ⎤⎦dt

−1

8

∫ = δ t + 3( )dt−1

8

∫ − 2 δ 4t( )dt−1

8

∫ = 0 − 2 ×1 / 4 δ t( )dt−1

8

∫ = −1 / 2

(b)

δ2 3t( )dt

1/ 2

5/ 2

∫ = δ 3t − 2n( )n=−∞

∞

∑ dt1/ 2

5/ 2

∫ =13

δ t − 2n / 3( )n=−∞

∞

∑ dt1/ 2

5/ 2

∫ =13

1+1+1⎡⎣ ⎤⎦ = 1




©M. J. Roberts - 3/16/11

Solutions 2-7

14. Graph the integral from negative infinity to time t of the functions in Figure E-14

which are zero for all time t < 0 .

This is the integral

g τ( )dτ−∞

t

∫ which, in geometrical terms, is the accumulated

area under the function g t( ) 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 = 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 second-degree 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)

t

11 2 3

12

g(t)

t

1

1 2 3

Figure E-14

g(t) dt g(t) dt

t

1

1 2 3

12 t

1

1 2 3

Even and Odd Functions 15. An even function

g t( ) is described over the time range 0 < t < 10 by

g t( ) =2t , 0 < t < 315− 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,

ddt

g t( ) = −ddt

g −t( )⇒ ddt

g t( )⎡

⎣⎢

⎤

⎦⎥

t=−6

= −ddt

g t( )⎡

⎣⎢

⎤

⎦⎥

t=6

= − −3( ) = 3 .

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

(a) g t( ) = 2t2 − 3t + 6




©M. J. Roberts - 3/16/11

Solutions 2-8

ge t( ) = 2t2 − 3t + 6 + 2 −t( )2

− 3 −t( ) + 62

=4t2 +12

2= 2t2 + 6

go t( ) = 2t2 − 3t + 6 − 2 −t( )2

+ 3 −t( ) − 62

=−6t2

= −3t

(b)

g t( ) = 20cos 40πt − π / 4( )

ge t( ) = 20cos 40πt − π / 4( ) + 20cos −40πt − π / 4( )

2

Using

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

ge t( ) =

20 cos 40πt( )cos −π / 4( ) − sin 40πt( )sin −π / 4( )⎡⎣ ⎤⎦+20 cos −40πt( )cos −π / 4( ) − sin −40πt( )sin − / 4( )⎡⎣ ⎤⎦

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪2

ge t( ) =

20 cos 40πt( )cos π / 4( ) + sin 40πt( )sin π / 4( )⎡⎣ ⎤⎦+20 cos 40πt( )cos π / 4( ) − sin 40πt( )sin π / 4( )⎡⎣ ⎤⎦

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪2

ge t( ) = 20cos π / 4( )cos 40πt( ) = 20 / 2( )cos 40πt( )

go t( ) = 20cos 40πt − π / 4( ) − 20cos −40πt − π / 4( )

2

Using

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

go t( ) =

20 cos 40πt( )cos −π / 4( ) − sin 40πt( )sin −π / 4( )⎡⎣ ⎤⎦−20 cos −40πt( )cos −π / 4( ) − sin −40πt( )sin −π / 4( )⎡⎣ ⎤⎦

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪2

go t( ) =

20 cos 40πt( )cos π / 4( ) + sin 40πt( )sin π / 4( )⎡⎣ ⎤⎦−20 cos 40πt( )cos π / 4( ) − sin 40πt( )sin π / 4( )⎡⎣ ⎤⎦

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪2

go t( ) = 20sin π / 4( )sin 40πt( ) = 20 / 2( )sin 40πt( )

(c) g t( ) = 2t2 − 3t + 6

1+ t

ge t( ) =

2t2 − 3t + 61+ t

+ 2t2 + 3t + 61− t

2




©M. J. Roberts - 3/16/11

Solutions 2-9

ge t( ) =

2t2 − 3t + 6( ) 1− t( ) + 2t2 + 3t + 6( ) 1+ t( )1+ t( ) 1− t( )

2

ge t( ) = 4t2 +12 + 6t2

2 1− t2( ) =6 + 5t2

1− t2

go t( ) =

2t2 − 3t + 61+ t

− 2t2 + 3t + 61− t

2

go t( ) =

2t2 − 3t + 6( ) 1− t( ) − 2t2 + 3t + 6( ) 1+ t( )1+ t( ) 1− t( )

2

go t( ) = −6t − 4t3 −12t

2 1− t2( ) = −t 2t2 + 91− t2

(d) g t( ) = t 2 − t2( ) 1+ 4t2( )

g t( ) = todd 2 − t2( )

even

1+ 4t2( )even

Therefore

g t( ) is odd,

ge t( ) = 0 and go t( ) = t 2 − t2( ) 1+ 4t2( )

(e)

g t( ) = t 2 − t( ) 1+ 4t( )

ge t( ) = t 2 − t( ) 1+ 4t( ) + −t( ) 2 + t( ) 1− 4t( )

2

ge t( ) = 7t2

go t( ) = t 2 − t( ) 1+ 4t( ) − −t( ) 2 + t( ) 1− 4t( )

2

go t( ) = t 2 − 4t2( )

17. Graph the even and odd parts of the functions in Figure E-17.

To graph the even part of graphically-defined functions like these, first graph

g −t( ) . Then add it (graphically, point by point) to

g t( ) and (graphically) divide

the sum by two. Then, to graph the odd part, subtract g −t( ) from

g t( )

(graphically) and divide the difference by two.

t

g(t)

1

1

t

g(t)

21

1

-1 Figure E-17




©M. J. Roberts - 3/16/11

Solutions 2-10

t

g (t)

1

1

t

g (t)

1

1

e

o

,

t

g (t)

21

1

-1

t

g (t)

21

1

-1

e

o

(a) (b)

18. Graph the indicated product or quotient

g t( ) of the functions in Figure E-18.

