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Sinusoidal Steady-state Analysis

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- 1. Network Analysis Chapter 2 e, Phasor, and Sinusoidal Steady-State Analysis Chien-Jung Li Department of Electronic Engineering National Taipei University of Technology
- 2. Department of Electronic Engineering, NTUT Compound Interest • 複利公式: 本金P, 年利率r, 一年複利n次, t年後本金加利息之總和為 = + 1 nt r S P n • Let P=1, r=1, and t=1 = + 1 1 n S n When n goes to infinite, S converges to 2.718… (= e) Let P=10萬, r/n=10%/12, t=1 S=11,0471 Let P=10萬, r/n=10%, and n=36, t=1 S=3,091,268 2/33
- 3. Department of Electronic Engineering, NTUT Development of Logarithm • Michael Stifel (1487-1567) • John Napier (1550-1617) • 利用對數而將乘法變成加法的特性，刻卜勒成功 計算了火星繞日的軌道。 ( )+ ∗ = = 2 52 5 7 m m m m ( )− = = 7 7 4 3 4 m m m m ( )− − = = = 2 2 3 1 3 1m m m mm − − − =⋯ ⋯3 2 1 0 1 2 3 , , , , 1, , , ,m m m m m m m 3/33
- 4. Department of Electronic Engineering, NTUT Definition of dB (分貝) • , where • Power gain • Voltage gain • Power (dBW) • Power (dBm) • Voltage (dBV) • Voltage (dBuV) ( )= ⋅10 logdB G ( )= aG b = ⋅ 2 1 10 log P P = ⋅ 2 1 20 log V V ( )= ⋅10 log 1-W P ( )= ⋅10 log 1-mW P ( )= ⋅20 log 1-Volt V ( )µ = ⋅20 log 1- V V 相對量 ((((比例,,,, 比值,,,, 無單位, dB), dB), dB), dB) 絕對量 ((((因相對於一絕對單位,,,, 因此可表示一絕對量.... 有單位,,,, 單位即為dBWdBWdBWdBW,,,, dBmdBmdBmdBm,,,, dBVdBVdBVdBV…)…)…)…) 4/33
- 5. Department of Electronic Engineering, NTUT In some textbooks, phasor may be represented as Euler’s Formula • Euler’s Formula cos sinjx e x j x= + ( ) ( ) ( ) { } { }ω φ φ ω ω φ + = ⋅ + = ⋅ = ⋅cos Re Re j t j j t p p pv t V t V e V e e φ φ= ⋅ = ∠ def j p pV V e V • Phasor (相量) Don’t be confused with VectorVectorVectorVector (向量) which is commonly denoted as A (How it comes?) 取實部 (即cosine部分) phasor A real sinusoidal signal v(t) that can be represented as: V V 5/33
- 6. Department of Electronic Engineering, NTUT Definition of e lim 1 n x n x e n→∞ = + 2 3 lim 1 1 1! 2! 3! n x n x x x x e n→∞ = + = + + + + … x jx= ( ) ( ) 2 3 1 1! 2! 3! jx jx jxjx e = + + + +… • Euler played a trick let , where 1j = − 1 lim 1 n n e n→∞ = + 6/33
- 7. Department of Electronic Engineering, NTUT • Since , , , How It Comes… 1j = − 2 1j = − 3 1j = − − 4 1j = = − + − + + − + − + … … 2 4 3 5 1 2! 4! 3! 5! x x x x j x 2 4 cos 1 2! 4! x x x = − + − +… 3 5 sin 3! 5! x x x x= − + − +… cos sinjx e x j x= + cos sinjx e x j x− = − cos 2 jx jx e e x − + = − − =sin 2 jx jx e e x j ( ) ( )= + + + +… 2 3 1 1! 2! 3! jx jx jxjx e • Use and we have (姊妹式) 7/33
- 8. Department of Electronic Engineering, NTUT Coordinate Systems x-axis y-axis x-axis y-axis P(r,θ) θ r P(x,y) 2 2 r x y= + 1 tan y x θ − = cosx r θ= siny r θ= Cartesian Coordinate System (笛卡兒座標系, 直角座標系) Polar Coordinate System (極坐標系) (x,0) (0,y) ( )cos ,0r θ ( )0, sinr θ Projection on x-axis Projection on y-axis 8/33
