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 />
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

()
g ( 2 − t ) = 7e ( ) = 7e
g ( t / 10 + 4 ) = 7e
g ( jt ) = 7e
g ( jt ) + g ( − jt )
e
= 7e

g 3 = 7e−9

(a)

−2 2−t −3

(b)

−t /5−11

(c)

− j 2t −3

(d)

− j 2t

−3

(e)

⎛
g⎜
⎝

(f)
2.

−7 + 2t

+ e j 2t
= 7e−3 cos 2t
2

( )

2

⎛ − jt − 3 ⎞
jt − 3 ⎞
+ g⎜
⎟
2 ⎠
⎝ 2 ⎟⎠
e− jt + e jt
=7
= 7 cos t
2
2

()

()

If g x = x 2 − 4x + 4 write out and simplify
(a)
(b)
(c)
(d)

()
g (u + v ) = (u + v )

g z = z 2 − 4z + 4
2

(

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

2

)

− 4 u + v + 4 = u 2 + v 2 + 2uv − 4u − 4v + 4
jt

2

(
− 4t + 4 ) − 4 ( t

j 2t

2

− 4e jt + 4 = e jt − 2
2

2

)

2

)

− 4t + 4 + 4

( ( ))

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

()

g 2 = 4−8+ 4 = 0

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’) ;

Solutions 2-1

Full file at />

Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed
©M. J. Roberts - 3/16/11

Full file at />⎡ 0.0247 +
⎢
⎢0.0920 +
⎢
1
⎢
0.0920
−
⎢
⎢⎣ 0.0247 −

4.

j0.155⎤
⎥
j0.289 ⎥
⎥
⎥
j0.289 ⎥
j0.155⎥⎦

Let two functions be defined by

(
(

)
)

⎧⎪1 , sin 20π t ≥ 0
x1 t = ⎨
and
⎩⎪−1 , sin 20π t < 0

()

( )
( )

⎧⎪t , sin 2π t ≥ 0
x2 t = ⎨
.
⎩⎪−t , sin 2π t < 0

()

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

-2

t

2
-2

Transformations of Functions

()

( )

() (

)

( )

For each function g t graph g −t , − g t , g t − 1 , and g 2t .

5.

(a)

(b)

g(t)

g(t)

4

3
-1

t

2

t

1
-3

g(-t)

g(-t)

4

-g(t)

-g(t)

3

t

-2

3

-1

t

1

g(t-1)

g(t-1)

g(2t)

4

3

4

t

3

-1

4

-3

1

1

t

2

1

t

2

g(2t)
3
-1
2

t

1

1
2
-3

-3

6.

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

()

( )

( )

()

( ) ( )

()

(

(

)

(a)

x t = 2 rect t / 4 , x −1 = 2 rect −1 / 4 = 2

(b)

x t = 5rect t / 2 sgn 2t , x 0.5 = 5rect 1 / 4 sgn 2 = 5

(c)

x t = 9 rect t / 10 sgn 3 t − 2

) ((

( )
))

(

()

) ()
(

) ( )

, x 1 = 9 rect 1 / 10 sgn −3 = −9

Solutions 2-2

Full file at />
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 />7.

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

((

()

) )

t0 and w in the functional transformation g 2 t = Ag1 t − t0 / w .
(a)

(a)

2
1
0
-1
-2

4

2
1
0
-1
-2

-4

4

2
1
0
-1
-2
-4

g2(t)

g1(t)

4

2
1
0
-1
-2
-4

-4

-2

0

t

2

-2

0

t

(b)

2

4

2

4

2

4

(b)

g2(t)

g1(t)

2
1
0
-1
-2
-4

-2

0

t

2

-2

0

t

(c)

(c)

g2(t)

g1(t)

2

1
0
-1
-2
-4

-2

0

t

2

-2

0

t

Figure E-7
Answers:

A = 2,t0 = 1, w = 1 , (b)

(a)

(c)
8.

A = −2,t0 = 0, w = 1 / 2 ,

A = −1 / 2,t0 = −1, w = 2

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

( (

()

))

t0 and a in the functional transformation g 2 t = Ag1 w t − t0 .

8

4

4

g2(t)

(a)

g1(t)

A = 2, t0 = 2, w = -2
8

0

-4

0
-4

-8
-10

-5

0

5

-8
-10

10

-5

0

5

10

t
t
Amplitude comparison yields A = 2 . Time scale comparison yields w = −2 .

( (

()

g 2 2 = 2 g1 −2 2 − t0

)) = 2 g (0) ⇒ −4 + 2t
1

0

= 0 ⇒ t0 = 2

8

4

4

g2(t)

(b)

g1(t)

A = 3, t0 = 2, w = 2
8

0

-4

0
-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 .

