Signals and Systems: Analysis of Signals Through Linear Systems - Solution manual

681 Pages·2003·6.02 MB·English

by  M.J. Roberts

M. J. Roberts - 7/12/03
Chapt er 2 - Mathematical Description of Signals
Solutions
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 +4 = 7e−5t−11
 
10
( )
(d) g jt = 7e−j2t−3
( ) ( )
(e) g jt +g −jt = 7e−3 e−j2t +ej2t = 7e−3cos(2t)
2 2
 jt−3 −jt−3
g  +g 
(f)  2   2  = 7e−jt +ejt = 7cos(t)
2 2
2. If g(x)= x2−4x +4 write out and simplify
(a) g(z)= z2−4z+4
(b) g(u+v)=(u+v)2−4(u+v)+4 = u2 +v2 +2uv−4u−4v +4
( ) ( ) ( )
(c) g ejt = ejt 2−4ejt +4 =ej2t −4ejt +4 = ejt −2 2
(d) g(g(t))=g(t2 −4t+4)=(t2 −4t+4)2−4(t2−4t+4)+4
g(g(t))= t4 −8t3 +20t2 −16t+4
(e) g(2)= 4−8+4 =0
3. What would be the numerical value of “g” in each of the following MATLAB
instructions?
(a) t = 3 ; g = sin(t) ; 0.1411
(b) x = 1:5 ; g = cos(pi*x) ; [-1,1,-1,1,-1]
(c) f = -1:0.5:1 ; w = 2*pi*f ; g = 1./(1+j*w) ;
Solutions 2-1
M. J. Roberts - 7/12/03
0.0247+ j0.155
 
0.0920+ j0.289
 
 
1
 
0.0920− j0.289
0.0247− j0.155
4. Let two functions be defined by
1 , sin(20πt)≥0 t , sin(2πt)≥0
x (t)= and x (t)= .
1 −1 , sin(20πt)<0 2 −t , sin(2πt)<0
Graph the product of these two functions versus time over the time range, −2< t<2.
x(t)
2
t
-2 2
-2
5. For each function, g(t), sketch g(−t), −g(t), g(t−1), and g(2t).
(a) (b)
g(t) g(t)
4 3
t -1 t
2 1
-3
g(-t) g(-t) -g(t) -g(t)
4 3 3
t -1 t t 1 t
-2 1 2 -1
-3 4 -3
g(t-1) g(t-1) g(2t) g(2t)
4 3 4 3
-1
2
t t t t
1 3 1 2 1 1
2
-3 -3
( )
6. A function, G f , is defined by
Solutions 2-2
M. J. Roberts - 7/12/03
( )  f 
G f =e−j2πf rect  .
 
2
( ) ( )
Graph the magnitude and phase of G f −10 +G f +10 over the range, −20< f <20.
G(f −10)+G(f +10)=e−j2π(f−10)rect f −10 +e−j2π(f+10)rect f +10
   
2 2
|G( f )|
1
f
-20 20
Phase of G( f )
π
f
-20 20
-π
7. Sketch the derivatives of these functions.
(All sketches at end.)
(a) g(t)=sinc(t) g′(t)= π2tcos(πt)−πsin(πt) = πtcos(πt)−sin(πt)
(πt)2 πt2
(b) g(t)=(1−e−t)u(t) g′(t)=e−t , t≥0=e−tu(t)
0 , t<0
(a) (b)
x(t) x(t)
1 1
t t
-4 4 -1 4
-1 -1

dx/dt dx/dt
1 1
t t
-4 4 -1 4
-1 -1
8. Sketch the integral from negative infinity to time, t, of these functions which are zero for
all time before time, t=0.
Solutions 2-3
M. J. Roberts - 7/12/03
g(t) g(t)
1 1
1 2 3
t t
1 1 2 3
2
∫ g(t) dt ∫ g(t) dt
1 1
1
2
t t
1 2 3 1 2 3
9. Find the even and odd parts of these functions.
(a) g(t)=2t2−3t+6
g (t)= 2t2 −3t+6+2(−t)2 −3(−t)+6 = 4t2 +12 =2t2 +6
e 2 2
g (t)= 2t2 −3t+6−2(−t)2 +3(−t)−6 = −6t =−3t
o 2 2
 π
(b) g(t)=20cos40πt− 
 
