Señales y sistemas michael j roberts 1ed solutions

LIBROS UNIVERISTARIOS
Y SOLUCIONARIOS DE
MUCHOS DE ESTOS LIBROS
LOS SOLUCIONARIOS
CONTIENEN TODOS LOS
EJERCICIOS DEL LIBRO
RESUELTOS Y EXPLICADOS
DE FORMA CLARA
VISITANOS PARA
DESARGALOS GRATIS.


M. J. Roberts - 7/12/03

Chapter 2 - Mathematical Description of Signals
Solutions
1. If g( t) = 7e −2 t − 3 write out and simplify
(a)

g( 3) = 7e −9

(b)

g(2 − t) = 7e −2( 2 − t ) − 3 = 7e −7 + 2 t

(c)

t
− −11

 t

5
g + 4 = 7e
 10


(d)

g( jt) = 7e − j 2 t − 3

(e)

(f)

g( jt) + g(− jt)
e − j 2t + e j 2t
= 7e −3
= 7e −3 cos(2 t)
2
2
 − jt − 3
 jt − 3
g

 + g
 2 
 2 
e − jt + e jt
=7

= 7 cos( t)
2
2

2. If g( x ) = x 2 − 4 x + 4 write out and simplify
(a)

g( z) = z 2 − 4 z + 4

(b)

g( u + v ) = ( u + v ) − 4 ( u + v ) + 4 = u 2 + v 2 + 2 uv − 4 u − 4 v + 4

(c)

g(e jt ) = (e jt ) − 4 e jt + 4 = e j 2 t − 4 e jt + 4 = (e jt − 2)

(d)

g(g( t)) = g( t 2 − 4 t + 4 ) = ( t 2 − 4 t + 4 ) − 4 ( t 2 − 4 t + 4 ) + 4

2

2

2

2

g(g( t)) = t 4 − 8 t 3 + 20 t 2 − 16 t + 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) ;

(b)

x = 1:5 ; g = cos(pi*x) ;

(c)

f = -1:0.5:1 ; w = 2*pi*f ; g = 1./(1+j*w) ;

0.1411
[-1,1,-1,1,-1]

Solutions 2-1


M. J. Roberts - 7/12/03

0.0247 +
0.0920 +



1

0.0920 −
0.0247 −

j 0.155 
j 0.289 



j 0.289 
j 0.155 

4. Let two functions be defined by
1 , sin(20πt) ≥ 0
x1 ( t) = 
−1 , sin(20πt) < 0

and

t , sin(2πt) ≥ 0
x 2 ( t) = 
.
− t , sin(2πt) < 0

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

-2

t

2
-2

5. For each function, g( t) , sketch g(− t) , − g( t) , g( t − 1) , and g(2t) .
(a)

(b)

g(t)

g(t)

4

3

t

2

-1

t

1
-3

g(-t)

g(-t)

4

-g(t)

-g(t)

3

t

-2

3

-1

t

1

2
4

-3

g(t-1)

g(t-1)

g(2t)

4

3

4

1

3

t

t

1

2

t

1
-1

t

-3

g(2t)
3
1

-3

6. A function, G( f ) , is defined by

Solutions 2-2

t

-1
2
1
2
-3

t


M. J. Roberts - 7/12/03

 f
G( f ) = e − j 2πf rect   .
 2
Graph the magnitude and phase of G( f − 10) + G( f + 10) over the range, −20 < f < 20 .
 f + 10 

 f − 10 
− j 2π f +10
G( f − 10) + G( f + 10) = e − j 2π ( f −10) rect 

 + e ( ) rect 
 2 
 2 
|G( f )|
1

-20

f

20

Phase of G( f )
π

-20

f

20
-π

7. Sketch the derivatives of these functions.
(All sketches at end.)

(a)

g( t) = sinc( t)

(b)

g( t) = (1 − e

−t

g′ ( t) =

π 2 t cos(πt) − π sin(πt) πt cos(πt) − sin(πt)
=
πt 2
(πt) 2
e − t , t ≥ 0 − t
g′ ( t) = 
 = e u( t)
0 , t < 0

) u(t)
(a)

(b)

x(t)

x(t)

1

1

-4

4

t

-1

-1

dx/dt

1

1
4

-1

t

-1

dx/dt
-4

4

t

-1

4

t

-1

8. Sketch the integral from negative infinity to time, t, of these functions which are zero for
all time before time, t = 0.