(a) (b)

t1

-11

-1

t1-1

1g(t)

Multiplication

1-11

-1

1-1

-1

1

t

t

g(t)Multiplication

g(t)

t1

-11

-1

g(t)

t1-1

1

-1

(c) (d)

11

-11 t

t

g(t)Multiplication

1

1

11 t

t

g(t)Multiplication

g(t)

t-1

-1 1

g(t)

t-1 1

1

(e) (f)

1-1

1

-1

1-1

1

...... t

t

g(t)Multiplication

1

1

-1

1-1

1

t

t

g(t)Multiplication

(e) (f)

g(t)

t1-1

1

-1

......

g(t)

t1

1

-1




©M. J. Roberts - 3/16/11

Solutions 2-11

(g) (h)

11

-11 t

t

g(t)Division

-1 -1 11

1

t

t

g(t)Division

g(t)

t

-1

g(t)

-1 1

1

t

Figure E-18

19. Use the properties of integrals of even and odd functions to evaluate these integrals

in the quickest way.

(a)

2 + t( )dt−1

1

∫ = 2even dt

−1

1

∫ + todd dt

−1

1

∫ = 2 2dt0

1

∫ = 4

(b)

4cos 10πt( ) + 8sin 5πt( )⎡⎣ ⎤⎦dt−1/ 20

1/ 20

∫ = 4cos 10πt( )even

dt

−1/ 20

1/ 20

∫ + 8sin 5πt( )odd

dt

−1/ 20

1/ 20

∫

4cos 10πt( ) + 8sin 5πt( )⎡⎣ ⎤⎦dt

−1/ 20

1/ 20

∫ = 8 cos 10πt( )dt0

1/ 20

∫ =8

10π

(c)

4 todd cos 10πt( )

even

odd

dt−1/ 20

1/ 20

∫ = 0

(d)

todd sin 10πt( )

odd

even

dt−1/10

1/10

∫ = 2 t sin 10πt( )dt0

1/10

∫ = 2 −tcos 10πt( )

10π0

1/10

+cos 10πt( )

10πdt

0

1/10

∫⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

todd sin 10πt( )

odd

even

dt−1/10

1/10

∫ == 21

100π+

sin 10πt( )10π( )2

0

1/10⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=

150π

(e)

e− t

even dt

−1

1

∫ = 2 e− t dt0

1

∫ = 2 e− tdt0

1

∫ = 2 −e− t⎡⎣ ⎤⎦0

1= 2 1− e−1( ) ≈ 1.264

(f)

todd e− t

even

odd

dt−1

1

∫ = 0

Periodic Signals 20. Find the fundamental period and fundamental frequency of each of these functions.

(a) g t( ) = 10cos 50πt( ) f0 = 25 Hz , T0 = 1 / 25 s

(b)

g t( ) = 10cos 50πt + π / 4( ) f0 = 25 Hz , T0 = 1 / 25 s




©M. J. Roberts - 3/16/11

Solutions 2-12

(c)

g t( ) = cos 50πt( ) + sin 15πt( )

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.

f0 = GCD 25,15 / 2( ) = 2.5 Hz , T0 = 1 / 2.5 = 0.4 s

(d)

g t( ) = cos 2πt( ) + sin 3πt( ) + cos 5πt − 3π / 4( )

f0 = GCD 1,3 / 2,5 / 2( ) = 1 / 2 Hz , T0 =

11 / 2

= 2 s

21. One period of a periodic signal

x t( ) with period T0 is graphed in Figure E-21.

Assuming x t( ) has a period T0 , what is the value of

x t( ) at time, t = 220ms ?

5ms 10ms 15ms 20mst

x(t)4321

-1-2-3-4

T0 Figure E-21

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 .

22. In Figure E-22 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)

Figure E-22

(a) f0 = 3 Hz and T0 = 1 / 3 s (b)

f0 = GCD 6,4( ) = 2 Hz and T0 = 1 / 2 s

(c) f0 = GCD 6,5( ) = 1 Hz and T0 = 1 s

Signal Energy and Power of Signals 23. Find the signal energy of these signals.




©M. J. Roberts - 3/16/11

Solutions 2-13

(a) x t( ) = 2rect t( )

Ex = 2rect t( ) 2

dt−∞

∞

∫ = 4 dt−1/ 2

1/ 2

∫ = 4

(b) x t( ) = A u t( ) − u t −10( )( )

Ex = A u t( ) − u t −10( )( ) 2

dt−∞

∞

∫ = A2 dt0

10

∫ = 10A2

(c)

x t( ) = u t( ) − u 10 − t( )

Ex = u t( ) − u 10 − t( ) 2

dt−∞

∞

∫ = dt−∞

0

∫ + dt10

∞

∫ → ∞

(d)

x t( ) = rect t( )cos 2πt( )

Ex = rect t( )cos 2πt( ) 2

dt−∞

∞

∫ = cos2 2πt( )dt−1/ 2

1/ 2

∫ =12

1+ cos 4πt( )( )dt−1/ 2

1/ 2

∫

Ex =12

dt−1/ 2

1/ 2

∫ + cos 4πt( )dt−1/ 2

1/ 2

∫=0

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=12

(e)

x t( ) = rect t( )cos 4πt( )

