signals and systems by m j roberts(solutions manual)_
TRANSCRIPT
-
M. J. Roberts - 7/12/03
Solutions 2-1
Chapter 2 - Mathematical Description of SignalsSolutions
1. If g t e t( ) = 7 2 3 write out and simplify(a) g 3 7 9( ) = e(b) g 2 7 72 2 3 7 2( ) = = ( ) +t e et t
(c) g t et
104 7 5
11+
=
(d) g jt e j t( ) = 7 2 3
(e) g g cosjt jt e e e e tj t j t( ) + ( )
=
+= ( )
27
27 23
2 23
(f)g g
cos
jt jte e
tjt jt
+
=
+= ( )
32
32
27
27
2. If g x x x( ) = +2 4 4 write out and simplify(a) g z z z( ) = +2 4 4(b) g u v u v u v u v uv u v+( ) = +( ) +( ) + = + + +2 2 24 4 2 4 4 4(c) g e e e e e ejt jt jt j t jt jt( ) = ( ) + = + = ( )2 2 24 4 4 4 2(d) g g gt t t t t t t( )( ) = +( ) = +( ) +( ) +2 2 2 24 4 4 4 4 4 4 4
g g t t t t t( )( ) = + +4 3 28 20 16 4(e) g 2 4 8 4 0( ) = + =
3. What would be the numerical 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](c) f = -1:0.5:1 ; w = 2*pi*f ; g = 1./(1+j*w) ;
-
M. J. Roberts - 7/12/03
Solutions 2-2
0 0247 0 1550 0920 0 289
10 0920 0 2890 0247 0 155
. .
. .
. .
. .
+
+
jj
jj
4. Let two functions be defined by
x, sin, sin1
1 20 01 20 0
tt
t( ) = ( )
( ) 0,
Strength = ( ) = ( ) = =
da a d a a1 1 1and for a < 0,
Strength = ( ) = ( ) = ( ) = =
da a d a d a a1 1 1 1Therefore for a > 0 and a < 0,
Strength = 1a
and a t ta
t t( )[ ] = ( )0 01 .40. Using the results of Exercise 39, show that
(a) comb ax
ax
n
an( ) =
=
1
From the comb definition, comb ax ax n( ) = ( )
.Then, using the property from Exercise 39, comb ax a x n
a ax
n
a( ) =
=
1 .(b) the average value of comb ax( ) is one, independent of the value of a
The period is 1/a. Therefore
comb comb combax
a
ax dx a ax dx a ax dxt
ta
a
a
a
a
( ) = ( ) = ( ) = ( )
+
110
01
12
12
12
12
comb ax x dx
a
a
( ) = ( ) =
1
2
12
1
-
M. J. Roberts - 7/12/03
Solutions 2-37
(c) a comb function of the form, 1a
ta
comb is a sequence of unit impulses spaced a units apart.
1 1a
ta a
a t an t ann n
comb = ( ) = ( )=
=
and (d) even though ata
t( ) = ( )1 , comb combaxa
x( ) ( )1
ax na
x nn n
( ) ( )=
=
1
1 1a
xn
a ax n
n n
( )=
=
QED
41. Sketch the generalized derivative of g sin rectt t t( ) = ( )3 2
.
Except at the discontinuities at t = 12
, the derivative is either zero, for t > 12
, or it
is the derivative of 32
sin t , 32 2
cost , for t >>Better figures needed
(a) comb cos cos cost t dt t n t dt t n t dtn n
( ) ( ) = ( ) ( ) = ( ) ( )
=
=
48 48 48 comb cos cost t dt n
n n
( ) ( ) = ( ) =
=
=
48 48 1
-
M. J. Roberts - 7/12/03
Solutions 2-40
(b) comb sint t dt( ) ( )
2comb sin sin sint t dt t n t dt n
n n
( ) ( ) = ( ) ( ) = ( ) =
=
=
2 2 2 0
(c) comb rectt t dt ( ) 24020
4
comb rect rect rectt t dt t n t dt nn n
( ) = ( ) ( ) = +( ) = =
=
24
4 2 4 4 2 4 00
20
0
20
(d) comb sinc sinc sinct t dt t n t dt nn n
( ) ( ) = ( ) ( ) = ( ) =
=
=
2
2
2
2
1
45. Sketch the derivatives of these functions.
(a) g sin sgnt t t( ) = ( ) ( )2 ( ) = ( )
g
sin , cossin , cos
tt t
t t2
2 2 02 2 0
(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
-
M. J. Roberts - 7/12/03
Solutions 2-41
46. Sketch the derivatives of these functions. Compare the average values of the magnitudesof the derivatives.
t
g (t)
......
-2 42 6 8 10 12
1
t
g (t)
......
-2 42 6 8 10 12
1
t
g (t)
......
-2 42 6 8 10 12
1
1 2 3
t
g' (t)
......
-2 42 6 8 10 12t
g' (t)
......
-2 42 6 8 10 12
1
t
g' (t)
......
-2 42 6 8 10 12
1 2 3
14
12
Average derivative is zero in each case.
47. A function, g t( ), has this description:It is zero for t < 5. It has a slope of 2 in the range, < < 5 2t . It has
the shape of a sine wave of unit amplitude and with a frequency of 14
Hz plus a constant inthe range, < 2 it decays exponentially toward zero with a time constant of2 seconds. It is continuous everywhere.
Write an exact mathematical description of this function.
g
,
,
sin ,
,
t
t
t t
tt
e tt
( ) =
<
< <
< < >
0 510 2 5 2
22 2
6 22
(a) Graph g t( ) in the range, <
-
M. J. Roberts - 7/12/03
Solutions 2-42
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
48. Find the even and odd parts of each of these CT functions.
(a) g sint t( ) = ( )10 20g sin sine t
t t( ) = ( ) + ( ) =10 20 10 202
0 , x sin sin sino tt t
t( ) = ( ) ( ) = ( )10 20 10 202
10 20
(b) g t t( ) = 20 3ge t
t t( ) = + ( ) =20 202
03 3
, go tt t
t( ) = ( ) =20 202
203 3
3
(c) x t t( ) = +8 7 2xe t
t tt( ) = + + + ( ) = +8 7 8 7
28 7
2 22 , xo t
t t( ) = + ( ) =8 7 8 72
02 2
(d) x t t( ) = +1xe t
t t( ) = + + + ( ) =1 12
1 , xo tt t
t( ) = + ( ) =1 12
(e) x t t( ) = 6ge t
t t( ) = + ( ) =6 62
0 , go tt t
t( ) = ( ) =6 62
6
(f) g cost t t( ) = ( )4 10
g cos cos cos cose tt t t t t t t t( ) = ( ) + ( ) ( ) = ( ) + ( ) ( ) =4 10 4 10
24 10 4 10
20
g cos cos cos cos coso tt t t t t t t t
t t( ) = ( ) ( ) ( ) = ( ) ( ) ( ) = ( )4 10 4 102
4 10 4 102
4 10
(g) g cost tt
( ) = ( )
g
cos cos cos cos
e t
tt
tt
tt
tt( ) =
( )+
( )
=
( )+
( )
=
2 2
0
-
M. J. Roberts - 7/12/03
Solutions 2-43
g
cos cos cos coscos
o t
tt
tt
tt
tt t
t( ) =
( )
( )
=
( )+
( )=
( )
2 2
(h) g sint tt
( ) = + ( )12 44
g
sin sin sin sinsin
e t
tt
tt
tt
tt t
t( ) =
+( )
+ +( )
=
+( )
+ +( )
= +( )12
44
12 44
2
12 44
12 44
212 4
4
g
sin sin sin sin
o t
tt
tt
tt
tt( ) =
+( )
( )
=
( )
( )=
12 44
12 44
2
44
44
20
(i) g cost t t( ) = +( ) ( )8 7 32
g cos cos cose tt t t t
t( ) = +( ) ( ) + ( ) ( ) = ( )8 7 32 8 7 322
8 32
g cos cos coso tt t t t
t t( ) = +( ) ( ) ( ) ( ) = ( )8 7 32 8 7 322
7 32
(j) g sint t t( ) = +( ) ( )8 7 322
gsin sin
e tt t t t( ) = +( ) ( ) + + ( )( ) ( ) =8 7 32 8 7 32
20
2 2
gsin sin
sino tt t t t
t t( ) = +( ) ( ) + ( )( ) ( ) = +( ) ( )8 7 32 8 7 322 8 7 322 2
2
49. Is there a function that is both even and odd simultaneously? Discuss.The only function that can be both odd and even simultaneously is the trivial signal,x t( ) = 0. Applying the definitions of even and odd functions,
x and xe ot s t t s t( ) = + = = ( ) ( ) = = = ( )0 02 00 0
20
proving that the signal is equal to both its even and odd parts and is therefore botheven and odd.
50. Find and sketch the even and odd parts of the CT function, x(t).
-
M. J. Roberts - 7/12/03
Solutions 2-44
t
x(t)
1-1-1
12
-2-3-4-52 3 4 5
t1
-1 -1
12
-2-3-4-5
2 3 4 5
x (t)e
t1
-1-1
12
-2-3-4-52 3 4 5
x (t)o
51. For each of the following signals decide whether it is periodic and, if it is, find theperiod.
(a) g sint t( ) = ( )28 400 Periodic. Fundamental frequency = 200 Hz, Period = 5 ms.(b) g cost t( ) = + ( )14 40 60 Periodic. Fundamental frequency = 30 Hz Period =
33.33...ms.(c) g cost t t( ) = ( )5 2 5000 Not periodic.(d) g sin cost t t( ) = ( ) + ( )28 400 12 500 Periodic. Two sinusoidal components with
periods of 5 ms and 4 ms. Least common multiple is 20 ms. Period of the overall signal is 20 ms.
(e) g sin cost t t( ) = ( ) ( )10 5 4 7 Periodic. The Periods of the two sinusoids are 25
s
and 27
s. Least common multiple is 2 . Period of the overall signal is 2 s .
(f) g sin sint t t( ) = ( ) + ( )4 3 3 3 Not periodic because least common multiple is infinite.52. The voltage illustrated in Figure E52 occurs in an analog-to-digital converter. Write a
mathematical description of it.
t (ms)-0.1 0.3
x(t)5
Signal in A/D Converter
Figure E52 Signal occurring in an A/D converter
-
M. J. Roberts - 7/12/03
Solutions 2-45
x tri tri .t t t( ) =
+
5 105 10
1 5 105 10
5
5
4
5
53. A signal occurring in a television set is illustrated in Figure E52. Write a mathematicaldescription of it.
t (s)-10 60
x(t)
-10
Signal in Television5
Figure E52 Signal occurring in a television set
x rect.
tt( ) =
10 2 5 105 10
6
6
54. The signal illustrated in Figure E54 is part of a binary-phase-shift-keyed (BPSK) binarydata transmission. Write a mathematical description of it.
t (ms)4
x(t)
-1
1
BPSK Signal
Figure E54 BPSK signal
x
sin rect . sin rect .
sin rect . sin rect .t
tt
tt
tt
tt
( ) =( )
( )
+ ( ) ( )
8000 0 5 1010
8000 1 5 1010
8000 2 5 1010
8000 3 5 1010
3
3
3
3
3
3
3
3
55. This signal illustrated in Figure E55 is the response of an RC lowpass filter to a sudden
change in excitation. Write a mathematical description of it.
