roberts chap 2 solutions
TRANSCRIPT
M. J. Roberts - 10/7/06
Solutions 2-1
Chapter 2 - Mathematical Description ofContinuous-Time Signals
SolutionsCT Functions
1. If g t( ) = 7e
2t 3 write out and simplify
(a) g 3( ) = 7e
9
(b) g 2 t( ) = 7e
2 2 t( ) 3= 7e
7+2t
(c) g t / 10 + 4( ) = 7e
t /5 11
(d) g jt( ) = 7e
j2t 3
(e)
g jt( ) + g jt( )2
= 7e3 e
j2t+ e
j2t
2= 7e
3 cos 2t( )
(f)
gjt 3
2+ g
jt 3
2
2= 7
ejt
+ ejt
2= 7cos t( )
2. If g x( ) = x
2 4x + 4 write out and simplify
(a) g z( ) = z
2 4z + 4
(b) g u + v( ) = u + v( )
2
4 u + v( ) + 4 = u2
+ v2
+ 2uv 4u 4v + 4
(c) g e
jt( ) = ejt( )
2
4ejt
+ 4 = ej2t 4e
jt+ 4 = e
jt 2( )2
(d) g g t( )( ) = g t
2 4t + 4( ) = t2 4t + 4( )
2
4 t2 4t + 4( ) + 4
g g t( )( ) = t
4 8t3+ 20t
2 16t + 4
(e) g 2( ) = 4 8 + 4 = 0
3. What would be the value of g in each of the following MATLABinstructions?
(a) t = 3 ; g = sin(t) ; 0.1411
(b) x = 1:5 ; g = cos(pi*x) ; [-1,1,-1,1,-1]
M. J. Roberts - 10/7/06
Solutions 2-2
(c) f = -1:0.5:1 ; w = 2*pi*f ; g = 1./(1+j*w’) ;
0.0247 + j0.155
0.0920 + j0.289
1
0.0920 j0.289
0.0247 j0.155
4. Let two functions be defined by
x1
t( ) =1 , sin 20 t( ) 0
1 , sin 20 t( ) < 0 and
x2
t( ) =t , sin 2 t( ) 0
t , sin 2 t( ) < 0 .
Graph the product of these two functions versus time over the time range, 2 < t < 2 .
t-2 2
x(t)
-2
2
Transformations of CT Functions
5. For each function, g t( ) , graph
g t( ) ,
g t( ) ,
g t 1( ) , and
g 2t( ) .
(a) (b)
t
g(t)
2
4
t
g(t)
1-1
3
-3
t
g(-t)
-2
4
t
g(-t)
1-1
3
-3
t
-g(t)
2
4
t
-g(t)
1
-1
3
-3
M. J. Roberts - 10/7/06
Solutions 2-3
t
g(t-1)
31
4
t
g(t-1)
1 2
3
-3
t
g(2t)
1
4
t
g(2t)
1
3
-32
12
-
6. Find the values of the following signals at the indicated times.
(a) x t( ) = 4 tri t( ) , x 1 / 2( ) = 4 tri 1 / 2( ) = 4 1 1 / 2( ) = 2
(b) x t( ) = 2rect t / 4( ) , x 1( ) = 2rect 1 / 4( ) = 2
(c) x t( ) = 10sinc t( ) , x 3 / 2( ) = 10
sin 3 / 2( )3 / 2
=2
3= 0.21221
(d) x t( ) = 5rect t / 2( )sgn 2t( ) , x 1( ) = 5rect 1 / 2( )sgn 2( ) = 2.5
(e) x t( ) = 2 tri 2 t 1( ) + 6rect t / 4( ) ,
x 3 / 2( ) = 2 tri 2 3 / 2 1( ) + 6rect 3 / 8( ) = 2 tri 1( ) + 6 = 6
(f) x t( ) = 9rect t / 10( )sgn 3 t 2( )( ) , x 1( ) = 9rect 1 / 10( )sgn 3( ) = 9
(g) x t( ) = 10sinc t + 2( ) / 4( ) , x -6( ) = 10sinc 1( ) = 0
7. For each pair of functions in Figure E-7 provide the values of the constants, A,
t0 and a in the functional transformation
g
2t( ) = Ag
1t t
0( ) / w( ) .
M. J. Roberts - 10/7/06
Solutions 2-4
-4 -2 0 2 4
-2-1012
t
g 1 (t)
(a)
-4 -2 0 2 4
-2-1012
t
g 2 (t)
(a)
-4 -2 0 2 4
-2-1012
t
g 1 (t)
(b)
-4 -2 0 2 4
-2-1012
t
g 2 (t)
(b)
-4 -2 0 2 4
-2-1012
t
g 1 (t)
(c)
-4 -2 0 2 4
-2-1012
tg 2 (
t)
(c)
Figure E-7
Answers: (a) A = 2,t
0= 1,w = 1 , (b)
A = 2,t
0= 0,w = 1 / 2 ,
(c) A = 1 / 2,t
0= 1,w = 2
8. For each pair of functions in Figure E-8 provide the values of the constants, A,
t0 and a in the functional transformation
g
2t( ) = Ag
1w t t
0( )( ) .
(a)
-10 -5 0 5 10-8
-4
0
4
8
t
g 1(t)
-10 -5 0 5 10-8
-4
0
4
8
A = 2, t0 = 2, w = -2
t
g 2(t)
(b) g 1(t)
A = 3, t0 = 2, w = 2
g 2(t)
-10 -5 0 5 10-8
-4
0
4
8
t-10 -5 0 5 10-8
-4
0
4
8
t
M. J. Roberts - 10/7/06
Solutions 2-5
(c) g 1(t)
A = 3, t0 = 3, w = 1/3
g 2(t)
-10 -5 0 5 10-8
-4
0
4
8
t-10 -5 0 5 10-8
-4
0
4
8
t
(d) g 1(t)
A = 2, t0 = 2, w = 1/3
g 2(t)
-10 -5 0 5 10-8
-4
0
4
8
t-10 -5 0 5 10-8
-4
0
4
8
t
(e) g 1(t)
A = 3, t0 = 2, w = 1/2
g 2(t)
-10 -5 0 5 10-8
-4
0
4
8
t-10 -5 0 5 10-8
-4
0
4
8
tFigure E-8
9. In Figure E-9 is plotted a CT function, g
1t( ) which is zero for all time outside the
range plotted. Let some other functions be defined by
g
2t( ) = 3g
12 t( ) ,
g
3t( ) = 2g
1t / 4( ) ,
g4
t( ) = g1
t 3
2
Find these values.
(a) g
21( ) = 3 (b)
g
31( ) = 3.5
(c)
g4
t( )g3
t( )t =2
=3
21( ) =
3
2(d)
g4
t( )dt
3
1
The function g
4t( ) 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.
M. J. Roberts - 10/7/06
Solutions 2-6
g4
t( )dt
3
1
= 2
t
g (t)
1-1-2-3-4 2 3 4
1
234
-4
-3-2-1
1
Figure E-9
10. A function, G f( ) , is defined by
G f( ) = e
j2 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 = 0of
width, 2, multiplied by a complex exponential. Therefore for frequencies greater thanone in magnitude it is zero. Its magnitude is simply the magnitude of the rectanglefunction because the magnitude of the complex exponential is one for any f.
e
j2 f= cos 2 f( ) + j sin 2 f( ) = cos 2 f( ) j sin 2 f( )
e
j2 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 therectangle function is zero between f = 1 and f = 1and undefined outside thatrange and the phase of a product is the sum of the phases. The phase of the complexexponential is
ej2 f
= cos 2 f( ) j sin 2 f( )( ) = tan 1sin 2 f( )cos 2 f( )
= tan 1sin 2 f( )cos 2 f( )
e
j2 f= tan 1 tan 2 f( )( )
The inverse tangent function is multiple-valued. Therefore there are multiple correctanswers for this phase. The simplest of them is found by choosing
ej2 f
= 2 f
M. J. Roberts - 10/7/06
Solutions 2-7
which is simply the coefficient of j in the original complex exponential expression.