- 9. Department of Electronic Engineering, NTUT Sine Waveform x-axis y-axis P(x,y) x y r θ θθ y θ 0 π/2 π 3π/2 2π Go along the circle, the projection on y-axis results in a sine wave. 9/33
- 10. Department of Electronic Engineering, NTUT x θ 0 π/2 π 3π/2 Cosine Waveform x-axis y-axis θ Go along the circle, the projection on x-axis results in a cosine wave. Sinusoidal waves relate to a CircleCircleCircleCircle very closely. Complete going along the circle to finish a cycle, and the angle θ rotates with 2π rads and you are back to the original starting-point and. Complete another cycle again, sinusoidal waveform in one period repeats again. Keep going along the circle, the waveform will periodically appear. 10/33
- 11. Department of Electronic Engineering, NTUT Complex Plan (I) It seems to be the same thing with x-y plan, right? • Carl Friedrich Gauss (1777-1855) defined the complex plan. He defined the unit length on ImImImIm-axis is equal to “j”. A complex Z=x+jy can be denoted as (x, yj) on the complex plan. (sometimes, ‘j’ may be written as ‘i’ which represent imaginary) Re-axis Im-axis Re-axis Im-axis P(r,θ) θ r P(x,yj) 2 2 r x y= + 1 tan y x θ − = cosx r θ= siny r θ= (x,0j) (0,yj) ( )cos ,0r θ ( )0, sinr θ ( )1j = − 11/33
- 12. Department of Electronic Engineering, NTUT Complex Plan (II) Re-axis Im-axis 1 Every time you multiply something by j, that thing will rotate 90 degrees. 1j = − 2 1j = − 3 1j = − − 4 1j = 1*j=j j j*j=-1 -1 -j -1*j=-j -j*j=1 (0.5,0.2j) (-0.2, 0.5j) (-0.5, -0.2j) (0.2, -0.5j) • Multiplying j by j and so on: 12/33
- 13. Department of Electronic Engineering, NTUT Sine Waveform Re-axis Im-axis P(x,y) x y r θ θθ y=rsinθ θ 0 π/2 π 3π/2 2π To see the cosine waveform, the same operation can be applied to trace out the projection on ReReReRe-axis. 13/33
- 14. Department of Electronic Engineering, NTUT Phasor Representation (I) – Sine Basis ( ) ( ) { } { }φ ω φ θ ω φ= + = =sin Im Imj j t j j sv t A t Ae e Ae e Re-axis Im-axis P(A,ф) y=Asin ф θ 0 π/2 π 3π/2 2π ф tθ ω= Given the phasor denoted as a point on the complex-plan, you should know it represents a sinusoidal signal. Keep this in mind, it is very very important! time-domain waveform 14/33
- 15. Department of Electronic Engineering, NTUT Phasor Representation (II) – Cosine Basis ( ) ( ) { } { }φ ω φ θ ω φ= + = =cos Re Rej j t j j sv t A t Ae e Ae e Re-axis Im-axis P(A,ф) y=Acos ф θ 0 π/2 π 3π/2 2π ф tθ ω= time-domain waveform 15/33
- 16. Department of Electronic Engineering, NTUT Phasor Representation (III) ( ) ( ) { }φ ω ω φ= + = 1 1 1 1 1sin Im j j t v t A t A e e Re-axis Im-axis P(A1,ф1) ф1 P(A2,ф2) P(A3,ф3) θ 0 π/2 π 3π/2 2π tθ ω= A1sin ф1 ( ) ( ) { }φ ω ω φ= + = 2 2 2 2 2sin Im j j t v t A t A e e ( ) ( ) { }φ ω ω φ= + = 3 3 3 3 3sin Im j j t v t A t A e e A2sin ф2 A3sin ф3 16/33