Ex = rect t( )cos 4πt( ) 2

dt−∞

∞

∫ = cos2 4πt( )dt−1/ 2

1/ 2

∫ =12

1+ cos 8πt( )( )dt−1/ 2

1/ 2

∫

Ex =12

dt−1/ 2

1/ 2

∫ + cos 8πt( )dt−1/ 2

1/ 2

∫=0

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=12

(f)

x t( ) = rect t( )sin 2πt( )

Ex = rect t( )sin 2πt( ) 2

dt−∞

∞

∫ = sin2 2πt( )dt−1/ 2

1/ 2

∫ =12

1− cos 4πt( )( )dt−1/ 2

1/ 2

∫

Ex =12

dt−1/ 2

1/ 2

∫ − cos 4πt( )dt−1/ 2

1/ 2

∫=0

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=12

24. A signal is described by

x t( ) = Arect t( ) + B rect t − 0.5( ) . What is its signal

energy?

Ex = Arect t( ) + B rect t − 0.5( ) 2

dt−∞

∞

∫

Since these are purely real functions,




©M. J. Roberts - 3/16/11

Solutions 2-14

Ex = Arect t( ) + B rect t − 0.5( )( )2

dt−∞

∞

∫

Ex = A2 rect2 t( ) + B2 rect2 t − 0.5( ) + 2AB rect t( )rect t − 0.5( )( )dt

−∞

∞

∫

Ex = A2 dt

−1/ 2

1/ 2

∫ + B2 dt0

1

∫ + 2AB dt0

1/ 2

∫ = A2 + B2 + AB

25. Find the average signal power of the periodic signal

x t( ) in Figure E-25.

1-1-2-3-4 2 3 4

1

-1-2-3

23

t

x(t)

Figure E-25

P =

1T0

x t( ) 2dt

t0

t0 +T0

∫ =13

x t( ) 2dt

−1

2

∫ =13

2t2

dt−1

1

∫ =43

t2dt−1

1

∫ =43

t3

3⎡

⎣⎢

⎤

⎦⎥−1

1

=89

26. Find the average signal power of these signals.

(a) x t( ) = A

Px = lim

T→∞

1T

A2

dt−T / 2

T / 2

∫ = limT→∞

A2

Tdt

−T / 2

T / 2

∫ = limT→∞

A2

TT = A2

(b) x t( ) = u t( )

Px = lim

T→∞

1T

u t( ) 2dt

−T / 2

T / 2

∫ = limT→∞

1T

dt0

T / 2

∫ = limT→∞

1T

T2=

12

(c)

x t( ) = Acos 2π f0t +θ( )

Px =

1T0

Acos 2π f0t +θ( ) 2dt

−T0 / 2

T0 / 2

∫ =A2

T0

cos2 2π f0t +θ( )dt−T0 / 2

T0 / 2

∫

Px =A2

2T0

1+ cos 4π f0t + 2θ( )( )dt−T0 / 2

T0 / 2

∫ =A2

2T0

t +sin 4π f0t + 2θ( )

4π f0

⎡

⎣⎢⎢

⎤

⎦⎥⎥−T0 / 2

T0 / 2

Px =A2

2T0

T0 +sin 4π f0T0 / 2 + 2θ( )

4π f0

−sin −4π f0T0 / 2 + 2θ( )

4π f0

=0

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=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

Acos 2π f0t( )

instead, which is somewhat easier algebraically. Exercises Without Answers in Text Signal Functions




©M. J. Roberts - 3/16/11

Solutions 2-15

27. Given the function definitions on the left, find the function values on the right.

(a) g t( ) = 100sin 200πt + π / 4( )

g 0.001( ) = 100sin 200π × 0.001+ π / 4( ) = 100sin π / 5+ π / 4( ) = 98.77

(b) g t( ) = 13− 4t + 6t2

g 2( ) = 13− 4 2( ) + 6 2( )2

= 29 (c)

g t( ) = −5e−2te− j2π t

g 1 / 4( ) = −5e−2/ 4e− j2π / 4 = −5e−1/ 2e− jπ / 2 = − j3.03

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

δ x( ) = lim

a→01 / a( )rect x/ a( ) , a > 0 .

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 δ 4x( ) = lim

a→01 / 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

δ −6x( ) = lim

a→01 / 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 the rectangle 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

δ bx( ) = lim

a→01 / a( )rect bx / a( ) for b

positive and for b negative ?

It is simply 1 / b .

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

δ a t − t0( )( ) = 1 / a( )δ t − t0( ) .

δ x( ) = 0 , x ≠ 0 ,

δ x( )dx

−∞

∞

∫ = 1

δ a t − t0( )⎡⎣ ⎤⎦ = 0 , where a t − t0( ) ≠ 0 or t ≠ t0

Strength = δ a t − t0( )⎡⎣ ⎤⎦dt

−∞

∞

∫

Let

a t − t0( ) = λ and ∴adt = dλ

Then, for a > 0,

Strength = δ λ( ) dλ

a−∞

∞

∫ =1a

δ λ( )dλ−∞

∞

∫ =1a=

1a

and for a < 0,




©M. J. Roberts - 3/16/11

Solutions 2-16

Strength = δ λ( ) dλ

a∞

−∞

∫ =1a

δ λ( )dλ∞

−∞

∫ = −1a

δ λ( )dλ−∞

∞

∫ = −1a=

1a

Therefore for a > 0 and a < 0,

Strength =

1a

and δ a t − t0( )⎡⎣ ⎤⎦ =

1aδ t − t0( ) .

30. Using the results of Exercise 29, show that

(a)

δ1 ax( ) = 1 / a( ) δ x − n / a( )

n=−∞

∞

∑

From the definition of the periodic impulse δ1 ax( ) = δ ax − n( )

−∞

∞

∑ .

