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EE202 Supplementary Materials for Self Study. Circuit Analysis Using Complex Impedance Passive Filters and Frequency Response. Acknowledgment. Dr. Furlani and Dr. Liu for lecture slides Ms. Colleen Bailey for homework and solution of complex impedance

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## EE202 Supplementary Materials for Self Study

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**EE202 Supplementary Materialsfor Self Study**Circuit Analysis Using Complex Impedance Passive Filters and Frequency Response**Acknowledgment**Dr. Furlani and Dr. Liu for lecture slides Ms. Colleen Bailey for homework and solution of complex impedance Textbook: Nilsson & Riedel, “Electric Circuits,” 8th edition**Steady-State Circuit Response to Sinusoidal Excitation -**Analysis Using Complex Impedance**Household Circuit Breaker Panel**240V Central Air 120V Lighting, Plugs, etc.**Single Frequency**Sinusoidal Signal**Sinusoidal Signal**Amplitude Peak-to-peak Root-mean-square Frequency Angular Frequency Period**Trigonometry Functions**Appendix F**Other Periodic Waveforms**Fundamental and Harmonics**Resistor Only Circuit**I=V/R, i(t)=v(t)/R Instantaneous Response**R-L Circuit**Transient Steady-state**Phase Shift**Time Delay or Phase Angle: t / T *2 or *360-degree**Phasor – Complex Number**Z Real(Z)+j Imag(Z) Y Imag(Z) tan-1(Y/X) X Real(Z) Reference**Observations**Single Frequency for All Variables Phasor Solution of Diff Eq. Algebraic equation Extremely simple Phase Delay between variables Physical Measurements Real part of complex variables v = Real{V}; i = Real{I}**Impedance in Series**Complex Impedance Resistance, Reactance**Example**=5000 rad/sec**Apply ZL=jL, ZC=1/j C**Zab=90+j(160-40)=90+j120=sqrt(902+1202)exp{jtan-1(120/90)} =150 53.13 degree I=750 30 deg / 150 53.13 deg = 5 -23.13 deg=5exp(-j23.13o)**Impedance in Parallel**Complex Admittance Conductance, Susceptance**Example**=200000 rad/sec**Apply ZL=jL, ZC=1/j C**Series: Use Z; Parallel: Use Y Y=0.2 36.87 deg; Z=5 -36.87 deg V=IZ=40 -36.87 deg**Kirchhoff’s Laws**Same Current at a Node Addition of current vectors (phasors) Voltage Around a Loop or Mesh Summation of voltage vectors (phasors)**Voltage divider**Vo=36.12-j18.84 (V)**Find VTh**Vx=100-I*10, Vx=I*(120-j40)-10*Vx; solve Vx and I VTH=10Vx+I*120=784-j288 (V)**Find ZTh**Calculate Ia Determine Vx Calculate Ib ZTh=VT/IT=91.2-j38.4 (Ohm)**Transformer**Time differentiation replaced by j**AC Sine Wave, Ideal Transformer**Voltage and Current Power Conserved**Transformer**• Power Applications • Convert voltage • vout=(N2/N1) vin • Signal Applications • Impedance transformation • Xab=(N1/N2)2 XL • Match source impedance with load to maximize power delivered to load**Frequency Response of Circuits**• Analysis Over a Range of Frequencies • Amplifier Uniformity • Filter Characteristics • Low pass filter • High pass filter • Bandpass filter • Equalizer**RC Filters**High Pass Low Pass

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