()

((

g 2 2 = 3g1 2 2 − t0

)) = 3g (0) ⇒ 4 − 2t
1

0

= 0 ⇒ t0 = 2

Solutions 2-3

Full file at />

Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed
©M. J. Roberts - 3/16/11

8

4

4

0

0

2

(c)

A = -3, t0= -6, w = 1/3

8

g (t)

g1(t)

Full file at />
-4

-4

-8
-10

-5

0

5

-8
-10

10

-5

0

5

10

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

(( ) (

()

g 2 0 = −3g1 1 / 3 0 − t0

)) = −3g ( 2) ⇒ −t
1

0

/ 3 = 2 ⇒ t0 = −6

OR
Amplitude comparison yields A = −3 . Time scale comparison yields w = −1 / 3 .
g 2 3 = −3g1 −1 / 3 3 − t0 = −3g1 0 ⇒ t0 / 3 − 1 = 0 ⇒ t0 = 3

((

()

))

)(

()

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

8

8
4

g2(t)

(d)

g1(t)

4
0
-4

0
-4

-8
-10

-5

0

5

-8
-10

10

-5

0

5

10

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

(( ) (

()

)) = −3g ( 2) ⇒ −t
1

0

/ 3 + 4 / 3 = 2 ⇒ t0 = −2

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

8

8

4

4

g2(t)

(e)

g1(t)

g 2 4 = −2 g1 1 / 3 4 − t0

0
-4

0

-4

-8
-10

-5

0

5

-8
-10

10

-5

0

5

10

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

()

(

g 2 t = 3g1 2 − t

)

,

()

( )

g 3 t = −2 g1 t / 4

⎛ t − 3⎞
g 4 t = g1 ⎜
⎝ 2 ⎟⎠

()

,

Find these values.

()

(a)

g 2 1 = −3

(c)

3
3
⎡ g 4 t g 3 t ⎤ = × −1 = −
⎣
⎦t = 2 2
2

() ()

(b)

( )

( )

g 3 −1 = −3.5
−1

(d)

∫ g (t ) dt
4

−3

()

The function g 4 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.
−1

∫ g (t ) dt = −2
4

−3

Solutions 2-4

Full file at />

Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed

©M. J. Roberts - 3/16/11

Full file at />g1(t)
4
3
2
1
-4

-3

-2

-1

-1
-2
-3
-4

1

2

3

4

t

Figure E-9
10.

( )

A function G f is defined by

( )

(

G f = e− j 2π f rect f / 2

)

(

.

)

(

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 = 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− j 2π f = cos −2π f + j sin −2π f = cos 2π f − j sin 2π f

(

)

(

)

)

e− j 2π f = cos 2 2π f + sin 2 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 = 1 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 2 π f −10

)

⎛ f − 10 ⎞
⎛ f + 10 ⎞
− j 2 π f +10
rect ⎜
+ e ( ) rect ⎜
⎝ 2 ⎟⎠
⎝ 2 ⎟⎠

Solutions 2-5

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Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed
©M. J. Roberts - 3/16/11

Full file at />|G( f )|
1

-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 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

() (

) ()

Graph the derivative of x t = 1 − e−t u t .

12.

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.

⎧e−t , t > 0
x′ t = ⎨
⎩⎪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 .

()

()

x(t)
1
-1

4

t

-1

dx/dt
1

-1

4

t

-1

13.

Find the numerical value of each integral.
(a)
8

(

)

8

( )

(

8

)

8

( )

()

∫ ⎡⎣δ t + 3 − 2δ 4t ⎤⎦ dt = ∫ δ t + 3 dt − 2 ∫ δ 4t dt = 0 − 2 × 1 / 4 ∫ δ t dt = −1 / 2

−1

−1

−1

−1

(b)
5/ 2

5/ 2

∞

1

5/ 2

∞

1

∫ δ (3t ) dt = ∫ ∑ δ (3t − 2n)dt = 3 ∫ ∑ δ (t − 2n / 3)dt = 3 ⎡⎣1 + 1 + 1⎤⎦ = 1

2

1/ 2

1/ 2 n= −∞

1/ 2 n= −∞

Solutions 2-6

Full file at />

Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed
©M. J. Roberts - 3/16/11

Full file at />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 .

∫ 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 = 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)

g(t)

1

1
1

2

3

t

1
2

1

2

3

1

2

3

t

Figure E-14
0 g(t) dt

0 g(t) dt

1

1

1
2

1

2

t

3

t

Even and Odd Functions
15.

()

An even function g t is described over the time range 0 < t < 10 by

⎧2t
, 0⎪
g t = ⎨15 − 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,
⎡d
⎤
⎡d
⎤
d
d
g t = − g −t ⇒ ⎢ g t ⎥
= − ⎢ g t ⎥ = − −3 = 3 .
dt
dt

⎣ dt
⎦t = −6
⎣ dt
⎦t =6

()

16.