Solutions 2-3


M. J. Roberts - 7/12/03

g(t)

g(t)

1

1
1

2

3

t

1
2

∫ g(t) dt

1

2

3

1

2

3

t

∫ g(t) dt

1

1

1

2

1

2

t

3

t

9. Find the even and odd parts of these functions.
(a)

g( t) = 2 t 2 − 3t + 6
2 t 2 − 3t + 6 + 2(− t) − 3(− t) + 6 4 t 2 + 12
g e ( t) =
=
= 2t 2 + 6
2
2
2

2 t 2 − 3t + 6 − 2(− t) + 3(− t) − 6 −6 t
g o ( t) =
=
= −3t
2
2

2

(b)

π

g( t) = 20 cos 40πt − 

4
π
π


20 cos 40πt −  + 20 cos −40πt − 



4
4
g e ( t) =
2

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

g e ( t) =


 
 π
 π


20 cos( 40πt) cos −  − sin( 40πt) sin −  
4
4 

 


+20 cos(−40πt) cos − π  − sin(−40πt) sin − π  
 
  


 4
 4  

2

 
π 
π   
20 cos( 40πt) cos  + sin( 40πt) sin   
4
4  
 


+20 cos( 40πt) cos π  − sin( 40πt) sin π  
 
  



 4
 4  

g e ( t) =
2

Solutions 2-4


M. J. Roberts - 7/12/03

20
π 
g e ( t) = 20 cos  cos( 40πt) =
cos( 40πt)
 4
2

π
π


20 cos 40πt −  − 20 cos −40πt − 


4
4
g o ( t) =

2
Using cos( z1 + z2 ) = cos( z1 ) cos( z2 ) − sin( z1 ) sin( z2 )

 
 π
 π

20 cos( 40πt) cos −  − sin( 40πt) sin −  
4
4 

 


−20 cos(−40πt) cos − π  − sin(−40πt) sin − π  
 
  


 4
 4  

g o ( t) =
2

g o ( t) =

 
π 
π   

20 cos( 40πt) cos  + sin( 40πt) sin   
4
4  
 


−20 cos( 40πt) cos π  − sin( 40πt) sin π  
 
  


 4
 4  

2

20
π 
g o ( t) = 20 sin  sin( 40πt) =
sin( 40πt)
 4
2
(c)

2 t 2 − 3t + 6
g( t) =
1+ t
2 t 2 − 3t + 6 2 t 2 + 3t + 6
+
1

1− t
+
t
g e ( t) =
2

(2t
g e ( t) =
g e ( t) =

2

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

4 t 2 + 12 + 6 t 2 6 + 5 t 2
=
1 − t2
2(1 − t 2 )

2 t 2 − 3t + 6 2 t 2 + 3t + 6
−
1
1− t
+
t
g o ( t) =
2

Solutions 2-5


M. J. Roberts - 7/12/03

(2t
g o ( t) =

2

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

−6 t − 4 t 3 − 12 t
2t 2 + 9
g o ( t) =
= −t
1 − t2
2(1 − t 2 )
sin(πt) sin(−πt)
+
t
π
−πt = sin(πt)
g e ( t) =
πt
2

(d)

g( t) = sinc( t)

(e)

g( t) = t(2 − t 2 )(1 + 4 t 2 )

g o ( t) = 0

g( t) = {t (2 − t 2 )(1 + 4 t 2 )
4 4
3
odd 12312
even

even

Therefore g( t) is odd, g e ( t) = 0 and g o ( t) = t(2 − t 2 )(1 + 4 t 2 )
(f)

g( t) = t(2 − t)(1 + 4 t)
g e ( t) =

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

g e ( t) = 7 t 2

g o ( t) =

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

g o ( t) = t(2 − 4 t 2 )