4
 π  π
20cos40πt−  +20cos−40πt− 
   
g (t)= 4 4
e 2
( ) ( ) ( ) ( ) ( )
Using cos z +z =cos z cos z −sin z sin z
1 2 1 2 1 2
   π  π 
20cos(40πt)cos−  −sin(40πt)sin−  
   
  4 4  
 
+20cos(−40πt)cos−π −sin(−40πt)sin−π
g (t)=   4 4 
e 2
  π π 
20cos(40πt)cos  +sin(40πt)sin  
   
  4 4  
 
+20cos(40πt)cosπ −sin(40πt)sinπ
g (t)=   4 4 
e 2
Solutions 2-4
M. J. Roberts - 7/12/03
g (t)=20cosπcos(40πt)= 20 cos(40πt)
e 4 2
 π  π
20cos40πt−  −20cos−40πt− 
   
g (t)= 4 4
o 2
( ) ( ) ( ) ( ) ( )
Using cos z +z =cos z cos z −sin z sin z
1 2 1 2 1 2
   π  π 
20cos(40πt)cos−  −sin(40πt)sin−  
   
  4 4  
 
−20cos(−40πt)cos−π −sin(−40πt)sin−π
g (t)=   4 4 
o 2
  π π 
20cos(40πt)cos  +sin(40πt)sin  
   
  4 4  
 
−20cos(40πt)cosπ −sin(40πt)sinπ
g (t)=   4 4 
o 2
g (t)=20sinπsin(40πt)= 20 sin(40πt)
o 4 2
(c) g(t)= 2t2 −3t+6
1+t
2t2 −3t+6 2t2 +3t+6
+
g (t)= 1+t 1−t
e 2
(2t2−3t+6)(1−t)+(2t2 +3t+6)(1+t)
g (t)= (1+t)(1−t)
e 2
g (t)= 4t2 +(12+)6t2 = 6+5t2
e 21−t2 1−t2
2t2 −3t+6 2t2 +3t+6
−
g (t)= 1+t 1−t
o 2
Solutions 2-5
M. J. Roberts - 7/12/03
(2t2−3t+6)(1−t)−(2t2 +3t+6)(1+t)
g (t)= (1+t)(1−t)
o 2
g (t)= −6t−(4t3 −)12t =−t2t2 +9
o 21−t2 1−t2
sin(πt) sin(−πt)
(d) g(t)=sinc(t) g (t)= πt + −πt = sin(πt) g (t)=0
e 2 πt o
(e) g(t)= t(2−t2)(1+4t2)
g(t)= {t (2−t2)(1+4t2)
12314243
odd
even even
Therefore g(t) is odd, g (t)=0 and g (t)= t(2−t2)(1+4t2)
e o
(f) g(t)= t(2−t)(1+4t)
g (t)= t(2−t)(1+4t)+(−t)(2+t)(1−4t) g (t)= 7t2
e 2 e
g (t)= t(2−t)(1+4t)−(−t)(2+t)(1−4t) g (t)= t(2−4t2)
o 2 o
10.Sketch the even and odd parts of these functions.
Solutions 2-6
M. J. Roberts - 7/12/03
g(t) g(t)
1 1
t t
1 1 2
-1
g (t) g (t)
e e
1 1
t t
1 1 2
-1
g (t) g (t)
o o
1 1
t t
1 1 2
-1
(a) (b)
( )
11.Sketch the indicated product or quotient, g t , of these functions.
(a) (b)
1 1
-1 -1
t t
1 1
-1 -1
g(t) g(t)
1
1
Multiplication Multiplication
-1
t t
-1 1 1
-1
g(t) g(t)
1 1
-1
t t
1 -1 1
-1 -1
Solutions 2-7
M. J. Roberts - 7/12/03
(c) (d)
1 1
t t
-1 1
g(t) g(t)
Multiplication Multiplication
1 1
t t
1 1
g(t) g(t)
-1 1
t
-1 1
t
-1 1
(e) (f)
1
... 1 ... t
1
t
-1 1 -1
g(t)
-1
g(t) 1
1 Multiplication
t
Multiplication 1
t
-1
-1 1
g(t)
g(t)
1
... ...
1
t
-1 1 t
1
-1 -1
(g) (h)
1 1
t t
-1 -1 -1 1
g(t) g(t)
π
Division Division
1
t t
1 1
g(t) g(t)
1
t
-1 -1 1 t
12. Use the properties of integrals of even and odd functions to evaluate these integrals in
the quickest way.
Solutions 2-8
M. J. Roberts - 7/12/03
1 1 1 1
(a) ∫(2+t)dt= ∫ 2{ dt+ ∫ {t dt=2∫2dt= 4

