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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 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