, M. J. Roberts - 7/12/03
Chapter 2 - Mathematical Description of Signals
Solutions
1. If g( t) = 7e −2 t − 3 write out and simplify
(a) g( 3) = 7e −9
(b) g(2 − t) = 7e −2( 2 − t ) − 3 = 7e −7 + 2 t
t
t
− −11
(c) g + 4 = 7e 5
10
(d) g( jt) = 7e − j 2 t − 3
g( jt) + g(− jt) e − j 2t + e j 2t
(e) = 7e −3 = 7e −3 cos(2 t)
2 2
jt − 3 − jt − 3
g + g
2 2 e − jt + e jt
(f) =7 = 7 cos( t)
2 2
2. If g( x ) = x 2 − 4 x + 4 write out and simplify
(a) g( z) = z 2 − 4 z + 4
g( u + v ) = ( u + v ) − 4 ( u + v ) + 4 = u 2 + v 2 + 2 uv − 4 u − 4 v + 4
2
(b)
g(e jt ) = (e jt ) − 4 e jt + 4 = e j 2 t − 4 e jt + 4 = (e jt − 2)
2 2
(c)
g(g( t)) = g( t 2 − 4 t + 4 ) = ( t 2 − 4 t + 4 ) − 4 ( t 2 − 4 t + 4 ) + 4
2
(d)
g(g( t)) = t 4 − 8 t 3 + 20 t 2 − 16 t + 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 + j 0.155
0.0920 + j 0.289
1
0.0920 − j 0.289
0.0247 − j 0.155
4. Let two functions be defined by
1 , sin(20πt) ≥ 0 t , sin(2πt) ≥ 0
x1 ( t) = and x 2 ( t) = .
−1 , sin(20πt) < 0 − 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
-1
2
t 1
t
-3
g(-t) g(-t) -g(t) -g(t)
4 3 3
-1 1
-2
t 1
t 2
t t
-1
-3 4 -3
g(t-1) g(t-1) g(2t) g(2t)
4 3 4 3
-1
2
1 3
t 1 2
t 1
t 1
t
2
-3 -3
6. A function, G( f ) , is defined by
Solutions 2-2
, M. J. Roberts - 7/12/03
f
G( f ) = e − j 2πf rect .
2
Graph the magnitude and phase of G( f − 10) + G( f + 10) over the range, −20 < f < 20 .
f − 10 f + 10
G( f − 10) + G( f + 10) = e − j 2π ( f −10) rect + e ( ) rect
− j 2π 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.)
π 2 t cos(πt) − π sin(πt) πt cos(πt) − sin(πt)
(a) g( t) = sinc( t) g′ ( t) = =
(πt) 2 πt 2
e − t , t ≥ 0 − t
(b) g( t) = (1 − e −t
) u(t) g′ ( t) = = e u( 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