ESE211 Lecture 3 Phasor Diagram of an RC Circuit V(t)=Vm sin(ωt) R Vi(t) C Vo(t) • Current is a reference in series circuit KVL: Vm = VR + VC V R ϕ Im VC V m ESE211 Lecture 3 Phasor Diagram of an RL Circuit V(t)=Vmsin(ωt) VR (t) R Vi(t) L Vo(t) KVL: Vm = VR + VL Vm VL ϕ VR Im ESE211 Lecture 3 Phasor Diagram of a Series RLC Circuit VC VL C L Vi R VR KVL: Vm = VC + VL + VR VL Vm VL-C VR Im VC • Voltages across capacitor and inductor compensate each other ESE211 Lecture 4 Integrating Circuit VR Vi(t) R Vo(t) C 22 VViR=+VC or VViR=+VC • Assume high frequency 1/ωC << R then VR >> VC VVRi≈ 11t t VOi=≈Idt Vdi t ∫∫ C 00RC Vi VV∝ dt Oi∫ • Accurate integrating can be obtained at high frequency that leads to a low output signal ESE211 Lecture 4 Differentiating Circuit VC Vi(t) C Vo(t) R 22 VViR=+VC or VViR=+VC • Assume low frequency 1/ωC >> R then VR << VC VR 1 t d Vii≈⇒Idt Iii=C V Vi VVCi≈ ∫ C 0 dt d VV==IR≈RCV ORi dt i d VOi∝ V dt • An accurate result can be achived at low frequency. ESE211 Lecture 4 Transient Processes in Passive Circuits RC Circuit without a Source • The circuit response is due only to the energy stored in the capacitor t=0 C R • The capacitor C is precharged to the voltage V0 Applying KCL: C dV/dt +V/R = 0 • This is the 1st order differential equation ESE211 Lecture 4 Solution of the Differential Equation dV/dt + V/RC = 0 dV/V = -dt/RC ln V = -t/RC +a V = A e -t/RC , A=e+a • e is the base of the natural logarithms, e = 2.718… • Continuity requires the initial condition: At t = 0 V(0) = V0 -t/RC V(t) = V0 e • The response governed by the elements themselves without external force is called a natural response ESE211 Lecture 4 The Time Constant V(t) V0 0.368 V0 0.05 V0 τ time 3τ • The time constant τ = RC is the rate at which the natural response decays to zero • Time τ is required to decay by a factor of 1/e=0.368 • After the time period of 3τ the transient process is considered to be completed ESE211 Lecture 4 Determination of the Time Constant V0 V V1 V/e τ t 1) From the solution of the equation By definition: as the time when the signal decreases by 1/e V(t+τ)/V(t) = 1/e The time t can be chosen arbitrarily 2) From the slope of the line at t=0 Differentiating the solution –t/τ dV/dt = -{V0/τ } e V1(t) = -{V0/τ }t +V0 V1(τ) = 0 ESE211 Lecture 4 Transient Response of an Integrating RC Circuits VR Vi VC KVL: Vi = VR + VC Vi t VO =VC VR t t ESE211 Lecture 4 Transient Response of a Differentiating RC Circuits VC Vi VR KVL: Vi = VC + VR Vi t VO = VR t VC ESE211 Lecture 5 Transfer Function Vi(t) Vo(t) Vi(t) = Vimcos(ωt) Vo(t) = Vomcos(ωt+ϕ) Vom(ω) ϕ(ω) jωt jϕ Vo(t) =Re { Vo e }, Vo = Vome V (ω) T (ω) = o Vi Amplitude response |T(ω)| = Vom(ω)/Vim Phase response ∠T(ω) = ϕ(ω) ESE211 Lecture 5 The Transfer Function of an Integrating RC Circuit R Vi(t) C Vo(t) 1 V o (ω) jωC 1 T(ω) = = = V 1 1+ jωRC i R + jωC 1 T (ω) = 1+ jωτ τ=RC ESE211 Lecture 5 Amplitude Response of the Integrating RC Circuit • The magnitude of the