( )

()

()

( )

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

()

g t = 2t 2 − 3t + 6
Solutions 2-7

Full file at />

Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed
©M. J. Roberts - 3/16/11

Full file at />
()

()

2t 2 − 3t + 6 − 2 −t

ge t =

go t =
(b)

( )

2t 2 − 3t + 6 + 2 −t

( )

2

( )

+ 3 −t − 6

2

()

(

()

( )

− 3 −t + 6

2

g t = 20cos 40π t − π / 4

ge t =

2

(

=

4t 2 + 12
= 2t 2 + 6
2

=

−6t
= −3t
2

)
)

(

20cos 40π t − π / 4 + 20cos −40π t − π / 4

)

2

(

)

( ) ( )

( ) ( )

Using cos z1 + z2 = cos z1 cos z2 − sin z1 sin z2 ,

(

) (
) ( ) (
) ⎫⎪
⎬
(
) (
) (
) ( )⎤⎦ ⎭⎪

(

) ( ) ( ) ( )
( ) ( ) ( ) ( )

⎧20 ⎡cos 40π t cos −π / 4 − sin 40π t sin −π / 4 ⎤
⎪ ⎣
⎦
⎨
⎪+20 ⎡⎣cos −40π t cos −π / 4 − sin −40π t sin − / 4
ge t = ⎩
2

()

⎧20 ⎡cos 40π t cos π / 4 + sin 40π t sin π / 4 ⎤ ⎫
⎪ ⎣
⎦ ⎪
⎨
⎬
⎡
⎪⎩+20 ⎣cos 40π t cos π / 4 − sin 40π t sin π / 4 ⎤⎦ ⎪⎭
ge t =
2

()

()

(

) (

(

)

)

(

g e t = 20cos π / 4 cos 40π t = 20 / 2 cos 40π t

()

go t =

(

)

(

20cos 40π t − π / 4 − 20cos −40π t − π / 4

)

)

2

(

)

( ) ( )

( ) ( )

Using cos z1 + z2 = cos z1 cos z2 − sin z1 sin z2 ,

(

) (
) ( ) (
) ⎫⎪
⎬
(
) (
) (
) (
)⎤⎦ ⎪⎭

(

) ( ) ( ) ( )
( ) ( ) ( ) ( )

⎧20 ⎡cos 40π t cos −π / 4 − sin 40π t sin −π / 4 ⎤
⎪ ⎣

⎦
⎨
⎡
⎪−20 ⎣cos −40π t cos −π / 4 − sin −40π t sin −π / 4
go t = ⎩
2

()

⎧20 ⎡cos 40π t cos π / 4 + sin 40π t sin π / 4 ⎤ ⎫
⎪ ⎣
⎦ ⎪
⎨
⎬
⎪−20 ⎡⎣cos 40π t cos π / 4 − sin 40π t sin π / 4 ⎤⎦ ⎭⎪
⎩
go t =
2

()

()

(

) (

)

(

)

(

g o t = 20sin π / 4 sin 40π t = 20 / 2 sin 40π t

(c)

()

g t =

)

2t 2 − 3t + 6
1+ t

2t 2 − 3t + 6 2t 2 + 3t + 6
+
1+ t
1− t
ge t =
2

()

Solutions 2-8

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Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed
©M. J. Roberts - 3/16/11

Full file at />
( 2t

2

)( ) (

(1 + t )(1 − t )

()

ge t =

()

ge t =

)( )

− 3t + 6 1 − t + 2t 2 + 3t + 6 1 + t
2

4t 2 + 12 + 6t 2

(

2 1− t2

)

=

6 + 5t 2
1− t2

2t 2 − 3t + 6 2t 2 + 3t + 6
−
1+ t
1− t
go t =
2

()

( 2t

2

)( ) (
(

()

()

(d)

)(

1+ t 1− t

go t =

go t =

)( )

− 3t + 6 1 − t − 2t 2 + 3t + 6 1 + t

)

2
−6t − 4t 3 − 12t

(

2 1− t2

() (

)

)(

g t = t 2 − t 2 1 + 4t 2

(

()

= −t

2t 2 + 9
1− t2

)

)(

)

g t = t 2 − t 2 1 + 4t 2
 

odd  
even

even

()

() (

()

)(

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

() (

(

)

)(

) ( )(

)(

)

)(

)

()

t 2 − t 1 + 4t + −t 2 + t 1 − 4t

()

t 2 − t 1 + 4t − −t 2 + t 1 − 4t

ge t =

go t =
17.