−1 −1even −1odd 0
(b)
1 1 1 1
2∫0 [4cos(10πt)+8sin(5πt)]dt= 2∫0 4cos(10πt)dt+ 2∫0 8sin(5πt)dt=82∫0cos(10πt)dt= 8
14243 14243 10π
−1 −1 even −1 odd 0
20 20 20
1
20
(c) ∫ 4 t{ cos(10πt)dt=0
14243
1 o1dd4243
− even
20
odd
1 1  1 
1
(d) 1∫0 t{ s1i4n(2104π3t)dt=21∫0tsin(10πt)dt=2−tcos1(01π0πt)10 +1∫0cos1(01π0πt)dt
−1 1odd42od4d 3 0  0 0 
10
even
1  1 
1∫0 t{ s1i4n(2104π3t)dt==21010π+ si(n1(01π0)π2t)10= 501π
−1 1odd42od4d 3  0 
10
even
1 1 1 [ ] ( )
(e) ∫e{−t dt=2∫e−tdt=2∫e−tdt=2 −e−t 1 =21−e−1 ≈1.264
0
−1even 0 0
1
(f) ∫ t{ e{−t dt=0
123
−1oddeven
odd
13.Find the fundamental period and fundamental frequency of each of these functions.
(a) g(t)=10cos(50πt) f =25 Hz , T = 1 s
0 0 25
(b) g(t)=10cos50πt+π f =25 Hz , T = 1 s
 4 0 0 25
(c) g(t)=cos(50πt)+sin(15πt)
 15 1
f =GCD25,  =2.5Hz , T = =0.4 s
0  2 0 2.5
(d) g(t)=cos(2πt)+sin(3πt)+cos5πt− 3π
 
4
Solutions 2-9
M. J. Roberts - 7/12/03
 3 5 1 1
f =GCD1, ,  = Hz , T = =2s
0  2 2 2 0 1
2
( )
14. Find the fundamental period and fundamental frequency of g t .
g(t) ... ...
t
1
(a) ... ... (b) +
g(t)
t
1
... ...
t
1
... ...
t
1
(c) +
g(t)
... ...
t
1
1
(a) f = 3Hz and T = s
0 0 3
(b) f =GCD(6,4) =2Hz and T = 1 s
0 0 2
(c) f =GCD(6,5) =1Hz and T =1s
0 0
15.Plot these DT functions.
[ ] 2πn 2π(n−2)
(a) x n = 4cos  −3sin  , −24≤ n <24
 12   8 
x[n]
7
n
-24 24
-7
n
[ ] −
(b) x n = 3ne 5 , −20≤ n<20
x[n]
6
n
-20 20
-6
[ ] n2
(c) x n =21  +14n3 , −5≤ n<5
 
2
Solutions 2-10