A more general solution would be e
j2 f= 2 f + 2n , n an integer . The
solution of the original problem is simply this solution except shifted up and downby 10 in f and added.
G f 10( ) + G f + 10( ) = ej2 f 10( )
rectf 10
2+ e
j2 f +10( )rect
f + 10
2
f -20 20
|G( f )|
1
f -20 20
Phase of G( f )
-π
π
11. Write an expression consisting of a summation of unit step functions to represent asignal which consists of rectangular pulses of width 6 ms and height 3 which occurat a uniform rate of 100 pulses per second with the leading edge of the first pulseoccurring at time t = 0 .
x t( ) = 3 u t 0.01n( ) u t 0.01n 0.006( )n=0
12. Write an expression consisting of a summation of triangle functions to represent aperiodic triangular wave whose maximum value is 5 and whose minimum value is 2with a period of 20 µs and a value of 5 at time t = 0 .
x t( ) = 3.5 + 1.5 tri 2 105t 2 10 5
n( )( ) tri 2 105t 10 5 2n 1( )( )( )
n=
Derivatives and Integrals of CT Functions
13. Graph the derivatives of these functions.
(All graphes at end.)
(a) g t( ) = sinc t( )
M. J. Roberts - 10/7/06
Solutions 2-8
g t( ) =
2t cos t( ) sin t( )
t( )2
=t cos t( ) sin t( )
t2
(b) g t( ) = 1 e
t( )u t( ) This function is constant zero for all time before time,
t = 0 , therefore its derivative during that time is zero. This function is aconstant minus a decaying exponential after time, t = 0 , and its derivative inthat time is therefore also a positive decaying exponential.
g t( ) =e
t , t > 0
0 , t < 0
Strictly speaking, its derivative is not defined at exactly t = 0 . Since the value of aphysical signal at a single point has no impact on any physical system (as long as itis finite) we can choose any finite value at time, t = 0 , without changing the effect ofthis signal on any physical system. If we choose 1/2, then we can write the derivativeas
g t( ) = e
t u t( ) .
t-4 4
x(t)
-1
1
t-4 4
dx/dt
-1
1
t-1 4
x(t)
-1
1
t-1 4
dx/dt
-1
1
(a) (b)
14. (a) If g t( ) = tri t / 2( ) what is the value of the first derivative of
g t 1( ) ,
d
dtg t 1( )( ) , at t = 2 ?
d
dtg t( )( ) =
0 , t < 2
1 / 2 , 2 < t < 0
1 / 2 , 0 < t < 2
0 , t > 2
d
dtg t 1( )( ) =
0 , t < 1
1 / 2 , 1 < t < 1
1 / 2 , 1 < t < 3
0 , t > 3
M. J. Roberts - 10/7/06
Solutions 2-9
At t = 2 ,
d
dtg t 1( )( ) = 1 / 2 .
(b) If g t( ) = sinc 2 t + 1( )( ) what is the value of
10g t / 10( ) at t = 4 ?
10g t / 10( ) = 10sinc 2 t / 10 + 1( )( )
At t = 4 , 10g t / 10( ) = 10sinc 2 4 / 10 + 1( )( ) = 10sinc 2.8( ) = 10
sin 2.8( )2.8
= 0.0668
15. Find the numerical value of each integral.
(a)
t + 3( ) 2 4t( ) dt
1
8
= t + 3( )dt
1
8
2 4t( )dt
1
8
= 0 2 1 / 4 t( )dt
1
8
= 1 / 2
(b)
2
3t( )dt
1/ 2
5/ 2
= 3t 2n( )n=
dt
1/ 2
5/ 2
=1
3t 2n / 3( )
n=
dt
1/ 2
5/ 2
=1
31+ 1+ 1 = 1
16. Graph the integral from negative infinity to time, t, of the functions in Figure E-16which are zero for all time t < 0 .
This is the integral,
g( )dt
, which, in geometrical terms, is the accumulated area
under the function, g t( ) , from time, to time, t. For the case of the two back-to-
back rectangular pulses, there is no accumulated area until after time, t = 0 , and thenin the time interval, 0 < t < 1 , the area accumulates linearly with time up to amaximum area of one at time, t = 1 . In the second time interval, 1 < t < 2 , the area islinearly 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 ofarea for t > 2.
In the second case of the triangular-shaped function, the area does not accumulatelinearly, 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 stopscompletely. The final value of the accumulated area must be the total area of thetriangle, which, in this case, is one.
g(t)
t
1
1 2 312
g(t)
t
1
1 2 3
Figure E-16
M. J. Roberts - 10/7/06
Solutions 2-10
∫ g(t) dt∫ g(t) dt
t
1
1 2 3
12
t
1
1 2 3
Even and Odd CT Functions
17. An even function, g(t), is described over the time range, 0 < t < 10 , by
g t( ) =
2t , 0 < t < 3
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
dtg t( ) =
d
dtg t( )
d
dtg t( )
t = 6
=d
dtg t( )
t =6
= 3( ) = 3 .
18. Find the even and odd parts of these functions.
(a) g t( ) = 2t
2 3t + 6
g
et( ) =
2t2 3t + 6 + 2 t( )
2
3 t( ) + 6
2=
4t2
+ 12
2= 2t
2+ 6
g
ot( ) =
2t2 3t + 6 2 t( )
2
+ 3 t( ) 6
2=
6t
2= 3t
(b) g t( ) = 20cos 40 t / 4( )
g
et( ) =
20cos 40 t / 4( ) + 20cos 40 t / 4( )2
Using cos z
1+ z
2( ) = cos z1( )cos z
2( ) sin z1( )sin z
2( ) ,
M. J. Roberts - 10/7/06
Solutions 2-11
g
et( ) =
20 cos 40 t( )cos / 4( ) sin 40 t( )sin / 4( )
+20 cos 40 t( )cos / 4( ) sin 40 t( )sin / 4( )2
g
et( ) =
20 cos 40 t( )cos / 4( ) + sin 40 t( )sin / 4( )
+20 cos 40 t( )cos / 4( ) sin 40 t( )sin / 4( )2
g
et( ) = 20cos / 4( )cos 40 t( ) = 20 / 2( )cos 40 t( )
g
ot( ) =
20cos 40 t / 4( ) 20cos 40 t / 4( )2
Using cos z
1+ z
2( ) = cos z1( )cos z
2( ) sin z1( )sin z
2( ) ,
g
ot( ) =
20 cos 40 t( )cos / 4( ) sin 40 t( )sin / 4( )
20 cos 40 t( )cos / 4( ) sin 40 t( )sin / 4( )2
g
ot( ) =
20 cos 40 t( )cos / 4( ) + sin 40 t( )sin / 4( )
20 cos 40 t( )cos / 4( ) sin 40 t( )sin / 4( )2
g
ot( ) = 20sin / 4( )sin 40 t( ) = 20 / 2( )sin 40 t( )
(c) g t( ) =
2t2 3t + 6
1+ t
g
et( ) =
2t2 3t + 6
1+ t+
2t2
+ 3t + 6
1 t
2
g
et( ) =
2t2 3t + 6( ) 1 t( ) + 2t
2+ 3t + 6( ) 1+ t( )
1+ t( ) 1 t( )2
M. J. Roberts - 10/7/06
Solutions 2-12
ge
t( ) =4t
2+ 12 + 6t
2
2 1 t2( )
=6 + 5t
2
1 t2
g
ot( ) =
2t2 3t + 6
1+ t
2t2
+ 3t + 6
1 t
2
g
ot( ) =
2t2 3t + 6( ) 1 t( ) 2t
2+ 3t + 6( ) 1+ t( )
1+ t( ) 1 t( )2
go
t( ) =6t 4t
3 12t
2 1 t2( )
= t2t
2+ 9
1 t2
(d) g t( ) = sinc t( )
g
et( ) =
sin t( ) / t + sin t( ) / t( )2
=sin t( )
t
g
ot( ) = 0
(e) g t( ) = t 2 t
2( ) 1+ 4t2( )
g t( ) = t
odd
2 t2( )
even
1+ 4t2( )
even
Therefore g t( ) is odd,
g
et( ) = 0 and g
ot( ) = t 2 t
2( ) 1+ 4t2( )
(f) g t( ) = t 2 t( ) 1+ 4t( )
g
et( ) =
t 2 t( ) 1+ 4t( ) + t( ) 2 + t( ) 1 4t( )2
ge
t( ) = 7t2
g
ot( ) =
t 2 t( ) 1+ 4t( ) t( ) 2 + t( ) 1 4t( )2
go
t( ) = t 2 4t2( )
19. Graph the even and odd parts of the functions in Figure E-19.
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
M. J. Roberts - 10/7/06
Solutions 2-13
two. Then, to graph the odd part, subtract g t( ) from
g t( ) (graphically) and divide
the difference by two.