t (ns)20
x(t)
-6
RC Filter Signal
-1.3333
4
Figure E55 Transient response of an RC filter
-
M. J. Roberts - 7/12/03
Solutions 2-46
x ut e tt
( ) = ( )
4 2 1 44
3
56. Find the signal energy of each of these signals:
(a) 2 rect ( )t , E t dt dt= ( )[ ] = =
2 4 4212
12
rect
(b) rect 8t( ) , E t dt dt= ( )[ ] = =
rect 8 182 116
116
(c) 34
rectt , E t dt dt=
= =
3 4 9 362
2
2
rect
(d) tri 2t( ) , E t dt t dt t t dt= ( )[ ] = ( ) = +( )
tri 2 1 2 1 2 2 22 212
12
2
12
12
E t t dt t t dt t t t t t t= + +( ) + +( ) = + + + +
=
1 4 4 1 4 4 2 4 3 2 4 3 13212
02
0
12
23
12
02
3
0
12
(e) 34
tri t , E t dt t dt t t dt= =
= +
3 4 9 1 4 9 1 2 4 42 2
4
4 2
4
4
tri
E t t dt t t dt t t t t t t= + + + +
= + +
+ +
=
9 1 2 16 1 2 16 9 4 48 4 48 242
4
0 2
0
4 2 3
4
0 2 3
0
4
(f) 2sin 200t( )E t dt t dt t dt= ( )[ ] = ( ) = ( )
2 200 4 200 4 12 12 4002 2sin sin sin E t t= + ( )
2 400400
cos
(g) t( ) (Hint: First find the signal energy of a signal which approaches animpulse some limit, then take the limit.)
t( ) = lima0
1a
rectta
E = lima0
1a
rectta
2dt
= lima0
1a2
rectta
dt
a
2
a
2 = lima0
a
a2
rect . sin rect . sin rect . sin rect .00 0 5 10 8000 1 5 10 8000 2 5 10 8000 3 53 3 3 t t t t t t t t) ( ) ( ) ( ) + ( ) ( ) ( ) (
-
M. J. Roberts - 7/12/03
Solutions 2-47
(h) x rectt ddt
t( ) = ( )( )ddt
t t trect( )( ) = + 12
12
E t t dtx = +
12 122
(i) x rect ramp rampt d t tt
( ) = ( ) = +
12 12
E t dt dtx = + +
122
12
12
12
finite infinite1 244 344 {
(j) x ut e tj t( ) = ( ) ( )1 8
E t dt e t dt e e dtxj t j t j t
= ( ) = ( ) =
( )
( ) +( ) x u2 1 8 2 1 8 1 80
E e dt ext
t
= =
=
20
2
0212
57. Find the average signal power of each of these signals:
(a) x sint t( ) = ( )2 200 This is a periodic function. Therefore
PT
t dtT
t dtxT
T
T
T
= ( )[ ] = ( )
1 2 200 4 12 12 40022
2
2
2
sin cos
PT
tt
TT T T T
xT
T
=
( )
=
( )+ +
( )
=
2 400400
22
200400 2
200400
22
2sin sin sin
For any sinusoid, the average signal power is half the square of the amplitude.
(b) x combt 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 theaverage power is the energy in one period, divided by the period, or infinite.
(c) x t e j t( ) = 100 This is a periodic function. Therefore
-
M. J. Roberts - 7/12/03
Solutions 2-48
PT
t dtT
e dt e e dtx Tj t
T
T
j t j t= ( ) = =
1 1 500
2
0
100 2
2
2100 100
1100
1100
00
0
x
P dtx = =
50 11
100
1100
58. Sketch these DT exponential and trigonometric functions.
(a) g cosn n[ ] = 4210 (b) g cos .n n[ ] = ( )4 2 2
(c) g cos .n n[ ] = ( )4 1 8 (d) g cos sinn n n[ ] = 22
63 2
6
(e) g nn
[ ] = 34
(f) g . sinn nn[ ] = ( ) 2 0 92
4
n-5 20
g[n]
-4
4
(a)
n-5 20
g[n]
-4
4
(b)
n-5 20
g[n]
-4
4
(c)
n-5 20
g[n]
-4
4
(d)
n-5 20
g[n]
-4
4
(e)
n-5 20
g[n]
-4
4
(f)
59. Sketch these DT singularity functions.
(a) g un n[ ] = +[ ]2 2 (b) g un n[ ] = [ ]5
(c) g rampn n[ ] = [ ]2 (d) g rampn n[ ] = 10 2
(e) g n n[ ] = [ ]7 1 (f) g n n[ ] = ( )[ ]7 2 1
-
M. J. Roberts - 7/12/03
Solutions 2-49
n-5 20
g[n]2
(a)
n-5 20
g[n]1
(b)
n-5 20
g[n]
-10
(c)
n-5 20
g[n]100
(d)
n-5 20
g[n]7
(e)
n-5 20
g[n]7
(f)
(g) g n n[ ] = 4
23
(h) g n n[ ] = 4
23
1
(i) g combn n[ ] = [ ]8 4 (j) g combn n[ ] = [ ]8 24(k) g rectn n[ ] = [ ]4 (l) g rectn n[ ] =
2 35
n-5 20
g[n]
-4
(g)
n-5 20
g[n]
-4
(h)
n-5 20
g[n]8
(i)
n-5 20
g[n]8
(j)
n-5 20
g[n]1
(k)
n-20 20
g[n]2
(l)
(m) g trin n[ ] = 5 (n) g sincnn[ ] = 4
(o) g sincn n[ ] = + 1
4
-
M. J. Roberts - 7/12/03
Solutions 2-50
n-10 10
g[n]1
(m)
n-20 20
g[n]
-1
1
(n)
n-20 20
g[n]
-1
1
(o)
(p) g drcl ,n n[ ] = 10 9
n-20 20
g[n]
-1
1
(p)
60. Sketch these combinations of DT functions.
(a) g u un n n[ ] = [ ] + [ ] (b) g u un n n[ ] = [ ] [ ]
(c) g cos combn n n[ ] = [ ]212 3 (d) g cos combn n n[ ] =
212 23
(e) g sin un e n nn
[ ] = [ ]
5 28
16 (f) g sin un n n[ ] = [ ]2
4
n-10 10
g[n]2
(a)
n-10
10
g[n]
-1
1
(b)
n-5 20
g[n]
-1
1
(c)
n-5 20
g[n]
-1
1
(d)
n-5 30
g[n]
-5
5
(e)
n-5
20
g[n]
-1
1
(f)
(g) g cos u cos un n n n n[ ] = +( ) +[ ]
[ ]2 1
121 2
12
-
M. J. Roberts - 7/12/03
Solutions 2-51
(h) g cos un m mm
n
[ ] = [ ]=
2120
(i) g comb combn m mm
n
[ ] = [ ] [ ]( )=
4 40
2
(j) g comb comb rectn m m mm
n
[ ] = [ ] + [ ]( ) [ ]=
4 3 4
(k) g comb combn n n[ ] = +[ ] [ ]2 21 (l) g n m mm
n
m
n
[ ] = [ ] [ ]=
+
=
1
n-5 20
g[n]
-1
1
(g)
n-5
20
g[n]
-2
2
(h)
n-5 20
g[n]1
(i)
n-10 10
g[n]6
(j)
n-5 20
g[n]
-1
1
(k)
n-5 10
g[n]1
(l)
61. Sketch the magnitude and phase of each function versus k.
(a) G sink k e jk
[ ] =
20 28
4
(b) G cos sinck k k[ ] = 2028 40
(c) G k k k k k k e jk
[ ] = +[ ] +[ ] + [ ] [ ] + [ ]( ) 8 2 4 2 4 8 8
-
M. J. Roberts - 7/12/03
Solutions 2-52
k-16 16
|G[k]|20
(a)
k-16 16
Phase of G[k]
-
k-16 16
|G[k]|20
(b)
k-16 16
Phase of G[k]
-
k-16 16
|G[k]|2
(c)
k-16 16
Phase of G[k]
-
62. Given the function definitions on the left, find the function values on the right.
(a)g n n e n[ ] = + 3 6
102 g .3 3 3 6
100 00000922 6[ ] = ( ) + = ( )e
(b)
g Ren jn
[ ] = +
12
g Re5 12
12
5
[ ] = +
=
j
(c)g n j n j n[ ] = ( ) + 2 10 42 g . .4 2 4 10 4 4 635 7 125 72[ ] = ( )( ) + ( ) = +j j j
63. Using MATLAB, for each function below plot the original function and the transformed function.
(a)
g
,
,
,
,
n
n
n n
n n
n
[ ] =
<
+ < >
5 05 3 0 4
23 4 841 8
2 g 3n[ ] vs. n
-
M. J. Roberts - 7/12/03
Solutions 2-53
n-10 20
g[n]
-10
50
(a)
n-10 20
g[3n]
-10
50
(b)g cos cosn n n[ ] = 10
220
24
4 2 1g n +( )[ ] vs. n
n40
g[n]
-10
10
(b)
n40
4g[2(n+1)]
-40
40
(c)g un e n
j n[ ] = [ ]8216
g n2
vs. n
-
M. J. Roberts - 7/12/03
Solutions 2-54
n-10 30
g[n]10
(c)
n-10 30
g[n/2]10
64. Given the graphical definition of a function, g[n], graph the indicated function(s), h[n]. (a)
n
g[n]
2 4 6 8-8 -6 -4 -2
2
4
6
-6
-4
-2
h gn n[ ] = [ ]2 4
g ,n n[ ] = >0 8
n
g[2n - 4]
2 4 6 8-8 -6 -4 -2
2
4
6
-6
-4
-2
(b)
n
g[n]
2 4 6 8-8 -6 -4 -2
2
4
6
-6
-4
-2
h n g n[ ] = 2
g ,n n[ ] = >0 8
-
M. J. Roberts - 7/12/03
Solutions 2-55
n
g[ ]
2 4 6 8-8 -6 -4 -2
2
4
6
-6
-4
-2
n2
(c)
n
g[n]
2 4 6 8-8 -6 -4 -2
2
4
6
-6
-4
-2
h n g n[ ] = 2
g n[ ] is periodic
n
g[ ]
2 4 6 8-8 -6 -4 -2
2
4
6
-6
-4
-2
n2
65. Sketch the accumulation from negative infinity to n of each of these DT functions.
(a) g cos un n n[ ] = ( ) [ ]2 (b) g cos un n n[ ] = ( ) [ ]4
n-5 20
Accumulation of g[n]20
(a)
n-5 20
Accumulation of g[n]20
(b)
66. Find and sketch the magnitude and phase of the even and odd parts of each of thisdiscrete-k function.
-
M. J. Roberts - 7/12/03
Solutions 2-56
G k j k[ ] = 10
1 4
Ge kj k j k
j k j k k[ ] =
++
=
( ) +( ) = +10
1 410
1 42
101 4 1 4
101 16 2
Go kj k j k j k
j k j kj k
k[ ] =
+=
( ) +( ) = +10
1 410
1 42
401 4 1 4
401 16 2
k-10 10
|Ge[k]|
10
k-10 10
Phase of Ge[k]
-
k-10 10
|Go[k]|
10
k-10 10
Phase of Go[k]
-
67. Find and sketch the even and odd parts of the DT function below.
n
g[n]
2 4 6 8-8 -6 -4 -2
2
4
6
-6
-4
-2
n
g [n]
2 4 6 8-8 -6 -4 -2
2
4
6
-6
-4
-2
n
g [n]
2 4 6 8-8 -6 -4 -2
2
4
6
-6
-4
-2
e o
68. Using MATLAB, plot each of these DT functions. If a function is periodic, find theperiod analytically and verify the period from the plot.
-
M. J. Roberts - 7/12/03
Solutions 2-57
(a) g sin sinn n n[ ] = = 32
64
Period is 4
(b) g sin cosn n n[ ] = + 2
310
3
g sin cos sin cosn n n n n n[ ] = + + = + = =
23
63
43
23
43
Period 3 Period 31 24 34 1 24 34
Period is 3
(c) g cos sinn n n[ ] = + 52
83 2
5
Period of 8 Period of 51 24 34 1 24 34
LCM of the periods is 40, therefore the period is 40.
n60
g[n]
-1
1
(a)
n60
g[n]
-2
2
(b)
n60
g[n]
-8
8
(c)
(d) g cosn n[ ] = 10 4 Not periodic.