)(

g t = t 2 − t 1 + 4t

()

g e t = 7t 2

2

(

)

)(

) ( )(

() (

g o t = t 2 − 4t 2

2

)

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

(graphically) and divide the difference by two.
g(t)

g(t)

1

1

1

t

1

2

t

-1

Figure E-17

Solutions 2-9

Full file at />
()

from g t


Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed
©M. J. Roberts - 3/16/11

Full file at />g e(t)

g e(t)

1

1

t

1

1

t

2

-1

g o(t)

g o(t)

1

1

t

1

1

t

2

-1

,

(a)

(b)

()

Graph the indicated product or quotient g t of the functions in Figure E-18.

18.

(a)

(b)
1
-1

1
-1
1

t

1

t

-1

g(t)

-1

g(t)

1

-1

Multiplication

t

1

1
-1

t

1

Multiplication

-1

g(t)

g(t)

1

1

-1

t

1

-1

-1

t

1
-1

(c)

(d)

1

1

t

-1

t

1

g(t)

g(t)
1

Multiplication

t

1

1

t

1

g(t)

g(t)
-1

Multiplication

1

t

-1

1
-1

(e)

t

1

(f)
1

1

...
-1

... t

1

1

t

-1

g(t)

-1

g(t)

1

-1

t

1

Multiplication

1

t

Multiplication

-1

(e)

(f)

g(t)

g(t)

1

...

...
-1

1

1

1

t

-1

1

t

-1

Solutions 2-10

Full file at />

Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed
©M. J. Roberts - 3/16/11

Full file at />(g)

(h)
1

1

t

-1

-1

g(t)

g(t)
1

/

Division

t

1

t

1

-1

Division

t

1

g(t)

g(t)

1

t
-1

-1

t

1

Figure E-18
19.

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

in the quickest way.
1

1

−1

−1 even

∫ ( 2 + t ) dt = ∫

(a)

1/ 20

∫

(b)

−1/ 20

1/ 20

∫

−1/ 20

∫

−1/ 20

1

−1 odd

0

(

)

( )

⎡ 4cos 10π t + 8sin 5π t ⎤ dt =
⎣
⎦

1/ 20

(

∫

1/ 20

)

( )

4cos 10π t dt + ∫ 8sin 5π t dt




−1/ 20
−1/ 20
even

(

)

(

odd

1/ 20

( )

⎡ 4cos 10π t + 8sin 5π t ⎤ dt = 8
⎣
⎦

1/ 20

(c)

1

2 dt + ∫ t dt = 2 ∫ 2dt = 4

8

∫ cos (10π t ) dt = 10π
0

)

4 t cos 10π t dt = 0


odd 
even




odd

1/10

(d)

∫

(

1/10

)

t sin 10π t dt = 2


odd

−1/10 odd

∫
0

(

⎡ cos 10π t
t sin 10π t dt = 2 ⎢ −t
⎢
10π
⎢⎣

(

)

)

1/10

1/10

+

0

∫

(

cos 10π t

0

10π

) dt ⎤⎥
⎥
⎥⎦

even

⎡
sin 10π t
⎢ 1
10π t dt == 2 ⎢
+
∫ oddt sin
2


100π
−1/10
10π

⎢
odd
⎣
1/10

(

)

(

(

)

)

even

1

(e)

1

1

(

0

⎤
1
⎥
⎥ = 50π
⎥
⎦

)

−t
−t
−1
∫ e dt = 2 ∫ e dt = 2 ∫ e dt = 2 ⎡⎣ −e ⎤⎦0 = 2 1 − e ≈ 1.264
−t

−1 even
1

(f)

1

1/10

∫

−t

0

0

−t

t e dt = 0
even
−1 odd

odd

Periodic Signals
20.

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

()

(

()

(

)

(a)

g t = 10cos 50π t

(b)

g t = 10cos 50π t + π / 4

f0 = 25 Hz , T0 = 1 / 25 s

)

f0 = 25 Hz , T0 = 1 / 25 s

Solutions 2-11

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Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed
©M. J. Roberts - 3/16/11

Full file at />
()

(

)

(

g t = cos 50π t + sin 15π t

(c)

)

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

( )

(

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

(d)

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

21.

)

1
=2s
1/ 2

()

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 ?
x(t)
4
3
2
1
-1
-2
-3
-4

5ms 10ms 15ms 20ms

t

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)

(a) ...

...
1

(b)

t

...

...

...
1

...

...

...