t
g(t)
1
1
t
g(t)
21
1
-1
Figure E-19
t
g (t)
1
1
t
g (t)
1
1
e
o
,
t
g (t)
21
1
-1
t
g (t)
21
1
-1
e
o
(a) (b)
20. Graph the indicated product or quotient g t( ) of the functions in Figure E-20.
t1
-1
1
-1
t1-1
1g(t)
Multiplication
t1
-1
1
-1
t1-1
-1
1
g(t)
g(t)g(t)
Multiplication
(a) (b)
t1
-1
1
-1
t1-1
1
-1
M. J. Roberts - 10/7/06
Solutions 2-14
t1
1
g(t)
Multiplication
(c)
t-1
1
t1
1
g(t)
g(t)
Multiplication
(d)
t1
1
t-1
-1 1
g(t)
t-1 1
1
(e) (f)
t1-1
1
-1
t1-1
1g(t)
Multiplication
...... t1
1
-1
t1
-1
1
g(t)
Multiplication
g(t)
t1-1
1
-1
......
g(t)
t1
1
-1
t1
1
g(t)
Division Division
(g)
t-1 -1 -1 1
1 1
g(t)
g(t)
(h)
t
1t
π
-1 1
1
t
g(t)
t
-1
Figure E-20
21. Use the properties of integrals of even and odd functions to evaluate these integrals inthe quickest way.
M. J. Roberts - 10/7/06
Solutions 2-15
(a)
2 + t( )dt
1
1
= 2even
dt
1
1
+ t
odd
dt
1
1
= 2 2dt
0
1
= 4
(b)
4cos 10 t( ) + 8sin 5 t( ) dt
1/ 20
1/ 20
= 4cos 10 t( )even
dt
1/ 20
1/ 20
+ 8sin 5 t( )odd
dt
1/ 20
1/ 20
4cos 10 t( ) + 8sin 5 t( ) dt
1/ 20
1/ 20
= 8 cos 10 t( )dt
0
1/ 20
=8
10
(c)
4 t
odd
cos 10 t( )even
odd
dt
1/ 20
1/ 20
= 0
(d)
t
odd
sin 10 t( )odd
even
dt
1/10
1/10
= 2 t sin 10 t( )dt
0
1/10
= 2 t
cos 10 t( )10
0
1/10
+cos 10 t( )
10dt
0
1/10
t
odd
sin 10 t( )odd
even
dt
1/10
1/10
== 21
100+
sin 10 t( )
10( )2
0
1/10
=1
50
(e)
et
even
dt
1
1
= 2 et
dt
0
1
= 2 etdt
0
1
= 2 et
0
1
= 2 1 e1( ) 1.264
(f)
t
odd
et
even
odd
dt
1
1
= 0
Periodic CT Functions
22. Find the fundamental period and fundamental frequency of each of these functions.
(a) g t( ) = 10cos 50 t( )
f0
= 25 Hz , T0
= 1 / 25 s
(b) g t( ) = 10cos 50 t + / 4( )
f0
= 25 Hz , T0
= 1 / 25 s
(c) g t( ) = cos 50 t( ) + sin 15 t( )
The fundamental period of the sum of two periodic signals is the least commonmultiple (LCM) of their two individual fundamental periods. The fundamentalfrequency of the sum of two periodic signals is the greatest common divisor (GCD)of their two individual fundamental frequencies.
M. J. Roberts - 10/7/06
Solutions 2-16
f
0= GCD 25,15 / 2( ) = 2.5 Hz , T
0= 1 / 2.5 = 0.4 s
(d) g t( ) = cos 2 t( ) + sin 3 t( ) + cos 5 t 3 / 4( )
f
0= GCD 1,3 / 2,5 / 2( ) = 1 / 2 Hz , T
0=
1
1 / 2= 2 s
23. One period of a periodic signal, x t( ) , with period,
T
0, is graphed in Figure E-23.
Assuming x t( ) has a period
T
0, what is the value of
x t( ) at time, t = 220ms ?
5ms 10ms 15ms 20mst
x(t)4321
-1-2-3-4
T0
Figure E-23
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 .
24. In Figure E-24 find the fundamental period and fundamental frequency of g t( ) .
g(t)
t......
1
t......
1
g(t)
t......
1
+(a) (b)
t......
1
g(t)
t......
1
+(c)
Figure E-24
(a) f
0= 3 Hz and T
0= 1 / 3 s
(b) f
0= GCD 6,4( ) = 2 Hz and T
0= 1 / 2 s
(c) f
0= GCD 6,5( ) = 1 Hz and T
0= 1 s
M. J. Roberts - 10/7/06
Solutions 2-17
Signal Energy and Power of CT Signals
25. Find the signal energy of these signals.
(a) x t( ) = 2rect t( )
Ex
= 2rect t( )2
dt = 4 dt
1/ 2
1/ 2
= 4
(b) x t( ) = A u t( ) u t 10( )( )
Ex
= A u t( ) u t 10( )( )2
dt = A2
dt
0
10
= 10A2
(c) x t( ) = u t( ) u 10 t( )
Ex
= u t( ) u 10 t( )2
dt = dt
0
+ dt
10
(d) x t( ) = rect t( )cos 2 t( )
Ex
= rect t( )cos 2 t( )2
dt = cos2 2 t( )dt
1/ 2
1/ 2
=1
21+ cos 4 t( )( )dt
1/ 2
1/ 2
Ex
=1
2dt
1/ 2
1/ 2
+ cos 4 t( )dt
1/ 2
1/ 2
=0
=1
2
(e) x t( ) = rect t( )cos 4 t( )
Ex
= rect t( )cos 4 t( )2
dt = cos2 4 t( )dt
1/ 2
1/ 2
=1
21+ cos 8 t( )( )dt
1/ 2
1/ 2
Ex
=1
2dt
1/ 2
1/ 2
+ cos 8 t( )dt
1/ 2
1/ 2
=0
=1
2
(f) x t( ) = rect t( )sin 2 t( )
Ex
= rect t( )sin 2 t( )2
dt = sin2 2 t( )dt
1/ 2
1/ 2
=1
21 cos 4 t( )( )dt
1/ 2
1/ 2
M. J. Roberts - 10/7/06
Solutions 2-18
Ex
=1
2dt
1/ 2
1/ 2
cos 4 t( )dt
1/ 2
1/ 2
=0
=1
2
26. A CT 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,
Ex
= Arect t( ) + B rect t 0.5( )( )2
dt
Ex
= A2 rect2
t( ) + B2 rect2
t 0.5( ) + 2AB rect t( )rect t 0.5( )( )dt
Ex
= A2
dt
1/ 2
1/ 2
+ B2
dt
0
1
+ 2AB dt
0
1/ 2
= A2
+ B2
+ AB
27. Find the average signal power of the periodic signal x(t) in Figure E-27.
1-1-2-3-4 2 3 4
1
-1
-2
-3
2
3
t
x(t)
Figure E-27
P =1
T0
x t( )2
dt
t0
t0
+T0
=1
3x t( )
2
dt
1
2
=1
32t
2
dt
1
1
=4
3t
2dt
1
1
=4
3
t3
31
1
=8
9
28. Find the average signal power of these signals.
(a) x t( ) = A
Px
= limT
1
TA
2
dt
T / 2
T / 2
= limT
A2
Tdt
T / 2
T / 2
= limT
A2
TT = A
2
M. J. Roberts - 10/7/06
Solutions 2-19
(b) x t( ) = u t( )
Px
= limT
1
Tu t( )
2
dt
T / 2
T / 2
= limT
1
Tdt
0
T / 2
= limT
1
T
T
2=
1
2
(c) x t( ) = Acos 2 f
0t +( )
Px
=1
T0
Acos 2 f0t +( )
2
dt
T0
/ 2
T0
/ 2
=A
2
T0
cos2 2 f0t +( )dt
T0
/ 2
T0
/ 2
Px
=A
2
2T0
1+ cos 4 f0t + 2( )( )dt
T0
/ 2
T0
/ 2
=A
2
2T0
t +sin 4 f
0t + 2( )
4 f0
T0
/ 2
T0
/ 2
Px
=A
2
2T0
T0
+sin 4 f
0T
0/ 2 + 2( )
4 f0
sin 4 f0T
0/ 2 + 2( )
4 f0
=0
=A
2
2
The average signal power of a periodic power signal is unaffected if it is shifted in
time. Therefore we could have found the average signal power of Acos 2 f
0t( )
instead, which is somewhat easier algebraically.