Then, using the property from Exercise 29 δ1 ax( ) = δ a x − n / a( )⎡⎣ ⎤⎦

−∞

∞

∑ =1a

δ x − n / a( )−∞

∞

∑ .

(b) Show that the average value of δ1 ax( ) is one, independent of the value of a

The period is 1 / a . Therefore

δ1 ax( ) =

11 / a

δ1 ax( )dxt0

t0 +1/ a

∫ = a δ1 ax( )dx−1/ 2a

1/ 2a

∫ = a δ ax( )dx−1/ 2a

1/ 2a

∫ Letting λ = ax

δ1 ax( ) = δ λ( )dλ

−1/ 2

1/ 2

∫ = 1

(c) Even though

δ at( ) = 1 / a( )δ t( ) ,

δ1 ax( ) ≠ 1 / a( )δ1 x( )

δ1 ax( ) = δ ax − n( )

n=−∞

∞

∑ ≠ 1 / a( ) δ x − n( )n=−∞

∞

∑ = 1 / a( )δ1 x( )

δ1 ax( ) ≠ 1 / a( )δ1 x( ) QED

Scaling and Shifting Functions 31. Graph these singularity and related functions.

(a) g t( ) = 2u 4 − t( ) (b)

g t( ) = u 2t( )

(c)

g t( ) = 5sgn t − 4( ) (d)

g t( ) = 1+ sgn 4 − t( )

(e)

g t( ) = 5ramp t +1( ) (f)

g t( ) = −3ramp 2t( )

(g)

g t( ) = 2δ t + 3( ) (h)

g t( ) = 6δ 3t + 9( )




©M. J. Roberts - 3/16/11

Solutions 2-17

g(t)

t4

2

4

2

(a)g(t)

t

1

(b)g(t)

t4

5

-5

(c)g(t)

t

(d)

-6

1

g(t)

t-1 1

10

(e)g(t)

t

(f)g(t)

t-3

2

(g) (h)g(t)

t-3

2

(i)

g t( ) = −4δ 2 t −1( )( ) (j)

g t( ) = 2δ1 t −1 / 2( )

(k)

g t( ) = 8δ1 4t( ) (l)

g t( ) = −6δ2 t +1( )

(m)

g t( ) = 2rect t / 3( ) (n)

g t( ) = 4rect t +1( ) / 2( )

(o)

g t( ) = −3rect t − 2( ) (p)

g t( ) = 0.1rect t − 3( ) / 4( )

g t( )

g t( ) g t( ) g t( ) g t( )

g t( ) g t( ) g t( )

t

t t

t

t tt

t

5 / 23 / 2

!3

0.1

531

(o) (p)

32. Graph these functions.

(a) g t( ) = u t( ) − u t −1( ) (b)

g t( ) = rect t −1 / 2( )

(c)

g t( ) = −4ramp t( )u t − 2( ) (d)

g t( ) = sgn t( )sin 2πt( )

(e)

g t( ) = 5e− t / 4 u t( ) (f)

g t( ) = rect t( )cos 2πt( )

(g)

g t( ) = −6rect t( )cos 3πt( ) (h)

g t( ) = u t +1 / 2( )ramp 1 / 2 − t( )




©M. J. Roberts - 3/16/11

Solutions 2-18

(a) (b) (c) (d)

g t( ) g t( ) g t( ) g t( )

g t( ) g t( ) g t( )g t( )

t

t

tt

tt t t

1 1

1

1

56

4

11

1

-1

-1

-8-16

1 / 21 / 2 1 / 2

!1 / 2!1 / 2 !1 / 2

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

(i)

g t( ) = rect t +1 / 2( ) − rect t −1 / 2( )

(j) g t( ) = δ λ +1( )

−∞

t

∫ − 2δ λ( ) + δ λ −1( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

dλ

(k) g t( ) = 2ramp t( )rect t −1( ) / 2( )

(l) g t( ) = 3rect t / 4( ) − 6rect t / 2( )

(i) (j) (k) (l)

g t( ) g t( )g t( ) g t( )

t t t t1

-1

1-1

1

-1 -3-2

1 12 2

2 3

-1 -1

33. Graph these functions.

(a) g t( ) = 3δ 3t( ) + 6δ 4 t − 2( )( )

Using the impulse scaling property,

g t( ) = δ t( ) + 3 / 2( )δ t − 2( )

t2

1

g(t)

32

(b) g t( ) = 2δ1 −t / 5( )

g t( ) = 2 δ −t / 5− n( )

n=−∞

∞

∑ = 10 δ t + 5n( )n=−∞

∞

∑ , 10

t

g(t)

5 10 15 20-10 -5

......

(c)

g t( ) = δ1 t( )rect t / 11( )




©M. J. Roberts - 3/16/11

Solutions 2-19

g t( ) = rect t / 11( ) δ t − n( )

n=−∞

∞

∑ = δ t − n( )n=−5

5

∑ ,

t

g(t)

1 2 3 4 5-5 -4 -3 -2 -1

1

(d) g t( ) = δ2 λ( ) − δ2 λ −1( )⎡⎣ ⎤⎦dλ

−∞

t

∫

t

g(t)

-1-2 21

1

3

34. 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.

g t( ) =0 , t < −5−10 − 2t , − 5 < t < −2sin πt / 2( ) , − 2 < t < 2

−6e− t / 2 , t > 2

⎧

⎨⎪⎪

⎩⎪⎪

(b) Graph

g t( ) in the range −10 < t < 10 .