10. Sketch the even and odd parts of these functions.

Solutions 2-6


M. J. Roberts - 7/12/03

g(t)

g(t)

1

1

t

1

1

2

t

-1

g e(t)

g e(t)

1

1

t

1

1

2

t

-1

g o(t)

g o(t)

1

1

t

1

1

2

t

-1

(a)

(b)

11. Sketch the indicated product or quotient, g( t) , of these functions.
(a)

(b)

1

1

-1
1

-1

t

1

-1

t

-1

g(t)

g(t)

1
Multiplication
-1

1

t

1

Multiplication

-1
1

t

-1

g(t)

g(t)

1

1

-1
1
-1

t

-1

1
-1

Solutions 2-7

t


M. J. Roberts - 7/12/03

(c)

(d)

1

1

t

-1

t

1

g(t)

g(t)

Multiplication

1
1

Multiplication

1

t

t

1

g(t)

g(t)
-1

1

t

-1

1
-1

t

1

(e)

(f)
1

1

...

...

-1

1

t

1

t

-1

g(t)

-1

g(t)

1

Multiplication

1
Multiplication
-1

t

1

t

1
-1

g(t)

g(t)

1

...

...
-1

1

1

t

t

1
-1

-1

(g)

(h)

1

1

t

-1

-1

-1

1

t

g(t)
Division

1
1

g(t)
π

Division

t

1

g(t)

t

g(t)
1

t
-1

-1

1

t

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

Solutions 2-8


M. J. Roberts - 7/12/03

1

1

−1

−1 even

∫ (2 + t)dt = ∫

(a)

1

1

−1 odd

0

2{ dt + ∫ {t dt = 2 ∫ 2 dt = 4

(b)
1
20

1
20

∫ [4 cos(10πt) + 8 sin(5πt)]dt = ∫

−

1
20

−

1
20

∫

(c)

1
−
20

1
20

4 cos(10πt) dt +
14243
even

1
20

1
20

−

1
20

odd

0

4 t{ cos(10πt) dt = 0
1424
3
odd
even
142
43
odd

1
1


10
10
cos(10πt) 
 cos(10πt)
(d) ∫ t{ sin(10πt) dt = 2 ∫ t sin(10πt) dt = 2 − t

+
dt 
1
424
3
10π 0 ∫0 10π
1 odd
0
odd


− 142
43


10
even
1
10

1
10

1


10
sin(10πt) 
1
 1

=
(10π3t)dt == 2
+
2
∫1 odd{t sin

1
424
100π
(10π ) 0  50π
odd
− 142

43

10
1
10

even

1

(e)

1

1

[

−t
−t
−t
−t
∫ e{ dt = 2∫ e dt = 2∫ e dt = 2 −e

−1 even

0

]

1
0

= 2(1 − e −1 ) ≈ 1.264

0

1

(f)

∫

−t
t{ e{
dt = 0
odd2

even
−1 1
3
odd

13. Find the fundamental period and fundamental frequency of each of these functions.
(a)

g( t) = 10 cos(50πt)

f 0 = 25 Hz , T0 =

1
s
25

(b)

π

g( t) = 10 cos 50πt + 

4

f 0 = 25 Hz , T0 =

1
s
25

(c)

g( t) = cos(50πt) + sin(15πt)
1
 15 
= 0.4 s
f 0 = GCD 25,  = 2.5 Hz , T0 =


2.5
2

(d)

8

(5π3t)dt = 8 ∫ cos(10πt)dt =
∫ 81sin
424
10π

3π 

g( t) = cos(2πt) + sin( 3πt) + cos 5πt − 

4

Solutions 2-9

M. J. Roberts - 7/12/03

1
 3 5 1
f 0 = GCD1, ,  = Hz , T0 = = 2 s
1
 2 2 2
2
14. Find the fundamental period and fundamental frequency of g( t) .

...

g(t)

(a) ...

...
1

t

...

t

1

(b)
...

...

+
t

1

...

...

t

1

(c)
...

...