CT Functions
29. Given the function definitions on the left, find the function values on the right.
(a) g t( ) = 100sin 200 t + / 4( )
g 0.001( ) = 100sin 200 0.001+ / 4( ) = 100sin / 5 + / 4( ) = 98.77
(b) g t( ) = 13 4t + 6t
2
g 2( ) = 13 4 2( ) + 6 2( )
2
= 29
(c) g t( ) = 5e
2te
j2 t
g 1 / 4( ) = 5e
2/ 4e
j2 / 4= 5e
1/ 2e
j / 2= j3.03
30. Let the CT unit impulse function be represented by the limit,
x( ) = lim
a 01 / a( ) tri x/ a( ) , a > 0 .
The function, 1 / a( ) tri x / a( ) has an area of one regardless of the value of a.
M. J. Roberts - 10/7/06
Solutions 2-20
(a) What is the area of the function,
4x( ) = lima 0
1 / a( ) tri 4x / a( ) ?
This is a triangle with the same height as 1 / a( ) tri x/ a( ) but 1/4 times the base
width. Therefore its area is 1/4 times as great or 1/4.
(b) What is the area of the function,
6x( ) = lima 0
1 / a( ) tri 6x / a( ) ?
This is a triangle with the same height as 1 / a( ) tri x/ a( ) but 1/6 times the base
width. (The fact that the factor is “-6” instead of “6” just means that the triangle isreversed in time which does not change its shape or area.) Therefore its area is 1/6times as great or 1/6.
(c) What is the area of the function,
bx( ) = lima 0
1 / a( ) tri bx / a( ) for b positive
and for b negative ?
It is simply 1 / b .
31. Using a change of variable and the definition of the unit impulse, prove that
a t t
0( )( ) = 1 / a( ) t t0( ) .
x( ) = 0 , x 0 ,
x( )d x = 1
a t t
0( ) = 0 , where a t t0( ) 0 or t t
0
Strength = a t t0( ) dt
Let
a t t
0( ) = and adt = d
Then, for a > 0,
Strength = ( )d
a=
1
a( )d =
1
a=
1
a
and for a < 0,
Strength = ( )d
a=
1
a( )d =
1
a( )d =
1
a=
1
a
Therefore for a > 0 and a < 0,
Strength =1
a
and
a t t0( ) =
1
a
t t0( ) .
32. Using the results of Exercise 31, show that
(a)
M. J. Roberts - 10/7/06
Solutions 2-21
1
ax( ) = 1 / a( ) x n / a( )n=
From the definition of the periodic impulse,
1ax( ) = ax n( ) .
Then, using the property from Exercise 31,
1ax( ) = a x n / a( ) =
1
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
1ax( ) =
1
1 / a1
ax( )dx
t0
t0
+1/ a
= a1
ax( )dx
1/ 2a
1/ 2a
= a ax( )dx
1/ 2a
1/ 2a
Letting = ax
1
ax( ) = ( )d
1/ 2
1/ 2
= 1
(d) Even though
at( ) = 1 / a( ) t( ) , 1
ax( ) 1 / a( ) 1x( )
1
ax( ) = ax n( )n=
1 / a( ) x n( )n=
= 1 / a( ) 1x( )
1ax( ) 1 / a( ) 1
x( ) QED
Scaling and Shifting Functions
33. Graph these CT singularity and related functions.
(a) g t( ) = 2u 4 t( ) (b)
g t( ) = u 2t( )
(c) g t( ) = 5sgn t 4( ) (d)
g t( ) = 1+ sgn 4 t( )
(e) g t( ) = 5ramp t + 1( ) (f)
g t( ) = 3ramp 2t( )
(g) g t( ) = 2 t + 3( ) (h)
g t( ) = 6 3t + 9( )
M. J. Roberts - 10/7/06
Solutions 2-22
g(t)
t4
2
4
2
(a)g(t)
t
1
(b)g(t)
t4
5
-5
(c)g(t)
t
(d)
-6
1
g(t)
t-1 1
10
(e)g(t)
t
(f)g(t)
t-3
2
(g) (h)g(t)
t-3
2
(i) g t( ) = 4 2 t 1( )( ) (j)
g t( ) = 2
1t 1 / 2( )
(k) g t( ) = 8
14t( ) (l)
g t( ) = 6
2t + 1( )
(m) g t( ) = 2rect t / 3( ) (n)
g t( ) = 4rect t + 1( ) / 2( )
(o) g t( ) = tri 4t( ) (p)
g t( ) = 6 tri t 1( ) / 2( )
g(t)
t
-2
1
1
-2
1-1 3
(i)g(t)
t
2
(j)g(t)
t
4
(k)g(t)
t
(l)
24
1
2
-6
-6
g(t)
t
(m)g(t)
t
(n)g(t)
t
(o) (p)g(t)
t
...... ......
......
1
41
41
23
23- -
1 3-1
(q) g t( ) = 5sinc t / 2( ) (r)
g t( ) = sinc 2 t + 1( )( )
(s) g t( ) = 10drcl t,4( ) (t)
g t( ) = 5drcl t / 4,7( )
M. J. Roberts - 10/7/06
Solutions 2-23
g(t)
t
5 -1
2
(q)g(t)
t
-1
(r)
t4
g(t)
-10
10
(s)
t8
g(t)
-1
5
(t)
(u) t
-3rect(t-2)
-3
32
52
(v)t
0.1
51 3
0.1rect t-34( )
(w) t-5 -13
-4tri 3+t2( )
-4
(x)
t
4sinc[5(t-3)]
4
-1 1 2 3 4 5 6
(y)t
4sinc(5t-3)
4
-1 1 2 3 4 5 6
34. Graph these CT functions.
(a) g t( ) = u t( ) u t 1( ) (b)
g t( ) = rect t 1 / 2( )
(c) g t( ) = 4ramp t( )u t 2( ) (d)
g t( ) = sgn t( )sin 2 t( )
M. J. Roberts - 10/7/06
Solutions 2-24
(e) g t( ) = 5e
t / 4 u t( ) (f) g t( ) = rect t( )cos 2 t( )
(g) g t( ) = 6rect t( )cos 3 t( ) (h)
g t( ) = rect t( ) tri t( )
g(t)
t1
1
1
1
-1
(a)g(t)
t
(b)g(t)
t2 4
-8
-16
(c)g(t)
t
(d)
1
-1
g(t)
t4
5
(e)g(t)
t
(f)g(t)
t
-6
6
(g) (h)g(t)
t
1
1
1
21
21-
21
21
21-
21
21-
(i) g t( ) = rect t( ) tri t + 1 / 2( ) (j)
g t( ) = u t + 1 / 2( )ramp 1 / 2 t( )
(k) g t( ) = tri2
t( ) (l) g t( ) = sinc2
t( )
(m) g t( ) = sinc t( ) (n)
g t( ) =
d
dttri t( )( )
(o) g t( ) = rect t + 1 / 2( ) rect t 1 / 2( )
(p)
g t( ) = + 1( )t
2 ( ) + 1( ) d
g(t)
t
1 1
-1 1
(i) (j)g(t)
t-1 1
(k)g(t)
t
(l)
1
1
-1-1
g(t)
t
1
1
(m)g(t)
t
(n) (o) (p)
1
21
21-
g(t)
t
1
21
21-
1
1
-1-1
g(t)
t
1
1
-1-1
g(t)
t
M. J. Roberts - 10/7/06
Solutions 2-25
(q) g t( ) = 3tri 2t / 3( ) + 3rect t / 3( )
32
32-
3
6
t
,
(r) g t( ) = 6 tri t / 3( )rect t / 3( )
32
32-
3
6
t
(s) 4sinc 2t( )sgn t( ) 1
212-
4
-4
t ,
(t) g t( ) = 2ramp t( )rect t 1( ) / 2( )
4
t2
(u) g t( ) = 4 tri t 2( ) / 2( )u 2 t( )
4
t2
,
(v) g t( ) = 3rect t / 4( ) 6rect t / 2( )
3
t21-2 -1
-3
(w) g t( ) = 10drcl t / 4,5( )rect t / 8( )
t-8 8
g(t)
-2
10
(g)
