ECEN 301Discussion #16 – Frequency Response1 Easier to Maintain Than to Retake Alma 59:9 9 And now as Moroni had supposed that there should be men sent to the city of Nephihah, to the assistance of the people to maintain that city, and knowing that it was easier to keep the city from falling into the hands of the Lamanites than to retake it from them, he supposed that they would easily maintain that city.

ECEN 301Discussion #16 – Frequency Response2 Lecture 16 – Frequency Response Back to AC Circuits and Phasors Frequency Response Filters

ECEN 301Discussion #16 – Frequency Response3 Frequency Response Frequency Response H(jω): a measure of how the voltage/current/impedance of a load responds to the voltage/current of a source

ECEN 301Discussion #16 – Frequency Response4 Frequency Response V L (jω) is a phase-shifted and amplitude-scaled version of V S (jω)

ECEN 301Discussion #16 – Frequency Response5 Frequency Response V L (jω) is a phase-shifted and amplitude-scaled version of V S (jω) Amplitude scaledPhase shifted

ECEN 301Discussion #16 – Frequency Response6 Frequency Response Example 1: compute the frequency response H V (jω) R 1 = 1kΩ, R L = 10kΩ, C = 10uF R 1 C v s (t) +–+– +RL–+RL–

ECEN 301Discussion #16 – Frequency Response7 Frequency Response Example 1: compute the frequency response H V (jω) R 1 = 1kΩ, R L = 10kΩ, C = 10uF R 1 C v s (t) +–+– +RL–+RL– 1.Note frequencies of AC sources Only one AC source so frequency response H V (jω) will be the function of a single frequency

ECEN 301Discussion #16 – Frequency Response8 Frequency Response Example 1: compute the frequency response H V (jω) R 1 = 1kΩ, R L = 10kΩ, C = 10uF R 1 C v s (t) +–+– +RL–+RL– 1.Note frequencies of AC sources 2.Convert to phasor domain Z 1 = R 1 Z LD =R L Z C =1/jωC VsVs +–+– ~

ECEN 301Discussion #16 – Frequency Response9 Frequency Response Example 1: compute the frequency response H V (jω) R 1 = 1kΩ, R L = 10kΩ, C = 10uF 1.Note frequencies of AC sources 2.Convert to phasor domain 3.Solve using network analysis Thévenin equivalent Z 1 = R 1 Z C =1/jωC

ECEN 301Discussion #16 – Frequency Response10 Frequency Response Example 1: compute the frequency response H V (jω) R 1 = 1kΩ, R L = 10kΩ, C = 10uF 1.Note frequencies of AC sources 2.Convert to phasor domain 3.Solve using network analysis Thévenin equivalent Z 1 = R 1 +VT–+VT– Z C =1/jωC VsVs +–+– ~

ECEN 301Discussion #16 – Frequency Response11 Frequency Response Example 1: compute the frequency response H V (jω) R 1 = 1kΩ, R L = 10kΩ, C = 10uF 1.Note frequencies of AC sources 2.Convert to phasor domain 3.Solve using network analysis Thévenin equivalent 4.Find an expression for the load voltage Z T Z LD VTVT +–+– ~ +VL–+VL–

ECEN 301Discussion #16 – Frequency Response12 Frequency Response Example 1: compute the frequency response H V (jω) R 1 = 1kΩ, R L = 10kΩ, C = 10uF 5.Find an expression for the frequency response Z T Z LD VTVT +–+– ~ +VL–+VL–

ECEN 301Discussion #16 – Frequency Response13 Frequency Response Example 1: compute the frequency response H V (jω) R 1 = 1kΩ, R L = 10kΩ, C = 10uF 5.Find an expression for the frequency response Z T Z LD VTVT +–+– ~ +VL–+VL–

ECEN 301Discussion #16 – Frequency Response14 Frequency Response Example 1: compute the frequency response H V (jω) R 1 = 1kΩ, R L = 10kΩ, C = 10uF 5.Find an expression for the frequency response Look at response for low frequencies (ω = 10) and high frequencies (ω = 10000) Z T Z LD VTVT +–+– ~ +VL–+VL– ω = 10 ω = Lowpass filter

Frequency Response Lowpass filter

ECEN 301Discussion #16 – Frequency Response16 Frequency Response Example 2: compute the frequency response H Z (jω) R 1 = 1kΩ, R L = 4kΩ, L = 2mH R 1 i s (t) +RL–+RL– L

ECEN 301Discussion #16 – Frequency Response17 Frequency Response Example 2: compute the frequency response H Z (jω) R 1 = 1kΩ, R L = 4kΩ, L = 2mH R 1 i s (t) +RL–+RL– L 1.Note frequencies of AC sources Only one AC source so frequency response H Z (jω) will be the function of a single frequency