+

g(t)

t

1

1
s

3

(a)

f 0 = 3 Hz and T0 =

(b)

f 0 = GCD(6, 4 ) = 2 Hz and T0 =

(c)

f 0 = GCD(6, 5) = 1 Hz and T0 = 1 s

1
s
2

15. Plot these DT functions.
(a)

 2π ( n − 2) 
 2πn 
x[ n ] = 4 cos
 − 3 sin

 12 


8

,

−24 ≤ n < 24

x[n]
7
-24

24

n

-7

(b)

x[ n ] = 3ne

−

n
5

,

−20 ≤ n < 20
x[n]
6

-20

20
-6
2

(c)

 n
x[ n ] = 21  + 14 n 3
 2

, −5 ≤ n < 5

Solutions 2-10

n

g(t)


M. J. Roberts - 7/12/03

x[n]
2000
-5

n

5

-2000

n

2

− 
 2πn 
 
16. Let x1[ n ] = 5 cos
 and x 2 [ n ] = −8e 6 . Plot the following combinations of those
 8 
two signals over the DT range, −20 ≤ n < 20 . If a signal has some defined and some
undefined values, just plot the defined values.

x[n]
40

-20

(a)

x[ n ] = x1[ n ] x 2 [ n ]

20

n

-40

x[n]
20
-20

(b)

x[ n ] = 4 x1[ n ] + 2 x 2 [ n ]

20

n

-40

x[n]
20
-20

(c)

x[ n ] = x1[2 n ] x 2 [ 3n ]

20

n

-40
x[n]
10000
-20

(d)

x[ n ] =

x1[2 n ]
x 2 [− n ]

20

n

-50000

x[n]
5
-20

(e)

n 
n 
x[ n ] = 2 x1   + 4 x 2  
2 
3

20

-40

Solutions 2-11

n


M. J. Roberts - 7/12/03

17. A function, g[ n ] is defined by

−2 , n < −4

g[ n ] = n , − 4 ≤ n < 1 .
4

, 1≤ n
n
n 
Sketch g[− n ], g[2 − n ], g[2n ] and g   .
2 
g[n]
4
-10

n

10
-4

g[- n]

g[2- n]

4

4
10

-10

10

n

-10

-4

-4

g[2n]

g[n/2]

4

n

4

-10

10

-10

n

10

-4

n

-4

18. Sketch the backward differences of these DT functions.

(a)

(b)

g[n]

g[n]

1

1

-4

20

n

-4

-1

20

n

-1

∆g[n-1]

∆g[n-1]

1

1

-4

20

n

-4

-1

20

n

-1

(c)
2

g[n] = (n/10)
4

-4

∆g[n-1]

20

n

0.5
-4
-0.25

20

n

19. Sketch the accumulation, g[ n ], from negative infinity to n of each of these DT functions.

Solutions 2-12


M. J. Roberts - 7/12/03

(a)

h[ n ] = δ [ n ]

(b)

h[ n ] = u[ n ]

(c)

 2πn 
h[ n ] = cos
 u[ n ]
 16 

(d)

 2πn 
h[ n ] = cos
 u[ n ]
 8 

(e)

 2πn 
h[ n ] = cos
 u[ n + 8]
 16 

(a)

-16

h[n]

h[n]

1

1

g[n]

16

n

(b)

-16

g[n]

1
-16

16

n

-16

16

(c)

16

1

n

(d)

g[n]

-16

-3

16

16

-1
3

n
-16

16

-3

h[n]
1

(e)

-16

-1

16

n

g[n]
3
-16

n

g[n]

3
-16

n

h[n]

1
-1

n

16

h[n]
-16

16

-3

16

n

20. Find and sketch the even and odd parts of these functions.
−

n
4

(a)

g[ n ] = u[ n ] − u[ n − 4 ]

(b)

g[ n ] = e

(c)

 2πn 
g[ n ] = cos

 4 

(d)

 2πn 
g[ n ] = sin
 u[ n ]
 4 

Solutions 2-13

u[ n ]

n

M. J. Roberts - 7/12/03

g[n]

g[n]

1
-10

-1

1
10

n

-10

-1

g [n]