35. Graph the following CT functions.
M. J. Roberts - 10/7/06
Solutions 2-26
(a) g t( ) = 3 3t( ) + 6 4 t 2( )( )
Using the impulse scaling property, g t( ) = t( ) + 3 / 2( ) t 2( )
t2
1
g(t)
32
(b) g t( ) = 2
1t / 5( )
g t( ) = 2 t / 5 n( )n=
= 10 t + 5n( )n=
, 10
t
g(t)
5 10 15 20-10 -5
......
(c) g t( ) =
1t( )rect t / 11( )
g t( ) = rect t / 11( ) t n( )n=
= t n( )n= 5
5
,
t
g(t)
1 2 3 4 5-5 -4 -3 -2 -1
1
(d) g t( ) = 5sinc t / 4( ) 2
t( )
g t( ) = 5sinc t / 4( ) t 2n( )n=
= 5 sinc n / 2( ) t 2n( )n=
t
g(t)
2 6 10
5
M. J. Roberts - 10/7/06
Solutions 2-27
(e)
g t( ) =2 ( ) 2
1( ) d
t
t
g(t)
-1-2 21
1
3
36. 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 andwith a frequency of 1 / 4 Hz plus a constant in the range, 2 < t < 2 . For t > 2 itdecays exponentially toward zero with a time constant of 2 seconds. It is continuouseverywhere.
(a) Write an exact mathematical description of this function.
g t( ) =
0 , t < 5
10 2t , 5 < t < 2
sin t / 2( ) , 2 < t < 2
6et / 2 , t > 2
(b) Graph g t( ) in the range, 10 < t < 10 .
(c) Graph g 2t( ) in the range, 10 < t < 10 .
(d) Graph 2g 3 t( ) in the range, 10 < t < 10 .
(e) Graph
2g t + 1( ) / 2( ) in the range, 10 < t < 10 .
M. J. Roberts - 10/7/06
Solutions 2-28
t-10 10
g(t)
-8
t-10 10
g(2t)
-8
t-10 10
2g(3- t)
-16 t-10 10
-2g(( t+1)/2)
16
37. Using MATLAB, for each function below plot the original function and the transformed function.
% Plotting functions and transformations of those functions
close all ;% (a) parttmin = -2 ; tmax = 2 ; N = 400 ;dt = (tmax - tmin)/N ; t = tmin + dt*[0:N]’ ;g0 = g322a(t) ; g1 = 5*g322a(2*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(‘5g(2t)’) ;
% (b) partfigure ;tmin = -3 ; tmax = 8 ; N = 100 ;dt = (tmax - tmin)/N ; t = tmin + dt*[0:N]’ ;g0 = g322b(t) ; g1 = -3*g322b(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)’) ;
% (c) partfigure ;tmin = 0 ; tmax = 96 ; N = 400 ;dt = (tmax - tmin)/N ; t = tmin + dt*[0:N]’ ;g0 = g322c(t) ; g1 = g322c(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)’) ;
% (d) part
M. J. Roberts - 10/7/06
Solutions 2-29
figure ;fmin = -20 ; fmax = 20 ; N = 200 ;df = (fmax - fmin)/N ; f = fmin + df*[0:N]’ ;G0 = G322d(f) ; G1 = abs(G322d(10*(f-10)) + G322d(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)g = (1-abs(t)).*(-1 < t & t < 1) ;y = 10*cos(20*pi*t).*g ;
function y = g322b(t)y = -2*(t <= -1) + 2*t.*(-1 < t & t <= 1) + ...(3-t.^2).*(1 < t & t <= 3) - 6*(t > 3) ;
function y = g322c(t)y = real(exp(j*pi*t) + exp(j*1.1*pi*t)) ;
function y = G322d(f)y = abs(5./(f.^2 - j*2 + 3)) ;
(a)
g t( ) = 10cos 20 t( ) tri t( )
5g 2t( ) vs. t
t-2 2
Original g(t)
-10
10
t-2 2
Transformed g(t)
-50
50
(b)
g t( ) =
2 , t < 1
2t , 1 < t < 1
3 t2 , 1 < t < 3
6 , t > 3
3g 4 t( ) vs. t
M. J. Roberts - 10/7/06
Solutions 2-30
t-4 8
Original g(t)
-6
2
t-4 8
Transformed g(t)
-10
20
(c)
g t( ) = Re e
j t+ e
j1.1 t( ) g t / 4( ) vs. t
t100
Original g(t)
-2
2
t100
Transformed g(t)
-2
2
(d)
G f( ) =5
f2
j2 + 3 G 10 f 10( )( ) + G 10 f + 10( )( ) vs. f
t-20 20
Original g(t)
1.5
t-20 20
Transformed g(t)
1.5
M. J. Roberts - 10/7/06
Solutions 2-31
38. The voltage illustrated in Figure E38 occurs in an analog-to-digital converter. Write amathematical description of it.
t (ms)-0.1 0.3
x(t)
5
Signal in A/D Converter
Figure E38 Signal occurring in an A/D converter
x t( ) = trit 5 10 5
5 10 5+ tri
t 1.5 10 4
5 10 5
39. A signal occurring in a television set is illustrated in Figure E39. Write amathematical description of it.
t (µs)-10 60
x(t)
-10
Signal in Television
5
Figure E39 Signal occurring in a television set
x t( ) = 10rectt 2.5 10 6
5 10 6
40. The signal illustrated in Figure E40 is part of a binary-phase-shift-keyed (BPSK)binary data transmission. Write a mathematical description of it.
t (ms)4
x(t)
-1
1
BPSK Signal
Figure E40 BPSK signal
M. J. Roberts - 10/7/06
Solutions 2-32
x t( ) =
sin 8000 t( )rectt 0.5 10 3
10 3sin 8000 t( )rect
t 1.5 10 3
10 3
+ sin 8000 t( )rectt 2.5 10 3
10 3sin 8000 t( )rect
t 3.5 10 3
10 3
41. The signal illustrated in Figure E41 is the response of an RC lowpass filter to asudden change in excitation. Write a mathematical description of it.
On a decaying exponential, a tangent line at any point intersects the final value one timeconstant later. Theconstant value before the decaying exponential is -4 V and the slope ofthe tangent line at 4 ns is -2.67V/4 ns or -2/3 V/ns.
t (ns)20
x(t)
-6
-4
RC Filter Signal
-1.3333
4
Figure E41 Transient response of an RC filter
x t( ) = 4 2 1 e
t 4( )/3( )u t 4( ) (times in ns)
42. Describe the signal in Figure E42 two ways.
...