- 17. Department of Electronic Engineering, NTUT Mathematical Operation j t j tde j e dt ω ω ω= ⋅ 1j t j t e dt e j ω ω ω = ⋅∫ ( ) ( )0 1 t v t i t dt C = ∫ ω = = ⋅ 1 CV I Z I j C ( ) ( )di t v t L dt = ω= ⋅ = ⋅LV j L I Z I ω = = 1 1 CZ j C sC ω= =LZ j L sL • LLLL and CCCC: from time-domain to phasor-domain analysis (s is the Laplace operator) ( )σ ω σ= + =, here let 0s j 17/33
- 18. Department of Electronic Engineering, NTUT Phasor Everywhere • 電路學、電子學: Phasor 常見為一個固定值 (亦可為變量) • 電磁學、微波工程: Phasor 常見為變動量, 隨傳播方向變化 • 通訊系統: Phasor 常見為變動量, 隨時間變化 此變動的phasor也經常被稱作複數波包(complex envelope)、波包 (envelope)，或帶通訊號的等效低通訊號(equivalent lowpass signal of the bandpass signal)。Phasor如果被拆成正交兩成分，常稱作I/Q訊 號，而在數位通訊裡表示I/Q訊號的複數平面(座標系)也被稱為星座 圖(constellation)。 • You will see “Phasor” many times in your E.E. life. It just appears with different names, and it is just a representation or an analysis technique. • Keep in mind that a phasor represents a signal, it’s like a head on your body. 18/33
- 19. Department of Electronic Engineering, NTUT Simple Relation Between Sine and Cosine • Sine CosineSine CosineSine CosineSine Cosine π/2 π 3π/2 2π sinθ θ 0 cosθ • Negative sine or cosineNegative sine or cosineNegative sine or cosineNegative sine or cosine ( )θ θ= +cos sin 90 ( )θ θ= −sin cos 90 ( )θ θ− = +cos cos 180 ( )θ θ− = +sin sin 180 Try to transform into sine-form:θ−cos ( ) ( ) ( )θ θ θ θ− = − + = + = −cos sin 90 sin 270 sin 90 19/33
- 20. Department of Electronic Engineering, NTUT Cosine as a Basis ( ) { }ω ω= =cos Re j t pv t V t Ve = ∠0pV V ( ) { }ωπ ω ω = = − = sin cos Re 2 j t p pv t V t V t Ve = ∠ − 90pV V ( ) ( ) { }ω ω ω π= − = + =cos cos Re j t p pv t V t V t Ve = ∠180pV V ( ) { }ωπ ω ω = − = + = sin cos Re 2 j t p pv t V t V t Ve = ∠90pV V cosinecosinecosinecosine sinesinesinesine negative cosinenegative cosinenegative cosinenegative cosine negative sinenegative sinenegative sinenegative sine Phasor Phasor Phasor Phasor 20/33
- 21. Department of Electronic Engineering, NTUT Sine as a Basis ( ) { }ω ω= =sin Im j t pv t V t Ve = ∠0pV V ( ) { }ωπ ω ω = = + = cos sin Im 2 j t p pv t V t V t Ve = ∠90pV V ( ) ( ) { }ω ω ω π= − = + =sin sin Im j t p pv t V t V t Ve = ∠180pV V ( ) { }ωπ ω ω = − = − = cos sin Im 2 j t p pv t V t V t Ve = ∠ − 90pV V Phasor Phasor Phasor Phasor cosinecosinecosinecosine sinesinesinesine negative cosinenegative cosinenegative cosinenegative cosine negative sinenegative sinenegative sinenegative sine 21/33