(e) g cos sinn n n[ ] = 32
72
6 (A trigonometric identity will be useful here.)
g sin sinn n n n n[ ] = + +
32
26
27
26
27
g sin sin sin sinn n n n n[ ] = +
=
32
1242
4042
32
4042
1242
The period is 42.
-
M. J. Roberts - 7/12/03
Solutions 2-58
n60
g[n]
-10
10
(d)
n60
g[n]
-4
4
(e)
69. Sketch the following DT functions.
(a) g n n n[ ] = [ ] + +[ ]5 2 3 1 n
53
2-1
g[n]
(b) g n n n[ ] = [ ] + ( )[ ]5 2 3 4 2 n
53
2
g[n]
(c) g u un n n[ ] = [ ] [ ]( )5 1 4n
2 4 6
g[n]
5
-5
(d) g rectn n[ ] = +[ ]8 14n
2-2-4-6 4 6
g[n]8
(e) g cosn n[ ] = 82
7
5 10 15 20
-10
-5
5
10
n
g[n]
-
M. J. Roberts - 7/12/03
Solutions 2-59
(f) g un e nn
[ ] = [ ]10 4 -2 2 4 6 8
-80
-60
-40
-20
n
g[n]
(g) g . un nn[ ] = ( ) [ ]10 1 284 -2 2 4 6 8
-80
-60
-40
-20
n
g[n]
(h) g un j nn
[ ] = [ ]4-4 -2 2 4 6 8
0.2
0.4
0.6
0.8
1
n
g[n]
(i) g ramp ramp rampn n n n[ ] = +[ ] [ ] + [ ]2 2 2
-4 -2 2 4 6 8
0.5
1
1.5
2
n
g[n]
(j) g rect combn n n[ ] = [ ] [ ]2 2-8 -6 -4 -2 2 4 6 8
0.2
0.4
0.6
0.8
1
n
g[n]
(k) g rect combn n n[ ] = [ ] +[ ]2 2 1
-
M. J. Roberts - 7/12/03
Solutions 2-60
-8 -6 -4 -2 2 4 6 8
0.2
0.4
0.6
0.8
1
n
g[n]
(l) g sin rectn n n[ ] = [ ]32
3 4
-15 -10 -5 5 10 15
-3
-2
-1
1
2
3
n
g[n]
(m) g cos un n n[ ] = 5
28 2
-5 5 10 15 20 25 30
-5
5
n
g[n]
70. Graph versus k , in the range, <
-
M. J. Roberts - 7/12/03
Solutions 2-61
(b) X sinck k e jk
[ ] =
2
24
k-20 20
|X[k]|1
k-20 20
Phase of X[k]
-
(c) X rectk k e jk
[ ] = [ ] 32
3
k-15 15
|X[k]|1
k-15 15
Phase of X[k]
-
(d) X kj k
[ ] =+
1
12
k-15 15
|X[k]|1
k-15 15
Phase of X[k]
-
-
M. J. Roberts - 7/12/03
Solutions 2-62
(e) X k jkj k
[ ] =+1
2
k-15 15
|X[k]|2
k-15 15
Phase of X[k]
-
(f) X combk k e jk
[ ] = [ ] 224
k-15 15
|X[k]|1
k-15 15
Phase of X[k]
-
71. Sketch the even and odd parts of these signals.
All plots below.
(a) x rectn n[ ] = +[ ]5 2
-
M. J. Roberts - 7/12/03
Solutions 2-63
n-10 10
xe[n]
-1
1
n-10 10
xo[n]
-1
1
(b) x combn n[ ] = [ ]3 1
n-10 10
xe[n]
-1
1
n-10 10
xo[n]
-1
1
(c) x cosn n[ ] = + 152
9 4
n-10 10
xe[n]
-20
20
n-10 10
xo[n]
-20
20
-
M. J. Roberts - 7/12/03
Solutions 2-64
(d) x sin rectn n n[ ] = [ ]2
415
n-10 10
xe[n]
-1
1
n-10 10
xo[n]
-1
1
72. What is the numerical value of each of the following accumulations?
(a) ramp nn
[ ] = + + + + ==
0
10
0 1 2 10 55L
(b) 12
1 12
120
6
6nn =
= + + +L .
Using
n
n
NN
N
=
==
0
1 111
,
, otherwise
12
1 12
1 12
1 112812
1272560
6
7
nn =
=
=
=
(c) u nn
n
[ ]=
2Using
n
n =
=
0, h . .h
t tt K e K e( ) = + 1 5 23 2 0 76Since the highest derivative of x is two less than the highest derivative ofy, the general solution is of the form,
h u. .t K e K e tt t( ) = +( ) ( ) 1 5 23 2 0 76Integrating the equation once from t = 0 to t = +0 ,
-
M. J. Roberts - 7/12/03
Solutions 3-28
( ) ( ) + ( ) ( )[ ] + ( ) = ( ) =+ +
+
+
h h h h h0 0 6 0 0 4 10
0
0
0
t dt t dt
We know that the impulse response cannot contain an impulse because itssecond derivative would be a triplet and there is no triplet excitation. Wealso know that the impulse response cannot be discontinuous at time, t = 0,because if it were the second derivative would be a doublet and there is nodoublet excitation. Therefore,
( ) ( ) = ( ) =+ +h h h0 0 1 0 1This requirement, along with the requirement that the solution be continuousat time, t = 0, leads to the two equations,
( ) = = [ ] = + =
+h . . . .. .0 1 5 23 0 76 5 23 0 761 5 23 2 0 76 0 1 2K e K e K Kt t
t
andh 0 0 1 2
+( ) = = +K K .Solving,
K1 0 2237= . and K2 0 2237= .
Then the total impulse response is
h . u. .t e e tt t( ) = ( ) ( ) 0 2237 0 76 5 23 .(c) 2 3( ) + ( ) = ( )y y xt t t
The homogeneous solution is yh ht
t K e( ) = 32
. The impulse response is of the form,
h ut K e t K tht
i( ) = ( ) + ( )32 .
Integrating from 0 to 0+ ,
2 0 0 3 0 0 00
0
h h h x x+ + ( ) ( )[ ] + ( ) = ( ) ( ) =
+
tor
2 0 3 2 3 0h +( ) + = + =K K Ki h iIntegrating a second time from 0 to 0+ ,
2 1 2 1 12
0
0
0
0
h t t K Ki i( ) = ( ) = = =
+
+
.
-
M. J. Roberts - 7/12/03
Solutions 3-29
Then Kh = 34
. Therefore
h ut e t tt( ) = ( ) + ( )3
412
32
(d) 4 9 2( ) + ( ) = ( ) + ( )y y x xt t t t
The homogeneous solution is yh ht
t K e( ) = 94
. The impulse response is of the form,
h ut K e t K tht
i( ) = ( ) + ( )32 .
Integrating once from 0 to 0+ ,
4 0 0 9 2 0 0 20
0
0
0
h h h x x+ + ( ) ( )[ ] + ( ) = ( ) + ( ) ( ) =
+
+
t tor
4 9 2K Kh i+ =
Integrating a second time from 0 to 0+ ,
4 1 4 1 14
0
0
h t K Ki i( ) = = =
+
.Then Kh =
116
. Therefore
h ut e t tt( ) = ( ) + ( )1
1614
94
20. Sketch g t( ).
(a) g rect rectt t t( ) = ( ) ( ) t
1
1 1
g(t)
(b) g rect rectt t t( ) = ( ) 2 t
g(t)
32
12
12
32
1
-
M. J. Roberts - 7/12/03
Solutions 3-30
(c) g rect rectt t t( ) = ( ) 1 2 t
g(t)
32
12
12
52
1
(d) g rect rect rect rectt t t t t( ) = ( ) + +( )[ ] ( ) + +( )[ ]5 5 4 4g rect * rect rect * rect
rect * rect rect * rect
t t t t t
t t t t
( ) = ( ) ( ) + +( ) ( )+ ( ) +( ) + +( ) +( )
5 4 5 45 4 5 4
t
rect(t-5)rect(t-4)1
8 9 10
t
rect(t+5)rect(t+4)1
-10 -9 -8
t
rect(t-5)rect(t+4)1
1 2
t
rect(t+5)rect(t-4)1
-2 -1
1
t
g(t)
-9 -1 1 9
21. Sketch these functions.
(a) g rectt t( ) = ( )4t
1
18
18-
g(t)
(b) g rect( )t t t( ) = ( )4 4t
4
18
18-
g(t)
(c) g rectt t t( ) = ( ) ( )4 4 2t
4
g(t)
2182-182+
(d) g rectt t t( ) = ( ) ( )4 4 2t
2
18
18-
g(t)
-
M. J. Roberts - 7/12/03
Solutions 3-31
(e) g rect combt t t( ) = ( ) ( )4t
1
18
18-
g(t)
1-1-2
......
(f) g rect combt t t( ) = ( ) ( )4 1t
1
18
18-
g(t)
1-1-2
......
(g) g rect combt t t( ) = ( ) ( )4 2t
21
18
18-
g(t)
1-1-2
......
(h) g rect combt t t( ) = ( ) ( )2 t
g(t)
1-1-2
......1
22. Plot these convolutions.
(a) g rect rect rectt t t t t t( ) = +( ) +( )[ ] =+
+ 2 2 12
21
2
t-4 1
g(t)
-1
1
(b) g rect tri rect tri trit t t t d t d( ) = ( ) ( ) = ( ) ( ) = ( )
12
12
-
M. J. Roberts - 7/12/03
Solutions 3-32
-4 4
rect() and tri(t- )1
t < -3/2
-4 4
rect() and tri(t- )1
-3/2 < t < -1/2
-4 4
rect() and tri(t- )1
-1/2 < t < 1/2
-4 4
rect() and tri(t- )1
1/2 < t < 3/2
-4 4
rect() and tri(t- )1
3/2 < t
If t < 32
, g t( ) = 0.
If < < 32
12
t ,
g t t d t d tt t t
( ) = = ( )( ) = +
>
+
+
+
1 1 2012
1
12
1 2
12
1
{
g t t t t t t( ) = + +( ) + +( ) +
1 1
21 1
2
122
12
2
2
g t t t( ) = + +2
232
98
If < 1 1
012
0
12
{ {
g t t d t d t tt
t
t
t
( ) = ( )( ) + ( )( ) = + + +
1 1 2 212
12 2
12
212
g t t t t t t t t t( ) = + + + + +
22 2
2
212 2
18
12
18 2 2
g t t( ) = 34
2
By symmetry, g gt t( ) = ( ) and
-
M. J. Roberts - 7/12/03
Solutions 3-33
g
,
,
,
t
t
t tt
t t
( ) =
>
+ <
0 05 0 2179 0 2179 0 .The response cannot have a discontinuity at zero. Integrating the differential equation oncefrom 0 to 0+ ,
( ) ( ) = ( ) =+ +h h h0 0 1 0 1 .Integrating the differential equation twice from 0 to 0+ ,
h h h0 0 0 0 0+ +( ) ( ) = ( ) =Therefore
h 0 0+( ) = =Kc and ( ) = + =+h . .0 0 05 0 2179 1K Kc s .Solving,
Kc = 0 and Ks = 4 589.and
h . sin . u.t e t tt( ) = ( ) ( )4 589 0 21790 05 .29. Find the impulse response of the system in Figure E29 and evaluate its BIBO stability.
y(t)x(t)23
18
Figure E29 A two-integrator system
-
M. J. Roberts - 7/12/03
Solutions 3-38
( ) = ( ) ( ) ( )y x y yt t t t23
18
The homogenoeus solution is
y cos . sin .ht
c st e K t K t( ) = ( ) + ( )[ ] 3 0 1179 0 1179implying that
h cos . sin . ,t e K t K t tt
c s( ) = ( ) + ( )[ ] > 3 0 1179 0 1179 0 .The response cannot have a discontinuity at zero. Integrating the differential equation oncefrom 0 to 0+ ,
( ) ( ) = ( ) =+ +h h h0 0 1 0 1 .Integrating the differential equation twice from 0 to 0+ ,
h h h0 0 0 0 0+ +( ) ( ) = ( ) =Therefore
h 0 0+( ) = =Kc and ( ) = + =+h .0 13 0 1179 1K Kc s .Solving,
Kc = 0 and Ks = 8 482.and
h . sin . ut e t tt
( ) = ( ) ( )8 482 0 11793 .30. Plot the amplitudes of the responses of the systems of Exercise 19 to the excitation, e j t ,
as a function of radian frequency, .
