Signals and systems analysis using transform methods and matlab 3rd edition roberts solutions manual

Solutions 2-1
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Signals and Systems Analysis Using Transform Methods and MATLAB 3rd
Edition Roberts SOLUTIONS MANUAL
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Chapter 2 - Mathematical Description of
Continuous-Time Signals
Solutions
Exercises With Answers in Text
Signal Functions
1. If gt 7e2t 3
write out and simplify
(a) g 3 7e9
8.6387 104
(b)
g2 t 7e22t 3
7e72t
(c) gt /10 4
7et/511
(d) g jt 7ej2t 3
(e)
gjtgjt
2
7e3 ej2t
ej2t
2
7e3
cos2t 0.3485 cos2t
g
jt 3 jt 3
g
jt jt
(f) 2 2
7
e e
7cost
2 2
2. If g x x2
4x 4 write out and simplify
(a) g z z2
4z 4
(b) gu v u v
2
4u v 4 u2
v2
2uv 4u 4v 4
(c) gejt
ejt
2
4ejt
4 ej2t
4ejt
4 ejt
2
2
(d) ggt gt2
4t 4 t2
4t 4
2
4t2
4t 4 4

Solutions 2-2
ggt t4
8t3
20t2
16t 4
(e) g2 4 8 4 0
3. Find the magnitudes and phases of these complex quantities.

Solutions 2-3
(a) e
3 j2.3
e
3 j2.3
e3
e j2.3
e3
cos2.3 jsin2.3
e
3 j2.3
e3
cos2
2.3 sin2
2.3 e3
0.0498
(b) e2 j6
e2 j6
e2
e j6
e2
7.3891
(c)
100
8 j13
100 100 100
6.5512
8 j13 8 j13 82
132
4. Let G f
j4 f
.
2 j7 f /11
(a) What value does the magnitude of this function approach as f approaches
positive infinity?
lim
j4 f j4 f 4 44
6.285
f 2 j7 f /11 j7 f /11 7 /11 7

Solutions 2-4
f 0
j /2
(b) What value (in radians) does the phase of this function approach as f
approaches zero from the positive side?
lim
f 0
j4 f 2 j7 f /11 lim j4 f 2 / 2 0 / 2
5. Let X f
jf
jf 10
(a) Find the magnitude X 4 and the angle in radians
X f
jf
X 4
j4 4e
0.3714ej1.19
jf 10 j4 10 10.77ej0.3805
X 4 0.3714
(b) What value (in radians) does approach as f approaches zero from
the positive side?
Shifting and Scaling
6. For each function gt graph gt, gt, gt 1, and g2t . (a)
(b)
g(t) g(t)
4 3
2
t 1
t
-3
g(-t) g(-t) -g(t) -g(t)
4 3
-2
t 1
t
-3
3
2
t
-1
t
4 -3

Solutions 2-5
(a) (a)
2 2
1 1
0 0
-1 -1
-2 -2
-4 -2
t
0 2 4 -4 -2
t
0 2 4
(b) (b)
2 2
1 1
0 0
-1 -1
-2 -2
-4 -2
t
0 2 4 -4 -2
t
0 2 4
(c) (c)
2 2
1 1
0 0
-1 -1
-2 -2
-4 -2 0 2 4 -4 -2 0 2 4
t t
g(t)g1(t)g1
(t)1
g(t)g(t)g(t)222
g(t-1) g(t-1) g(2t) g(2t)
4
1 3
t
3
1 2
t
-3
4 3
- 1
1
t 1
t
2
-3
7. Find the values of the following signals at the indicated times.
(a) xt 2rectt / 4 , x1 2rect1/ 4 2
(b) xt 5rectt / 2sgn2t , x0.5 5rect1/ 4sgn1 5
(c) xt 9rectt /10sgn3t 2 , x1 9rect1/10sgn 3 9
8. For each pair of functions in Figure E-8 provide the values of the constants A,
t0
and w in the functional transformation g2
t Ag1
t t0
/ w .
Figure E-8
Answers: (a) A 2,t0 1,w 1 , (b) A 2,t0 0,w 1/ 2 , (c) A
1/ 2,t0 1,w 2
9. For each pair of functions in Figure E-9 provide the values of the constants A,

Solutions 2-6
-10 -5 0 5 10 -10 -5 0 5 10
t t
-10 -5 0 5 10 -10 -5 0 5 10
t t
0
0
g(t)g(t)g(t)111
g(t)g(t)g(t)222
t0
and a in the functional transformation g2 t Ag1 wt t0
.
A = 2, t0
= 2, w = -2
8 8
4 4
(a)
0 0
-4 -4
-8 -8
Amplitude comparison yields A 2. Time scale comparison yields w 2.
g2 2 2g1 22 t0 2g1 0 4 2t0 0 t0 2
A = 3, t = 2, w = 2
(b)
8 8
4 4
0 0
-4 -4
-8 -8
Amplitude comparison yields A 3. Time scale comparison yields w 2.
g2 2 3g1 22 t0 3g1 0 4 2t0 0 t0 2
A = -3, t = -6, w = 1/3
8 8
4 4
(c)
0 0
-4 -4
-8
-10 -5 0 5 10
-8
-10 -5 0 5 10
t t
Amplitude comparison yields A 3. Time scale comparison yields w 1/ 3.
g2 0 3g1 1/ 30 t0 3g1 2 t0 / 3 2 t0 6
OR
g2
3 3g1
1/ 33 t0
3g1
0 t0
/ 3 1 0 t0
3