transfer function is 1 T (ω) = 1+ ω 2τ 2 • The Integrating RC circuit is a LOW-PASS filter 1.0 0.707 | in /V o 0.5 |V 0.0 -3 -2 -1 0 1 2 3 0.001ω 0.01ω 0.1ω ω 10ω 100ω 1000ω ο ο ο ο ο ο ο Frequency (ω) 0 100 slope -20 dB/dec -1 -3 dB -20 | 10 ] in /V [dB o 0 |V -2 -40 10 V 10-3 -3 -2 -1 0 1 2 3 -60 0.001ω 0.01ω 0.1ω ω 10ω 100ω 1000ω ο ο ο ο ο ο ο Frequency (ω) ESE211 Lecture 5 Phase Response of the Integrating RC Circuit The angle of the transfer function is Im[]T(ω) ∠T (ω) = arctan Re[]T(ω) 1− jωτ ∠T(ω) = ∠ = −arctan(ωτ ) (1+ jωτ )(1− jωτ ) 0 o V -45 -90 -3 -2 -1 0 1 2 3 0.001 ωο 0.01ω ο 0.1ω ο ωο 10 ωο 100 ωο 1000 ωο Frequency ( ω) ESE211 Lecture 5 The Transfer Function of a Differentiating RC Circuit C Vi(t) R Vo(t) V (ω) I(ω) ⋅ R T(ω) = o = Vi I(ω) ⋅ X RC (ω) R R jωRC T (ω) = = = X (ω) 1 1+ jωRC RC R + jωC jωτ (1− jωτ ) ωτ (ωτ + j) T (ω) = 2 2 = 2 2 1+ω τ 1+ω τ τ = RC ESE211 Lecture 5 Amplitude response of the Differentiating RC Circuit • Amplitude response of the RC circuit is the magnitude of T(ω) ωτ (ωτ + j) ωτ T(ω) = 2 2 = 1+ ω τ 1+ ω 2τ 2 • The Differentiating RC circuit is a HIGH-PASS filter 1.0 | V =0.707 V in 0 in V o/ 0.5 |V 0.0 -3 -2 -1 0 1 2 3 0.001 0.01 0.1 ω 10 100ω 1000ω ωο ωο ωο ο ωο ο ο Frequency ( ω) 0 100 slope -1 -20 dB/dec -20 ] | 10 -3 dB point at ω in or 0 /V -6 dB/oct [dB o 0 V |V 10-2 -40 10-3 -3 -2 -1 0 1 2 3 -60 0.001ω 0.01ω 0.1ω ω 10 100 1000 ο ο ο ο ωο ωο ω Frequency ( ω) ο ESE211 Lecture 5 Phase response of the Differentiating RC Circuit • Phase response of the RC circuit is the phase angle of T(ω) ωτ (ωτ + j) ∠T (ω) = ∠ 1+ ω 2τ 2 1 ∠T(ω) = arctan ωτ 90 0 V 45 0 -3 -2 -1 0 1 2 3 0.001 0.01 0.1 ω 10 100ω 1000ω ωο ωο ωο ο ωο ο ο Frequency ( ω) ESE211 Lecture 5 Frequency Response of a Series RLC Circuit VC VL C L Vi R VR KVL: Vi = VC + VL + VR VL VL-C Vi VR Im VC • Voltages across capacitor and inductor compensate each other ESE211 Lecture 5 Amplitude Response of a Series RLC Circuit R R |V 0 /V IN |= = 2 1 1 jωL + + R R 2 + ωL − jωC ωC 1.0 V =0.707 V | 0 in in /V 0.5 o |V 0.0-3 -2 -1 0 1 2 3 100 1000 0.001ωο 0.01ωο 0.1ωο ωο 10ωο ωο ωο Frequency ( ω) 1.5 | in /V L 1.0 |V |, |VC/Vin| |VL/Vin| in /V C 0.5 |V 0.0 -3 -2 -1 0 1 2 3 0.001 0.01 0.1 ω ωο ωο ωο ο 10ωο 100ωο 1000ω Frequency (ω) ο ESE211 Lecture 5 Amplitude and Phase Response of the Ladder RC Circuit R R Vin C C Vo 0 o V -90 -180 -3 -2 -1 0 1 2 3 0.001ωο 0.01ωο 0.1ωο ωο 10ωο 100 ωο 1000 ωο Frequency ( ) ω 0 100 -2 -40 | 10 ] in /V [dB o slope 0 |V -4 -80 10 V -40 dB/dec 10-6 -3 -2 -1 0 1 2 3 -120 0.001ω 0.01ω 0.1ω ω 10ω 100ω 1000ω ο ο ο ο ο ο ο Frequency (ω) ESE211 Lecture 5 Amplitude and Phase Response of a Wien-Bridge Circuit R C −1 1 + jωC ZOUT R T(ω) = = −1 Vin R C Vo ZTOTAL 1 1 + jωC + R + R jωC 100 50 0 o V -50 -100 -3 -2 -1 0 1 2 3 0.001 0.01 0.1 ω 10 100ω 1000ω ωο ωο ωο ο ωο ο ο Frequency (ω ) 0.33 0.24 V =0.707 V | 0 max n i 5 BandWidth /V o |V 0.0-3 -2 -1 0 1 2 3 100 1000 0.001ωο 0.01ωο 0.1ωο ωο 10ωο ωο ωο Frequency ( ω) .