(a) ( ) + ( ) = ( )y y xt t t5y p
j tt Ke( ) =
K j= +1
5
-10 10
|K|0.2
(b) ( ) + ( ) + ( ) = ( )y y y xt t t t6 4
K j= +1
4 62
-
M. J. Roberts - 7/12/03
Solutions 3-39
-5 5
|K|0.25
(c) 2 3( ) + ( ) = ( )y y xt t t
Kj
j= +
2 3
-5 5
|K|0.5
(d) 4 9 2( ) + ( ) = ( ) + ( )y y x xt t t t
Kj
j=+
+
24 9
-5 5
|K|0.25
31. Plot the responses of the systems of Exercise 19 to a unit-step excitation.
(a) h ut e tt( ) = ( )5
h h u ,
( ) = ( ) = ( ) = = [ ] = ( ) > 1 5 50
50
515
15
1 0t d e d e d e e tt t t
t t
h ,
( )
-
M. J. Roberts - 7/12/03
Solutions 3-40
(b) h . u. .t e e tt t( ) = ( ) ( ) 0 2237 0 76 5 23
h h . u . ,. . . .
( ) = ( ) = ( ) ( ) = ( ) > 1 0 76 5 23 0 76 5 230
0 2237 0 2237 0t d e e d e e d tt t t
h . . . ,. .
( ) = [ ] [ ]{ } >1 0 76 0 5 23 00 2237 1 316 0 1912 0t e e tt t t th . . . ,. .
( ) = ( ) ( )[ ] >1 0 76 5 230 2237 1 316 1 0 1912 1 0t e e tt th . . . u. .
( ) = ( ) ( )[ ] ( )1 0 76 5 230 2237 1 316 1 0 1912 1t e e tt t
t5
h- 1(t)
0.25
(c) h ut e t tt( ) = ( ) + ( )34
12
32
h h u ,
( ) = ( ) = ( ) + ( ) = > 1
32
32
0
34
12
12
34
0t d e d e d tt t t
h ,
( ) = + = +
>1
32
0
321
234
23
12
12
1 0t e e tt
t
h u
( ) = ( )1321
2t e t
t
t5
h- 1(t)
0.5
(d) h ut t e tt( ) = ( ) ( )14
116
94
h h u ,
( ) = ( ) = ( ) + ( ) = > 1
94
94
0
116
14
14
116
0t d e d e d tt t t
-
M. J. Roberts - 7/12/03
Solutions 3-41
h ,
( ) = + = +
>1
94
0
941
41
1649
14
136
1 0t e e tt
t
h u
( ) = +
( )1941
4136
1t e tt
t5
h- 1(t)
0.25
32. A CT system is described by the block diagram in Figure E32.
y(t)x(t) 1414
34
Figure E32 A CT system
Classify the system as to homogeneity, additivity, linearity, time-invariance, stability,causality, memory, and invertibility.
4 3( ) + ( ) + ( ) = ( )y y y xt t t tHomogeneity:Let x g1 t t( ) = ( ). Then 4 31 1 1( ) + ( ) + ( ) = ( )y y y gt t t t .Let x g2 t K t( ) = ( ) . Then 4 32 2 2( ) + ( ) + ( ) = ( )y y y gt t t K t .If we multiply the first equation by K, we get
4 31 1 1K t K t K t K t( ) + ( ) + ( ) = ( )y y y gTherefore
4 3 4 31 1 1 2 2 2K t K t K t t t t( ) + ( ) + ( ) = ( ) + ( ) + ( )y y y y y yThis can only be true for all time if y y2 1t K t( ) = ( ) .Homogeneous
Additivity:Let x g1 t t( ) = ( ). Then 4 31 1 1( ) + ( ) + ( ) = ( )y y y gt t t t .
-
M. J. Roberts - 7/12/03
Solutions 3-42
Let x h2 t t( ) = ( ) . Then 4 32 2 2( ) + ( ) + ( ) = ( )y y y ht t t t . Let x g h3 t t t( ) = ( ) + ( ) . Then 4 33 3 3( ) + ( ) + ( ) = ( ) + ( )y y y g ht t t t tAdding the first two equations,
4 31 2 1 2 1 2( ) + ( )[ ] + ( ) + ( )[ ] + ( ) + ( )[ ] = ( ) + ( )y y y y y y g ht t t t t t t tTherefore
4 3 4 33 3 3 1 2 1 2 1 2( ) + ( ) + ( ) = ( ) + ( )[ ] + ( ) + ( )[ ] + ( ) + ( )[ ]y y y y y y y y yt t t t t t t t t4 3 4 33 3 3 1 2 1 2 1 2( ) + ( ) + ( ) = ( ) + ( )[ ] + ( ) + ( )[ ] + ( ) + ( )[ ]y y y y y y y y yt t t t t t t t t
This can only be true for all time if y y y3 1 2t t t( ) = ( ) + ( ).Additive
Since it is homogeneous and additive, it is also linear.
It is also incrementally linear.
It is not statically non-linear because it is linear.
Time Invariance:Let x g1 t t( ) = ( ). Then 4 31 1 1( ) + ( ) + ( ) = ( )y y y gt t t t .Let x g2 0t t t( ) = ( ) . Then 4 32 2 2 0( ) + ( ) + ( ) = ( )y y y gt t t t t .The first equation can be written as
4 31 0 1 0 1 0 0 ( ) + ( ) + ( ) = ( )y y y gt t t t t t t tTherefore
4 3 4 31 0 1 0 1 0 2 2 2 ( ) + ( ) + ( ) = ( ) + ( ) + ( )y y y y y yt t t t t t t t tThis can only be true for all time if y y2 1 0t t t( ) = ( ) .Time Invariant
Stability:
The eigenvalues are = -0.1250 + 0.4841 = -0.1250 - 0.4841
1
2
jj
Therefore the homogeneous solution is of the form,
y -0.1250 + 0.4841 -0.1250 - 0.4841t K e K ej t j t( ) = +( ) ( )1 2 .If there is no excitation, but the zero-excitation response is not zero, the response will decayto zero as time increases. Since the particular solution has the same form as the excitationplus all its unique derivatives, the response to any bounded input will be a bounded output.Stable
-
M. J. Roberts - 7/12/03
Solutions 3-43
Causality:
The system equation can be rewritten as
y x y yt d d d d dt t t
( ) = ( ) ( ) ( )
14 31 1 2 1 1 1 1 22 2
So the response at any time, t t= 0 , depends on the excitation at times, t t< 0 and not on anyfuture values.Causal
Memory:The response at any time, t t= 0 , depends on the excitation at times, t t< 0 .System has memory.
Invertibility:
The system equation,4 3( ) + ( ) + ( ) = ( )y y y xt t t t
expresses the excitation in terms of the response and its derivatives. Therefore the excitationis uniquely determined by the response.Invertible.
33. A system has a response that is the cube of its excitation. Classify the system as tohomogeneity, additivity, linearity, time-invariance, stability, causality, memory, andinvertibility.
y xt t( ) = ( )3Homogeneity:Let x g1 t t( ) = ( ). Then y g1 3t t( ) = ( ).Let x g2 t K t( ) = ( ) . Then y g g g2 3 3 3 1 3t K t K t Ky t K t( ) = ( )[ ] = ( ) ( ) = ( ) .Not homogeneous
Additivity:Let x g1 t t( ) = ( ). Then y g1 3t t( ) = ( ).Let x h2 t t( ) = ( ) . Then y h2 3t t( ) = ( ) . Let x g h3 t t t( ) = ( ) + ( ) . Then y g h g h g h g h y y3
3 3 3 2 21 23 3t t t t t t t t t t t( ) = ( ) + ( )[ ] = ( ) + ( ) + ( ) ( ) + ( ) ( ) ( ) + ( )
Not additive
Since it is not homogeneous and not additive, it is also linear.
It is also not incrementally linear.
It is statically non-linear because it is non-linear without memory (lack of memory provenbelow).
-
M. J. Roberts - 7/12/03
Solutions 3-44
Time Invariance:Let x g1 t t( ) = ( ). Then y g1 3t t( ) = ( ).Let x g2 0t t t( ) = ( ) . Then y g y2
30 1 0t t t t t( ) = ( ) = ( ) .
Time Invariant
Stability:If x t( ) is bounded then y xt t( ) = ( )3 is also bounded.Stable
Causality:The response at any time, t t= 0 , depends only on the excitation at time, t t= 0 and not onany future values.Causal
Memory:The response at any time, t t= 0 , depends only on the excitation at time, t t= 0 and not onany past values.System has no memory.
Invertibility:
Solve y xt t( ) = ( )3 for x t( ) . x yt t( ) = ( )13
. The cube root operation is multiple valued.Therefore the system is not invertible.
34. A CT system is described by the differential equation,
t t t t( ) ( ) = ( )y y x8 .Classify the system as to linearity, time-invariance and stability.
Homogeneity:Let x g1 t t( ) = ( ). Then t t t t( ) ( ) = ( )y y g1 18 .Let x g2 t K t( ) = ( ) . Then t t t K t ( ) ( ) = ( )y y g2 28 .If we multiply the first equation by K, we get
tK t K t K t( ) ( ) = ( )y y g1 18Therefore
tK t K t t t t( ) ( ) = ( ) ( )y y y y1 1 2 28 8This can only be true for all time if y y2 1t K t( ) = ( ) .Homogeneous
Additivity:Let x g1 t t( ) = ( ). Then t t t t( ) ( ) = ( )y y g1 18 .Let x h2 t t( ) = ( ) . Then t t t t ( ) ( ) = ( )y y h2 28 . Let x g h3 t t t( ) = ( ) + ( ) . Then t t t t t ( ) ( ) = ( ) + ( )y y g h3 38Adding the first two equations,
-
M. J. Roberts - 7/12/03
Solutions 3-45
t t t t t t t( ) + ( )[ ] ( ) + ( )[ ] = ( ) + ( )y y y y g h1 2 1 28Therefore
t t t t t t t t( ) + ( )[ ] ( ) + ( )[ ] = ( ) ( )y y y y y y1 2 1 2 3 38 8t t t t t t t ty y y y y y1 2 1 2 3 38 8( ) + ( )[ ] ( ) + ( )[ ] = ( ) ( )
This can only be true for all time if y y y3 1 2t t t( ) = ( ) + ( ).Additive
Since it is homogeneous and additive, it is also linear.
It is also incrementally linear.
It is not statically non-linear because it is linear.
Time Invariance:Let x g1 t t( ) = ( ). Then t t t t( ) ( ) = ( )y y g1 18 .Let x g2 0t t t( ) = ( ) . Then t t t t t ( ) ( ) = ( )y y g2 2 08 .The first equation can be written as
t t t t t t t t( ) ( ) ( ) = ( )0 1 0 1 0 08y y gTherefore
t t t t t t t t t( ) ( ) ( ) = ( ) ( )0 1 0 1 0 2 28 8y y y yThis equation is not satisfied if y y2 1 0t t t( ) = ( ) therefore y y2 1 0t t t( ) ( ) .Time Variant
Stability:
The homogeneous solution to the differential equation is of the form,
ty t y t( ) = ( )8To satisfy this equation the derivative of y times t must be of the same functionalform as y itself. This is satisfied by a homogeneous solution of the form,
y t Kt( ) = 8If there is no excitation, but the zero-excitation response is not zero, the response willincrease without bound as time increases.Unstable
Causality:
The system equation can be rewritten as
-
M. J. Roberts - 7/12/03
Solutions 3-46
y x yt d dt t
( ) = ( ) + ( )
8So the response at any time, t t= 0 , depends on the excitation at times, t t< 0 and not on anyfuture values.Causal
Memory:The response at any time, t t= 0 , depends on the excitation at times, t t< 0 .System has memory.
