2장 Sinusoids (4)_회전 페이저 해석/ 페이저 합/ 오일러 공식/ 복소 진폭/ 사인파의 합/ 페이저 합 규칙 & 증명/ 시간-이동을 위상으로 변환

Rotating phasor interpretation

• Consider $Z_{3}=Z_{1}Z_{2}$, where $Z_{1}=r_{1}e^{j\theta _{1}}$ and $Z_{2}=r_{2}e^{j\theta _{2}}$
• $Z_{3}=r_{1}r_{2}e^{j(\theta _{1}+\theta _{2})}$

$Z_{3}$은 $Z_{1}$과 $Z_{2}$을 곱한 것.

$z(t) = Xe^{j\omega t}\;\;\;\;\;\;\;\; X = Ae^{j\varphi }$

• Where $z(t)$ is the product of the complex number $X$ (called complex amplitude) and $e^{j\omega t}$
• The complex amplitude is also called a phasor.
• The complex number $X$ can be represented as a vector in the complex plane, where the vector's magnitude. $(\left|X \right|=A)$ is the amplitude, and vector's angle $(\angle X=\varphi )$

X : complex amplitude (복소 진폭)
복소 진폭은 phasor (페이저)라고도 불림.
복소수 $X$는 벡터의 크기가 있는 복소수 평면에서 벡터로 표현될 수 있음.
$(\left|X \right|=A)$는 진폭, 벡터의 각도 $(\angle X=\varphi )$

• The complex exponential signal can be written as $z(t) = Xe^{j\omega t}=Ae^{j\varphi}e^{j\omega t} =Ae^{j\theta (t)}$
  where $\theta (t) = \omega t+\varphi$ (Radians)
• Now, if $t$ increases, the complex vector $z(t)$ will be simple rotate at a constant rate, determined by the radian frequency $\omega$.
• Thus, the complex exponential signal is rotating phasor.

$t$가 증가하면 복소 벡터 $z(t)$는 라디안 주파수 $\omega$에 의해 결정되는 일정한 속도로 단순 회전함.
즉 $X$에 $e^{j\omega t}$를 곱하면 X가 $\omega$만큼 회전함.
따라서, 복소 지수 신호는 회전 페이저임.

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Rotating phasors

• The plot fig(a) shows the relationship between a single complex rotating phasor and the cosine signal $z(t) = e^{j(t-\frac{\pi}{4})}$ at the specific time $t=1.5\pi$

특정 시간 $t=225^{\circ}$을 줌. $cos225^{\circ}$가 됨. 그래프에서 답 보면 -0.7정도가 됨.// 그림은 단일 복소 회전 페이저와 코사인 신호 $z(t) = e^{j(t-\frac{\pi}{4})}$ 사이의 관계를 보여준다

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PHASOR Addition

• Phasors = Complex Amplitude
  • Complex Numbers represent Sinusoids

$z(t)=Xe^{j\omega t}=(Ae^{j\varphi})e^{j\omega t}$

• Develop the ABSTRACTION :
  • Adding Sinusoids = Complex Addition
  • PHASOR ADDITION THEOREM

분홍색과 빨간색 더하기

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AVOID Trigonometry

• Algebra, even complex, is EASIER !!!
• Can you recall $cos(\theta_{1}+\theta_{2})$ ?
• Use : real part of $e^{j(\theta_{1}+\theta_{2})}=cos(\theta_{1}+\theta_{2})$

$e^{j(\theta_{1}+\theta_{2})}=e^{j\theta_{1}}e^{j\theta_{2}}$
$=(cos\theta_{1}+jsin\theta_{1})(cos\theta_{2}+jsin\theta_{2})$

$=$$(cos\theta_{1}cos\theta_{2}-sin\theta_{1}sin\theta_{2})$$+j(...)$

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Euler's FORMULA

• Complex Exponential
  • Real part is cosine.
  • Imaginary part is sine.
  • Magnitude is one.

$e^{j\omega t}=cos(\omega t)+jsin(\omega t)$

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Real & Imaginary Part Plots

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POP QUIZ : Complex Amp

• Fine the COMPLEX AMPLITUDE for :

$x(t) = \sqrt{3}cos(77\pi t+0.5\pi)$

• Use EULER's FORMULA :

$x(t)=\Re \left\{\sqrt{3}e^{j(77\pi t+0.5\pi)} \right\}=\Re \left\{\sqrt{3}e^{j0.5\pi}e^{j77\pi t} \right\}$

$X=\sqrt{3}e^{j0.5\pi}$

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WANT to ADD SINUSOIDS

• ALL SINUSOIDS have SAME PREQUENCY
• HOW to GET $\left\{Amp, Phase \right\}$ of RESULT ?