Solutions 2-7
g
1
(t)g
1
(t)
g
2
(t)g
2
(t)
A = -2, t0
= -2, w = 1/3
8 8
4 4
(d)
0 0
-4 -4
-8
-10 -5 0 5 10
-8
-10 -5 0 5 10
t t
g2
4 2g1
1/ 34 t0
3g1
2 t0
/ 3 4 / 3 2 t0
2
A = 3, t0 = -2, w = 1/2
8 8
4 4
(e)
0 0
-4 -4
-8
-10 -5 0 5 10
-8
-10 -5 0 5 10
t t
g2
0 3g1
1/ 20 t0
3g1
1 t0
/ 2 1 t0
2
Figure E-9
10. In Figure E-10 is plotted a function g1 t which is zero for all time outside the
range plotted. Let some other functions be defined by
t 3
g2 t 3g1 2 t , g3 t 2g1 t / 4 , g4 t g1  2
Find these values.
(a) g2 1 3 (b) g3 1 3.5
3 3
1
(c) g4 tg3 tt2
2
1
2
(d) g4 t dt
3
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.
1
g4 t dt 2
3

Solutions 2-7
1
g (t)
-4 -3
4
3
2
1
-2 -1
-1
-2
-3
-4
1 2 3 4 t
11. A function G f is defined by
Figure E-10
G f ej2 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.
ej2 f
cos2 f j sin2 f cos2 f jsin2 f 
ej2 f
cos2
2 f sin2
2 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
ej2 f
cos2 f jsin2 f tan1 sin2f
tan1 sin2f
cos2 f  cos2 f
ej2 f
tan1
tan2 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
ej2 f
2 f

Solutions 2-8
1 2
which is simply the coefficient of j in the original complex exponential expression.
A more general solution would be e j 2 f
2 f 2n , 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 j2 f 10
rect
f 10
e j2 f 10
rect
f 10
2 2
12. Let x t 3rectt 1 / 6 and x t rampt ut ut 4 .1 2 
(a) Graph them in the time range 10 t 10. Put scale numbers on the
vertical axis so that actual numerical values could be read from the
graph.
x (t)1
x (t)2
6 6
-10 10
-6
-10 10
-6
(b) Graph xt x 2t x t / 2 in the time range 10 t 10. Put
scale numbers on the vertical axis so that actual numerical values
could be read from the graph.

Solutions 2-9

5
11
x 2t 3rect2t 1 / 6 and x t / 2 rampt / 2 ut / 2 ut / 2 41 2
x 2t 3rectt 1/ 2 / 3 and x

t / 2 1/ 2rampt ut ut 81 2 
x (2t)-x1 2
(t/2)
8
-10 10
-8
13. Write an expression consisting of a summation of unit step functions to represent a
signal which consists of rectangular pulses of width 6 ms and height 3 which occur
at a uniform rate of 100 pulses per second with the leading edge of the first pulse
occurring at time t 0 .

xt 3
n0
ut 0.01n ut 0.01n 0.006
14. Find the strengths of the following impulses.
(a) 34t
34t 3
1
t Strength is 3 / 4
4
(b) 53t 1
53t 1
5
t 1 Strength is
5
3 3
15. Find the strengths and spacing between the impulses in the periodic impulse
9 5t.
9 5t 9 5t 11k
9 
t 11k / 511 k
k
The strengths are all -9/5 and the spacing between them is 11/5.

Solutions 2-10
1
4
1
4

t
Derivatives and Integrals of Functions
16. Graph the derivative of xt 1 et
ut.
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 a positive
decaying exponential.
xt
e , 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
xt et
ut .
x(t)
t
-1
-1
dx/dt
t
-1
-1
Alternate Solution using the chain rule of differentiation and the fact that the
impulse occurs at time t 0 .
d
xt 1 et
t et
ut et
ut
dt 14243
0 for t0
17. Find the numerical value of each integral.
11 4
(a) u 4 t dt dt 6
2 2
8
(b) t 3 2 4t dt
1

Solutions 2-11
3
2
4
8 8 8
t 3 2 4t dt t 3dt 2 4t dt
1 1 1
8 8
t 3 2 4t dt 0 2 1/ 4 t dt 1/ 2
(c)
1
5/2
2 3t dt
1/2
5/2 5/2 
1
1
5/2
2 3t dt  3t 2n dt
3
 t 2n / 3dt
1/2 1/2 n
5/2
1/2 n
1
2 3t dt 1 1 1 1
1/2
(d) t 4 ramp2t dt ramp24 ramp8 8
10 10
10
(e) ramp2t 4 dt 2t 4 dt t2
4t 100 40 4 8 64
3 2
(f)
82
3sin200t t 7 dt 0 (Impulse does not occur between 11 and 82.)
11
(g)
5
sin t / 20 dt = 0 Odd function integrated over
5
symmetrical limits.
10 10
(h) 39t2
 t 1dt 39 t2
t 1 4k dt4 2 2
k
10
39t2
 10
t 1dt 39 t2
t 1 t 5 t 9 dt4 
2 2
10
39t2

2
t 1dt 3912
52
92
4173
(i)

e18t
ut 10t 2 dt .

Solutions 2-12


t

e18t
ut 10t 2 dt e18t
ut 10t 1/ 5dt

Using the scaling property of the impulse,
e18t
ut 10t 2 dt
1

e18t
ut t 1/ 5dt
10 
Using the sampling property of the impulse,
e18t
ut 10t 2 dt
1
e18/5
0.002732
10
9 9
(j) 9t 4 / 5dt 45 t 4 dt 45
2 2
3 3
(k) 53t 4 dt
5
t 4 dt 0
6
36
(l)

ramp3t t 4 dt ramp3 4 12
(m)
17
3 t cos2t / 3dt
1
The impulses in the periodic impulse occur at ...-9,-6,-3,0,3,6,9,12,15,18,...
At each of these points the cosine value is the same, one, because its
period is 3 seconds also. So the integral value is simple the sum of the
strengths of the impulse that occur in the time range, 1 to 17 seconds
17
3 t cos2t / 3dt 5
1
18. Graph the integral from negative infinity to time t of the functions in Figure E-18
which are zero for all time t 0.
This is the integral g d which, in geometrical terms, is the accumulated

area under the function gt 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

Solutions 2-13
dt
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 2 3
t
1
2
1
t
1 2 3
Figure E-18
g(t) dt g(t) dt
1
1
2
t
1 2 3
1
t
1 2 3
19. If 4ut 5
d
xt, what is the function xt? xt 4rampt 5
Generalized Derivative
20. The generalized derivative of 18rect
t 2 consists of two impulses.
3
Find their numerical locations and strengths.
Impulse #1: Location is t = 0.5 and strength is 18.
Impulse #2: Location is t = 3.5 and strength is -18.
Even and Odd Functions
21. Classify the following functions as even, odd or neither .
(a) cos2t trit 1 Neither