-1

n

-10

-1

-10

10

n

-10

-1

10

-10

-1

10

10

n

10

n

1

n

-10

-1

go[n]

go[n]

1
-1

n

ge[n]

-1

-10

10

1

n

ge[n]
1

n

g[n]

-1

-10

10

o
1

g[n]
1

n

g [n]

o
1

-1

10

1
10

g [n]
-10

n

g [n]
e

e
1

-10

10

10

n

1

-10

-1

21. Sketch g[ n ].
(a)

(b)

g1[n]

g1[n]

1
-10

1
10

-4

n

20

-1

g2[n]

g2[n]

g[n]

1

g[n]

1

-10
10

n

Multiplication
-4

-1

20

(d)

1

1
20

n

-10

-1

10

n

-1

g2[n]

g[n]

g[n]

1
20
-1

Multiplication

g1[n]

g1[n]

-4

n

-1

(c)

-4

n

-1

n

g[n]

1

Multiplication
-10

10
-1

Solutions 2-14

n

Multiplication


M. J. Roberts - 7/12/03

(a)

(b)

g[n]

g[n]

1

1

-10
10

n

-4

-1

20

n

-1

(c)

(d)

g[n]

g[n]

1

1

-4

20

n

-10

-1

10

n

-1

22. Find the fundamental DT period and fundamental DT frequency of these functions.
(a)

 2πn 
g[ n ] = cos

 10 

N 0 = 10 , F0 =

1
10

(b)

 πn 
g[ n ] = cos 
 10 

N 0 = 20 , F0 =

1
20

(c)

 2πn 
 2πn 
g[ n ] = cos
 + cos

 5 
 7 

N 0 = 35 , F0 =

1
35

(d)

g[ n ] = e

j

N 0 = 20 , F0 =

1
20

g[ n ] = e

−j

N 0 = 12 , F0 =

1
12

(e)

2πn
20

2πn
3

+e

−j

+e

2πn
20

−j

2πn
4

23. Graph the following functions and determine from the graphs the fundamental period of
each one (if it is periodic).
(a)

 2πn 
 2πn 
g[ n ] = 5 sin
 + 8 cos

 4 
 6 

(b)

 14πn 
 7πn 
g[ n ] = 5 sin
 + 8 cos

 12 
 8 

(c)

πn
−j 

g[ n ] = Re e jπn + e 3 



(d)

n
−j 

g[ n ] = Re e jn + e 3 



Solutions 2-15


M. J. Roberts - 7/12/03

(a)

(b)

g[n]

g[n]

12

12

-24

n

24

-24

n

24

-12

-12

N0 = 12

N0 = 24

(c)

(d)

g[n]

g[n]

2

2

24

-24

n

-24

n

24
-2

-2

N0 = 6

Not Periodic

24. Find the signal energy of these signals.

(a)

∞

x( t) = 2 rect ( t)

∫ 2 rect (t)

Ex =

−∞

(b)

dt = 4 ∫ dt = 4
−

∞

∫

−∞

10

A(u( t) − u( t − 10)) dt = A 2 ∫ dt = 10 A 2
2

0

∞

x( t) = u( t) − u(10 − t)

Ex =

∫ u(t) − u(10 − t)

−∞

(d)

2

dt =

0

∞

−∞

10

∫ dt + ∫ dt → ∞

x( t) = rect ( t) cos(2πt)
∞

Ex =

∫ rect (t) cos(2πt)

−∞

2

1
2

dt =

1
2

∫ cos (2πt)dt = 2 ∫ (1 + cos(4πt))dt
2

−

1

1
2



1

 12
2
 1

1
E x =  ∫ dt + ∫ cos( 4πt) dt  =
2
2 1
1

−
−
2
2
14
4244
3


=0

(e)

1
2

x( t) = A(u( t) − u( t − 10))
Ex =

(c)

2

1

2

x( t) = rect ( t) cos( 4πt)

Solutions 2-16

−

1
2


M. J. Roberts - 7/12/03

∞

Ex =

∫ rect (t) cos(4πt)