ECEN 301Discussion #16 – Frequency Response18 Frequency Response Example 2: compute the frequency response H Z (jω) R 1 = 1kΩ, R L = 4kΩ, L = 2mH R 1 i s (t) +RL–+RL– L 1.Note frequencies of AC sources 2.Convert to phasor domain I s (jω) Z LD = R L Z L = jωL Z 1 =R 1 +VL–+VL–

ECEN 301Discussion #16 – Frequency Response19 Frequency Response Example 2: compute the frequency response H Z (jω) R 1 = 1kΩ, R L = 4kΩ, L = 2mH 1.Note frequencies of AC sources 2.Convert to phasor domain 3.Solve using network analysis Current divider & Ohm’s Law I s (jω) Z LD = R L Z L = jωL Z 1 =R 1 +VL–+VL– I L (jω)

ECEN 301Discussion #16 – Frequency Response20 Frequency Response Example 2: compute the frequency response H Z (jω) R 1 = 1kΩ, R L = 4kΩ, L = 2mH I s (jω) Z LD = R L Z L = jωL Z 1 =R 1 +VL–+VL– I L (jω) 4.Find an expression for the frequency response

ECEN 301Discussion #16 – Frequency Response21 Frequency Response Example 2: compute the frequency response H Z (jω) R 1 = 1kΩ, R L = 4kΩ, L = 2mH I s (jω) Z LD = R L Z L = jωL Z 1 =R 1 +VL–+VL– I L (jω) 4.Find an expression for the frequency response Look at response for low frequencies (ω = 10) and high frequencies (ω = 10000) ω = 10 ω = Lowpass filter

ECEN 301Discussion #16 – Frequency Response22 1 st and 2 nd Order RLC Filters Graphing in Frequency Domain Filter Orders Resonant Frequencies Basic Filters

ECEN 301Discussion #16 – Frequency Response23 Frequency Domain Graphing in the frequency domain: helpful in order to understand filters -ω-ωω ππ cos(ω 0 t) π[δ(ω – ω 0 ) + δ(ω – ω 0 )] Time domainFrequency domain Fourier Transform

ECEN 301Discussion #16 – Frequency Response24 Frequency Domain Graphing in the frequency domain: helpful in order to understand filters -WW 1 sinc(ω 0 t) Time domainFrequency domain W/π π/W Fourier Transform

ECEN 301Discussion #16 – Frequency Response25 Basic Filters Electric circuit filter: attenuates (reduces) or eliminates signals at unwanted frequencies Low-pass High-pass Band-pass Band-stop ωω ω ω

ECEN 301Discussion #16 – Frequency Response26 Filter Orders Higher filter orders provide a higher quality filter

ECEN 301Discussion #13 – Phasors27 Impedance +L–+L– +C–+C– Re Im -π/2 π/2 R -1/ωC ωLωL ZRZR ZCZC ZLZL Phasor domain +R–+R– V s (jω) +–+– ~ + V Z (jω) – I(jω) Z Impedance of resistors, inductors, and capacitors

ECEN 301Discussion #13 – Phasors28 Impedance +C–+C– Re Im -π/2 -1/ωC ZCZC Impedance of capacitors +C–+C– +C–+C–

ECEN 301Discussion #13 – Phasors29 Impedance +L–+L– Re Im π/2 ωLωL ZLZL Impedance of inductors +L–+L– +L–+L–

ECEN 301Discussion #16 – Frequency Response30 Resonant Frequency Resonant Frequency (ω n ): the frequency at which capacitive impedance and inductive impedance are equal and opposite (in 2 nd order filters) C + v i (t) – + v o (t) – L C + v i (t) – + v o (t) – L Impedances in seriesImpedances in parallel

ECEN 301Discussion #16 – Frequency Response31 Resonant Frequency Resonant Frequency (ω n ): the frequency at which capacitive impedance and inductive impedance are equal and opposite (in 2 nd order filters) Z L =jωL + V i (jω) – + V o (jω) – Z C =1/jωC Impedances in series

ECEN 301Discussion #16 – Frequency Response32 Resonant Frequency Resonant Frequency (ω n ): the frequency at which capacitive impedance and inductive impedance are equal and opposite (in 2 nd order filters) + V i (jω) – + V o (jω) – Z EQ =0 Impedances in series

ECEN 301Discussion #16 – Frequency Response33 Resonant Frequency Resonant Frequency (ω n ): the frequency at which capacitive impedance and inductive impedance are equal and opposite (in 2 nd order filters) Impedances in parallel Z L =jωL + V i (jω) – + V o (jω) – Z C =1/jωC