Solutions 2-14

15
Cosine is even but the shifted triangle is neither even nor odd which
means it has a non-zero even part and a non-zero odd part. So the product
also has a non-zero even part and a non-zero odd part.
(b) sin2t rectt / 5 Odd
Sine is odd and rectangle is even. Therefore the product is odd.
22. An even function gt is described over the time range 0 t 10 by
2t , 0 t 3
gt 3t , 3 t 7 .
2 , 7 t 10
(a) What is the value of gt at time t 5?
Since gt is even, gt gt g5 g5 15 3 5 0 . (b) What is the
value of the first derivative of g(t) at time t 6?
Since gt is even,
d
gt
d
gt
d
gt
 d
gt

3 3.
dt dt dt t6 dt t6
23. Find the even and odd parts of these functions.
(a) gt 2t2
3t 6
ge t  2t2
3t 6 2t
2
3t6

2
2t2
3t 6 2t
2
3t6
4t2
12
2
6t
2t2
6
go t  3t
2 2
(b) gt 20cos 40t / 4
20cos40t / 420cos40t / 4

ge t 
2
Using cos z1 z2 cos z1 cos z2 sinz1 sin z2
,

Solutions 2-15
2
20

cos 40t cos / 4 sin 40tsin / 4 

ge t 
20 cos40tcos / 4sin40tsin/ 4
2
20


ge t 
20 cos40tcos / 4sin40tsin / 4
2
ge
t 20cos / 4 cos 40t 20 / 2cos 40t
20cos40t / 420cos40t / 4

go t 
2
Using cos z1
z2
cos z1
cos z2
sinz1
sin z2
,
20


go t 
20 cos40tcos/ 4sin40tsin/ 4
2
20


go t 
20 cos40tcos / 4sin40tsin / 4
2
go
t 20sin / 4sin 40t 20 / 2sin 40t
(c) gt
2t 3t 6
1 t
2t2
3t 6 2t2
3t 6
ge t 1t 1t
2
2t2
3t 61 t 2t2
3t 61 t 1 t1 tge t 
2

Solutions 2-16
{
2
2
2
ge t  4t2
12 6t2
21 t2

6 5t2

1 t2
2t2
3t 6 2t2
3t 6
go t 1t 1t
2
2t2
3t 61 t 2t2
3t 61 t 1 t1 tgo t 
2
go t  6t 4t3
12t
21 t2

t
2t2
9
1 t2
(d) gt t2 t2
1 4t2

gt t 2 t2
1 4t2

odd 123 14243
even even
Therefore gt is odd, ge t 0 and go t t2 t 1 4t2
(e) gt t2 t1 4t
t2 t14tt2 t14t
ge t 
2
t2 t14tt2 t14t

ge t 7t
2
go
t 
2
go t t2 4t
(f) gt
20 4t 7t
1 t
ge t 
20 4t2
7t
1 t
20 4t2
7t

1 t
2
40 8t2
1 t
2
20 4t2

1 t

Solutions 2-17
1
1
1
1
1
,
go t 
20 4t2
7t
1 t
20 4t
2
7t

1 t
2
14t
1 t
2
7t
1 t
24. Graph the even and odd parts of the functions in Figure E-24.
To graph the even part of graphically-defined functions like these, first graph
gt . Then add it (graphically, point by point) to gt and (graphically) divide the
sum by two. Then, to graph the odd part, subtract gt from gt
(graphically) and divide the difference by two.
(a) (b)
g(t) g(t)
1
t t1 2
-1
Figure E-24
ge
(t) ge
(t)
1
t 1 2
t
-1
g (t)o
go
(t)
1
1
t 1 2
t
-1
(a) (b)
25. Graph the indicated product or quotient gt of the functions in Figure E-25.

Solutions 2-18
1
-1
1
1
1
1
1
1
-1 1
-1 1
-1 1
1
(a) (b)
1
-1
1
-1
1
-1 1
t
g(t)
Multiplication
t
1
-1
t1
-1
1
-1
t1
-1
g(t)
Multiplication
g(t) g(t)
1
-1
t
1
-1
1
t-1 1
-1
(c) (d)
t t
g(t) g(t)
Multiplication
t
Multiplication
t
g(t) g(t)
t
t
(e) (f)

1
-1 1
-1
1
-1 1

t
t
g(t)
Multiplication
1
1
t
-1
g(t)
Multiplication
1
t
-1

g(t)
1
-1 1
-1

t
g(t)
1
1
t
-1

Solutions 2-19
1
-1
1
1
{
0
1
(g) (h)
t
g(t)
1
-1 -1 1
t

g(t)
Division Division
t
1
t
g(t)
-1
t
Figure E-25
g(t)
1
-1 1
t
26. Use the properties of integrals of even and odd functions to evaluate these integrals
in the quickest way.
1 1 1 1
(a) 2 t dt  2{ dt  t{ dt 2 2dt 4
1 1 even 1 odd 0
(b)
1/20
 4cos10t 8sin 5t dt 
1/20
 4cos10t dt 
4243
1/20
 8sin 5t dt
142431/20
1/20
1/20
1
1/20
even 1/20 odd