2

1
2

1
2

∫ cos (4πt)dt = 2 ∫ (1 + cos(8πt))dt

dt =

1

2

−∞

−

1
2

−

1
2



1

 12
2
 1
1
E x =  ∫ dt + ∫ cos(8πt) dt  =
2
2 1
1


−
−
2
2
14
4244
3


=0

(f)

x( t) = rect ( t) sin(2πt)
∞

Ex =

∫ rect (t) sin(2πt)

2

1
2

∫ sin (2πt)dt = 2 ∫ (1 − cos(4πt))dt

dt =

−∞

1
2

1

2

−

1
2

−

1
2



1

 12
2
 1
1
E x =  ∫ dt − ∫ cos( 4πt) dt  =
2
2 1

1

−
−
2
2
14
4244
3


=0

(g)

x[ n ] = A rect N 0 [ n ]
∞

E x = ∑ A rect N 0 [ n ] = A
−∞

(h)

x[ n ] = Aδ [ n ]

2

N0

2

∑ (1) = (2N

−N0

0

+ 1) A 2

x[ n ] = Aδ [ n ]
∞

0

E x = ∑ Aδ [ n ] = A 2 ∑ (1) = A 2
2

−∞

(i)

(j)
(k)

0

∞

∞

x[ n ] = comb N 0 [ n ]

E x = ∑ comb N 0 [ n ] =

x[ n ] = ramp[ n ]

E x = ∑ ramp[ n ] = ∑ n 2 → ∞

2

−∞

∞

−∞

2

∞
0

x[ n ] = ramp[ n ] − 2 ramp[ n − 4 ] + ramp[ n − 8]

Solutions 2-17

∑ (1) → ∞

−∞
n = mN 0

M. J. Roberts - 7/12/03
∞

E x = ∑ ramp[ n ] − 2 ramp[ n − 4 ] + ramp[ n − 8] = (0 2 + 12 + 2 2 + 32 + 4 2 + 32 + 2 2 + 12 + 0 2 )
2

−∞

E x = 1 + 4 + 9 + 16 + 9 + 4 + 1 = 44
25. Find the signal power of these signals.
x( t) = A

(a)

1
T →∞ T

x( t) = u( t)

(b)

Px = lim

∫
−

T
2

T
2 2

A
A dt = lim
T →∞ T
2

T
2

∫ u(t)
−

2

A2
T = A2
∫T dt = Tlim
→∞ T

−

2
T
2

1
1T 1
dt = lim

=
∫
T →∞ T
T →∞ T 2
2
0

dt = lim

T
2

x( t) = A cos(2πf 0 t + θ )

(c)

Px =

1
Px = lim
T →∞ T

T
2

1
T0

T0
2

A cos(2πf 0 t + θ ) dt =

∫

−

2

T0
2

2

A
T0

T0
2

∫ cos (2πf t + θ )dt
2

0

−

T0
2

T0
2

T0

A  sin( 4πf 0 t + 2θ )  2
A
1 + cos( 4πf 0 t + 2θ )) dt =
Px =
(
t +

∫
2T0 T0
2T0 
4πf 0
 − T0
−
2

2

2

2

T
T






sin 4πf 0 0 + 2θ sin −4πf 0 0 + 2θ 





A2
A
2
2


=
T
Px =
+
−
2T0  0
4πf 0
4πf 0
2
14444444
24444444
3


=0

2

∞

(d)

x( t) = A ∑ rect ( t − 2 n )
n =−∞

1
Px =
T0

(e)

T0
2

∫

−

T0
2

∞

2

2 1

A
A ∑ rect ( t − 2 n ) dt =
2
n =−∞

1
2 2

A
∫ rect (t) dt = 2
−1

∞
 1

x( t) = 2 A − + ∑ rect ( t − 2 n ) 
 2 n =−∞


Solutions 2-18

2

A2
∫1 dt = 2

−

2

M. J. Roberts - 7/12/03

1
Px =
T0

T0
2

∫

T0
2

−

2

∞
 1

4 A2
1
− + rect ( t) dt
2 A − + ∑ rect ( t − 2 n )  dt =
∫
2 −1 2
 2 n =−∞



2

1

2

1

2

1

1
1
Px = 2 A ∫ − + rect ( t) dt = 4 A 2 ∫ − + rect ( t) dt
2
2
−1
0
2


 12
2
2
1
 1 
2   1

Px = 4 A ∫   dt + ∫  −  dt  = A 2
 2
 2
1

0


2
(f)