ECEN 301Discussion #16 – Frequency Response34 Resonant Frequency Resonant Frequency (ω n ): the frequency at which capacitive impedance and inductive impedance are equal and opposite (in 2 nd order filters) + V i (jω) – + V o (jω) – Z EQ =∞ Impedances in parallel

ECEN 301Discussion #16 – Frequency Response35 Low-Pass Filters Low-pass Filters: only allow signals under the cutoff frequency (ω 0 ) to pass Low-pass ω1ω1 ω0ω0 ω2ω2 ω3ω3 H L (jω) R C + v i (t) – + v o (t) – R C + v i (t) – + v o (t) – L 1 st Order 2 nd Order

ECEN 301Discussion #16 – Frequency Response36 Low-Pass Filters 1 st Order Low-pass Filters: Z R + V i (jω) – + V o (jω) – Z C =1/jωC Z R + V i (jω) – + V o (jω) – ω 1 : ω → 0 ω 3 : ω → ∞ Low-pass ω1ω1 ω0ω0 ω3ω3 Z R + V i (jω) – + V o (jω) –

ECEN 301Discussion #16 – Frequency Response37 Low-Pass Filters 2 nd Order Low-pass Filters: Z R + V i (jω) – + V o (jω) – Z C =1/jωC Z L =jωL ω 1 : ω → 0 Z R + V i (jω) – + V o (jω) – ω 2 : ω = ω n ω 3 : ω → ∞ Z R + V i (jω) – + V o (jω) – Low-pass ω1ω1 ω0ω0 ω2ω2 ω3ω3 Z R + V i (jω) – + V o (jω) – Resonant frequency

ECEN 301Discussion #16 – Frequency Response38 High-Pass Filters High-pass Filters: only allow signals above the cutoff frequency (ω 0 ) to pass H H (jω) High-pass ω1ω1 ω0ω0 ω2ω2 ω3ω3 + v i (t) – + v o (t) – 1 st Order R C R + v i (t) – + v o (t) – 2 nd Order C L

ECEN 301Discussion #16 – Frequency Response39 High-Pass Filters 1 st Order High-pass Filters: Z C =1/jωC + V i (jω) – + V o (jω) – Z R =R Z R + V i (jω) – + V o (jω) – ω 1 : ω → 0 ω 3 : ω → ∞ High-pass ω1ω1 ω0ω0 ω3ω3 Z R + V i (jω) – + V o (jω) –

ECEN 301Discussion #16 – Frequency Response40 High-Pass Filters 2 nd Order High-pass Filters: Z R + V i (jω) – + V o (jω) – Z C =1/jωC Z L =jωL ω 1 : ω → 0 ω 2 : ω = ω n ω 3 : ω → ∞ Z R + V i (jω) – + V o (jω) – High-pass ω1ω1 ω0ω0 ω2ω2 ω3ω3 Z R + V i (jω) – + V o (jω) – Z R + V i (jω) – + V o (jω) – Resonant frequency

ECEN 301Discussion #16 – Frequency Response41 Band-Pass Filters Band-pass Filters: only allow signals between the passband (ω a to ω b ) to pass H B (jω) Band-pass ω1ω1 ωaωa ω2ω2 ω3ω3 ωbωb + v i (t) – + v o (t) – 2 nd Order R C L

ECEN 301Discussion #16 – Frequency Response42 Band-Pass Filters 2 nd Order Band-pass Filters: Z R + V i (jω) – + V o (jω) – Z C =1/jωC Z L =jωL ω 1 : ω → 0 ω 2 : ω = ω n ω 3 : ω → ∞ Band-pass ω1ω1 ωaωa ω2ω2 ω3ω3 ωbωb Z R + V i (jω) – + V o (jω) – + V i (jω) – Z R + V o (jω) – Z R + V i (jω) – + V o (jω) – Resonant frequency

ECEN 301Discussion #16 – Frequency Response43 Band-Stop Filters Band-stop Filters: allow signals except those between the passband (ω a to ω b ) to pass H N (jω) + v i (t) – + v o (t) – 2 nd Order R C L Band-stop ω1ω1 ωaωa ω2ω2 ω3ω3 ωbωb

ECEN 301Discussion #16 – Frequency Response44 Band-Stop Filters 2 nd Order Band-stop Filters: ω 1 : ω → 0 ω 2 : ω = ω n ω 3 : ω → ∞ Z R + V i (jω) – + V o (jω) – Z C =1/jωC Z L =jωL Band-stop ω1ω1 ωaωa ω2ω2 ω3ω3 ωbωb Z R + V i (jω) – + V o (jω) – Z R + V i (jω) – + V o (jω) – Z R + V i (jω) – + V o (jω) – Resonant frequency