(c) Graph

g 2t( ) in the range −10 < t < 10 .

(d) Graph

2g 3− t( ) in the range −10 < t < 10 .

(e) Graph

−2g t +1( ) / 2( ) in the range −10 < t < 10 .

t-10 10

g(t)

-8

t-10 10

g(2t)

-8

t-10 10

2g(3- t)

-16 t-10 10

-2g(( t+1)/2)16

35. Using MATLAB, for each function below plot the original function and the

transformed function.

% Plotting functions and transformations of those functions




©M. J. Roberts - 3/16/11

Solutions 2-20

% (a) part figure ; tmin = -3 ; tmax = 8 ; N = 100 ; dt = (tmax - tmin)/N ; t = tmin + dt*[0:N]’ ; g0 = g322a(t) ; g1 = -3*g322a(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)’) ; % (b) part figure ; tmin = 0 ; tmax = 96 ; N = 400 ; dt = (tmax - tmin)/N ; t = tmin + dt*[0:N]’ ; g0 = g322b(t) ; g1 = g322b(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)’) ; % (c) part figure ; fmin = -20 ; fmax = 20 ; N = 200 ; df = (fmax - fmin)/N ; f = fmin + df*[0:N]’ ; G0 = G322c(f) ; G1 = abs(G322c(10*(f-10)) + G322c(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)

y = -2*(t <= -1) + 2*t.*(-1 < t & t <= 1) + ... (3-t.^2).*(1 < t & t <= 3) - 6*(t > 3) ;

function y = g322b(t)

y = real(exp(j*pi*t) + exp(j*1.1*pi*t)) ;

function y = G322c(f) y = abs(5./(f.^2 - j*2 + 3)) ;

(a)

g t( ) =−2 , t < −12t , −1< t < 13− t2 , 1< t < 3−6 , t > 3

⎧

⎨⎪⎪

⎩⎪⎪

−3g 4 − t( ) vs. t

t-4 8

Original g(t)

-6

2

t-4 8

Transformed g(t)

-10

20

(b)

g t( ) = Re e jπ t + e j1.1π t( )

g t / 4( ) vs. t




©M. J. Roberts - 3/16/11

Solutions 2-21

t100

Original g(t)

-2

2

t100

Transformed g(t)

-2

2

(c)

G f( ) = 5

f 2 − j2 + 3

G 10 f −10( )( ) + G 10 f +10( )( ) vs. f

t-20 20

Original g(t)1.5

t-20 20

Transformed g(t)1.5

36. A signal occurring in a television set is illustrated in Figure E36. Write a

mathematical description of it.

t (µs)-10 60

x(t)

-10

Signal in Television5

Figure E36 Signal occurring in a television set

x t( ) = −10rect

t − 2.5×10−6

5×10−6

⎛

⎝⎜⎞

⎠⎟

37. The signal illustrated in Figure E37 is part of a binary-phase-shift-keyed (BPSK)

binary data transmission. Write a mathematical description of it.

t (ms)4

x(t)

-1

1

BPSK Signal

Figure E37 BPSK signal




©M. J. Roberts - 3/16/11

Solutions 2-22

x t( ) =sin 8000πt( )rect

t − 0.5×10−3

10−3

⎛

⎝⎜⎞

⎠⎟− sin 8000πt( )rect

t −1.5×10−3

10−3

⎛

⎝⎜⎞

⎠⎟

+ sin 8000πt( )rectt − 2.5×10−3

10−3

⎛

⎝⎜⎞

⎠⎟− sin 8000πt( )rect

t − 3.5×10−3

10−3

⎛

⎝⎜⎞

⎠⎟

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

38. The signal illustrated in Figure E38 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.

t (ns)20

x(t)

-6

-4

RC Filter Signal

-1.3333

4

Figure E38 Transient response of an RC filter

x t( ) = −4 − 2 1− e− t−4( )/3( )u t − 4( ) (times in ns)

39. Describe the signal in Figure E39 as a ramp function minus a summation of step

functions.

...4

15

x(t)

t

Figure E39

x t( ) = 3.75ramp t( ) −15 u t − 4n( )

n=1

∞

∑

40. Mathematically describe the signal in Figure E-40 .

......9

9

x(t)

t

Semicircle

Figure E-40

The semicircle centered at t = 0 is the top half of a circle defined by

x2 t( ) + t2 = 81

Therefore

x t( ) = 81− t2 , − 9 < t < 9 .

This one period of this periodic function. The other periods are just shifted versions.




©M. J. Roberts - 3/16/11

Solutions 2-23

x t( ) = rect

t −18n18

⎛⎝⎜

⎞⎠⎟

81− t −18n( )2

n=−∞

∞

∑

(The rectangle function avoids the problem of imaginary values for the square roots of negative numbers.)

41. Let two signals be defined by

x1 t( ) = 1 , cos 2π t( ) ≥ 00 , cos 2πt( ) < 0

⎧⎨⎪

⎩⎪ and x2 t( ) = sin 2πt / 10( ) ,

Plot these products over the time range, −5 < t < 5 .