Invertibility:
The system equation,t t t t( ) ( ) = ( )y y x8
expresses the excitation in terms of the response and its derivatives. Therefore the excitationis uniquely determined by the response.Invertible.
35. A CT system is described by the equation,
y xt d
t
( ) = ( )
3 .Classify the system as to time-invariance, stability and invertibility.
Homogeneity:
Let x g1 t t( ) = ( ). Then y g13
t d
t
( ) = ( )
.Let x g2 t K t( ) = ( ) . Then y g g y2
3 3
1t K d K d K t
t t
( ) = ( ) = ( ) = ( )
.Homogeneous
Additivity:
Let x g1 t t( ) = ( ). Then y g13
t d
t
( ) = ( )
.Let x h2 t t( ) = ( ) . Then y h2
3
t d
t
( ) = ( )
. Let x g h3 t t t( ) = ( ) + ( ) .
-
M. J. Roberts - 7/12/03
Solutions 3-47
Then y g h g h y y33 3 3
1 2t d d d t t
t t t
( ) = ( ) + ( )[ ] = ( ) + ( ) = ( ) + ( )
Additive
Since it is both homogeneous and additive, it is also linear.
It is also incrementally linear, since any linear system is incrementally linear.
It is not statically non-linear because it is linear.
Time Invariance:
Let x g1 t t( ) = ( ). Then y g13
t d
t
( ) = ( )
.Let x g2 0t t t( ) = ( ) . Then y g g y g2 0
3 3
1 0
300
t t d u du t t d
t tt
t t
( ) = ( ) = ( ) ( ) = ( )
.Time Variant
Stability:
If x t( ) is a constant, K, then y t Kd K dt t
( ) = =
3 3 and, as t , y t( ) increases withoutbound.Unstable
Causality:The response at time, t = 3, depends partially on the excitation at time t = 1 which is inthe future.Not causal
Memory:The response at any time, t t= 0 , depends partially on excitations in the past, t t< 0 .System has memory.
Invertibility:
Differentiate both sides of y xt d
t
( ) = ( )
3 w.r.t. t yielding ( ) = y xt t3 . Then it followsthat x yt t( ) = ( )3 .Invertible.
36. A CT system is described by the equation,
-
M. J. Roberts - 7/12/03
Solutions 3-48
y xt dt
( ) = ( )
+ 3 .Classify the system as to linearity, causality and invertibility.
Homogeneity:
Let x g1 t t( ) = ( ). Then y g13
t dt
( ) = ( )
+ .Let x g2 t K t( ) = ( ) . Then y g g y2
3 3
1t K d K d K tt t
( ) = ( ) = ( ) = ( )
+
+ .Homogeneous
Additivity:
Let x g1 t t( ) = ( ). Then y g13
t dt
( ) = ( )
+ .Let x h2 t t( ) = ( ) . Then y h2
3
t dt
( ) = ( )
+ . Let x g h3 t t t( ) = ( ) + ( ) . Then y g h g h y y3
3 3 3
1 2t d d d t tt t t
( ) = ( ) + ( )[ ] = ( ) + ( ) = ( ) + ( )
+
+
+ Additive
Since it is both homogeneous and additive, it is also linear.
It is also incrementally linear, since any linear system is incrementally linear.
It is not statically non-linear because it is linear.
Time Invariance:
Let x g1 t t( ) = ( ). Then y g13
t dt
( ) = ( )
+ .Let x g2 0t t t( ) = ( ) . Then y g g y2 0
3 3
1 0
0
t t d u du t tt t t
( ) = ( ) = ( ) = ( )
+
+
.Time Invariant
Stability:
If x t( ) is a constant, K , then y t Kd K dt t
( ) = =
+
+ 3 3 and, as t , y t( ) increaseswithout bound.Unstable
Causality:The response at any time, t t= 0 , depends partially on the excitation at times t t t0 0 3< < +which are in the future.
-
M. J. Roberts - 7/12/03
Solutions 3-49
Not causal
Memory:The response at any time, t t= 0 , depends partially on excitations in the past, t t< 0 .System has memory.
Invertibility:
Differentiate both sides of y xt dt
( ) = ( )
+ 3 w.r.t. t yielding ( ) = +( )y xt t 3 . Then itfollows that x yt t( ) = ( )3 .Invertible.
37. Show that the system described by y Re xt t( ) = ( )( ) is additive but not homogeneous.(Remember, if the excitation is multiplied by any complex constant and the system ishomogeneous, the response must be multiplied by that same complex constant.)
y Re xt t( ) = ( )( )Homogeneity:
Let x g h1 t t j t( ) = ( ) + ( ), where g t( ) and h t( ) are both real-valued functions.Then y Re g h g1 t t j t t( ) = ( ) + ( )( ) = ( ) .Let x g h2 t K jK t j tr i( ) = +( ) ( ) + ( )[ ], where Kr and Ki are both real constants.Then
y Re g h2 t K jK t j tr i( ) = +( ) ( ) + ( )[ ]( )y Re g h g h g h2 t K t K t jK t jK t K t K tr i i r r i( ) = ( ) ( ) + ( ) + ( )( ) = ( ) ( )
If we multiply the first equation by K jKr i+ , we get
K jK t K jK t j tr i r i+( ) ( ) = +( ) ( ) + ( )( )y Re g h1K jK t K jK tr i r i+( ) ( ) = +( ) ( )y g1
Thereforey y2 1t K jK tr i( ) +( ) ( )
unless Ki = 0.
Not homogeneous
Additivity:
Let x g h1 1 1t t j t( ) = ( ) + ( ), where g1 t( ) and h1 t( ) are both real-valued functions.Then y Re g h g1 1 1 1t t j t t( ) = ( ) + ( )( ) = ( ) .Let x g h2 2 2t t j t( ) = ( ) + ( ) , where g2 t( ) and h2 t( ) are both real-valued functions.Then y Re g h g2 2 2 2t t j t t( ) = ( ) + ( )( ) = ( ) .
-
M. J. Roberts - 7/12/03
Solutions 3-50
Let x g h g h3 1 1 2 2t t j t t j t( ) = ( ) + ( ) + ( ) + ( ) .Then y Re g h g h g g y y3 1 1 2 2 1 2 1 2t t j t t j t t t t t( ) = ( ) + ( ) + ( ) + ( )( ) = ( ) + ( ) = ( ) + ( ).Additive
38. Graph the magnitude and phase of the complex-sinusoidal response of the systemdescribed by
( ) + ( ) = y yt t e j ft2 2as a function of cyclic frequency, f.
This is a complex exponential excitation which has been applied for all time. Theresponse is the particular solution of the form,
y t Ke j ft( ) = 2where K is a complex constant. Substituting the solution form into the equation,
+ = j fKe Ke ej ft j ft j ft2 22 2 2 or
K j f= 12
11
.
Therefore y t j f ej ft( ) =
12
11
2
-4 -3 -2 -1 0 1 2 3 40
0.1
0.2
0.3
0.4
0.5
|K|
Exercise 38
-4 -3 -2 -1 0 1 2 3 4-1.5
-1
-0.5
0
0.5
1
1.5
Frequency, f (Hz)
Phas
e of
K
39. A DT system is described by
y xn mm
n
[ ] = ( )=
+1 .Classify this system as to time invariance, BIBO stability and invertibility.
-
M. J. Roberts - 7/12/03
Solutions 3-51
Homogeneity:
Let x g1 n n[ ] = [ ]. Then y g11
n mm
n
[ ] = [ ]=
+Let x g2 n K n[ ] = [ ]. Then y g g y2
1 1
1n K m K m K nm
n
m
n
[ ] = [ ] = [ ] = [ ]=
+
=
+ .Homogeneous.
Let x g1 n n[ ] = [ ]. Then y g11
n mm
n
[ ] = [ ]=
+Let x h2 n n[ ] = [ ]. Then y h2
1
n mm
n
[ ] = [ ]=
+Let x g h3 n n n[ ] = [ ] + [ ]. Then y g h g h y y3
1 1 1
1 2n m m m m n nm
n
m
n
m
n
[ ] = [ ] + [ ]( ) = [ ] + [ ] = [ ] + [ ]=
+
=
+
=
+ .Additive.
Since the system is homogeneous and additive it is also linear.
The system is also incrementally linear because it is linear.
The system is not statically non-linear because it is linear.
Time Invariance:
Let x g1 n n[ ] = [ ]. Then y g11
n mm
n
[ ] = [ ]=
+ .Let x g2 0n n n[ ] = [ ]. Then y g2 0
1
n m nm
n
[ ] = [ ]=
+ .The first equation can be rewritten as
y g g y1 01
0
1
2
0
n n m q n nm
n n
q
n
[ ] = [ ] = [ ] = [ ]=
+
=
+ Time invariant
Stability:If the excitation is a constant, the response increases without bound.Unstable
Causality:At any discrete time, n n= 0 , the response depends on the excitation at the next discrete timein the future.Not causal.
Memory:At any discrete time, n n= 0 , the response depends on the excitation at that discrete time andprevious discrete times.System has memory.
-
M. J. Roberts - 7/12/03
Solutions 3-52
Invertibility:Inverting the functional relationship,
y xn mm
n
[ ] = [ ]=
+1 .Invertible.Taking the first backward difference of both sides of the original system equation,
y y x xn n m mm
n
m
n
[ ] [ ] = [ ] [ ]=
+
=
+ 1 1 1 1
x y yn n n+[ ] = [ ] [ ]1 1The excitation is uniquely determined by the response.Invertible.
40. A DT system is described by
n n n ny y x[ ] [ ] = [ ]8 1 .Classify this system as to time invariance, BIBO stability and invertibility.
Homogeneity:
Let x g1 n n[ ] = [ ]. Then n n n ny y g1 18 1[ ] [ ] = [ ]Let x g2 n K n[ ] = [ ]. Then n n n K ny y g2 28 1[ ] [ ] = [ ]Multiply the first equation by K. nK n K n K ny y g1 18 1[ ] [ ] = [ ]Then, equating results,
nK n K n n n ny y y y1 1 2 28 1 8 1[ ] [ ] = [ ] [ ]If this equation is to be satisfied for all n,
y y2 1n K n[ ] = [ ].Homogeneous.
Additivity:
Let x g1 n n[ ] = [ ]. Then n n n ny y g1 18 1[ ] [ ] = [ ]Let x h2 n n[ ] = [ ]. Then n n n ny y h2 28 1[ ] [ ] = [ ]Let x g h3 n n n[ ] = [ ] + [ ]. Then n n n n ny y g h3 38 1[ ] [ ] = [ ] + [ ]Add the two first two equations.
n n n n n n ny y y y g h1 2 1 28 1 1[ ] + [ ]( ) [ ] + [ ]( ) = [ ] + [ ]Then, equating results,
n n n n n n n ny y y y y y1 2 1 2 3 38 1 1 8 1[ ] + [ ]( ) [ ] + [ ]( ) = [ ] [ ]If this equation is to be satisfied for all n,
-
M. J. Roberts - 7/12/03
Solutions 3-53
y y y3 1 2n n n[ ] = [ ] + [ ].Additive.
Since the system is both homogeneous and additive, it is linear.
Since the system is linear it is also incrementally linear.
Since the system is linear, it is not statically non-linear.
