ECE 231 -9

# ECE 231 -9 - ECE-231 Circuits and Systems I Fall 2011...

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Unformatted text preview: ECE-231 Circuits and Systems I Fall 2011 Session 9 Dr. Stewart Personick Office: ECEC Room 331 [email protected] Review of the last portion of Session 8 Substituting Complex Exponentials for Sine and Cosine Waveforms If i(t) = A e j2 π ft and i(t) flows through a resistor, then the voltage produced across the resistor is: Vr(t) = R x i(t) = R x A e j2 π ft Substituting Complex Exponentials for Sine and Cosine Waveforms If i(t) = A e j2 π ft and i(t) flows through an inductor, then the voltage produced across the inductor is: VL(t) = L x d[ i(t) ]/dt = L x j 2 π f x A e j2 π ft Substituting Complex Exponentials for Sine and Cosine Waveforms If i(t) = A e j2 π ft and i(t) flows through a capacitor, then the voltage produced across the capacitor is {since i(t) = C d[VC(t)]/dt}: VC(t) = (1/C) x the integral of i(t) dt = (1/C) x (1/ j 2 π f) x A e j2 π ft Using complex exponentials to solve for currents and voltages in circuits Circuit containing resistors, inductors and capacitors +- Vs(t) = A e j2 π ft = A cos (2 π ft) + j A sin (2 π ft) Using complex exponentials to solve for currents and voltages in circuits Circuit containing resistors, inductors and capacitors +- Vs(t) = A e j2 π ft = A cos (2 π ft) + j A sin (2 π ft) First solve for the node voltages and mesh currents using Vs(t) : For example: Vab(t) = Γ e j2 π ft and i1(t) = Η e j2 π ft Then take the real part of each result to obtain: Vab(t) = Re { Γ e j2 π ft} and i1(t) =Re { Η e j2 π ft} Vs(t) Using sine waves to solve for currents and voltages in RLC networks (example) +- Vs(t) = A e j2 π ft a Vs(t) = Vab(t) + Vbc(t) + Vcd(t) [i.e. Kirchhoff’s voltage law] A e j2 π ft = j2 π f x L x Λ x e j2 π ft + 1/[j2 π f x C] x Λ x e j2 π ft + R x Λ x e j2 π ft A = j2 π f x L x Λ + 1/[j2 π f x C] x Λ + R x Λ b i(t) = Λ x e j2 π ft- + Vab(t) c d Solve for the unknown: Λ Using sine waves to solve for currents and voltages in RLC networks (example) +- Vs(t) = A e j2 π ft a Vs(t) = Vab(t) + Vbc(t) + Vcd(t) A e j2 π ft = j2 π f x L x Λ x e j2 π ft + 1/[j2 π f x C] x Λ x e j2 π ft + R x Λ x e j2 π ft A = j2 π f x L x Λ + 1/[j2 π f x C] x Λ + R x Λ b i(t) = Λ x e j2 π ft- + Vab(t) c d Using sine waves to solve for currents and voltages in RLC networks (example) +- a Vs(t) = Vab(t) + Vbc(t) + Vcd(t) A e j2 π ft = j2 π f x L x Λ x e j2 π ft + 1/[j2 π f x C] x Λ x e j2 π ft + R x Λ x e j2 π ft A = j2 π f x L x Λ + 1/[j2 π f x C] x Λ + R x Λ b i(t) = Re { Λ x e j2 π ft }- + Vab(t) c d Vs(t) =Re {A e j2 π ft } = A cos(2 π ft) New materials Example: R-L-C circuit +- Vs(t) = Re{ Aej2 π ft) } volts = A cos(2 π f t) volts a i(t) = Re { i(t) } = Re { Λ x e j2 π ft} Λ = A / { j2 π fL + 1/[j2 π fC] + R } = A / { ZL + Zc + R } Vcd(t)= Re {1000 Ω x i(t )} = Re {1000 Ω x A exp(j2 π ft) volts / [j2 π fL - j/2 π fC +1050 Ω ]} L = 1H XL = j2 π fL R = 1000 Ω d i(t) C = 1 μ F XC = -j/ (2 π fC) Vcd(t) = Vcd cos(2 π ft + ϑ ); where Vcd and ϑ both depend upon the value of f c R = 50 Ω b Impedance of a 1H Inductor and...
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## This note was uploaded on 03/01/2012 for the course ECE 231 taught by Professor Pietrucha during the Spring '08 term at NJIT.

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ECE 231 -9 - ECE-231 Circuits and Systems I Fall 2011...

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