1/20
4cos10t 8sin5t dt 8 
0
cos10t dt
8
10
(c)
1/10
1/20
1/20
4 t cos10t dt 0
odd
14243
1442ev
4en
43
odd
1/10
cos10t
1/10 1/10
cos10t
(d) t{ sin10t dt 2 t sin10t dt 2t dt
1/10 odd
14243
0
10 0
10
142od
4d
3
even
1/10
1 sin10t
 1/10 
1
t{ sin10t dt 2 
2
1/10 odd
14243 100 10  50
142od4d 3 0

even
1 1 1
(e) e
t
dt 2 e
t
dt 2 et
dt 2et

21 e1
1.264{ 
1 even 0
0
0

Solutions 2-20
{
0
5
(f)
1
t{ et
dt 0
1 o{dd even
odd
Periodic Signals
27. Find the fundamental period and fundamental frequency of each of these functions.
(a) gt 10cos50t f0 25 Hz , T0 1/ 25 s
(b) gt 10cos50t / 4 f0
25 Hz , T0
1/ 25 s
(c) gt cos50t sin15t
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 GCD25,15 / 2 2.5 Hz , T0 1/ 2.5 0.4 s
(d) gt cos2t sin 3t cos5t 3 / 4
f0 GCD1,3/ 2,5 / 2 1/ 2 Hz , T0 
(e) gt 3sin20t 8cos 4t
1
1/ 2
2 s
10 2 2
f0
GCD , Hz , T / 2 s
(f) gt 10sin20t 7cos10t
f GCD
10
,  0 Hz , T , gt is not periodic0 0
(g) gt 3cos2000t 8sin2500t
f0
GCD1000,1250 250 Hz , T0
4 ms

Solutions 2-21
(h)
g1 t is periodic with fundamental period T01 15s
g2 t is periodic with fundamental period T02 40s
T LCM15s,40s 120s , f
1
8333
1
Hz0 0
120s 3
28. Find a function of continuous time t for which the two successive transformations
t t and t t 1 leave the function unchanged.
cos2nt , 
one.
1/n
t , n an integer. Any even periodic function with a period of
29. One period of a periodic signal xt with period T0
is graphed in Figure E-29.
Assuming xt has a period T0
, what is the value of xt at time, t 220ms?
x(t)
4
3
2
1
-1
-2
-3
-4
t5ms 10ms 15ms 20ms
T0
Figure E-29
Since the function is periodic with period 15 ms,
x220ms x220ms n 15ms where n is any integer. If we choose n 14 we get
x220ms x220ms 14 15ms x220ms 210ms x10ms 2 .
Signal Energy and Power of Signals
30. Find the signal energy of each of these signals.
 1/2
2
(a) xt 2rectt Ex 
2rectt dt 4 1/2
dt 4
(b) xt Aut ut 10

Solutions 2-22
2


Ex 
Aut ut 10 dt A2
10
dt 10A2
0
(c) xt ut u10 t
0 
2
Ex  ut u10 t dt dt dt
10
(d) xt rectt cos2t

2 1/2
2 1
1/2
Ex 
rectt cos 2t dt  1/2
cos 2t dt 
2 1/2
1 cos 4t dt
(e) xt rectt cos 4t

Ex 
rectt cos 4t
2
dt 
1/2
1/2
cos2
4t dt
1
2
1/2
1/2
1 cos8t dt

1
1/2

1/2
1
Ex
2
 dt  cos8t dt 
21/2 11/2442443
0
(f) xt recttsin2t

Ex 

recttsin2t
2
dt 
1/2

1/2
sin2
2t dt
1
2
1/2

1/2
1 cos 4t dt

Solutions 2-23
E
2
2
0
(g) xt 

1
t 1 , t 1 

0 , otherwise
1 1
2
x
t 1
1
3
dt 2
0
1
t 1
2
dt 2
0

t2
2t 1dt
2
t 
E 2 t2
t
2
1
1 1
x 
3 0
3 3
31. A signal is described by xt Arectt Brectt 0.5 . What is its signal energy?

Ex 
Arectt Brectt 0.5
2
dt
Since these are purely real functions,

Ex Arectt Brectt 0.5 dt

Ex A rect2

t B2
rect2 t 0.5 2ABrectt rectt 0.5dt

1/2 1
2 2
1/2
2 2
Ex A dt B dt 2AB dt A B AB
1/2 0 0
32. Find the average signal power of the periodic signal xt in Figure E-32.
x(t)
3
2
1
-4 -3 -2 -1
-1
-2
-3
1 2 3 4
Figure E-32
t0 T0 2 1 1 3 1
P
1
xt
2
dt
1
xt
2
dt
1
2t
2
dt
4
t2
dt
4 t 8
T0 t
31
31
3 1
3 3 1
9

Solutions 2-24
0
T T 4 4
 3
1
2 2
1
2
0
5

33. Find the average signal power of each of these signals.
(a) xt A Px lim
T /2
 A
2
dt lim
A
T /2
 dt lim
A
T A2
T T T /2
T T T /2
T T
(b) xt ut Px lim
T /2
 ut
2
dt lim
1
T /2
1 T 1
dt lim
T T T /2
(c) xt Acos2 f0t 
T T 0
T T 2 2
P
1
x
T
T0 /2
 Acos2 f0
t  A2
dt 
T
T0 /2
 cos2 2 f0
t dt
0 T0 /2 0 T0 /2
A2 T0 /2
A2 
sin4f t 2
T0 /2
Px
2T
1 cos 4 f t 2dt 
2T
t
0
0 T0 /2 0 4 f0 T /2

A2  sin4f T / 2 2

sin4f T / 2 2 A2
0 0 0 0
Px
2T
T0 
4 f

4 f 20
144440
44442444440
4443
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 Acos2 f0t instead, which is somewhat easier algebraically.
(d) xt is periodic with fundamental period four and one fundamental period is
described by xt t1 t , 1 t 5 .
P
1
xt
2
dt
1
tt 1
2
dt
1
5
t4
2t3
t2
dtx 
0
0
1 1
5 4 3
5
P
1 t t t  1 3125 625 125 1 1 1x
4 5 2
4 51
2 3 5 2 3
88.5333
(e) xt is periodic with fundamental period six. This signal is described
over one fundamental period by