Time Invariance:
Let x g1 n n[ ] = [ ]. Then n n n ny y g1 18 1[ ] [ ] = [ ]Let x g2 0n n n[ ] = [ ]. Then n n n n ny y g2 2 08 1[ ] [ ] = [ ]We can re-write the first equation as
n n n n n n n n( ) [ ] [ ] = [ ]0 1 0 1 0 08 1y y gThen, equating results,
n n n n n n n n n( ) [ ] [ ] = [ ] [ ]0 1 0 1 0 2 28 1 8 1y y y yThis equation cannot be satisfied for all n, therefore
y y2 1 0n n n[ ] [ ].Time Variant.
Stability:If x is bounded, so is y.Stable.
Causality:We can rearrange the system equation into
y g y1 18 1
nn n
n[ ] = [ ] + [ ]
showing that the response at time, n, depends on the excitation at time, n, and the responseat a previous time.Causal.
Memory:The response depends on past values of the response.The system has memory.
Invertibility:The original system equation, n n n ny y x[ ] [ ] = [ ]8 1 , expresses the excitation in terms ofthe response.Invertible.
-
M. J. Roberts - 7/12/03
Solutions 3-54
41. A DT system is described byy xn n[ ] = [ ] .
Classify this system as to linearity, BIBO stability, memory and invertibility.
Homogeneity:Let x g1 n n[ ] = [ ]. Then y g1 n n[ ] = [ ]Let x g2 n K n[ ] = [ ]. Then y g g2 n K n K n[ ] = [ ] = [ ] .Multiplying the first equation by K,
K n K n ny g y1 2[ ] = [ ] [ ] .Not homogeneous.
Let x g1 n n[ ] = [ ]. Then y g1 n n[ ] = [ ]Let x h2 n n[ ] = [ ]. Then y h2 n n[ ] = [ ]Let x g h3 n n n[ ] = [ ] + [ ]. Then y g h y y3 1 2n n n n n[ ] = [ ] + [ ] [ ] + [ ] .Not additive.
Since the system is not homogeneous and not additive it is also not linear.
The system is also not incrementally linear.
The system is statically non-linear because of the square-root relationship between excitationand response.
Time Invariance:
Let x g1 n n[ ] = [ ]. Then y g1 n n[ ] = [ ] .Let x g2 0n n n[ ] = [ ]. Then y g2 0n n n[ ] = [ ] .The first equation can be rewritten as
y g y1 0 0 2n n n n n[ ] = [ ] = [ ]Time invariant
Stability:If the excitation is bounded, the response is bounded.Stable
Causality:At any discrete time, n n= 0 , the response depends only on the excitation at that same time.Causal.
Memory:At any discrete time, n n= 0 , the response depends only on the excitation at that same time.System has no memory.
Invertibility:
-
M. J. Roberts - 7/12/03
Solutions 3-55
Inverting the functional relationship,x yn n[ ] = [ ]2 .
Invertible.
42. Graph the magnitude and phase of the complex-sinusoidal response of the systemdescribed by
y yn n e j n[ ] + [ ] = 12
1
as a function of .
The equation can be written as
y yn n e jn n[ ] + [ ] = ( ) =12 1
where = e j
The particular solution has the form,y n K n[ ] = .
Substituting the solution form into the equation,
K Kn n n + =12
1
or
K K + =12
.
Solving for K,
Ke
e
j
j=
+=
+
12
12
.
Therefore y n e
e
j
jn[ ] =
+
12
.
-
M. J. Roberts - 7/12/03
Solutions 3-56
F -2 2
|K|2
F -2 2
Phase of K
-
43. Find the impulse response, h n[ ], of the system in Figure E43.
D
y[n]x[n]0.9
2
Figure E43 DT system block diagram
y x . yn n n[ ] = [ ] + [ ]2 0 9 1or
y . y xn n n[ ] [ ] = [ ]0 9 1 2The homogeneous solution (for n 0) is of the form,
y n Khn[ ] =
therefore the characteristic equation is
K Khn
hn =0 9 01. .
and the eigenvalue is = 0 9. and, therefore, y .n Khn[ ] = ( )0 9
We can find an initial condition to evaluate the constant, Kh , by directly solving thedifference equation for n = 0.
y x . y0 2 0 0 9 1 2[ ] = [ ] + [ ] = .Therefore
2 0 9 20= ( ) =K Kh h. .Therefore the total solution is
y .n n[ ] = ( )2 0 9
-
M. J. Roberts - 7/12/03
Solutions 3-57
which is the impulse response.
44. Find the impulse responses of these systems.
(a) 3 4 1 2 1y y y x xn n n n n[ ] + [ ] + [ ] = [ ] + [ ]This can be written as
y x x y yn n n n n[ ] = [ ] + [ ] [ ] [ ]( )13 1 4 1 2 .The homogeneous solution is
yh hn
hnn K K[ ] = +1 1 2 2
where
1 223
13,
= .
The impulse response is the response to a single unit impulse at n = 0 plusthe response to another single unit impulse at n = 1. The response to a single unit impulse atn = 0 is
h u0 1 213
1n K K nhn
hn[ ] = + ( )
[ ]
subject to the initial conditions, h0 013[ ] = and h0 1
49[ ] = (found from the recursion
relation). Solving for the constants,
K Kh h1 216
12
= =, .
Therefore
h u016
13
12
1n nn
n[ ] = + ( )
[ ]and
h u un n nn
n
n
n[ ] = + ( )
[ ] + + ( )
[ ]
16
13
12
1 16
13
12
1 11
1
or
h u u u un n n n nn
n[ ] = [ ] [ ]( ) + ( ) [ ] [ ]( )16
13
3 1 12
1 1
(b) 52
6 1 10 2y y y xn n n n[ ] + [ ] + [ ] = [ ]The homogeneous solution is
-
M. J. Roberts - 7/12/03
Solutions 3-58
yh hn
hnn K K[ ] = +1 1 2 2
where1 2 1 2 1 6, . .= j .
h . . . . un K j K j nh n h n[ ] = +( ) + +( )( ) [ ]1 21 2 1 6 1 2 1 6h 0 2
5[ ] = and h 12425[ ] =
Solving for the constants,
K j K jh h1 20 2 0 15 0 2 0 15= + = . . , . .
h . . . . . . . . un j j j j nn n[ ] = +( ) +( ) + ( ) ( )( ) [ ]0 2 0 15 1 2 1 6 0 2 0 15 1 2 1 6h . . . . u. . . .n j e j e nj n j n[ ] = +( ) + ( )( ) [ ]+( ) ( )0 2 0 15 0 2 0 150 693 2 214 0 693 2 214
h . . . . u. . . . .n e e e j e j e nn j n j n j n j n[ ] = + + ( ) [ ] 0 693 2 214 2 214 2 214 2 2140 2 0 2 0 15 0 15h . cos . . sin . un n n nn[ ] = ( ) ( )( ) [ ]2 0 4 2 2143 0 3 2 2143
45. Plot g n[ ]. Use the MATLAB conv function if needed.
(a) g rect sinn n n[ ] = [ ] 12
9
n-10
10
g[n]
-3
3
......
G drcl , comb combF F j F F( ) = ( ) +
3 3 2
19
19
G drcl , comb drcl , combF j F F F F( ) = ( ) + ( )
32
3 19
3 19
G drcl , comb drcl , combF j F F( ) = +
32
19
3 19
19
3 19
Since the Dirichlet function is even,
G drcl , comb drcl , combF j F F( ) = +
32
19
3 19
19
3 19
-
M. J. Roberts - 7/12/03
Solutions 3-59
g drcl , sin . sinn n n[ ] = = 319
3 29
2 5321 29
(b) g rect sinn n n[ ] = [ ] 22
9
n-1010
g[n]
-3
3
......
G drcl , comb combF F j F F( ) = ( ) +
5 5 2
19
19
G drcl , comb drcl , combF j F F F F( ) = ( ) + ( )
52
5 19
5 19
G drcl , comb drcl , combF j F F( ) = +
52
19
5 19
19
5 19
Since the Dirichlet function is even,
G drcl , comb drcl , combF j F F( ) = +
52
19
5 19
19
5 19
g drcl , sin . sinn n n[ ] = = 519
5 29
2 8794 29
(c) g rect sinn n n[ ] = [ ] 42
9
n-10 10
g[n]
-3
3
G drcl , comb combF F j F F( ) = ( ) +
9 9 2
19
19
G drcl , comb drcl , combF j F F F F( ) = ( ) + ( )
92
9 19
9 19
G drcl , comb drcl , combF j F F( ) = +
92
19
9 19
19
9 19
Since the Dirichlet function is even,
-
M. J. Roberts - 7/12/03
Solutions 3-60
G drcl , comb drcl , combF j F F( ) = +
92
19
9 19
19
9 19
g drcl , sinn n[ ] = =919
9 29
0
(d) g rect rect combn n n n[ ] = [ ] [ ] [ ]3 3 14
-30 -20 -10 0 10 20 30
7
n
g[n]
... ...
rect rect rect rect rect tri3 3 3 3 33
3
77
n n m n m n mn
m m
[ ] [ ] = [ ] [ ] = [ ] = =
=
g rect rect comb tri combn n n n n n[ ] = [ ] [ ] [ ] = [ ]3 3 14 147 7g tri trin n n m n m
m m
[ ] = [ ] =
=
=
7 7 14 7147
(e) g rect rect combn n n n[ ] = [ ] [ ] [ ]3 3 7
7
n
g[n]
......
rect rect rect rect rect tri3 3 3 3 33
3
77
n n m n m n mn
m m
[ ] [ ] = [ ] [ ] = [ ] = =
=
g rect rect comb tri combn n n n n n[ ] = [ ] [ ] [ ] = [ ]3 3 7 77 7
-
M. J. Roberts - 7/12/03
Solutions 3-61
g tri trin n n m n mm m
[ ] = [ ] = =
=
=
7 7 7 77
77
(f) g cos un n nn
[ ] = [ ]227
78
10 15 20 25 30 35
-0.1
0
0.1
n
g[n]
(g) gsinc sinc
n
n n
[ ] =
42 2
42 2
-40 -30 -20 -10 10 20 30 40
0.5
g[n]
n
46. Find the impulse responses of the subsystems in Figure E 46 and then convolve them tofind the impulse response of the cascade connection of the two subsystems. You mayfind this formula for the summation of a finite series useful,
n
n
NN
N
=
==
0
1 111
1
,
,
.
D
y [n]x[n] 1
D
y [n]245
Figure E 46 Two cascaded subsystems
y x y1 1 1n n n[ ] = [ ] [ ] and y y y2 1 245 1n n n[ ] = [ ] [ ]
-
M. J. Roberts - 7/12/03
Solutions 3-62
h u1 1n nn[ ] = ( ) [ ] and h u2 45n n
n
[ ] = [ ]
h h h u u u un n n n n m n mnn
m
n m
m
[ ] = [ ] [ ] = ( ) [ ] [ ] = ( ) [ ] [ ]
=
1 2 1 45 145
h un n mmn m
m
m
n m
m
n
[ ] = ( ) [ ] = ( )
=
=
1 45 1450 0
h nn
m
m
m
n n m
m
n
[ ] = ( ) =
= =
45 145
45
540 0
h nn
n
n n n
n[ ] =
=
= ( )
+
+45
1 54
1 54
4 45
1 54
4 45
54
1
1
1
h n nn
n
n
[ ] = ( )
= ( )
+
4 1 45
54
5 1 1 45
1
47. For the system of Exercise 43, let the excitation, x n[ ], be a unit-amplitude complexsinusoid of DT cyclic frequency, F. Plot the amplitude of the response complexsinusoid versus F over the range, <
-
M. J. Roberts - 7/12/03
Solutions 3-63
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
-1
1
5
10
15
20
F
-0.8 -0.6 -0.4 -0.2 0
0.2 0.4 0.6 0.8
-1.5
-1
-0.5
0.5
1
1.5
F
|K|
Phase of K
48. In the second-order DT system below what is the relationship between a, b and c thatensures that the system is stable?
