Signals and Systems - Electrical Engineering

460 C H A P T E R 8: Discrete-Time Signals and Systems

To find the power we use the trigonometric identity (or Euler’s equation) cos^2 (θ)=0.5( 1 +

cos( 2 θ)), and so replacingx 1 [n], we have

Px=

4

2 N 0

0.5





N∑ 0 − 1

n= 0

1 +

N∑ 0 − 1

n= 0

cos(0.2πn−π/ 2 )



 N 0 = 20

=

2

40

[20+0]= 1

where the sum of the cosine is zero, as we are adding the values of the periodic cosine over a

period.

2. For 0 =3.2,x 2 [n]=2 cos(3.2n/ 10 −π/ 4 )forn≥0 and zero otherwise. The signal now does
   not repeat periodically aftern=0, as the frequency 3.2/10 (which equals the rational 32/100)
   cannot be expressed as 2πm/N(which due toπis an irrational value) for integersmandN.
   Thus, in this case we do not have a close form for the power, so we can simply say that the
   power is

Px= lim

N→∞

1

2 N+ 1

∑N

n=−N

∣

∣x 2 [n]

∣

∣^2

and conjecture that because the corresponding analog signal has finite power, so wouldx 2 [n].

Thus, we use MATLAB to compute the power for both cases.

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% Example 8.7 --- Power

%%%%%%%%%%%%%%%%%%%%%%%

clear all;clf

n = 0:100000;

x2 = 2∗cos(0.1∗n∗3.2−pi/4); % non-periodic for positive n

x1 = 2∗cos(0.1∗n∗pi−pi/4); % periodic for positive n

N = length(x1)

Px1 = sum(x1.ˆ2)/(2∗N+1) % power of x1

Px2 = sum(x2.ˆ2)/(2∗N+1) % power of x2

P1 = sum(x1(1:20).ˆ2)/(20); %power in period of x1

The signalx 1 [n] in the script has unit power and so does the signalx 2 [n] when we consider

100,001 samples. The two signals and their squared magnitudes are shown in Figure 8.1. n

nExample 8.8

Determine if a discrete-time exponential

x[n]= 2 (0.5)n n≥ 0

and zero otherwise, has finite energy, finite power, or both.