Solutions 2-25



xt rect
t 2
4rect
t 4 , 0 t 6.
3 2
The signal can be described in the time period 0 t 6 by
0 , 0 t 1/ 2
1 , 1/ 2 t 3

xt 3 , 3 t
7
2
4 , 7 / 2 t 5
0 , 5 t 6
1 2 1 2 5 2 1 2 3 2 2.5 4.5 24
Px
6
0 1
2
3
2
4
2
0
2
1 

5.167
6
Exercises Without Answers in Text
Signal Functions
34. Let the unit impulse function be represented by the limit,
x lim1/ arectx/ a , a 0 .
a0
The function 1/ a rect x / a has an area of one regardless of the value of a. (a)
What is the area of the function 4x lim1/ a rect 4x / a ?
a0
This is a rectangle with the same height as 1/ arectx/ a but 1/4 times the base
width. Therefore its area is 1/4 times as great or 1/4.
(b) What is the area of the function 6x lim1/ a rect6x / a ?
a0
This is a rectangle with the same height as 1/ arectx/ 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 bx lim1/ a rectbx / a for b
a0
positive and for b negative ?
It is simply 1/ b .

Solutions 2-26
a
35. Using a change of variable and the definition of the unit impulse, prove that
at t0 1/ a t t0 .
x 0 , x 0 , x dx 1
at t0 0 , where at t0 0 or t t0
Let
Strength 

at t0 dt
Then, for a > 0,
at t0 and adt d

Strength
d 1
d
1 1
and for a < 0,
a

a a a
Strength
d 1
d
1
d
1 1
a a 
Therefore for a > 0 and a < 0,
a a a
Strength
1
and at t
1
t t .
a
0
a
0
36. Using the results of Exercise 35,

(a) Show that 1 ax 1/ a x n / a
n

From the definition of the periodic impulse 1 ax ax n.

Then, using the property from Exercise 35
1
1 ax 

a x n / a x n / a .
(b) Show that the average value of 1 ax is one, independent of the value of a
The period is 1/ a. Therefore
1 ax 1
1/ a
t0 1/a
t0
1 ax dx a
1/2a
1/2a
1 ax dx a
1/2a
1/2a
axdx

Solutions 2-27
Letting ax
1 ax 
1/2
d 1
1/2
(c) Show that even though at 1/ a t, 1 ax 1/ a 1 x

1 ax ax n 1/ a x n 1/ a 1 x
n n
1 ax 1/ a 1 x QED
Scaling and Shifting
37. A signal is zero for all time before t 2, rises linearly from 0 to 3 between t
2 and t 4 and is zero for all time after that. This signal can be expressed in the
form xt Arect
t t01
tri
t t02
. Find the numerical values of the
constants.
w1 w2
xt 3rect
t 1
tri
t 4
6 6
38. Let xt 3et /4
ut 1 and let yt 4 x5t .
(a) What is the smallest value of t for which yt is not zero?
yt 4 x5t 12e5t/4
u5t 1
Smallest t for which yt 0 occurs where 5t 1 0 or t 1/ 5.
(b) What is the maximum value of yt over all time?
maxyt 9.3456
(c) What is the minimum value of yt over all time?
minyt 0

Solutions 2-28
2
4
2
4
1
(d) What is the value of y1?
yt 12e5t /4
u5t 1 y1 12e5/4
u 4 12e5 /4
3.4381
(e) What is the largest value of t for which yt 2?
yt 2 12e5t/4
e5t/4
1/ 6 5t / 4 ln1/ 6 1.7918
t 41.7918 / 5 1.4334
39. Graph these singularity and related functions.
(a) gt 2u 4 t (b) gt u2t
(c) gt 5sgnt 4 (d) gt 1 sgn 4 t (e) gt 5
rampt 1 (f) gt 3ramp2t (g) gt 2t 3
(h) gt 6 3t 9
(a) (b) (c) (d)
g(t) g(t) gt g(t)
t
g(t)
10
-1 1
t
1
g(t)
-6
5
t
t
g(t)
2
t
-3
t
-3
t
g(t)
2
t
(e) (f) (g) (h)
(i) gt 4 2t 1 (j) gt 21 t 1/ 2
(k)
(m)
gt 81 4t
gt 2rectt / 3
(l)
(n)
gt 62 t 1
gt 4rectt 1/ 2

Solutions 2-29
4
3 5
2 2
Graph these functions.
(a)
(c)
gt ut ut 1
gt 4rampt ut 2
(b)
(d)
gt rectt 1/ 2
gt sgntsin2t
(e)
(g)
gt 5et/4
ut
gt 6rectt cos 3t
(f)
(h)
gt rectt cos2t
gt ut 1/ 2ramp1/ 2 t
(o) gt 3rectt 2 (p) gt 0.1rectt 3 / 4
g(t)
1
-2
g(t)
t
2
1
...
t
g(t)
1
4
...
...
g(t)
-6
... t
g(t)
2
t
- 3 3
2 2
g(t)
-2
g(t)
t -3
g(t)
0.1
t
1 3 5
40.