...
1

+
t

t

1

(c)

t

1

+

g(t)

t

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.
Solutions 2-12

Full file at />
g(t)


Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed
©M. J. Roberts - 3/16/11

Full file at />∞

()

()

()

( () (

x t = 2 rect t

(a)

x t = A u t − u t − 10

(b)
∞

Ex =

∫ ( () (

A u t − u t − 10

−∞

2

() () (

x t = u t − u 10 − t

(c)
∞

Ex =

))

∫ u (t ) − u (10 − t )

2

1/ 2

1/ 2

(

( ))

1/ 2

(

( ))

(

( ))

(

)

dt = 4

∫

dt = 4

−1/ 2

10

0

)
0

∞

−∞

10

∫ dt + ∫ dt → ∞

dt =

() ( )

x t = rect t cos 2π t

(d)
∞

Ex =

−∞

))

2

2 rect t

dt = A2 ∫ dt = 10 A2

−∞

()

()

∫

Ex =

∫

() ( )

rect t cos 2π t

−∞

2

1/ 2

∫

dt =

( )

1
1 + cos 4π t dt
2 −1/∫ 2

( )

1
1 + cos 8π t dt
2 −1/∫ 2

cos 2 2π t dt =

−1/ 2

⎡
⎤
1/ 2
1/ 2
⎥ 1
1⎢
E x = ⎢ ∫ dt + ∫ cos 4π t dt ⎥ =

2 ⎢ −1/ 2
2
−1/ 2

⎥
⎢⎣
⎥⎦
=0

( )

()

∞

Ex =

() ( )

x t = rect t cos 4π t

(e)

∫

() ( )

rect t cos 4π t

−∞

2

1/ 2

dt =

∫

cos 2 4π t dt =

−1/ 2

⎡
⎤
1/ 2
1/ 2
⎥ 1
1⎢
E x = ⎢ ∫ dt + ∫ cos 8π t dt ⎥ =
2 ⎢ −1/ 2
2
−1/ 2

⎥
⎢⎣
⎥
⎦
=0

( )

()

∞

Ex =

() ( )

x t = rect t sin 2π t

(f)

∫

−∞

() ( )

rect t sin 2π t

2

1/ 2

dt =

∫

1/ 2

( )

sin 2 2π t dt =

−1/ 2

1
1 − cos 4π t dt
2 −1/∫ 2

⎡
⎤
1/ 2
1/ 2
⎥ 1
1⎢
E x = ⎢ ∫ dt − ∫ cos 4π t dt ⎥ =
2 ⎢ −1/ 2
2
−1/ 2

⎥
⎢⎣
⎥
⎦
=0

( )

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,

Solutions 2-13

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©M. J. Roberts - 3/16/11

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Ex =

∫ ( Arect (t ) + B rect (t − 0.5))

2

dt

−∞
∞

Ex =

∫ (A

2

()

−∞

1/ 2

E x = A2

∫

−1/ 2

25.

(

)

))

() (

rect 2 t + B 2 rect 2 t − 0.5 + 2 AB rect t rect t − 0.5 dt
1

1/ 2

0

0

dt + B 2 ∫ dt + 2 AB ∫ dt = A2 + B 2 + AB

()

Find the average signal power of the periodic signal x t in Figure E-25.
x(t)
3
2
1
-4 -3 -2 -1

-1

1

2

3

4

t

-2
-3

Figure E-25
1
P=
T0

26.

t0 +T0

∫ x (t )

2

t0

2

1
dt = ∫ x t
3 −1

()

1

2

1
1
2
1
4
4 ⎡ t3 ⎤
8
dt = ∫ 2t dt = ∫ t 2 dt = ⎢ ⎥ =
3 −1
3 −1

3 ⎣ 3 ⎦ −1 9

Find the average signal power of these signals.
T /2

Px = lim

() ()

Px = lim

x t =A

(b)

x t =u t

(c)

()

T /2

1
u t
T →∞ T ∫
−T / 2

(

x t = Acos 2π f0t + θ
T /2

Px =

T /2

2
1
A2
A2
A dt = lim
dt = lim
T = A2
∫
∫
T →∞ T
T →∞ T
T →∞ T
−T / 2
−T / 2

()

(a)

1 0
Acos 2π f0t + θ
T0 −T∫ / 2

(

)

()

2

1
T →∞ T

dt = lim

T /2

T →∞

0

)

2

T /2

dt =

0

A2 0

cos 2 2π f0t + θ dt
T0 −T∫ / 2

(

)

0

(

T /2
A2 0
A2 ⎡ sin 4π f0t + 2θ
Px =
1
+
cos
4
π
f
t
+
2
θ
dt
=
⎢t +
0
2T0 −T∫ / 2

2T0 ⎢⎣
4π f0
0

(

A2
Px =
2T0

(

1T 1
=
2 2

∫ dt = lim T

))

) ⎤⎥

T0 / 2

⎥⎦ −T / 2
0

⎡
⎤
⎢

sin 4π f0T0 / 2 + 2θ
sin −4π f0T0 / 2 + 2θ ⎥ A2
−
⎢T0 +
⎥=
4π f0
4π f0
2
⎢
 ⎥
⎢⎣
⎥
⎦
=0

(

)

(

)

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
Solutions 2-14

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)


Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed
©M. J. Roberts - 3/16/11

Full file at />27.