(a)

x1 2t( )x2 −t( ) (b)

x1 t / 5( )x2 20t( )

(c)

x1 t / 5( )x2 20 t +1( )( ) (d)

x1 t − 2( ) / 5( )x2 20t( )

t-5 5

x1(t)x2(t)

-1

1

(a)

t-5 5

x1(t)x2(t)

-1

1

(b)

t-5 5

x1(t)x2(t)

-1

1

(c)

t-5 5

x1(t)x2(t)

-1

1

(d)

42. Given the graphical definition of a function in Figure E-42, graph the indicated transformation(s).

(a)

-2-2 2 3 4 5 611

2

-2

t

g(t)

g t( )→ g 2t( )g t( )→ −3g −t( )

g t( ) = 0 , t > 6 or t < −2

(b)




©M. J. Roberts - 3/16/11

Solutions 2-24

-2-2 21 3 4 5 6

12

-2

t

g(t)

t → t + 4

g t( )→ −2g t −1( ) / 2( )

g t( ) is periodic with fundamental period, 4

Figure E-42

(a)

The transformation g t( )→ g 2t( ) simply compresses the time scale by a factor of

2. The transformation g t( )→ −3g −t( ) time inverts the signal, amplitude inverts

the signal and then multiplies the amplitude by 3.

-2-2 2 4 6

12

-2

t

g(2t)

-2-2-4 2 4 6

36

-6

t

-3g(-t)

(b)

-2-2 21 3 4 5 6 7

12

-2

t

g(t + 4)

-2-2 21 3 4 5 6 7 8

24

-4

t

-2g( )t -12

43. For each pair of functions graphed in Figure E-43 determine what transformation

has been done and write a correct functional expression for the transformed function.

(a)




©M. J. Roberts - 3/16/11

Solutions 2-25

-2-2 21 3 4 5 6

2

-1t

g(t)

-1 1 2 3 4-2-3-4

2

-1t

(b)

-2-2 21 3 4 5 6

2

t

g(t)

-2-2 21 3 4 5 6-1t

Figure E-43 In (b), assuming

g t( ) is periodic with fundamental period 2 find two different

transformations which yield the same result (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 two successive transformations is then

t → − t − 2( ) = 2 − t . Therefore the

transformation is

g t( )→ g 2 − t( )

(b)

g t( )→ − 1 / 2( )g t +1( ) or g t( )→ − 1 / 2( )g t −1( )

44. Write a function of continuous time t for which the two successive changes t → −t

and t → t −1 leave the function unchanged. cos 2πt( ) ,

δ1 t( ) , etc...

(Any even periodic function with a period of one.) 45. Graph the magnitude and phase of each function versus f.

(a) G f( ) = jf

1+ jf / 10

f -100 100

|G( f )|10

f -100 100

- !

!

!G f( )




©M. J. Roberts - 3/16/11

Solutions 2-26

(b) G f( ) = rect

f −1000100

⎛⎝⎜

⎞⎠⎟+ rect

f +1000100

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥e− jπ f /500

(c) G f( ) = 1

250 − f 2 + j3 f

(b) (c)

f -1100 1100

|G( f )|1

f -1100 1100

- !

!

!G f( )

f -50 50

|G( f )|0.02

f -50 50

- !

!

!G f( )

46. Graph versus f , in the range −4 < f < 4 the magnitude and phase of

(a) X f( ) = 5rect 2 f( )e+ j2π f

5

! / 2"! / 2

1 / 4"1 / 4

!X f( )

X f( )

f

f

(b) X f( ) = j5δ f + 2( ) − j5δ f − 2( )

f

f

X f( )

!X f( )2!2

" / 2

!" / 2

5

(c) X f( ) = 1 / 2( )δ1/ 4 f( )e− jπ f

X f( ) = e− jπ f

2δ f − n

4⎛⎝⎜

⎞⎠⎟n=−∞

∞

∑

f

f

X f( )

!X f( )1!1

4"

!4"

1 / 2

Generalized Derivative




©M. J. Roberts - 3/16/11

Solutions 2-27

47. Graph the generalized derivative of g t( ) = 3sin πt / 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 πt / 2( ) ,

3π / 2( )cos πt / 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 .

t

ddt (g(t))

32

32

Alternate solution:

g t( ) = 3sin πt / 2( ) u t +1 / 2( ) − u t −1 / 2( )⎡⎣ ⎤⎦

ddt

g t( )( ) = 3sin πt / 2( ) δ t +1 / 2( ) − δ t −1 / 2( )⎡⎣ ⎤⎦ + 3π / 2( )cos πt / 2( ) u t +1 / 2( ) − u t −1 / 2( )⎡⎣ ⎤⎦

ddt

g t( )( ) = 3sin −π / 4( )δ t +1 / 2( ) − 3sin π / 4( )δ t −1 / 2( )⎡⎣ ⎤⎦ + 3π / 2( )cos πt / 2( )rect t( )

ddt

g t( )( ) = −3 2 / 2 δ t +1 / 2( ) + δ t −1 / 2( )⎡⎣ ⎤⎦ + 3π / 2( )cos πt / 2( )rect t( )

Derivatives and Integrals of Functions 48. What is the numerical value of each of the following integrals?

(a) δ t( )cos 48πt( )dt

−∞

∞

∫ = cos 0( ) = 1 ,

(b) δ t − 5( )cos πt( )dt

−∞

∞

∫ = cos 5π( ) = −1

(c) δ t − 8( )rect t / 16( )dt

0

20

∫ = rect 8 / 16( ) = 1 / 2

49. What is the numerical value of each of the following integrals?

(a) δ1 t( )cos 48πt( )dt

−∞

∞

∫ = δ t − n( )n=−∞

∞

∑ cos 48πt( )dt−∞

∞

∫ = δ t − n( )cos 48πt( )dt−∞

∞

∫n=−∞

∞

∑

δ1 t( )cos 48πt( )dt

−∞

∞

∫ = cos 48nπ( )n=−∞

∞

∑ = 1n=−∞

∞

∑ → ∞




©M. J. Roberts - 3/16/11

Solutions 2-28

t-0.1 0.1

x(t)