- 22. Department of Electronic Engineering, NTUT Addition of Sinusoidal A basic property of sinusoidal functions is that the sum of an arbitrary number of sinusoids of the same frequency is equivalent to a single sinusoid of the given frequency. It must be emphasized that all sinusoids must be of the same frequency. ( ) ( )ω θ= +sinpv t V t θ= ∠1 1 1pV V θ= ∠2 2 2pV V θ= ∠n pn nV V = + + +⋯1 2 nV V V V ( ) ( ) ( ) ( )ω θ ω θ ω θ= + + + + + +⋯1 1 2 2sin sin sinp p pn nv t V t V t V t ( )1v t ( )2v t ( )nv t 22/33
- 23. Department of Electronic Engineering, NTUT Example ( ) ( ) ( )= +0 1 2v t v t v t ( ) ( )= −1 20cos 100 120v t t ( ) ( )= − +2 15sin 100 60v t t = ∠ − = −1 20 30 17.3205 10V j = ∠ − = − −2 15 120 7.5 12.9904V j ( ) ( )= − + − −0 17.3205 10 7.5 12.9904V j j ( ) ( )= −0 25sin 100 66.87v t t = − = ∠ −9.8205 22.9904 25 66.87j = ∠ − = − −1 20 120 10 17.321V j = ∠ = − +2 15 150 12.9904 7.5V j ( ) ( )= − − + − +0 10 17.321 12.9904 7.5V j j = − − = ∠22.9904 9.8205 25 203.13j ( ) ( )= +0 25cos 100 203.13v t t ( )= −25sin 100 66.87t Choose the basis you like, and the results are identical. andFor calculate use sine function as a basis use cosine function as a basis 23/33
- 24. Department of Electronic Engineering, NTUT Steady-state Impedance = = + V Z R jX I • Steady-state impedance resistance reactance = = + I Y G jB Z • Steady-state admittance conductance susceptance = +30 40Z j = Ω30R = Ω40X = = − + 1 0.012 0.016 30 40 Y j j = 0.012G S = −0.016X S 24/33
- 25. Department of Electronic Engineering, NTUT Conversion to Phasor-domain ( )i t ( )v t V I RR ( )i t ( )v t ( )i t ( )v t C L ω 1 j C V I ωj LV I = ⋅V R I ω = ⋅ 1 V I j C ω= ⋅V j L I V I V I V I V and I are in-phase V lags I by 90o V leads I by 90o R C L 25/33
- 26. Department of Electronic Engineering, NTUT Frequency Response Frequency-independent All pass Frequency-dependent High-pass Frequency-dependent Low-pass V I R ω 1 j C V I ωj LV I = + =Z R jX R ω = + = 1 Z R jX C ω π= 2 f ω π= 2 f ω π= 2 f ω= + =Z R jX L 26/33
- 27. Department of Electronic Engineering, NTUT Calculate the Impedance (I) ω 1 j C V • Calculate the impedance of a 0.01-uF capacitor at (a) f=50Hz (b) 1kHz (c) 1MHz ( )π − = + = + = − Ω ⋅ × 6 1 0 318.309 k 2 50 0.01 10 Z R jX j j = − Ω318.309 kX = Ω318.309 kZ I (a) f = 50 Hz ( )π − = + = + = − Ω × ⋅ ×3 6 1 0 15.92 k 2 1 10 0.01 10 Z R jX j j = − Ω15.92 kX = Ω15.92 kZ (b) f = 1 kHz ( )π − = + = + = − Ω × ⋅ ×6 6 1 0 15.92 2 1 10 0.01 10 Z R jX j j = − Ω15.92X = Ω15.92Z (c) f = 1 MHz = 0.01 µFC 27/33
- 28. Department of Electronic Engineering, NTUT Calculate the Impedance (II) • Calculate the impedance of a 100-mH inductor at (a) f=50Hz (b) 1kHz (c) 1MHz ( )π − = + = + ⋅ × = Ω3 0 2 50 100 10 31.42Z R jX j j = Ω31.42X = Ω31.42Z (a) f = 50 Hz ( )π − = + = + × ⋅ × = Ω3 3 0 2 1 10 100 10 628.32Z R jX j j = Ω628.32X = Ω628.32Z (b) f = 1 kHz ( )π − = + = + × ⋅ × = Ω6 3 0 2 1 10 100 10 628.32 kZ R jX j j = Ω628.32 kX = Ω628.32 kZ (c) f = 1 MHz ωj LV I = 100 mHL 28/33