4
15
x(t)
t
Figure E42
(a) As a ramp function minus a summation of step functions.
x t( ) = 3.75ramp t( ) 15 u t 4n( )n=0
(b) As a summation of products of triangle functions and rectangle functions.
x t( ) = 15 rectt 2 4n
4tri
t 4 4n
4n=0
8000 t( )rect t 0.5 10 3( ) sin 8000 t( )rect t 1.5 10 3( ) + sin 8000 t( )rect t 2.5 10 3( ) sin 8000 t( )rect t 3.5 (
M. J. Roberts - 10/7/06
Solutions 2-33
43. Mathematically describe the signal in Figure E-43 .
......9
9
x(t)
t
Semicircle
Figure E-43
The semicircle centered at t = 0 is the top half of a circle defined by
x2
t( ) + t2
= 81
Therefore
x t( ) = 81 t
2 , 9 < t < 9 .
This one period of this periodic function. The other periods are just shifted versions.
x t( ) = rectt 18n
1881 t 18n( )
2
n=
(The rectangle function avoids the problem of imaginary values for the square rootsof negative numbers.)
44. Let two signals be defined by
x1
t( ) =1 , cos 2 t( ) 1
0 , cos 2 t( ) < 1and
x
2t( ) = sin 2 t / 10( ) .
Plot these products over the time range, 5 < t < 5 .
(a) x
12t( )x
2t( ) (b)
x
1t / 5( )x
220t( )
(c) x
1t / 5( )x
220 t + 1( )( ) (d)
x
1t 2( ) / 5( )x
220t( )
M. J. Roberts - 10/7/06
Solutions 2-34
t-5 5
x1(t)x
2(t)
-1
1
(a)
t-5 5
x1(t)x
2(t)
-1
1
(b)
t-5 5
x1(t)x
2(t)
-1
1
(c)
t-5 5
x1(t)x
2(t)
-1
1
(d)
45. Given the graphical definition of a function in Figure E-45, graph the indicatedtransformation(s).
(a)
-2-2 2 3 4 5 61
12
-2
t
g(t)
g t( ) g 2t( )g t( ) 3g t( )
g t( ) = 0 , t > 6 or t < 2
(b)
-2-2 21 3 4 5 6
12
-2
t
g(t)
t t + 4
g t( ) 2g t 1( ) / 2( )
g t( ) is periodic with fundamental period, 4
Figure E-45
(a)
M. J. Roberts - 10/7/06
Solutions 2-35
The transformation, g t( ) g 2t( ) , simply compresses the time scale by a factor of 2.
The transformation g t( ) 3g t( ) time inverts the signal, amplitude inverts the
signal and then multiplies the amplitude by 3.
-2-2 2 4 6
12
-2
t
g(2t)
-2-2-4 2 4 6
36
-6
t
-3g(-t)
(b)
-2-2 21 3 4 5 6 7
12
-2
t
g(t + 4)
-2-2 21 3 4 5 6 7 8
24
-4
t
-2g( )t -12
46. For each pair of functions graphed in Figure E-46 determine what transformation hasbeen done and write a correct functional expression for the transformed function.
(a)
M. J. Roberts - 10/7/06
Solutions 2-36
-2-2 21 3 4 5 6
2
-1t
g(t)
-1 1 2 3 4-2-3-4
2
-1t
(b)
-2-2 21 3 4 5 6
2
t
g(t)
-2-2 21 3 4 5 6-1t
In (b), assuming g t( ) is periodic with fundamental period, 2, find two different
transformations which yield the same result
Figure E-46(a)It should be visually obvious that the transformed signal has been time inverted andtime shifted. By identifying a few corresponding points on both curves we see thatafter the time inversion the shift is to the right by 2. This corresponds to twosuccessive transformations, t t , followed by t t 2 . The overall effect of the
two successive transformations is then t t 2( ) = 2 t . Therefore the
transformation is
g t( ) g 2 t( )
(b) g t( ) 1 / 2( )g t + 1( ) or g t( ) 1 / 2( )g t 1( )
47. Let a function be defined by g t( ) = tri t( ) . Below are four other functions based on
this function. All of them are zero for large negative values of t.
g
1t( ) = 5g 2 t( ) / 6( )
g
2t( ) = 7g 3t( ) 4g t 4( )
g
3t( ) = g t + 2( ) 4g t + 4( ) / 3( )
g
4t( ) = 5g t( )g t 1 / 2( )
(a) Which of these transformed functions is the first to become non-zero
(becomes non-zero at the earliest time)? g
3t( )
(b) Which of these transformed functions is the last to go back to zero and stay
there? g
1t( )
M. J. Roberts - 10/7/06
Solutions 2-37
(c) Which of these transformed functions has a maximum value that is greaterthan all the other maximum values of all the other transformed functions?
g
2t( )
(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? g
1t( )
48. Name a function of continuous time t for which the two successive transformations
t t and t t 1 leave the function unchanged. cos 2 t( ) ,
1t( ) , etc...
(Any even periodic function with a period of one.)
49. Graph the magnitude and phase of each function versus f.
(a) G f( ) = sinc f( )e
j f /8
This function is a ratio of two functions, jf and 1 + jf/10. The magnitude of the ratiois the ratio of the magnitudes. At very low values of f , the ratio approaches 0because the numerator approaches 0 and the denominator approaches 1. At veryhigh values of f the denominator is approximately jf/10 and the magnitude of the ratioapproaches 10. All these statements are equally true for positive and negative f.Therefore the magnitude is an even function of f.
The phase of the ratio is the phase of the numerator minus the phase of thedenominator. For any positive f , the phase of the numerator is the phase of j times apositive constant. That is some number on the positive imaginary axis in thecomplex plane. So the phase is / 2 radians or 90°. For very small positive f, thedenominator is approximately just the real number, 1, whose phase is 0. Thereforefor very small positive f approaching 0 the phase approaches / 2 . For very largepositive f , the phase of the denominator approaches / 2 also and the differencebetween the numerator and denominator phases approaches 0. The behavior fornegative f is similar except that the phase of the numerator is now / 2 . So thephase for negative f is exactly the negative of the phase for the corresponding positivef. That is, the phase is an odd function of f.
(b)
G f( ) =jf
1+ jf / 10
M. J. Roberts - 10/7/06
Solutions 2-38
f -16 16
|G( f )|
1
(a)
f -16 16
Phase of G( f )
- π
π
f -100 100
|G( f )|
10
(b)
f -100 100
Phase of G( f )
- π
π
(c)
G f( ) = rectf 1000
100+ rect
f + 1000
100e
j f /500
(d)
G f( ) =1
250 f2
+ j3 f
f -1100 1100
|G( f )|
1
(c)
f -1100 1100
Phase of G( f )
- π
π
f -50 50
|G( f )|
0.02
(d)
f -50 50
Phase of G( f )
- π
π
(e) G f( ) =
0.01f( )sinc 25 f( )e
j f /50
100
M. J. Roberts - 10/7/06
Solutions 2-39
f -0.2 0.2
|G( f )|
0.01
(e)
f -0.2 0.2
Phase of G( f )
- π
π
50. Graph versus f , in the range, 4 < f < 4 , the magnitude and phase of
(a) X f( ) = sinc f( )
The phase in this plot is the phase of a purely real function. If we only plotted purelyreal functions we would not need to graph magnitude and phase separately. A simplereal plot of the function would be sufficient and clearer. But most transforms that wewill later graph are complex functions and magnitude and phase plots are good waysof representing them. Since this function is purely real its value always lies on thereal axis of the complex plane. When it is positive the simplest phase answer is 0.When it is negative the simplest phase answer is either positive or negative radians.Later, in the study of transform methods applied to systems, we will find that wealways have an even magnitude and an odd phase. For that reason, it is consistentand logical to choose phase values so as to make the plot an odd function. Here that
is done by making the phase for negative values of sinc f( ) be for positive f and
for negative f.
f
|X( f )|
X( f )1-1-2-3-4 2 3 4
f 1
-1-2-3-42 3 4
π
-π
1
M. J. Roberts - 10/7/06
Solutions 2-40
(b) X f( ) = 2sinc f( )e
j4 f
f
|X( f )|
1-1-2-3-4 2 3 4
f 1
-1-2-3-42 3 4
4π-4π
X( f )
2
π
(c) X f( ) = 5rect 2 f( )e
+ j2 f
f
|X( f )|
f
X( f )
5
14
14
π2
π2
(d) X f( ) = 10sinc2
f / 4( )f
|X( f )|
1-1-2-3-4 2 3 4
f 1-1-2-3-4 2 3 4
X( f )
10
(e) X f( ) = j5 f + 2( ) j5 f 2( )
f
|X( f )|
1-1-2-3-4 2 3 4
f 1-1-2-3-4 2 3 4
X( f )
5
π2
π2
(f) X f( ) = 1 / 2( ) 1/ 4
f( )ej f
M. J. Roberts - 10/7/06
Solutions 2-41
X f( ) =e
j f
2f
n
4n=
f
|X( f )|
f 1-1-2-3-4 2 3 4
X( f )
π
-π-2π-3π-4π
2π3π4π
14
14
34
12
12
12
34 1-1
... ...