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

()

(

g t = 100sin 200π t + π / 4

(

)

)

(

)

(

)

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

()

(b)

g t = 13 − 4t + 6t 2

(c)

g t = −5e−2t e− j 2π t

()

() ()

2

g 2 = 13 − 4 2 + 6 2 = 29

()

(

)

g 1 / 4 = −5e−2/ 4 e− j 2π / 4 = −5e−1/ 2 e− jπ / 2 = − j3.03
28.

Let the unit impulse function be represented by the limit,

()

(

)

(

)

δ x = lim 1 / a rect x/ a , a > 0 .

(

a→0

) (

)

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

( ) ( ) ( )

This is a rectangle with the same height as (1 / a ) rect ( x/ a ) but 1/4 times the base
(a)

What is the area of the function δ 4x = lim 1 / a rect 4x / a ?
a→0

width. Therefore its area is 1/4 times as great or 1/4.

( ) (
)
This is a rectangle with the same height as (1 / a ) rect ( x/ a ) but 1/6 times the base
(b)

(

)

What is the area of the function δ −6x = lim 1 / a rect −6x / a ?
a→0

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 1 / a rect bx / a for b
a→0

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 − t )

()

∞

∫ δ ( x ) dx = 1

δ x =0 , x≠0 ,

(

.

0

−∞

)

(

)

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

Strength =

∫ δ ⎡⎣ a (t − t )⎤⎦ dt
0

−∞

Let

(

)

a t − t0 = λ and ∴ adt = d λ
Then, for a > 0,
∞

Strength =

−∞

∞

( ) daλ = 1a ∫ δ ( λ ) d λ = 1a =

∫δ λ

−∞

1
a

and for a < 0,
Solutions 2-15

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Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed
©M. J. Roberts - 3/16/11

Full file at />−∞

∫ δ (λ )

Strength =

∞

dλ 1
=
a
a

−∞

∞

1

1

∫ δ (λ ) dλ = − a ∫ δ (λ ) dλ = − a =
∞

−∞

1
a

Therefore for a > 0 and a < 0,

1
1
and δ ⎡⎣ a t − t0 ⎤⎦ = δ t − t0
a
a

(

Strength =
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 .
−∞

∞
1
Then, using the property from Exercise 29 δ1 ax = ∑ δ ⎡⎣ a x − n / a ⎤⎦ =
a
−∞

( )

(b)

(

)

( )

1
1/ a

( )

δ1 ax =

t0 +1/ a

1/ 2a

∫ ( )

1/ 2a

∫ ( )

δ1 ax dx = a

−1/ 2a

t0

∫ δ ( ax ) dx

δ1 ax dx = a

−1/ 2a

Letting λ = ax
1/ 2

( )

δ1 ax =

( ) (

∫ δ (λ ) dλ = 1

−1/ 2

) () ( ) (

) ()

Even though δ at = 1 / a δ t , δ1 ax ≠ 1 / a δ1 x

( ) ∑ δ ( ax − n) ≠ (1 / a ) ∑ δ ( x − n) = (1 / a )δ ( x )

δ1 ax =

∞

∞

n= −∞

n= −∞

( ) (

1

) ()

δ1 ax ≠ 1 / a δ1 x

QED

Scaling and Shifting Functions
31.

−∞

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

The period is 1 / a . Therefore

(c)

∞

∑δ ( x − n / a) .

Graph these singularity and related functions.

()

(

)

(a)

g t = 2u 4 − t

(c)

g t = 5s gn t − 4

(e)

g t = 5ramp t + 1

(g)

g t = 2δ t + 3

()

(

()

()

(

(

)
)

)

() ( )

(b)

g t = u 2t

(d)

g t = 1 + s gn 4 − t

(f)

g t = −3ramp 2t

(h)

g t = 6δ 3t + 9

()

(

()

( )

()

(

)

)

Solutions 2-16

Full file at />

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©M. J. Roberts - 3/16/11

Full file at />(a)

(c)

(b)

g(t)

(d)

g(t)

g(t)

g(t)

5
2

1
t

4

t

(e)

(g)

(h)

g(t)

10

1

1

-1

()

( ( ))

()

( )

g t = −4δ 2 t − 1

(k)

g t = 8δ1 4t

(m)

g t = 2 rect t / 3

(o)

g t = −3rect t − 2

()

( )

()

(

()

g(t)

2

2

t

t

-3

(i)

)

()

t

-3

(

)

(j)

g t = 2δ1 t − 1 / 2

(l)

g t = −6δ 2 t + 1

(n)

g t = 4 rect t + 1 / 2

(p)

g t = 0.1rect t − 3 / 4

()

g t

g(t)

-6

t

t

4

-5

(f)
g(t)

2

t

4

()

(

()

(( ) )
(( ) )

()

()

g t

)

()

g t

g t

t

t
t

t

()

(p)

(o)

()

g t

()

()

g t

g t

g t

0.1

3/ 2 5/ 2
t

t

t

1

−3

32.