-1

1

(b) δ1 t( )sin 2πt( )dt

−∞

∞

∫

δ1 t( )sin 2πt( )dt

−∞

∞

∫ = δ t − n( )sin 2πt( )dt−∞

∞

∫n=−∞

∞

∑ = sin 2nπ( )n=−∞

∞

∑ = 0

t-2 2

x(t)

-1

1

(c) 4 δ4 t − 2( )rect t( )dt

0

20

∫

4 δ4 t − 2( )rect t( )dt

0

20

∫ = 4 δ t − 2 − 4n( )rect t( )dt0

20

∫n=−∞

∞

∑ = 4 rect 2 + 4n( )n=−∞

∞

∑ = 0

t-8 8

x(t)4

50. Graph the time derivatives of these functions.

(a) g t( ) = sin 2πt( )sgn t( )

′g t( ) = 2π−cos 2πt( ) , t < 0

cos 2πt( ) , t ≥ 0

⎧⎨⎪

⎩⎪

(b) g t( ) = cos 2πt( )

′g t( ) = 2πsin 2πt( ) , cos 2πt( ) < 0

− sin 2πt( ) , cos 2πt( ) > 0

⎧⎨⎪

⎩⎪

(a) (b)

t-4 4

x(t)

-1

1

t-4 4

dx/dt

-6

6

t-1 1

x(t)

-1

1

t-1 1

dx/dt

-6

6

Even and Odd Functions

51. Graph the even and odd parts of these signals.

(a) x t( ) = rect t −1( )




©M. J. Roberts - 3/16/11

Solutions 2-29

xe t( ) = rect t −1( ) + rect t +1( )

2 ,

xo t( ) = rect t −1( ) − rect t +1( )

2 >>>need graph

(b)

x t( ) = 2sin 4πt − π / 4( )rect t( )

xe t( ) = −2sin π / 4( )cos 4πt( )rect t( ) ,

xo t( ) = 2cos π / 4( )sin 4πt( )rect t( )

t-10 10

xe(t)

-4

4

(c)

t-10 10

xo(t)

-4

4

t-1 1

xe(t)

-2

2

(d)

t-1 1

xo(t)

-2

2

52. Find the even and odd parts of each of these functions.

(a) g t( ) = 10sin 20πt( )

ge t( ) = 10sin 20πt( ) +10sin −20πt( )

2= 0 ,

go t( ) = 10sin 20πt( ) −10sin −20πt( )

2= 10sin 20πt( )

(b)

g t( ) = 20t3

ge t( ) = 20t3 + 20 −t( )3

2= 0 ,

go t( ) = 20t3 − 20 −t( )3

2= 20t3

(c) x t( ) = 8 + 7t2

xe t( ) = 8 + 7t2 + 8 + 7 −t( )2

2= 8 + 7t2 ,

xo t( ) = 8 + 7t2 − 8 − 7 −t( )2

2= 0

(d) x t( ) = 1+ t

xe t( ) = 1+ t +1+ −t( )

2= 1 ,

xo t( ) = 1+ t −1− −t( )

2= t

(e) x t( ) = 6t

ge t( ) = 6t + 6 −t( )

2= 0 ,

go t( ) = 6t − 6 −t( )

2= 6t

(f) g t( ) = 4t cos 10πt( )

ge t( ) = 4t cos 10πt( ) + 4 −t( )cos −10πt( )

2=

4t cos 10πt( ) + 4 −t( )cos 10πt( )2

= 0

go t( ) = 4t cos 10πt( ) − 4 −t( )cos −10πt( )

2=

4t cos 10πt( ) − 4 −t( )cos 10πt( )2

= 4t cos 10πt( )




©M. J. Roberts - 3/16/11

Solutions 2-30

(g) g t( ) = cos πt( )

πt

ge t( ) =

cos πt( )πt

+cos −πt( )

−πt2

=

cos πt( )πt

+cos πt( )−πt

2= 0

go t( ) =

cos πt( )πt

−cos −πt( )

−πt2

=

cos πt( )πt

+cos πt( )

πt2

=cos πt( )

πt

(h) g t( ) = 12 +

sin 4πt( )4πt

ge t( ) =

12 +sin 4πt( )

4πt+12 +

sin −4πt( )−4πt

2=

12 +sin 4πt( )

4πt+12 +

sin 4πt( )4πt

2= 12 +

sin 4πt( )4πt

go t( ) =

12 +sin 4πt( )

4πt−12 −

sin −4πt( )−4πt

2=

sin 4πt( )4πt

−sin 4πt( )

4πt2

= 0

(i) g t( ) = 8 + 7t( )cos 32πt( )

ge t( ) = 8 + 7t( )cos 32πt( ) + 8 − 7t( )cos −32πt( )

2= 8cos 32πt( )

go t( ) = 8 + 7t( )cos 32πt( ) − 8 − 7t( )cos −32πt( )

2= 7t cos 32πt( )

(j)

g t( ) = 8 + 7t2( )sin 32πt( )

ge t( ) =

8 + 7t2( )sin 32πt( ) + 8 + 7 −t( )2( )sin −32πt( )2

= 0

go t( ) =

8 + 7t2( )sin 32πt( ) − 8 + 7 −t( )2( )sin −32πt( )2

= 8 + 7t2( )sin 32πt( )

53. 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( ) = 0 . Applying the definitions of even and odd functions,

xe t( ) = 0 + 0

2= 0 = x t( ) and xo t( ) = 0 − 0

2= 0 = x t( )

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




©M. J. Roberts - 3/16/11

Solutions 2-31

t

x(t)

1-1-1

12

-2-3-4-52 3 4 5

Figure E-54

t1

-1 -1

12

-2-3-4-5

2 3 4 5

x (t)e

t1

-1-1

12

-2-3-4-5

2 3 4 5

x (t)o

Periodic Functions

55. For each of the following signals decide whether it is periodic and, if it is, find the

fundamental period.