Solutions 2-30
3
-1 1
(a) (b) (c) (d)
gt gt gt gt
1 1
-8
t t
1 1 -16
t 1
t
-1 1
gt
gt gt gt
1
6
5 1/ 2 1/ 2
t t t
4
t -1
1/ 2 1/ 2 1/ 2 1/ 2
(e) (f) (g) (h)
(i) gt rectt 1/ 2 rectt 1/ 2
t 
(j) gt 1 2 1d

(k) gt 2rampt rectt 1/ 2
(l) gt 3rectt / 4 6rectt / 2
(i) (j) (k) (l)
gt gt
gt gt 2
1 1
1
t
1
t t t
-1 -1 2 -2 2
-1 -1 -3
41. A continuous-time signal xt is defined by the graph below. Let
yt 4xt 3 and let zt 8xt / 4 Find the numerical values.
(a) y 3 4 x 3 3 4x6 4 6 24
(b) z4 8x4 / 4 8x1 8 3 24
(c)
d
zt
dt t10
d
8xt / 4
dt t10
8
d
xt / 4
dt t10

Solutions 2-31
d
zt
8
d
xt
dt t10/4
8 3
6
dt t10 4 4
x(t)
10
t
-10 10
-10
42. Find the numerical values of
(a) ramp32 rect2 /10 ramp6 rect1/ 5 6 1 6
(b)
(c)

3t 4 cos t /10 dt 3cos 4 /10 0.9271

d
2sgnt / 5 rampt 8
dt t13
d
2sgnt / 5 rampt 8 2sgnt / 5 ut 8 2rampt 8 2 t
dt
d
2sgnt / 5 rampt 8

2sgn13 / 5 u5 2ramp5 2 13 2
dt t13
14243 { 14243 {
1 1 5 0
43. Let a function be defined by gt trit . Below are four other functions based on this
function. All of them are zero for large negative values of t.
g1 t 5g
2 t 
g t 7g 3t 4gt 4
6
2
t 4

1
g3 t gt 2 4g
3
g t 5gtg t 4
2
(a) Which of these transformed functions is the first to become non-zero
(becomes non-zero at the earliest time)? g3 t

Solutions 2-32
x(t)
PhaseofX(f)|X(f)|
f
(b) Which of these transformed functions is the last to go back to zero and
stay there? g1 t
(c) Which of these transformed functions has a maximum value that is greater
than all the other maximum values of all the other transformed functions?
(“Greater than” in the strict mathematical sense of “more positive than”.
For example, 2 5.) g2 t
(d) Which of these transformed functions has a minimum value that is less
than all the other minimum values of all the other transformed functions?
g1 t
44. (a) Write a functional description of the time-domain energy signal in Figure
E-44 as the product of two functions of t.
Energy Signal
1
0.5
a
f
0
b
-0.5
-1
-1 -0.5 0 0.5 1
t
Figure E-44
xt recttsin2t
(b) Write a functional description of the frequency-domain signal
as the sum of two functions of f.
X f sinc f j / 2 f 1 f 1
X f j / 2 sinc f 1 j / 2sinc f 1
(c) Find the numerical values of a and b.
a is at the first positive-frequency null of sinc f 1 which occurs at
f 2 . Therefore a = 2.

Solutions 2-33


The phase values are all either / 2 or / 2 because of the
j / 2 factor. Therefore b / 2.
45. A function gt 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 , t 5
10 2t , 5 t 2
gt 
sin t / 2 , 2 t 2
6et/2
, t 2
(b) Graph gt in the range 10 t 10.
g(t)
t-10 10
-8
(c) Graph g2t in the range 10 t 10.
g(2t)
t-10 10
-8
(d) Graph 2g 3 t in the range 10 t 10.
2g(3- t)
t-10 10
-16

Solutions 2-34
16



(e) Graph 2gt 1/ 2 in the range 10 t 10.
-2g(( t+1)/2)
t-10 10
46. A signal occurring in a television set is illustrated in Figure E46. Write a
mathematical description of it.
x(t)
Signal in Television
5
-10 60 t (s)
-10
Figure E46 Signal occurring in a television set
t 2.5 106 
x t 10rect
5 106
47. The signal illustrated in Figure E47 is part of a binary-phase-shift-keyed (BPSK)
binary data transmission. Write a mathematical description of it.
x(t)
BPSK Signal
1
t (ms)4
-1
Figure E47 BPSK signal
t 0.5 10 t 1.5 10 
sin 8000t rect sin 8000t rect
xt  3
103 
3
103
3

3
sin8000t rect
t 2.5 10
sin8000t rect
t 3.5 10
103
103

Solutions 2-35
15
4
48. The signal illustrated in Figure E48 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.
x(t)
RC Filter Signal
4
20
-1.3333
-4
-6
Figure E48 Transient response of an RC filter
xt 4 21 et 4/3
ut 4 (times in ns)
49. Describe the signal in Figure E49 as a ramp function minus a summation of step
functions.
x(t)
Figure E49

...
t
xt 3.75 rampt 15ut 4n
n1
50. Mathematically describe the signal in Figure E-50 .
...
9
Semicircle
9
...
Figure E-50
The semicircle centered at t 0 is the top half of a circle defined by
x2
t t2
81
Therefore

Solutions 2-36
1
5
1
5
1
5
1
5

18
xt 81 t2
, 9 t 9 .
This one period of this periodic function. The other periods are just shifted
versions.
xt rect
t 18n

81 t 18n
2

n
(The rectangle function avoids the problem of imaginary values for the square
roots of negative numbers.)
51. Let two signals be defined by
1 , cos 2t 0
x1 t 

0 , cos2t 0 and x2 t sin2t /10 ,
Graph these products over the time range, 5 t 5.
(a) x1 2t x2 t (b) x1 t / 5x2 20t
(c) x1 t / 5x2 20t 1 (d) x1 t 2 / 5 x2 20t
x (t)x (-t) x (t)x (t)1 2 1 2
t t-5 -5
-1
x (t)x (t)
-1
x (t)x (t)
1 2
1 2
-5
t -5
t
-1 -1
52. Given the graphical definitions of functions in Figure E-52, graph the indicated
transformations.
(a)

Solutions 2-37
g(t)
2
1
1
t-2 2 3 4 5 6
-2
gt g2t
gt 3gt
gt 0 , t 6 or t 2
(b)
x(t)
3
2
1
-4 -3 -2 -1
-1
-2
-3
1 2 3 4
t t 4
gt 2gt 1/ 2
gt is periodic with fundamental period, 4
Figure E-52
(a)
The transformation gt g2t simply compresses the time scale by a factor of
2. The transformation gt 3gt time inverts the signal, amplitude inverts the
signal and then multiplies the amplitude by 3.
g(2t)
2
1
t-2 2 4 6
-2
-3g(-t)
(b)
6
3
-4 -2
-6
2g(t + 4)
t2 4 6
2
1
-2 1
-2
t2 3 4 5 6 7
-2g( t -1 )2
4
2
-2 1
-4
t2 3 4 5 6 7 8