Generalized Derivative
51. Graph the generalized derivative of g t( ) = 3sin t / 2( )rect t( ) .
Except at the discontinuities at t = ±1 / 2 , the derivative is either zero, for t > 1 / 2 , or
it is the derivative of 3sin t / 2( ) ,
3 / 2( )cos t / 2( ) , for
t < 1 / 2 . At the
discontinuities the generalized derivative is an impulse whose strength is thedifference between the limit approached from above and the limit approached from
below. In both cases that strength is 3 / 2 .
t
ddt
(g(t))
3π2
32
Alternate solution:
g t( ) = 3sin t / 2( ) u t + 1 / 2( ) u t 1 / 2( )
d
dtg t( )( ) = 3sin t / 2( ) t + 1 / 2( ) t 1 / 2( ) + 3 / 2( )cos t / 2( ) u t + 1 / 2( ) u t 1 / 2( )
d
dtg t( )( ) = 3sin / 4( ) t + 1 / 2( ) 3sin / 4( ) t 1 / 2( ) + 3 / 2( )cos t / 2( )rect t( )
d
dtg t( )( ) = 3 2 / 2 t + 1 / 2( ) + t 1 / 2( ) + 3 / 2( )cos t / 2( )rect t( )
M. J. Roberts - 10/7/06
Solutions 2-42
Derivatives and Integrals of CT Functions
52. What is the numerical value of each of the following integrals?
(a)
t( )cos 48 t( )dt = cos 0( ) = 1 , (b)
t 5( )cos t( )dt = cos 5( ) = 1
(c)
t 8( ) tri t / 32( )dt
0
20
= tri 8 / 32( ) =3
4
(d)
t 8( )rect t / 16( )dt
0
20
= rect 8 / 16( ) = 1 / 2
(e)
t 1.5( )sinc t( )dt
2
2
= sinc 1.5( ) =sin 3 / 2( )
3 / 2=
2
3
(f)
t 1.5( )sinc 4t( )dt
2
2
= sinc 4 1.5( ) =sin6
6= 0
53. What is the numerical value of each of the following integrals?
(a)
1t( )cos 48 t( )dt = t n( )
n=
cos 48 t( )dt = t n( )cos 48 t( )dt
n=
1
t( )cos 48 t( )dt = cos 48n( )n=
= 1n=
t-0.1 0.1
x(t)
-1
1
(b)
1t( )sin 2 t( )dt
1
t( )sin 2 t( )dt = t n( )sin 2 t( )dt
n=
= sin 2n( )n=
= 0
M. J. Roberts - 10/7/06
Solutions 2-43
t-2 2
x(t)
-1
1
(c)
44
t 2( )rect t( )dt
0
20
44
t 2( )rect t( )dt
0
20
= 4 t 2 4n( )rect t( )dt
0
20
n=
= 4 rect 2 + 4n( )n=
= 0
t-8 8
x(t)4
(d)
1t( )sinc t( )dt
2
2
= t n( )sinc t( )dt
2
2
n=
= sinc n( )n=
= 1
t-8 8
x(t)4
54. Graph the derivatives of these functions.
(a) g t( ) = sin 2 t( )sgn t( )
g t( ) = 2cos 2 t( ) , t < 0
cos 2 t( ) , t 0
(b) g t( ) = 2 tri t / 2( ) 1
g t( ) =
1+ t , 2 < t < 0
1 t , 0 < t < 2
0 , otherwise
g t( ) =
1 , 2 < t < 0
1 , 0 < t < 2
0 , otherwise
(c) g t( ) = cos 2 t( )
g t( ) = 2sin 2 t( ) , cos 2 t( ) < 0
sin 2 t( ) , cos 2 t( ) > 0
M. J. Roberts - 10/7/06
Solutions 2-44
(a) (b) (c)
t-4 4
x(t)
-1
1
t-4 4
dx/dt
-6
6
t-4 4
x(t)
-1
1
t-4 4
dx/dt
-1
1
t-1 1
x(t)
-1
1
t-1 1
dx/dt
-6
6
Even and Odd CT Functions
55. Graph the even and odd parts of these CT signals.
(a) x t( ) = rect t 1( )
x
et( ) =
rect t 1( ) + rect t + 1( )2
, x
ot( ) =
rect t 1( ) rect t + 1( )2
(b) x t( ) = tri t 3 / 4( ) + tri t + 3 / 4( )
x
et( ) = tri t 3 / 4( ) + tri t + 3 / 4( ) ,
x
0t( ) = 0
t-3 3
xe(t)
-1
1
(a)
t-3 3
xo(t)
-1
1
t-3 3
xe(t)
-1
1
(b)
t-3 3
xo(t)
-1
1
(c) x t( ) = 4sinc t 1( ) / 2( )
xe
t( ) =4sinc t 1( ) / 2( ) + 4sinc t 1( ) / 2( )
2= 2
sin t 1( ) / 2( )t 1( ) / 2
+sin t 1( ) / 2( )
t 1( ) / 2
M. J. Roberts - 10/7/06
Solutions 2-45
xe
t( ) = 2
sint
2 2
t 1( ) / 2+
sint
2 2
t 1( ) / 2= 4
sint
2 2
t 1( )
sint
2 2
t + 1( )
Using
sin x + y( ) = sin x( )cos y( ) + cos x( )sin y( )
xe
t( ) = 4
sin t / 2( )cos / 2( )=0
+ cos t / 2( )sin / 2( )= 1
t 1( )
sin t / 2( )cos / 2( )=0
+ cos t / 2( )sin / 2( )= 1
t + 1( )
xe
t( ) = 4cos t / 2( )
t + 1( )
cos t / 2( )t 1( )
= 4cost
2
t 1( ) t + 1( )t + 1( ) t 1( )
xe
t( ) = 4cost
2
2
t2 1( )
= 8cos t / 2( )
t2 1( )
xe
t( ) = 8cos t / 2( )
t2 1( )
Similarly,
xo
t( ) = 8t
cos t / 2( )t
2 1( )
(d) x t( ) = 2sin 4 t / 4( )rect t( )
x
et( ) = 2sin / 4( )cos 4 t( )rect t( ) ,
x
ot( ) = 2cos / 4( )sin 4 t( )rect t( )
M. J. Roberts - 10/7/06
Solutions 2-46
t-10 10
xe(t)
-4
4
(c)
t-10 10
xo(t)
-4
4
t-1 1
xe(t)
-2
2
(d)
t-1 1
xo(t)
