COMPUTING

Solutions to End−of−Chapter Exercises1.orG( jω)The magnitude of (1) isand the phase angle or argument, isV outR=R---------------------------V + 1 ⁄ jωC inV out jωRC jωRC + ω 2 R 2 C 2 ωRC( j + ωRC)= ---------- = ----------------------- = ---------------------------------------- = --------------------------------------V in 1 + jωRC 1 + ω 2 R 2 C 2 1 + ω 2 R 2 C 2ωRC 1 + ω 2 R 2 C 2 ∠atan( 1 ⁄ ( ωRC)) 1= ------------------------------------------------------------------------------------------- = -------------------------------------------- ∠atan( 1 ⁄ ( ωRC))1 + ω 2 R 2 C 21+1⁄( ω 2 R 2 C 2 )G( jω)1= --------------------------------------------1+1⁄( ω 2 R 2 C 2 )(2)(1)θ= arg{ G( jω)} =atan( 1 ⁄ ωRC)(3)We can obtain a quick sketch for the magnitude G( jω) versus ω by evaluating (2) at ω = 0 ,ω = 1⁄RC, and ω→∞. Thus,As ω → 0 ,For ω = 1⁄RC,and as ω→∞,G( jω) ≅ 0G( jω) = 1⁄ 2 = 0.707G( jω) ≅ 1We will use the MATLAB script below to plot Gjω ( ) versus radian frequency ω . This is shownon the plot below where, for convenience, we let RC = 1.w=0:0.02:100; RC=1; magGs=1./sqrt(1+1./(w.*RC).^2); semilogx(w,magGs); gridWe can also obtain a quick sketch for the phase angle, i.e., θ = arg{ G( jω)} versus ω , by evaluating(3) at ω = 0 , ω = 1⁄RC, ω = – 1 ⁄ RC, ω →–∞, and ω→∞. Thus,as ω → 0 ,For ω = 1⁄RC,For ω = – 1 ⁄ RC,θ ≅– atan0≅ 0°θ = – atan1= – 45°θ = – atan( – 1)= 45°Electronic Devices and Amplifier Circuits with MATLAB Computing, Second EditionCopyright © Orchard Publications1−27