Solutions 2-38
53. For each pair of functions graphed in Figure E-53 determine what transformation
has been done and write a correct functional expression for the transformed
function.
(a)
g(t)
2
-2 -1
t1 2 3 4 5 6
2
-4 -3 -2 -1-1
t
1 2 3 4
(b)
g(t)
2
-2 1
t2 3 4 5 6 -2
-1
t1 2 3 4 5 6
Figure E-53
In (b), assuming gt 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
gt g2 t
(b) gt 1/ 2gt 1 or gt 1/ 2gt 1
54. Graph the magnitude and phase of each function versus f.
(a) G f
jf
1 jf /10

Solutions 2-39
 
(b) G f rect
f 1000
rect
f 1000
e j f /500
100 100
(c) G f
1
250 f 2
j3f
(b) (c)
|G( f )|
1
f-20 20
Phase of G( f )

f-20 20
-
Generalized Derivative
55. Graph the generalized derivative of gt 3sin t / 2 rectt .
Except at the discontinuities at t 1/ 2, the derivative is either zero, for t 1/ 2,
or it is the derivative of 3sin t / 2, 3 / 2cos t / 2, for t 1/ 2. At the

Solutions 2-40



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 .
d (g(t))
dt
3
2
t
3
2
Alternate solution:
gt 3sin t / 2 ut 1/ 2 ut 1/ 2
d
gt
 3sint / 2 t 1/ 2 t 1/ 2 
dt 3 / 2cost / 2 ut 1/ 2 ut 1/ 2
d
gt
 3sin / 4 t 1/ 2 3sin / 4 t 1/ 2

dt 3 / 2cost / 2rectt 
d
gt 
dt
3 2 / 2 t 1/ 2 t 1/ 2 
3
/ 2cost / 2 rectt
56. Find the generalized derivative of the function described by xt
4 , t 3
.
7t , t 3
xt 17 t 3
0 , t 3
17 t 3 7ut 3
7 , t 3

Derivatives and Integrals of Functions
57. What is the numerical value of each of the following integrals?
(a)

t cos 48tdt cos0 1 ,
(b)

t 5 cos tdt cos5 1

Solutions 2-41
1
4
1
1
6
4
6
1
4
(c)
20
t 8rectt /16 dt rect8 /16 1/ 2
0
(d)
22
8e4t
t 2 dt 8e8
23,848
8
(e)
82
3sin200t t 7 dt 0
11
10
(f) 39t2

2
t 1dt 3912
52
92
4173
58. Graph the time derivatives of these functions.
cos2t , t 0
(a) gt sin2tsgnt gt 2 

cos2t , t 0
(b) gt cos2t gt 2
sin2t , cos2t 0
sin2t , cos2t 0
(a) (b)
x(t) x(t)
t
-4
-1
dx/dt
t
-4
-6
-1
t
-1
dx/dt
-1
t
-6
Even and Odd Functions
59. Graph the even and odd parts of these signals.
(a) xt rectt 1
rectt 1rectt 1
rectt 1rectt 1
xe t 
2
, xo t 
2

xo(t)xe(t)xo(t)xe(t)
Exercise 4.1.4 (a) - Even and Odd Parts of rect(t-1)
1
0
-1
-3 -2 -1 0 1 2 3
1
0
-1
-3 -2 -1 0 1 2 3
t
(b) xt 2sin 4t / 4 rectt
xe t 2sin / 4cos 4t rectt ,xo t 2cos / 4sin 4t rectt
Exercise 4.1.4 (d) - Even and Odd Parts of 2sin(4*pi*t-pi/4)*rect(t)
2
0
-2
-1 -0.5 0 0.5 1
2
0
-2
-1 -0.5 0 0.5 1
t
60. 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, xt 0. Applying the definitions of even and odd functions,

xe t 0 0
2
0 xt and xo t  0 0
2
0 xt
proving that the signal is equal to both its even and odd parts and is therefore both
even and odd.
61. Find and graph the even and odd parts of the function xt in Figure E-61
x(t)
2
1
-5 -4 -3 -2 -1
-1
1 2 3 4 5
t
xe(t) xo(t)
2
1
-5 -4 -3 -2 -1
-1
1 2 3 4 5
t
2
1
-5 -4 -3 -2 -1
-1
1 2 3 4 5
t
Figure E-61
Periodic Functions
62. For each of the following signals decide whether it is periodic and, if it is, find the
fundamental period.
(a) gt 28sin 400t Periodic. Fundamental frequency = 200 Hz, Period = 5 ms. (b)
gt 14 40cos60t Periodic. Fundamental frequency = 30 Hz
Period = 33.33...ms.
(c) gt 5t 2cos5000t Not periodic.
(d) gt 28sin 400t 12cos500t 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) gt 10sin5t 4cos 7t Periodic. The Periods of the two sinusoids are 2 / 5 s
and 2 / 7 s. Least common multiple is 2 . Period of the overall signal is 2 s.
(f) gt 4sin 3t 3sin 3t Not periodic because least common multiple is infinite.
63. Is a constant a periodic signal? Explain why it is or is not periodic and, if it is
periodic what is its fundamental period?

2

2
4
4


 a
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
64. Find the signal energy of each of these signals.