(g)

(h)

x[ n ] = A

1
Px = lim
N →∞ 2 N

x[ n ] = u[ n ]

Px = lim

x[ n ] = A

N −1

∑

n =− N
N −1

1
N →∞ 2 N

∑

n =− N

A2
A = lim
N →∞ 2 N
2

N −1

A2
(1) = Nlim
(2N ) = A 2
∑
→∞ 2 N
n =− N

1
N →∞ 2 N

u[ n ] = lim
2

N −1

N →∞

n =0

∞

∑ rect [n − 8m]

m =−∞

2

N −1

2

∞
A2 7 ∞
1 0
8
Px =
A
rect
n
−
m
=

[
]
∑ ∑ 2
∑ ∑ rect 2[n − 8m]
2 N 0 n =− N 0 m =−∞
2 × 8 n =−8 m =−∞

Px =

(i)

(j)

1

N

∑ (1) = lim 2N = 2

2

2
7
 10 A 2 5 A 2
A 2  −6
1
1
1
(
)

+
(
)
+
(
)
∑
∑ ∑= 6  = 2 × 8 = 8
2 × 8 n =−8
n =−2
n

2
1
1
comb N 0 [ n ] =
∑
N 0 n = N0
N0

x[ n ] = comb N 0 [ n ]

Px =

x[ n ] = ramp[ n ]

1
Px = lim
N →∞ 2 N

N −1

∑

n =− N

ramp[ n ]

2

1
= lim
N →∞ 2 N

N −1

∑n

2

→∞

n =0

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, ∆t , of the plot. Explain why the plots
look the way they do.
(a)

∆t =

1
24

(b)

∆t =

1
12

(c)

∆t =

1
4

(d)

∆t =

1
2

(e)

∆t =

2

3

(f)

∆t =

5
6

Solutions 2-19


M. J. Roberts - 7/12/03

(g)

∆t = 1
∆t = 1/24

∆t = 1/12

x(t)

x(t)

1

1

10

t

-1

10

t

-1

∆t = 1/4

∆t = 1/2

x(t)

x(t)

1

1

10

t

-1

10

t

-1

∆t = 2/3

∆t = 5/6

x(t)

x(t)

1

1

10

t

-1

10

t

-1

∆t = 1

x(t)
1

10

t

-1

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

(b)

π

g( t) = 100 sin 200πt + 

4
π
π π 

g(0.001) = 100 sin 200π × 0.001 +  = 100 sin +  = 98.77

 5 4
4
g( t) = 13 − 4 t + 6 t 2

g(2) = 13 − 4 (2) + 6(2) = 29
2

(c)
g( t) = −5e −2 t e − j 2πt

π

−2
− j 2π
−
−j
 1
4
= −5e 2 e 2 = − j 3.03
g  = −5e 4 e
 4
1

1

28. Sketch these CT exponential and trigonometric functions.

Solutions 2-20

1


M. J. Roberts - 7/12/03

(a)

g( t) = 10 cos(100πt)

(c)

g( t) = 5e

−

t
10

(b)

g( t) = 40 cos(60πt) + 20 sin(60πt)

(d)

g( t) = 5e 2 cos(2πt)

−

(a)

t

(b)

g(t)

g(t)

10

60

0.04

t

t

0.066667

-10

-60

(c)

g(t)

(d)

g(t)

5

5

8

t

50

t

-5

29. Sketch these CT singularity and related functions.
(a)

g( t) = 2 u( 4 − t)

(b)

g( t) = u(2 t)

(c)

g( t) = 5 sgn( t − 4 )

(d)

g( t) = 1 + sgn( 4 − t)

(e)

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

(f)

g( t) = −3 ramp(2 t)

(g)

g( t) = 2δ ( t + 3)

(h)

g( t) = 6δ ( 3t + 9)