3

5

Graph these functions.

() () (

)

(a)

g t = u t − u t −1

(c)

g t = −4 ramp t u t − 2

(e)

g t = 5e−t / 4 u t

(g)

g t = −6 rect t cos 3π t

)

()

(

)

()

() ( )

()

() ( )

(b)

g t = rect t − 1 / 2

(d)

g t = s gn t sin 2π t

()

() (

()

()

(f)

g t = rect t cos 2π t

()

() ( )

(h)

g t = u t + 1 / 2 ramp 1 / 2 − t

() (

)

(

Solutions 2-17

Full file at />
)

t


Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed

©M. J. Roberts - 3/16/11

Full file at />(a)

(b)

()

()

g t
1

g t

t

t

t

-1

()

(

-1

)

(

(k)

g t = 2 ramp t rect t − 1 / 2

(

)

( ) (

(j)

()

1

t

-1

-1

(h)

⎤
⎥⎦

(l)

()

g t

()

3

2
1

t

1/ 2

−1 / 2

g t

g t
1

t

(k)

()

g t

1

1/ 2

)⎥ d λ

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

(i)

()

)

⎡
g t = ⎢ ∫ δ λ + 1 − 2δ λ + δ λ − 1
⎢⎣ −∞

()

1

g t
1

(g)

(j)

t

1
t

(f)

g t = rect t + 1 / 2 − rect t − 1 / 2

-1

t

−1 / 2

(i)

(l)

33.

1/ 2

−1 / 2
4
(e)

()

g t
6

g t

5

g t

()

()

()

()

-8
t
-16

1

g t

(d)

g t

1
1

(c)

t

t

2

-1

-2 -1 1
-3

2

t

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

)

g(t)
3
2
1
t

2
(b)

()

(

)

g t = 2δ1 −t / 5

g(t)

∞

()

(

∞

)

(

)

g t = 2 ∑ δ −t / 5 − n = 10 ∑ δ t + 5n ,
n= −∞

n= −∞

10
...
-10 -5

(c)

()

() (

...

5 10 15 20

)

g t = δ1 t rect t / 11

Solutions 2-18

Full file at />
t


Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed
©M. J. Roberts - 3/16/11

Full file at />
g(t)

()

(

∞

5

n= −∞

n= −5

) ∑ δ (t − n) = ∑ δ (t − n)

g t = rect t / 11

,

1

-5 -4 -3 -2 -1

1 2 3 4 5

t

t

(d)

( ) ∫ ⎡⎣δ ( λ ) − δ ( λ − 1)⎤⎦ d λ

g t =

2

2

−∞

g(t)
1

-2 -1

34.

1

2

t

3

()

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.
⎧0
,
⎪
⎪−10 − 2t ,
g t =⎨
⎪sin π t / 2
⎪−6e−t / 2 ,

⎩

()

(

)

, −2t>2

()

(b)

Graph g t in the range −10 < t < 10 .

(c)

Graph g 2t in the range −10 < t < 10 .

(d)

Graph 2 g 3 − t in the range −10 < t < 10 .

(e)

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

( )

(

)

(( ) )
g(t)

-10

g(2t)
10

t

-10

-8

-10

t

-2g(( t+1)/2)
10

-16

10

-8

2g(3- t)

35.

t < −5
− 5 < t < −2

t

16

-10

10

t

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

Plotting functions and transformations of those functions

Solutions 2-19

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Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed
©M. J. Roberts - 3/16/11

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%
(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)

⎧−2 , t < −1
⎪
⎪2t , − 1 < t < 1
g t =⎨
2
⎪3 − t , 1 < t < 3
⎪⎩−6 , t > 3

()

(

−3g 4 − t

)

vs. t

Original g(t)
2
-4

8

t

-6

Transformed g(t)
20

-4

8

t

-10

(b)

()

(

g t = Re e jπ t + e j1.1π t

)

( )

g t/4

vs. t

Solutions 2-20

Full file at />

Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed
©M. J. Roberts - 3/16/11

Full file at />Original g(t)
2

t

100
-2

Transformed g(t)
2

t

100
-2

(c)

( )

G f =

( (

5
f − j2 + 3

))

( (

G 10 f − 10 + G 10 f + 10

2

))

vs. f

Original g(t)

1.5

-20

20

t

Transformed g(t)
1.5

-20

36.

