Introduction to signals - Continuous and Discrete Time Signals and Systems - page 54

(v) x 5[ k ] = exp( j(7 π k / 4)) + cos(4 π k / 7 + π );
(vi) x 6[ k ] = sin(3 π k / 8) cos(63 π k / 64).
1.7 Determine if the following CT signals are energy or power signals or
neither. Calculate the energy and power of the signals in each case:
(i) x 1( t ) = cos( π t ) sin(3 π t );

cos(3 π t ) − 3 ≤ t ≤ 3;

(v) x 5( t ) =

(ii) x 2( t ) = exp( − 2 t );
(iii) x 3( t ) = exp( − j2 t );
(iv) x 4( t ) = exp( − 2 t ) u ( t );

0 elsewhere;

0 ≤ t

≤ 2

t

(vi) x 6( t ) = 4 − t 2 ≤ t ≤ 4

0 elsewhere .

1.8 Repeat Problem 1.7 for the following DT sequences:
(i) x 1[ k ] = cos

π k
4

3 π k
8

sin ;

3 π k
16

cos − 10 ≤ k ≤ 0

(ii) x 2[ k ] =

0 elsewhere;

(iii) x 3[ k ] = ( − 1) k ;
(iv) x 4[ k ] = exp( j( π k / 2 + π/ 8));

2 k 0 ≤ k ≤ 10

1 1 ≤ k ≤ 15
0 elsewhere.
1.9 Show that the average power of the CT periodic signal x ( t ) =

(v) x 5[ k ] =

A sin( ω 0 t +

θ ), with real-valued coefficient A ,isgivenby A 2 / 2.
1.10 Show that the average power of the CT signal y ( t ) = A 1 sin( ω 1 t + φ 1 ) +
A 2 sin( ω 2 t + φ 2 ), with real-valued coefficients A 1 and A 2 ,isgivenby

A 1

+ A 2
2

ω 1 = ω 2

2

P y =

A 1
2

+ A 2
2

+ A 1 A 2 cos( φ 1 − φ 2 ) ω 1 = ω 2 .

1.11 Show that the average power of the CT periodic signal x ( t ) =

D exp[j( ω 0 t + θ )] is given by D 2 .
1.12 Show that the average power of the following CT signal:

N

D n e j ω n t ,ω p x ( t ) = = ω r if p = r ,

n = 1

for 1 ≤ p , r ≤ N , is given by

N

D n 2 . P x =

n = 1

1.13 Calculate the average power of the periodic function shown in Fig. P1.13
and defined as

− 2 m − 1 − 2 m

12 < t ≤ 2

x ( t )

=

t = (0 , 1]

− 2 m − 2 − 2 m − 1

02 < t ≤ 2

+

m ∈ Z and x ( t ) = x ( t + 1) .