(a)

(c)

(b)

g(t)

(d)

g(t)

g(t)

g(t)

5
2

1
t

4

t

(e)

g(t)

10

-1

(i)

1

1

t

(h)
g(t)

g(t)

2

2

t

-6

g( t) = −4δ (2( t − 1))

-3

(j)

t

4

-5

(g)

(f)
g(t)

2

t

4

t

-3

 1
g( t) = 2 comb t − 
 2

Solutions 2-21

t


M. J. Roberts - 7/12/03

(k)

g( t) = 8 comb( 4 t)

(l)

 t + 1
g( t) = −3 comb

 2 

(m)

 t
g( t) = 2 rect  

 3

(n)

 t + 1
g( t) = 4 rect 

 2 

(o)

g( t) = tri( 4 t)

(p)

 t − 1
g( t) = −6 tri

 2 

(i)

(k)

(j)

g(t)
1

2

2

t

...

-2

...
1

(m)

...

...
t

t

t

-2

(p)

g(t)

g(t)

-1

-1
4

t

1
4

-6

(q)

 t
g( t) = 5 sinc 
 2

(r)

g( t) = − sinc(2( t + 1))

(s)

g( t) = −10 drcl( t, 4 )

(t)

t 
g( t) = 5 drcl , 7

4 
(r)

(q)
g(t)

g(t)
-1

5

t

t

-1

2

(s)

(t)

g(t)

g(t)

10

5

4
-10

3

...

-6

(o)
1

4

3
2

t

1
4

g(t)

2

g(t)
-1 1

...

(n)

g(t)

-3
2

(l)

g(t)

g(t)

t
-1

Solutions 2-22

8

t

1

3

t

t


M. J. Roberts - 7/12/03

( )

0.1rect t-3
4

-3rect(t-2)
3
2

5
2

0.1

t

(u)

(v)

-3

1

( )

-4tri 3+t
2

3 5

t

4sinc[5(t-3)]
4

(w)

-5 3 -1

t

(x)

-4

-1

1 2 3 4 5 6

4sinc(5t-3)
4
(y)

-1

1 2 3 4 5 6

t

30. Sketch these combinations of CT functions.
(a)

g( t) = u( t) − u( t − 1)

(b)

 1
g( t) = rect  t − 
 2

(c)

g( t) = −4 ramp( t) u( t − 2)

(d)

g( t) = sgn( t) sin(2πt)

(e)

g( t) = 5e

(f)

g( t) = rect ( t) cos(2πt)

(g)

g( t) = −6 rect ( t) cos( 3πt)

(h)

g( t) = rect ( t) tri( t)

−

t
4

u( t)

Solutions 2-23

t


M. J. Roberts - 7/12/03

g(t)

(a)

1

(c)

(b)

g(t)

(d)

g(t)

g(t)

1

2

t

1

(e)

t
-1

(g)

g(t)
1

-1
2

5
t

4

1

-16

(f)

g(t)

t

-8

t

1

4

1
2

t

-1

(h)

g(t)
6

-1
2

-6

1

g(t)
1
1
2

1
2

t
-1
2

t

1

2

(i)

 1
g( t) = rect ( t) tri t + 
 2

(j)

 1
1 
g( t) = u t +  ramp − t
 2
2 

(k)

g( t) = tri2 ( t)

(l)

g( t) = sinc 2 ( t)

(m)

g( t) = sinc( t)

(n)

g( t) =

(o)

 1
 1
g( t) = rect  t +  − rect  t − 
 2
 2

(p)

t

g( t) =  ∫ δ (λ + 1) − 2δ (λ ) + δ (λ − 1) dλ
−∞


(i)

(k)

(j)

g(t)
1

1
2

t

t

1
2

t

-1

(o)

(p)

g(t)

g(t)

g(t)

1

1

1

1

1

-1
-1
1

1

-1

(n)

g(t)

g(t)
1

1

-1
2

(m)

(l)

g(t)

g(t)
1

-1

2

d
(tri(t))
dt

t

1
-1

t

-1

t

Solutions 2-24

1
-1
-1

t

1

t