-2
2
56. Find the even and odd parts of each of these CT functions.
(a) g t( ) = 10sin 20 t( )
g
et( ) =
10sin 20 t( ) + 10sin 20 t( )2
= 0 , x
ot( ) =
10sin 20 t( ) 10sin 20 t( )2
= 10sin 20 t( )
(b) g t( ) = 20t
3
g
et( ) =
20t3+ 20 t( )
3
2= 0 ,
g
ot( ) =
20t3 20 t( )
3
2= 20t
3
(c) x t( ) = 8 + 7t
2
x
et( ) =
8 + 7t2
+ 8 + 7 t( )2
2= 8 + 7t
2 , x
ot( ) =
8 + 7t2 8 7 t( )
2
2= 0
(d) x t( ) = 1+ t
x
et( ) =
1+ t + 1+ t( )2
= 1 , x
ot( ) =
1+ t 1 t( )2
= t
(e) x t( ) = 6t
g
et( ) =
6t + 6 t( )2
= 0 , g
ot( ) =
6t 6 t( )2
= 6t
(f) g t( ) = 4t cos 10 t( )
g
et( ) =
4t cos 10 t( ) + 4 t( )cos 10 t( )2
=4t cos 10 t( ) + 4 t( )cos 10 t( )
2= 0
M. J. Roberts - 10/7/06
Solutions 2-47
g
ot( ) =
4t cos 10 t( ) 4 t( )cos 10 t( )2
=4t cos 10 t( ) 4 t( )cos 10 t( )
2= 4t cos 10 t( )
(g) g t( ) =
cos t( )t
g
et( ) =
cos t( )t
+cos t( )
t
2=
cos t( )t
+cos t( )
t
2= 0
g
ot( ) =
cos t( )t
cos t( )t
2=
cos t( )t
+cos t( )
t
2=
cos t( )t
(h) g t( ) = 12 +
sin 4 t( )4 t
g
et( ) =
12 +sin 4 t( )
4 t+ 12 +
sin 4 t( )4 t
2=
12 +sin 4 t( )
4 t+ 12 +
sin 4 t( )4 t
2= 12 +
sin 4 t( )4 t
g
ot( ) =
12 +sin 4 t( )
4 t12
sin 4 t( )4 t
2=
sin 4 t( )4 t
sin 4 t( )4 t
2= 0
(i) g t( ) = 8 + 7t( )cos 32 t( )
g
et( ) =
8 + 7t( )cos 32 t( ) + 8 7t( )cos 32 t( )2
= 8cos 32 t( )
g
ot( ) =
8 + 7t( )cos 32 t( ) 8 7t( )cos 32 t( )2
= 7t cos 32 t( )
(j) g t( ) = 8 + 7t
2( )sin 32 t( )
g
et( ) =
8 + 7t2( )sin 32 t( ) + 8 + 7 t( )
2( )sin 32 t( )
2= 0
g
ot( ) =
8 + 7t2( )sin 32 t( ) 8 + 7 t( )
2( )sin 32 t( )
2= 8 + 7t
2( )sin 32 t( )
57. Is there a function that is both even and odd simultaneously? Discuss.
M. J. Roberts - 10/7/06
Solutions 2-48
The only function that can be both odd and even simultaneously is the trivial signal,
x t( ) = 0 . Applying the definitions of even and odd functions,
x
et( ) =
0 + 0
2= 0 = x t( ) and x
ot( ) =
0 0
2= 0 = x t( )
proving that the signal is equal to both its even and odd parts and is therefore botheven and odd.
58. Find and graph the even and odd parts of the CT function x t( ) in Figure E-58
t
x(t)
1
-1-1
12
-2-3-4-5
2 3 4 5
Figure E-58
t1
-1 -1
12
-2-3-4-5
2 3 4 5
x (t)e
t1
-1-1
12
-2-3-4-5
2 3 4 5
x (t)o
Periodic CT Functions
59. For each of the following signals decide whether it is periodic and, if it is, find theperiod.
(a) g t( ) = 28sin 400 t( ) Periodic. Fundamental frequency = 200 Hz, Period = 5 ms.
(b) g t( ) = 14 + 40cos 60 t( ) Periodic. Fundamental frequency = 30 Hz Period =
33.33...ms.
(c) g t( ) = 5t 2cos 5000 t( ) Not periodic.
(d) g t( ) = 28sin 400 t( ) + 12cos 500 t( ) 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.
M. J. Roberts - 10/7/06
Solutions 2-49
(e) g t( ) = 10sin 5t( ) 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) g t( ) = 4sin 3t( ) + 3sin 3t( ) Not periodic because least common multiple is infinite.
Signal Energy and Power of CT Signals
60. Find the signal energy of each of these signals.
(a) 2rect t( ) ,
E = 2rect t( )2
dt = 4 dt
1/ 2
1/ 2
= 4
(b) rect 8t( ) ,
E = rect 8t( )2
dt = dt
1/16
1/16
=1
8
(c)
3rectt
4 ,
E = 3rectt
4
2
dt = 9 dt
2
2
= 36
(d) tri 2t( ) ,
Keep in mind that the square of a rectangle function is another rectangle function but thesquare of a triangle function is not another triangle function. Use the definition of thetriangle function and the fact that a triangle function is even to reduce the work.
E = tri 2t( )2
dt = 1 2t( )2
dt
1/ 2
1/ 2
= 1 2 2t + 2t2( )dt
1/ 2
1/ 2
E = 1+ 4t + 4t2( )dt
1/ 2
0
+ 1 4t + 4t2( )dt
0
1/ 2
= t + 2t2
+ 4t
3
31/ 2
0
+ t 2t2
+ 4t
3
30
1/ 2
=1
3
(e)
3trit
4 ,
E = 3trit
4
2
dt = 9 1t
4
2
dt
4
4
= 9 1 2t
4+
t
4
2
dt
4
4
E = 9 1+t
2+
t2
16dt
4
0
+ 1t
2+
t2
16dt
0
4
= 9 t +t
2
4+
t3
484
0
+ tt
2
4+
t3
480
4
= 24
(f) 2sin 200 t( )
E = 2sin 200 t( )2
dt = 4 sin2 200 t( )dt = 41
2
1
2sin 400 t( ) dt
E = 2 t +cos 400 t( )
400
M. J. Roberts - 10/7/06
Solutions 2-50
(g)
t( ) (Hint: First find the signal energy of a signal which approaches an
impulse some limit, then take the limit.)
t( ) = lim
a 01 / a( )rect t / a( )
E = lima 0
1
arect
t
a
2
dt = lima 0
1
a2
rectt
adt
a / 2
a / 2
= lima 0
a
a2
(h) x t( ) =
d
dtrect t( )( )
d
dtrect t( )( ) = t + 1 / 2( ) t 1 / 2( )
Ex
= t + 1 / 2( ) t 1 / 2( )2
dt
(i)
x t( ) = rect( )d
t
= ramp t + 1 / 2( ) ramp t 1 / 2( )
Ex
= t + 1 / 2( )2
dt
1/ 2
1/ 2
finite
+ dt
1/ 2
infinite
(j) x t( ) = e
1 j8( )tu t( )
Ex
= x t( )2
dt = e1 j8( )t
u t( )2
dt = e1 j8( )t
e1+ j8( )t
dt
0
Ex
= e2t
dt
0
=e
2t
20
=1
2
61. Find the average signal power of each of these signals:
(a) x t( ) = 2sin 200 t( ) This is a periodic function. Therefore
Px
=1
T2sin 200 t( )
2
dt
T / 2
T / 2
=4
T
1
2
1
2cos 400 t( ) dt
T / 2
T / 2
M. J. Roberts - 10/7/06
Solutions 2-51
Px
=2
Tt
sin 400 t( )400
T / 2
T / 2
=2
T
T
2
sin 200 T( )400
+T
2+
sin 200 T( )400
= 2
For any sinusoid, the average signal power is half the square of the amplitude.
(b) x t( ) =
1t( ) 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 averagepower is the energy in one period, divided by the period, or infinite.
(c) x t( ) = e
j100 t This is a periodic function. Therefore
Px
=1
T0
x t( )2
dtT
0
=1
T0
ej100 t
2
dt
T0
/ 2
T0
/ 2
= 50 ej100 t
ej100 t
dt
1/100
1/100
Px
= 50 dt
1/100
1/100
= 1
62. A CT signal, x, is periodic with fundamental period, T
0= 6 . This signal is described
over the time period, 0 < t < 6 , by
rect t 2( ) / 3( ) 4rect t 4( ) / 2( ) .
What is the signal power of this signal?The signal, x, can be described in the time period, 0 < t < 6 , by
x t( ) =
0 , 0 < t < 1 / 2
1 , 1 / 2 < t < 3
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 thefundamental period.
P =1
602 1
2+ 12 5
2+ 3( )
2 1
2+ 4( )
2 3
2+ 02 1 =
2.5 + 4.5 + 24
6= 5.167