2 we can represent almost any waveform vst as a sum

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2. We can represent almost any waveform, Vs(t), as a sum of sine waves and cosine waves (Fourier analysis) 3. Why would we want to represent a waveform Vs(t) as a sum of sine waves and cosine waves?
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Using sine waves to solve for currents and voltages in RLC networks 1. From superposition, we can solve for i(t) and Vab(t) by representing Vs(t) as the sum of a number of voltage waveforms Vs(t) = V1(t) + V2(t) + …. + Vn(t) , and solving for the waveforms i(t) and Vab(t) that are produced by individually applying each of: V1(t), V2(t), etc. 2. We can represent almost any waveform, Vs(t), as a sum of sine waves and cosine waves (Fourier analysis) 3. Why would we want to represent a waveform Vs(t) as a sum of sine waves and cosine waves? 4. Because: If Vs(t) is a sine wave, at frequency: f; then the currents and voltages produced in an RLC circuit will also be sine waves , at the same frequency: f
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Using sine waves to solve for currents and voltages in RLC networks (example) + - Vs(t) = Vs sin(2 π ft) a b i(t) - + Vab(t) In this simple case Vs(t) = Vab(t) i(t) = C (Farads) x d [Vab(t)]/dt = C (Farads) x d [Vs(t)]/dt Therefore : i(t)= C (Farads) x 2 π f x Vs cos(2 π ft) = C (Farads) x 2 π f x Vs sin(2 π ft + π /2 ) C
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Using sine waves to solve for currents and voltages in RLC networks (example) + - a In this simple case Vs(t) = Vab(t); and Vab(t) = L x d[i(t)]/dt d[i(t)]/dt= (1/L) x Vab(t) = (1/L) x Vs sin(2 π ft) Next: Integrate both sides of the above equation, with respect to t i(t) = -(1/2 π f) x (1/L) x Vs cos (2 π ft ) = (1/2 π f) x (1/L) x Vs sin (2 π ft - π /2 ) b i(t) - + Vab(t) L Vs(t) = Vs sin(2 π ft)
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Complex Numbers X = A + j B; where j is the square root of -1 A = “the real part of X ” = Re{ X } B= “the imaginary part of X ” = Im{ X }
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Complex Numbers Im ( X ) = B Re ( X ) = A X = A + jB; where j is the square root of -1 A = the real part of X = Re{ X } B= the imaginary part of X = Im{ X } X
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Complex Numbers Im ( X ) = B Re ( X ) = A X = A + jB; where j is the square root of -1 A = the real part of X = Re{ X } B= the imaginary part of X = Im{ X } X Φ If the length of the blue arrow is C, and the angle (in radians) is Φ , then: (from trigonometry) A= C cos ( Φ ), and B = C sin ( Φ )
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Substituting Complex Exponentials for Sine and Cosine Waveforms Identity: e j x = cos (x) + j sin (x) ; where x is a real number When we calculate cos(x) and sin(x), we assume that
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  • Fall '08
  • PIETRUCHA
  • Volt, Square wave, Complex number, Sine wave, Sine Waves, Euler's formula

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