D
x[n] y[n]
b
a
Dc
1
yx y y
nn b n c n
a[ ] = [ ] [ ] + [ ]( )1 2
Stability is determined by the eigenvalues of the homogeneous solution.
a n b n c ny y y[ ] + [ ] + [ ] =1 2 0The eigenvalues are
1 2
2 42,
=
b b aca
For stability the magnitudes of all the eigenvalues must be less than one. Therefore
+ 0,
then find an expression for the current, i t( ), for time, t > 0.R = 2 1 C = 3 F
V = 10 V R = 6 2t = 0i(t)
i (t)v (t)+-
s i (t)
s
C
C
i i it t ts C( ) = ( ) + ( ) , is st VR( ) = 1 , i vC Ct Cddt
t( ) = ( )( )
v iC Ct t R( ) + ( ) =2 0 , v vC Ct R C ddt t( ) + ( )( ) =2 0
-
M. J. Roberts - 7/12/03
Solutions 3-71
vCtt Ke( ) = , = 1
2R C, vC
t
R Ct Ke( ) = 2
vC
t
R CKe K K0 10 102( ) = = = = , vCt
R Ct e( ) = 10 2
i vC Ct
R Ct
R Ct C ddt
tR C
CeR
e( ) = ( )( ) = = 10 102 2
2 2
i t VR R
e est
R Ct
( ) = + = + 1 2
1810 5 53
2
56. The water tank in Figure E56 is filled by an inflow, x t( ) , and is emptied by an outflow,y t( ) . The outflow is controlled by a valve which offers resistance, R , to the flow ofwater out of the tank. The water depth in the tank is d t( ) and the surface area of thewater is A, independent of depth (cylindrical tank). The outflow is related to the waterdepth (head) by
y dt tR
( ) = ( ) .
The tank is 1.5 m high with a 1m diameter and the valve resistance is 10 sm2
.
(a) Write the differential equation for the water depth in terms of the tank dimensionsand valve resistance.
(b) If the inflow is 0 05. m3
s, at what water depth will the inflow and outflow rates be
equal, making the water depth constant?
(c) Find an expression for the depth of water versus time after 1 3m of water is dumpedinto an empty tank.
(d) If the tank is initially empty at time, t = 0, and the inflow is a constant 0 2. m3
s after
time, t = 0, at what time will the tank start to overflow?
Inflow, x(t)
d(t)
Outflow, y(t)
R
Surface area, A
Valve
-
M. J. Roberts - 7/12/03
Solutions 3-72
Figure E56 Water tank with inflow and outflow
(a)y dt t
R( ) = ( )
ddt
A t t td x y( ) = ( ) ( )volume123
A t t tR
( ) = ( ) ( )d x d
A t tR
t( ) + ( ) = ( )d d x
(b) For the water height to be constant, ( ) =d t 0. Then
d x . .t R t( ) = ( ) = =10 0 05 0 5sm
m
sm2
3
(c) Dumping 1 3m of water into an empty tank is exciting this system with a unit impulseof water inflow. The impulse response, h t( ), of the system is the solution of
A t tR
t( ) + ( ) = ( )h h The solution is
h ut Ke tt
AR( ) = ( ) .We can find K by finding the initial water depth in response to 1 3m being suddenly dumpedin. The surface area is 0 7854 2. m . Therefore the initial depth is 1.273 m and
h . ut e tt
AR( ) = ( )1 273 .(d) The response to a step of flow is the convolution of the impulse response with thestep excitation.
d h x . u . ut t t e t tt
AR( ) = ( ) ( ) = ( ) ( )1 273 0 2
d . u u . ut e t d e t dAR AR( ) = ( ) ( ) = ( )
0 2546 0 25460
For t < 0, d t( ) = 0For t > 0,
-
M. J. Roberts - 7/12/03
Solutions 3-73
d . . .t e d AR e AR eARt
AR
t t
AR( ) = = =
0 2546 0 2546 0 2546 10 0
For all time,
d u.t e tt
( ) = ( )
2 1 7 854 .
Solving for a depth of 1.5 m,
1 5 2 1 0 25 1 3867 854
10 8867 854 7 854. . ..
.
. .
=
= =
e ett t
s
57. The suspension of a car can be modeled by the mass-spring-dashpot system of FigureE57 Let the mass, m, of the car be 1500 kg, let the spring constant, Ks, be 75000
Nm
and let the shock absorber (dashpot) viscosity coefficient, Kd , be 20000N sm
.
At a certain length, d0, of the spring, it is unstretched and uncompressed and exerts noforce. Let that length be 0.6 m.
(a) What is the distance, y xt t( ) ( ), when the car is at rest?(b) Define a new variable z y xt t t( ) = ( ) ( ) constant such that, when the system is atrest, z t( ) = 0 and write a describing equation in z and x which describes an LTI system.Then find the impulse response.
(c) The effect of the car striking a curb can be modeled by letting the road surface heightchange discontinuously by the height of the curb, hc. Let hc = 0 15. m. Graph z t( ) versustime after the car strikes a curb.
Automobile Chassis
y(t)
x(t)
Spring ShockAbsorber
Figure E57 Car suspension model
Using the basic principle, F ma= , we can write
K t t d K ddt
t t mg m ts dy x y x y( ) ( ) [ ] + ( ) ( )[ ] + = ( )0
-
M. J. Roberts - 7/12/03
Solutions 3-74
orm t K t K t K t K t K d mgd s d s s( ) + ( ) + ( ) = ( ) + ( ) + y y y x x 0 .
(a) At rest all the derivatives are zero and
K t t d mgs y x( ) ( ) ( ) + =0 0 .Solving,
y x . . .t t K d mgK
s
s
( ) ( ) = = =0 75000 0 6 1500 9 875000
0 404 m
(b) The describing equation is
m t K t K t K t K t K d mgd s d s s( ) + ( ) + ( ) = ( ) + ( ) + y y y x x 0 .which can be rewritten as
m t K t t K t t K d mgd s s( ) + ( ) ( )[ ] + ( ) ( )[ ] + =y y x y x 0 0or
m t K t t K t t d mgKd s s
( ) + ( ) ( )[ ] + ( ) ( ) + =y y x y x 0 0
Let z y xt t t d mgKs
( ) = ( ) ( ) +0 . Then ( ) = ( ) + ( )y z xt t t and
m t t K t K td s( ) + ( )[ ] + ( ) + ( ) =z x z z 0or
m t K t K t m td s( ) + ( ) + ( ) = ( )z z z xThis equation is in a form which describes an LTI system. We can find its impulseresponse. After time, t = 0, the impulse response is the homogenous solution. Theeigenvalues are
1 22 2
2
42 2 4
6 667 2 357,
. .=
= =
K K mKm
Km
Km
Km
jd d s d d s .
The homogeneous solution is
h . . . .t K e K e K e K eht
ht
hj t
hj t( ) = + = + +( ) ( )1 2 1 6 667 2 357 2 6 667 2 3571 2 .
Since the system is underdamped another (equivalent) form of homogeneous solution willbe more convenient,
h cos . sin ..t e K t K tt h h( ) = ( ) + ( )[ ]6 667 1 22 357 2 357 .The impulse response can have a discontinuity at t = 0 and an impulse but no higher-ordersingularity there. Therefore the general form of the impulse response is
-
M. J. Roberts - 7/12/03
Solutions 3-75
h cos . sin . u.t K t e K t K t tt h h( ) = ( ) + ( ) + ( )[ ] ( ) 6 667 1 22 357 2 357Integrating both sides of the describing equation,
m K K t dtd s( ) ( )( ) + ( ) ( )( ) + ( ) =+ +
+
h h h h h0 0 0 0 00
0
.
(The integral of the doublet, which is the derivative of the impulse excitation, is zero.)Since the impulse response and all its derivatives are zero before time, t = 0, it follows thenthat
m K K t dtd s( ) + ( ) + ( ) =+ +
+
h h h0 0 00
0
andm K K K K K Kh h d h s +( ) + + =6 667 2 357 01 2 1. . .
Integrating the describing equation a second time,
m K t dtdh h0 00
0+( ) + ( ) =
+
or
mK K Kh d1 0+ = .
Integrating the describing equation a third time,
m t dt mh( ) =
+
0
0
ormK m K= = 1 .
Solving for the other two constants, K Km
hd
1 = and
mKm
K KKm
Kd h dd
s + + =6 667 2 357 02. .
or
K
Km
Km
Km
h
s d d
2
2
2 6 667
2 357=
+ .
.
Thereforeh . cos . . sin . u.t t e t t tt( ) = ( ) + ( ) ( )[ ] ( ) 6 667 13 333 2 357 16 497 2 357
(c) The response to a step of size 0.15 is then the convolution,
Km
d
-
M. J. Roberts - 7/12/03
Solutions 3-76
z . u ht t t( ) = ( ) ( )0 15or
z . . cos . . sin . u u.t e t d( ) = ( ) + ( ) ( )[ ] ( ){ } ( )
0 15 13 333 2 357 16 497 2 3576 667
z . . cos . . sin . u.t e t d( ) = ( ) + ( ) ( )[ ]{ } ( )
0 15 13 333 2 357 16 497 2 3576 6670
For t < 0, z t( ) = 0.For t > 0,
using
e bx dx ea b
a bx b bxaxax
sin sin cos( ) =+
( ) ( )[ ] 2 2e bx dx e
a ba bx b bxax
ax
cos cos sin( ) =+
( ) + ( )[ ] 2 2we get
z . u .
. . cos . . sin .
. . sin . . cos .
.
.
t t
e
e
t
( ) = ( ) + ( ) + ( )[ ]
( ) ( )[ ]
0 15 0 1513 333
506 667 2 357 2 357 2 357
16 49750
6 667 2 357 2 357 2 357
6 667
6 667
0
or
z . u .
. . cos . . sin .
. . sin . . cos .
.
.
.
.
.
.
t t
et t
et t
t
t
( ) = ( ) +
( ) + ( )[ ] ( ) ( )[ ]
+
0 15 0 15
13 33350
6 667 2 357 2 357 2 357
16 49750
6 667 2 357 2 357 2 357
13 333 6 66750
16 497 2 35750
6 667
6 667
z . u . . sin . cos . u.t t e t t tt( ) = ( ) + ( ) ( )[ ] +{ } ( )0 15 0 15 2 812 2 357 2 357 13 333or
z . . sin . cos . u.t e t t tt( ) = ( ) ( )[ ] ( )0 15 2 812 2 357 2 3573 333
t2
z(t)
-0.2
0.1
58. As derived in the text, a simple pendulum is approximately described for small angles, , by the differential equation,
-
M. J. Roberts - 7/12/03
Solutions 3-77
mL t mg t t( ) + ( ) ( ) xwhere m is the mass of the pendulum, L is the length of the massless rigid rod supportingthe mass and is the angular deviation of the pendulum from vertical.
(a) Find the general form of the impulse response of this system.
mL t mg t t( ) + ( ) ( )h h The form of the homogeneous solution is
h hj g
Lt
h
j gL
tt K e K e t( ) = +
( )
1 2 u
or, more conveniently,
h h ht KgL
t KgL
t t( ) = +
( )1 2cos sin u .
There can be no discontinuity or impulse in the impulse response therefore this is also theimpulse response. Integrate the differential equation once through zero.
mLgL
K KLg mL m gLh h2 2
1 1 1 1
= =
Now integrate again through zero.
Kh1 0= .
Therefore
h sin utm gL
gL
t t( ) = ( )
1 1
(b) If the mass is 2 kg and the rod length is 0.5 m, at what cyclic frequency will thependulum oscillate?
The cyclic frequency is 12
9 812
0 704
.
.= . The mass is irrelevant.
59. Pharmacokinetics is the study of how drugs are absorbed into, distributed through,metabolized by and excreted from the human body. Some drug processes can beapproximately modeled by a one compartment model of the body in which V is thevolume of the compartment, C t( ) is the drug concentration in that compartment, ke is arate constant for excretion of the drug from the compartment and k0 is the infusion rateat which the drug enters the compartment.
