2장 Sinusoids (2)_위상 변위와 시간 변위 / 사인파 샘플링

Phase shift and time shift

• The phase shift parameter $\varphi $ determines the time locations of maxima and minima.

$5cos(0.3\pi t+1.2\pi)$

위상 이동 매개 변수는 최댓값과 최솟값의 시간 위치를 결정함.

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TIME - SHIFT

(a) s(t)의 식

(b)의 t 값 범위 구하기

(b) s(t-2)의 식

• In a mathematical formula, we can replace $t$ with $t-t_{m}$

$x(t-t_{m})=Acos(\omega (t-t_{m}))$

• Then the $t=0$ point moves to $t=t_{m}$
• Peak value of $cos(\omega (t-t_{m}))$ is now at $t=t_{m}$

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TIME-SHIFTED SINUSOID

$x(t+4)=5cos(0.3\pi(t+4)) =5cos(0.3\pi (t-(-4)))$

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PHASE ↔ TIME-SHIFT

• Equate the formulas :                          (좌변 : 시간)        (우변: 위상)

$Acos(\omega (t-t_{m})) = Acos(\omega t+\varphi )$

• and we obtain: $-\omega t_{m}=\varphi $
• or, $t_{m}=-\frac{\varphi }{\omega }$

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SINUSOID from a PLOT

• Measure the period, T
  • Between peaks or zero crossings
  • Compute frequency : $\omega = \frac{2\pi}{T}$
• Measure time of a peak : $t_{m}$
  • Compute phase : $\phi=-\omega t_{m} $
• Measure height of positive peak: A

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$(A,\;\omega , \;\phi )$ from a PLOT$

A = 5

$T=\frac{0.01sec}{1\;period}=\frac{1}{100}$     →      $\omega = \frac{2\pi}{T} =\frac{2\pi}{0.01}=200\pi$

$t_{m}=-0.00125sec$      →      $\varphi = -\omega t_{m}=-(200\pi)(t_{m}) =0.25\pi$

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PHASE is AMBIGUOUS

• The cosine signal is periodic
  • period is 2$\pi$

$Acos(\omega t+\varphi +2\pi)=Acos(\omega t+\varphi )$

• Thus adding any multiple of $2\pi$ leaves $x(t)$ unchanged

if $t_{m} = -\frac{\varphi }{\omega }$, then

$t_{m_{2}} = \frac{-(\varphi +2\pi)}{\omega }=-\frac{\varphi }{\omega }-\frac{2\pi}{\omega }=t_{m}-T$

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Sampling and plotting sinusoids

$x(t) = 20cos(2\pi(40)t-0.4\pi)$

• Sampling period → $Ts=0.005$ sec.
• There are five samples/cycle.
• Straight lines rather than the smooth cosine function → $Ts=0.0025$ sec.
• The choice of $Ts$ depends on the frequency of the cosine signal.

Ts의 선택은 코사인 신호의 주파수에 따라 달라짐

(a) -0.03s ~ 0.045s 이므로 주기는 0.045 - (-0.03) = 0.075 이고 샘플링이 16개 이므로 샘플링 주기는 0.075/16 = 약 0.005 sec이다. 매끄러운 코사인 함수보다는 직선 →
(b) 한 주기당 5개의 샘플이 있으므로 0.005*5 = 0.0025 따라서 주기가 0.0025sec이다.