(a) 2rectt , E 


2rectt
2
1/2
dt 4 
1/2
1/16
dt 4
1
(b) rect8t , E 

rect8t

dt 
2

1/16
dt 
8
2
(c) 3rect
t 
, E

3rect
t 
dt 9 dt 36 
2
(d) 2sin200t

2 1 1
E 
2sin200t dt 4 sin2
200t dt 4




sin 400t dt
2
E 2t
cos400t
400

(e) t (Hint: First find the signal energy of a signal which approaches an
impulse some limit, then take the limit.)
t lim1/ a rectt / aa0
2
E lim
1
rect
t 
dt lim
1
a/2
rect
t
dt lim
a
a0 a a 
(f) xt
d
rectt
dt
a0 a2 
a/2
a0 a2
d
rectt t 1/ 2 t 1/ 2
dt

Ex t 1/ 2 t 1/ 2

2
dt





2

t
(g) xt rect d rampt 1/ 2 rampt 1/ 2

Ex 
1/2
t 1/ 2
2

dt dt
11/2
442443
finite
(h) xt e1j8 t
ut
1
{/2
infinite

2 1j8 t
2

1 j8 t 1 j8 t
Ex 
xt dt e

ut dt e e dt
0

Ex e 2t e2t
1
dt
2 20 0
(i) xt 2rectt / 4 3rect
t 1
4
2 t 1
Ex 

xt dt 

2rectt / 4 3rect dt
4
The function can be expressed as
2 , 2 t 1 xt 1 ,
1 t 2
3 , 2 t 3
1 2 3
x

E 4 dt dt 9 dt 4 3 9 16
Alternate Solution:
2 1 2
Since the function is real, the square of its magnitude equals its square.

Ex
2rectt / 4 3rect
2
t 1 
dt
4 
Ex
 
2rectt / 4
2
dt
 
12rectt / 4rect
t 1 
dt 

3rect
2
t 1 
dt


4 4

 
4 4
2
2


E 4 rectt / 4 dt 12 rectt / 4rect
t 1
dt 9 rect
t 1
dtx 
2 2 3
x

E 4 dt 12 dt 9 dt 16 36 36 16
2 1 1
(j) xt 3rect
t 1
tri
t 4
6 6
From a sketch it is easily observed that xt
t 2
2
, 2 t 4.
2
4
t 2
2 4
Ex 

xt dt 
2
dt 1/ 4 t
2
4t 4dt
t3
Ex 1/ 4
3
2t2
4
4t 1/ 4 64
3
32 16 
8
8 8
3
 18

(k) xt 5e4t
ut

2
2

8t e8t
Ex
 xt dt 25 e dt 25
8
 25 / 8 or 3.125
0 0
(l) A signal xt has the following description:
1. It is zero for all time t 4.
2. It is a straight line from the point t 4,x 0 to the point
t 4,x 4 .
3. It is a straight line from the point t 4,x 4
t 3,x 0.
4. It is zero for all time t 3.
to the point
The only non-zero values of xt lie on a straight line between t 4 and t 3.
The signal energy of this signal is therefore the same as this signal shifted
to the left so that it starts its non-zero values at t 7 and goes to zero at t 0
. Such a signal would be described in its non-zero range by
xt 4 / 7t , -7 t 0

1

1
Its signal energy is
0
Ex 4 / 7t
0
2
dt 16 / 49 t3 / 3dt 1/ 316 / 49 343 37.333
7 7
65. An even continuous-time energy signal xt is described in positive time by
xt
3ut 5ut 4 11ut 7 , 0 t 10
.
0 , t 10
Another continuous-time energy signal yt is described by yt 3x2t 2. (a)
Find the signal energy Ex
of xt .
(b) Find the signal energy Ey
of yt .

Ey 
yt
2

dt 
3x 2t 2
2

dt 9 x2

2t 2 dt
Let 2t 2 d 2dt.
66. Find the average signal power of each of these signals:
(a) xt 2sin200t This is a periodic function. Therefore
1
T /2
2 4
T /2
Px
T T /2
2sin200t dt 
T T /2
cos 400t dt
2 2
2 

sin400t
T /2

2 T

sin200T T sin200T 
2Px
T
t
 400 T /2

T 2

400 2
400
For any sinusoid, the average signal power is half the square of the amplitude.

0
0
x
x[t]
P 
2
4
(b) xt 1 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) xt ej100 t
This is a periodic function. Therefore
1
x
T T
xt
2
dt  1
T0 /2
 ej100 t 2
dt 50
1/100
 ej100t
e j100 t
dt
0
Px 50
0
1/100
1/100
dt 1
T0 T /2 1/100
(d) A periodic continuous-time signal with fundamental period 12 described
over one fundamental period by
xt 3rect
t 3
4 rectt / 2 1 , 6 t 6
4
One Fundamental Period
5
0
-5
-10 -5 0 5 10
t
Px
1/T0
xt
 6
dt 1/12 3rect
t 3 2
4 rectt / 2 1 dt
T 6
1 1 1 
36 32 48
Px 1/12 9 dt 16 dt 24 dt  5 / 3 1.667
5 3 3
12
(e) xt 3sgn2t 4 xt
2
9 P 9
67. A signal x is periodic with fundamental period T0 6 . This signal is described
over the time period 0 t 6 by


rectt 2 / 3 4 rectt 4 / 2 .
What is the signal power of this signal?
The signal x can be described in the time period 0 t 6 by
0 , 0 t 1/ 2
1 , 1/ 2 t 3
xt 3 , 3 t 7 / 2
4 , 7 / 2 t 5

0 , 5 t 6
The signal power is the signal energy in one fundamental period divided by the
fundamental period.
2.5 4.5 24
P
1
02 1
12 5
3
2 1
4
2 3
02
1

5.167
6 2 2 2 2 6
Signals and Systems Analysis Using Transform Methods and MATLAB
3rd Edition Roberts SOLUTIONS MANUAL
Full download at:
http://testbanklive.com/download/signals-and-systems-analysis-using-
transform-methods-and-matlab-3rd-edition-roberts-solutions-manual/
Signals and Systems
J. Roberts
M J Roberts 2003 Solutions Manual
Analysis Using Transform
Analysis Using Transform Methods & MATLAB
Signals and Systems Analysis Using Transform
Signals and Systems 2nd Edition Solutions Manual
Solutions Manual

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