-
M. J. Roberts - 7/12/03
Solutions 3-78
(a) Write a differential equation in which the infusion rate is the excitation and the drugconcentration is the response.
(b) Let the parameter values be ke = 0 4 1. hr , V = 20 l and k0 200=mghr
(where l is the
symbol for liter). If the initial drug concentration is C 0 10( ) = mgl
, plot the drugconcentration as a function of time (in hours) for the first 10 hours of infusion. Find thesolution as the sum of the zero-excitation response and the zero-state response.
(a) The differential equation is
V ddt
t k Vk teC C( )( ) = ( )0or
ddt
t k t kVe
C C( )( ) + ( ) = 0(b) The eigenvalue is 0.4 and the zero-excitation response is
C .t e t( ) = 10 0 4 mgl
(t in hours).
The impulse response (to a unit impulse of infusion rate) is h u.
te
tt
( ) = ( )0 4
20mg
l. The
step response to an infusion rate of k0 200=mghr
is then C .t e t( ) = ( )25 1 0 4 mgl . The sumof the two responses is
C . . .t e e et t t( ) = + ( )( ) = ( ) 10 25 1 25 150 4 0 4 0 4mgl mgl .
t (hours)10
C (t)25
60. At the beginning of the year 2000, the country, Freedonia, had a population, p, of 100million people. The birth rate is 4% per annum and the death rate is 2% per annum,compounded daily. That is, the births and deaths occur every day at a uniform fractionof the current population and the next day the number of births and deaths changesbecause the population changed the previous day. For example, every day the number
of people who die is the fraction, 0 02365.
, of the total population at the end of the previousday (neglect leap-year effects). Every day 275 immigrants enter Freedonia.
(a) Write a difference equation for the population at the beginning of the nth day afterJanuary 1, 2000 with the immigration rate as the excitation of the system.
-
M. J. Roberts - 7/12/03
Solutions 3-79
(b) By finding the zero-exctiation and zero-state responses of the system determine thepopulation of Freedonia be at the beginning of the year 2050.
(a) The difference equation is
p p . p . pn n n n+[ ] = [ ] + [ ] [ ] +1 0 043650 02365
275
p . pn n+[ ] + ( ) [ ] =1 1 5 48 10 2755The eigenvalue is 1 5 48 10 1 00005485+ =. . and the zero-excitation response is
p .n n[ ] = ( )10 1 00005488
The impulse response is h . un nn[ ] = ( ) [ ]1 0000548 . The response to the immigration rate isthe convolution of the impulse response with the immigration rate, 275u n[ ], or
p . u . u . .n n nn mm
n
[ ] = ( ) [ ] [ ] = ( )=
1 0000548 273 9 273 9 1 00005480
1
Using the summation formular for a geometric series,
p ..
u.
.
u . . un n n nn n
n[ ] =
[ ] = [ ] = ( ) [ ]2751 1 00005481 1 0000548 2751 0000548 1
0 00005485018248 2 1 0000548 1
The total solution isp . . . u ,n n nn n[ ] = ( ) + ( ) [ ] 10 1 0000548 5018248 2 1 0000548 1 08
p . . . ,n nn[ ] = ( ) 1 05 10 1 0000548 5018248 2 08(b) The beginning of the year 2050 is the 18250th day.
p . . . , ,18250 1 05 10 1 0000548 5018248 2 280 420 0008 18250[ ] = ( ) =61. A car rolling on a hill can be modeled as shown in Figure E61. The excitation is the
force, f t( ), for which a positive value represents accelerating the car forward with themotor and a negative value represents slowing the car by braking action. As it rolls, thecar experiences drag due to various frictional phenomena which can be approximatelymodeled by a coefficient, k f , which multiplies the cars velocity to produce a forcewhich tends to slow the car when it moves in either direction. The mass of the car is mand gravity acts on it at all times tending to make it roll down the hill in the absence ofother forces. Let the mass, m, of the car be 1000 kg, let the friction coefficient, k f , be
5 N sm
and let the angle, , be 12
.
-
M. J. Roberts - 7/12/03
Solutions 3-80
(a) Write a differential equation for this system with the force, f t( ), as the excitation andthe position of the car, y t( ) , as the response.
(b) If the nose of the car is initially at position, y 0 0( ) = , with an initial velocity,( )[ ] =
=
y tt 0
10 ms
, and no applied acceleration or braking force, graph the velocity of the car,
( )y t , for positive time.(c) If a constant force, f t( ), of 200 N is applied to the car what is its terminal velocity ?
mgsin()
f(t)y(t)
Figure E61 Car on an inclined plane
(a) Summing forces,f sin y yt mg k t m tf( ) ( ) ( ) = ( )
orm t k t mg tf( ) + ( ) + ( ) = ( )y y sin f
(b) The zero-excitation response can be found by setting the force, f t( ), to zero yieldingm t k t mgf( ) + ( ) = ( )y y sin
The homogeneous solution is yh h hkm
tt K K e
f
( ) = + 1 2 . The particular solution must be in theform of a linear function of t, to satisfy the differential equation. Choosing the form,
y p pt K t( ) =and solving, K mg
kp f= ( )sin . Then the total zero-excitation response is
y sint K K e mgk
th h
km
t
f
f
( ) = + ( )1 2 Using the initial conditions,
y 0 0 1 2( ) = = +K Kh hand
( ) = = ( )y sin0 10 2km
Kmgk
fh
f .
Solving,
-
M. J. Roberts - 7/12/03
Solutions 3-81
Km
kg
m
kh f f2
2
5 510 1 0146 10 2000 1 0346 10=
( ) = = sin . .
and Kh151 0346 10= .
y . .t e tt
( ) =
1 0346 10 1 507 285 200
( ) = = =
+
y . . . . .t e e et t t1 0346 10
200507 28 517 28 507 28 517 28 1 10
5200 200 200
t1000
y(t)
-550
(c) The differential equation is
m t k t mg tf( ) + ( ) + ( ) = ( )y y sin fWe can re-write the equation as
m t k t t mgf( ) + ( ) = ( ) ( )y y f sin treating the force due to gravity as part of the excitation. Then the impulse response is thesolution of
m t k t tf( ) + ( ) = ( )h h which is of the form,
h ut K K e th hkm
tf
( ) = + ( )
1 2 .
Integrating both sides of the differential equation through t = 0 we get
m k mkm
K k K Kff
h f h h( ) + ( ) = = + +( )+ +h h0 0 1 2 1 2
Integrating a second time yields,
m m K Kh hh 0 0 1 2+( ) = = +( ) .
Solving,k K
Kf h
h
01 1
10
1
2
=
we get
-
M. J. Roberts - 7/12/03
Solutions 3-82
Kk
Kkh f
hf
1 21 1
= = , .
So the impulse response is
h ut ek
t
km
t
f
f
( ) = ( )
1 .
Now, if we say that the force, f t( ), is a step of size, 200 N, the excitation of the system isx u sint t mg( ) = ( ) ( )200 .
But this is going to cause a problem. The problem is that the term, ( )mgsin , is aconstant, therefore presumed to have acted on the system for all time before time, t = 0.The implication from that is that the position at time, t = 0, is at infinity. Since we are onlyinterested in the final velocity, not position, we can assume that the car was held in place aty t( ) = 0 until the force was applied and gravity was allowed to act on the car. That makesthe excitation,
x sin ut mg t( ) = ( )[ ] ( )200 and the response is
y x h sin u ut t t mg t ek
t
km
t
f
f
( ) = ( ) ( ) = ( )[ ] ( ) ( )
200 1
or
y sint mgk
e df
km
t f
( ) = ( )
200 10
or
y sin sint mgk
m
ke
mgk
tm
ke
m
kf f
km
t
f f
km
t
f
f f
( ) = ( ) +
=
( )+
200 200
0
The terminal velocity is the derivative of position as time approaches infinity which, in thiscase is
+( ) = ( ) = = y sin . .200 200 2536 435
467 3mgkf
ms
.
Obviously a force of 200 N is insufficient to move the car forward and its terminal velocityis negative indicating it is rolling backward down the hill.
62. A block of aluminum is heated to a temperature of 100 C. It is then dropped into aflowing stream of water which is held at a constant temperature of 10C. After 10seconds the temperature of the ball is 60C. (Aluminum is such a good heat conductorthat its temperature is essentially uniform throughout its volume during the coolingprocess.) The rate of cooling is proportional to the temperature difference between theball and the water.
(a) Write a differential equation for this system with the temperature of the water as theexcitation and the temperature of the block as the response.
-
M. J. Roberts - 7/12/03
Solutions 3-83
(b) Compute the time constant of the system.(c) Find the impulse response of the system and, from it, the step response.(d) If the same block is cooled to 0 C and dropped into a flowing stream of water at 80C, at time, t = 0, at what time will the temperature of the block reach 75C?
(a) The controlling differential equation isddt
t K ta w aT T T( ) = ( )( )or
1K
ddt
t ta a wT T T( ) + ( ) =where Ta is the temperature of the aluminum ball and Tw is the temperature of the water.The solution is
TaKtt e( ) = +90 10
(b) We can find the constant, K, by using the temperature after 10 seconds,
T .aKe K10 60 90 10 0 058810( ) = = + = .
(c) The impulse response is the solution of the equation,1K
ddt
t t th h( ) + ( ) = ( ) .The form of the solution is
h ut K e thKt( ) = ( )
Integrating both sides of the differential equation through t = 0,
1 0 1K
KK
K Kh hh+( ) = = =
Thereforeh u . u.t Ke t e tKt t( ) = ( ) = ( ) 0 0588 0 0588 .
The unit step response is the integral of the impulse response,
h u.
( ) = ( ) ( )1 0 05881t e tt .(d) The response is the response to a step of 80 C .
T t t e tat( ) = ( ) = ( ) ( )
80 80 110 0588h u. .
To find the time at which the temperature is 75 C, t75, solve
T t e tat
750 0588
7575 80 1 75( ) = = ( ) ( ) . u .
-
M. J. Roberts - 7/12/03
Solutions 3-84
Solving,t75 47 153= . .
63. A well-stirred vat has been fed for a long time by two streams of liquid, fresh water at0.2 cubic meters per second and concentrated blue dye at 0.1 cubic meters per second.The vat contains 10 cubic meters of this mixture and the mixture is being drawn from thevat at a rate of 0.3 cubic meters per second to maintain a constant volume. The blue dyeis suddenly changed to red dye at the same flow rate. At what time after the switch doesthe mixture drawn from the vat contain a ratio of red to blue dye of 99:1?
Let the concentration of red dye be denoted by C tr ( ) and the concentration of bluedye be denoted by C tb ( ). The concentration of water is constant throughout at 23. Therates of change of the dye concentrations are governed by
ddt
VC t C t fb b draw( )( ) = ( )ddt
VC t f C t fr r r draw( )( ) = ( )where V is the constant volume, 10 cubic meters, fdraw is the flow rate of the draw from thevat and fr is the flow rate of red dye into the tank. Solving the two differential equations,
C t ebf
Vtdraw( ) = 1
3and
C t erf
Vtdraw( ) =
13
1 .
Then the ratio of red to blue dye concentration is
C tC t
e
e
e
e
er
b
fV
t
fV
t
fV
t
fV
t
fV
t
draw
draw
draw
draw
draw( )( ) =
=
=
13
1
13
1 1 .
Setting that ratio to 99 and solving for t99,
99 1 153 50 310
9999
= =e tt
.
. seconds
64. Some large auditoriums have a noticeable echo or reverberation. While a littlereverberation is desirable, too much is undesirable. Let the response of an auditorium toan acoustic impulse of sound be
h t e t nnn
( ) = =
50 .
-
M. J. Roberts - 7/12/03
Solutions 3-85
We would like to design a signal processing system that will remove the effects ofreverberation. In later chapters on transform theory we will be able to show that thecompensating system that can remove the reverberations has an impulse response of theform,
h gcn
t n