20

t

A signal occurring in a television set is illustrated in Figure E36.
mathematical description of it.

Write a

Signal in Television
x(t)

5

-10

60

t (µs)

-10

Figure E36 Signal occurring in a television set

⎛ t − 2.5 × 10−6 ⎞
x t = −10 rect ⎜
⎟
⎝ 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.
BPSK Signal
x(t)
1

4

t (ms)

-1

Figure E37 BPSK signal
Solutions 2-21

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Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed
©M. J. Roberts - 3/16/11

Full file at />
⎡
⎛ t − 0.5 × 10−3 ⎞
⎛ t − 1.5 × 10−3 ⎞ ⎤
−
sin
8000
π
t
rect
⎢sin 8000π t rect ⎜
⎟
⎜
⎟ ⎥
10−3
10−3
⎝
⎠
⎝
⎠ ⎥
⎢
x t =⎢

−3
−3 ⎥
⎛
⎞
⎛
⎞
⎢ + sin 8000π t rect t − 2.5 × 10 − sin 8000π t rect t − 3.5 × 10 ⎥
⎜
⎟
⎜
⎟⎥
−3
−3
⎢⎣
10
10
⎝
⎠
⎝
⎠⎦

()

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.

RC Filter Signal
x(t)

4

t (ns)

20

-1.3333
-4
-6

Figure E38 Transient response of an RC filter

(

()

)

(

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

39.

)

(times in ns)

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

x(t)
15

...
t

4
Figure E39

()

()

∞

(

x t = 3.75ramp t − 15∑ u t − 4n
40.

n=1

)

Mathematically describe the signal in Figure E-40 .

x(t)

...

Semicircle

9

...
9

t

Figure E-40
The semicircle centered at t = 0 is the top half of a circle defined by

()

x 2 t + t 2 = 81
Therefore

()

x t = 81 − t 2 , − 9 < t < 9 .

This one period of this periodic function. The other periods are just shifted
versions.
Solutions 2-22

Full file at />

Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed
©M. J. Roberts - 3/16/11

Full file at />⎛

⎞

( ) ∑ rect ⎜⎝ t −1818n ⎟⎠
∞

x t =

(

81 − t − 18n

n= −∞

)

2

(The rectangle function avoids the problem of imaginary values for the square
roots of negative numbers.)
41.

Let two signals be defined by

⎪⎧1 , cos ( 2π t ) ≥ 0
x1 ( t ) = ⎨
⎩⎪0 , cos ( 2π t ) < 0

and x 2 ( t ) = sin ( 2π t / 10 ) ,

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

( ) ( )

(a)

x1 2t x 2 −t

(c)

x1 t / 5 x 2 20 t + 1

( ) ( (

))

((

) ) ( )

x1 t − 2 / 5 x 2 20t

(d)

(a)

(b)

x1(t)x2(t)

x1(t)x2(t)

1

1

-5

5

t

-5

5

-1

(c)

(d)

x1(t)x2(t)

x1(t)x2(t)

-5

t

-1

1

1

5

t

-5

5

-1

42.

( ) ( )

x1 t / 5 x 2 20t

(b)

t

-1

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

g(t)
2
1

() ( )

g ( t ) → −3g ( −t )
g t → g 2t

1

-2

2 3 4 5 6

t

-2

()

g t = 0 , t > 6 or t < −2

(b)

Solutions 2-23

Full file at />

Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed
©M. J. Roberts - 3/16/11

Full file at />g(t)
2
1
-2

()

t→t+4

1 2 3 4 5 6

t

(( ) )

()

g t → −2 g t − 1 / 2

-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.

g(2t)
2
1
-2

2

4

6

2

4

6

t

-2

-3g(-t)
6
3
-4

-2

t

-6
(b)

g(t + 4)
2
1
-2

1 2 3 4 5 6 7

t

-2

-2g( t -1 )
2
4
2
-2

1 2 3 4 5 6 7 8

t

-4
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)
Solutions 2-24

Full file at />

Solution Manual for Signals and Systems Analysis Using Transform Methods and MATLAB 2nd Ed
©M. J. Roberts - 3/16/11

Full file at />g(t)
2

2
-2

t

1 2 3 4 5 6

-1

-4 -3 -2 -1 -1

1 2 3 4

t

(b)

g(t)
2
-2

t

1 2 3 4 5 6

-2

1 2 3 4 5 6

-1

t

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

(b)
44.

()

()

(

) (

g t →g 2−t

(

)

)

()

(

) (

)

)

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

Write a function of continuous time t for which the two successive changes t → −t
and t → t − 1 leave the function unchanged. cos 2π t , 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
|G( f )|
10

-100

100

f

( )

G f
π

-100

100

f

-π

Solutions 2-25

Full file at />