- 29. Department of Electronic Engineering, NTUT Calculate the Impedance (III) • Calculate the impedance of following circuit at (a) f=50Hz (b) 1kHz (c) 1MHz ( ) ( ) π − = + = + = − Ω ⋅ × 6 1 200 0.2 318.309 k 2 50 0.01 10 Z R jX j j = Ω318.309 kZ (a) f = 50 Hz ( ) ( ) π − = + = + = − Ω × ⋅ ×3 6 1 200 0.2 15.92 k 2 1 10 0.01 10 Z R jX j j = Ω15.92 kZ (b) f = 1 kHz ( ) ( ) π − = + = + = − Ω × ⋅ ×6 6 1 200 200 15.92 2 1 10 0.01 10 Z R jX j j = Ω200.63Z (c) f = 1 MHz ω 1 j C = 0.01 µFC R = Ω200R = ∠ − Ω318.309k 89.96Z = ∠ − Ω15.92k 89.26Z = ∠ Ω200.63 -4.55Z 29/33
- 30. Department of Electronic Engineering, NTUT Calculate the Impedance (IV) • Calculate the impedance of following circuit at (a) f=50Hz (b) 1kHz (c) 1MHz ( ) ( )π − = + = + ⋅ × = + Ω3 200 2 50 100 10 200 31.42Z R jX j j = Ω202.45Z (a) f = 50 Hz ( ) ( )π − = + = + × ⋅ × = + Ω3 3 200 2 1 10 100 10 200 628.32Z R jX j j = Ω659.38Z (b) f = 1 kHz ( ) ( )π − = + = + × ⋅ × = + Ω6 3 200 2 1 10 100 10 0.2 628.32 kZ R jX j j = Ω628.32 kZ (c) f = 1 MHz ωj L = 100 mHL R = Ω200R = ∠ Ω202.45 8.93Z = ∠ Ω659.38 72.34Z = ∠ Ω628.32 k 89.98Z 30/33
- 31. Department of Electronic Engineering, NTUT Power in AC Circuits ( ) ( )ω φ= +sinpi t I t ( ) ( )ω φ θ= + +sinpv t V t Instantaneous power absorbed by the circuit: ( ) ( ) ( ) ( ) ( )ω φ θ ω φ= = + + +sin sinp pp t v t i t V I t t ( ) ( ) ( )= =∫ ∫0 0 1 1T T P p t dt v t i t dt T T Average power: ( ) ( )= − − + 1 1 sin sin cos cos 2 2 A B A B A B Steady-state AC circuit ( )i t ( )v t ( ) ( )ω φ θ ω φ= + + +∫0 1 sin sin T p pV I t t dt T 31/33
- 32. Department of Electronic Engineering, NTUT Power in AC Circuits Average power: ( )θ ω φ θ = − + + ∫ ∫0 0 cos cos 2 2 2 T Tp pV I dt t dt T ( ) ( ) { } θ θ θ ∗ = = = = 0 cos 1 cos cos Re 2 2 2 2 T p p p p p pV I V I V I t T VI T T Steady-state AC circuit ( )i t ( )v t ( ) ( )ω φ θ ω φ= + + +∫0 1 sin sin T p pP V I t t dt T V Iθ φ ( )φ θ+ = j pV V e φ∗ − = j pI I e θ =* j p pVI V I e { } θ=* Re cosp pVI V I 32/33
- 33. Department of Electronic Engineering, NTUT Root-Mean-Square (RMS) Value θ θ θ= = = cos cos cos 22 2 p p p p rms rms V I V I P V I (RMS value is also called the effective value) When the circuit contains L and C, the current and voltage may not be in-phase (they can be in-phase if effects of L and C cancelled at the given frequency), and hence the apparent power may not be totally absorbed by the circuit. Define RMS voltage and current as = 2 p rms V V = 2 p rms I I power factor (PF) is define as the power factor (功率因子/因素)θ≤ ≤0 cos 1 ×Actual power = Apparent power Power factor θ= cosrms rmsV I 33/33

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