$x_{1}(t)=1.7cos(2\pi(10)t+70\pi/180)$

$x_{2}(t)=1.9cos(2\pi(10)t+200\pi/180)$

$x_{3}(t)= x_{1}(t)+x_{2}(t)=1.532cos(2\pi(10)t+141.79\pi/180)$

주파수가 10Hz로 똑같은 신호 $x_{1}(t)$, $x_{2}(t)$
$x_{1}(t)$랑 $x_{2}(t)$를 더하면?
$z=\sqrt{x^{2}+y^{2}}$
노란색 박스에 써넣은 공식 이용.
1.  $x_{1}(t)$과 $x_{2}(t)$를 위상으로 나타내기.
$X_{1}=A_{1}e^{j\varphi _{1}}=1.7e^{j70\pi /180}$
$X_{2}=A_{2}e^{j\varphi _{2}}=1.9e^{j200\pi /180}$
2. 두 위상을 직교좌표계 형태로 변환.
$X_{1}=1.7(cos(70\pi/180)+jsin(70\pi/180))=0.5814+j1.5975$
$X_{2}=1.9(cos(200\pi/180)+jsin(200\pi/180))=-1.7854-j0.6498$
3. 실수부와 허수부 더하기.
$X_{3}=X_{1}+X_{2}$
      $=(0.5814+j1.5975)+(-1.7854-j0.6498)$
      $=-1.204+j0.9477$
4. 극좌표계로 다시 변환.
$A=\sqrt{\left [ 1.7cos(70\pi/180)+1.9cos(200\pi/180) \right ]^{2}+\left [ 1.7sin(70\pi/180)+1.9sin(200\pi/180) \right ]^{2}}$
$\phi=tan^{-1}\left ( \frac{y}{x} \right )=141.79^{\circ}$
$X_{3} = 1.532e^{j141.79\pi/180}$

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PHASOR ADDITION RULE

$x(t) = \sum_{k=1}^{N}A_{k}cos(\omega_{0}t+\phi _{k})=Acos(\omega_{0}t+\phi )$

Get the new complex amplitude by complex addition

$\sum_{k=1}^{N}A_{k}e^{j\phi _{k}}=Ae^{j\phi }$

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Phasor Addition Proof

$\sum_{k=1}^{N}A_{k}cos(\omega_{0}t+\phi _{k})=\sum_{k=1}^{N}\Re \left\{A_{k}e^{j(\omega_{0}t+\phi _{k})} \right\}$

                                         $=\Re \left\{\sum_{k=1}^{N} A_{k}e^{j\phi _{k}} e^{j\omega_{0}t} \right\}$

                                         $=\Re \left\{\left ( \sum_{k=1}^{N}A_{k}e^{j\phi _{k}} \right )e^{j\omega_{0}t} \right\}$

                                         $=\Re \left\{ (Ae^{j\phi })e^{j\omega_{0}t}\right\}=Acos(\omega_{0}t+\phi )$

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POP QUIZ : Add Sinusoids

• ADD THESE 2 SINUSOIDS :

$x_{1}(t)=cos(77\pi t)$

$x_2(t)=\sqrt{3}cos(77\pi t+0.5\pi)$

• COMPLEX ADDITION :

$1e^{j0}+\sqrt{3}e^{j0.5\pi}$

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POP QUIZ (answer)

• COMPLEX ADDITION :

• CONVERT back to cosine form :

$x_{3}(t) = 2cos(77\pi t+\frac{\pi}{3})$

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ADD SINUSOIDS

• Sum Sinusoid has SAME Frequency

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ADD SINUSOIDS EXAMPLE

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Convert Time-Shift to Phase

• Measure peak times:
  • ${\color{Blue} t_{m1}}=-0.0195,\;\;{\color{Blue} t_{m2}}=-0.0556,\;\;{\color{Blue} t_{m3}}=-0.0394$
• Conver to phase($t=0.1$)
  • ${\color{Blue} \phi_{1}}=-\omega {\color{Blue} t_{m1}}=-2\pi (\frac{{\color{Blue} t_{m1}}}{T})=\frac{70\pi}{180}$
  • ${\color{Blue} \phi _{2}}=\frac{200\pi}{180}$
• Amplitudes
  • $A_{1}=1.7,\;\;A_{2}=1.9,\;\;A_{3}=1.532$

$f_{0}=10Hz,\;\;T_{0}=0.1sec$
$t_{m1}=-\frac{\phi T_{0}}{2\pi}=\frac{\phi }{\omega}=\frac{-0.39}{20\pi}=-0.0195$

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Phasor Add: Numerical

• Convert Polar to Cartesian
  • $X_{1}=0.5814+{\color{Blue} j}1.597$
  • $X_{2}=-1.785-{\color{Blue} j}0.6498$
  • sum$=X_{3}=-1.204+{\color{Blue} j}0.9476$
• Convert back to Polar
  • $X_{3}=1.532$ at angle $\frac{141.79\pi}{180}$
  • This is the sum

왼쪽부터 $X_{2}$, $X_{3}$, $X_{1}$