신호및 시스템 개정 2판 솔루션 Simon S. Haykin [signals and systems 2ed haykin] (구매자 판매중지)

CHAPTER 1
1.1 to 1.41 - part of text

1.42

(a) Periodic: Fundamental period = 0.5s (b) Nonperiodic (c) Periodic Fundamental period = 3s (d) Periodic Fundamental period = 2 samples (e) Nonperiodic (f) Periodic: Fundamental period = 10 samples (g) Nonperiodic (h) Nonperiodic (i) Periodic: Fundamental period = 1 sample

l.43

π 2 y ( t ) = ? 3 cos ? 200t + -- ? ? ? ? 6? ?
2 π = 9 cos ? 200t + -- ? ? 6?

9 π = -- cos ? 400t + -- ? 1 ? 2 3? 9 (a) DC component = -2 9 π (b) Sinusoidal component = -- cos ? 400t + -- ? ? 2 3? 9 Amplitude = -2

1

200 Fundamental frequency = -------- Hz π

1.44

The RMS value of sinusoidal x(t) is A ? 2 . Hence, the average power of x(t) in a 1-ohm resistor is ( A ? 2 ) = A2/2.
2

1.45

Let N denote the fundamental period of x[N]. which is de?ned by 2π N = ----Ω The average power of x[n] is therefore 1 2 P = --- ∑ x [ n ] N 1 = --N
n=0 N -1 N -1

∑A
n=0

2

A = ----N

n=0 2 N -1

2 2πn cos ? --------- + φ? ? N ? 2…(생략)