(a) g t( ) = 28sin 400πt( ) Periodic. Fundamental frequency = 200 Hz, Period = 5 ms.

(b) g t( ) = 14 + 40cos 60πt( ) Periodic. Fundamental frequency = 30 Hz Period = 33.33...ms.

(c) g t( ) = 5t − 2cos 5000πt( ) Not periodic.

(d) g t( ) = 28sin 400πt( ) +12cos 500πt( ) 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.

(e) g t( ) = 10sin 5t( ) − 4cos 7t( ) Periodic. The Periods of the two sinusoids are 2π / 5 s

and 2π / 7 s. Least common multiple is 2π . Period of the overall signal is 2π s. (f)

g t( ) = 4sin 3t( ) + 3sin 3t( ) Not periodic because least common multiple is infinite.

56. 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) 2rect −t( ) ,

E = 2rect −t( )⎡⎣ ⎤⎦

2dt

−∞

∞

∫ = 4 dt−1/ 2

1/ 2

∫ = 4

(b) rect 8t( ) ,

E = rect 8t( )⎡⎣ ⎤⎦

2dt

−∞

∞

∫ = dt−1/16

1/16

∫ =18

(c) 3rect

t4

⎛⎝⎜

⎞⎠⎟

, E = 3rect

t4

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

2

dt−∞

∞

∫ = 9 dt−2

2

∫ = 36




©M. J. Roberts - 3/16/11

Solutions 2-32

(d) 2sin 200πt( )

E = 2sin 200πt( )⎡⎣ ⎤⎦

2dt

−∞

∞

∫ = 4 sin2 200πt( )dt−∞

∞

∫ = 412−

12

sin 400πt( )⎡

⎣⎢

⎤

⎦⎥dt

−∞

∞

∫

E = 2 t +

cos 400πt( )400π

⎡

⎣⎢⎢

⎤

⎦⎥⎥−∞

∞

→ ∞

(e)

δ t( ) (Hint: First find the signal energy of a signal which approaches an impulse some limit, then take the limit.)

δ t( ) = lim

a→01 / a( )rect t / a( )

E = lim

a→0

1a

rectta

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

2

dt−∞

∞

∫ = lima→0

1a2 rect

ta

⎛⎝⎜

⎞⎠⎟

dt−a / 2

a / 2

∫ = lima→0

aa2 →∞

(f) x t( ) = d

dtrect t( )( )

ddt

rect t( )( ) = δ t +1 / 2( ) − δ t −1 / 2( )

Ex = δ t +1 / 2( ) − δ t −1 / 2( )⎡⎣ ⎤⎦

2dt

−∞

∞

∫ → ∞

(g) x t( ) = rect λ( )dλ

−∞

t

∫ = ramp t +1 / 2( ) − ramp t −1 / 2( )

Ex = t +1 / 2( )2dt

−1/ 2

1/ 2

∫finite

+ dt

1/ 2

∞

∫infinite

→∞

(h)

x t( ) = e −1− j8π( )t u t( )

Ex = x t( ) 2

dt−∞

∞

∫ = e −1− j8π( )t u t( ) 2dt

−∞

∞

∫ = e −1− j8π( )te −1+ j8π( )tdt0

∞

∫

Ex = e−2tdt

0

∞

∫ =e−2t

−2⎡

⎣⎢

⎤

⎦⎥

0

∞

=12

58. Find the average signal power of each of these signals:

(a) x t( ) = 2sin 200πt( ) This is a periodic function. Therefore

Px =

1T

2sin 200πt( )⎡⎣ ⎤⎦2

dt−T / 2

T / 2

∫ =4T

12−

12

cos 400πt( )⎡

⎣⎢

⎤

⎦⎥dt

−T / 2

T / 2

∫

Px =

2T

t −sin 400πt( )

400π

⎡

⎣⎢⎢

⎤

⎦⎥⎥−T / 2

T / 2

=2T

T2−

sin 200πT( )400π

+T2+

sin −200πT( )400π

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 2

For any sinusoid, the average signal power is half the square of the amplitude.




©M. J. Roberts - 3/16/11

Solutions 2-33

(b) x t( ) = δ1 t( ) This is a periodic signal whose period, T, is 1. Between

−T / 2 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)

x t( ) = e j100π t This is a periodic function. Therefore

Px =

1T0

x t( ) 2dt

T0∫ =

1T0

e j100π t 2dt

−T0 / 2

T0 / 2

∫ = 50 e j100π te− j100π tdt−1/100

1/100

∫

Px = 50 dt

−1/100

1/100

∫ = 1

59. A 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( ) − 4rect 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

x t( ) =

0 , 0 < t < 1 / 21 , 1 / 2 < t < 3−3 , 3 < t < 7 / 2−4 , 7 / 2 < t < 50 , 5 < t < 6

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

The signal power is the signal energy in one fundamental period divided by the fundamental period.

P =

16

02 ×12+12 ×

52+ −3( )2

×12+ −4( )2

×32+ 02 ×1

⎛⎝⎜

⎞⎠⎟=

2.5+ 4.5+ 246

= 5.167




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Chapter 2 - Mathematical Description of Continuous-Time ...€¦ · !M. J. Roberts - 3/16/11 Solutions 2-1 Chapter 2 - Mathematical Description of Continuous-Time Signals Solutions