Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8EE201 Lecture 35 P. 1Frequency ResponseDefinition: Frequency response of a circuit is the graph of the ratio of the phasor output to the phasor input as a function of frequency.Example: plot the frequency response of the circuit below.1 0.01 F+_Vin+_VoutStep 1: Find impedance of capacitor and resistor pair. Zin = R + (1/jC) = 1 + (1/j0.01)EE201 Lecture 35 P. 2Step 2: Use voltage division to establish output to input ratio.V outV in=11+ (1/j0.01)= j0.011+j0.01H(j)=Magnitude 10.51 10 100 1000Phase(degrees)806040201 10 100 1000Frequency(rad/sec)“High Pass”FilterEE201 Lecture 35 P. 3Two plots, one for magnitude and one for phase angle, are needed because H(j) is a complex quantity.Important frequencies for RC circuits. = 0 and = H(j0) = 0 90ºH(j) = 1 0ºFor magnitude:|H(j)| 1 as |H(j)| 0 as 0For phase: H(j) 0o as H(j) 90 º as 0EE201 Lecture 35 P. 41 1H+_Vin+_VoutExample: Plot the frequency response for the circuit below.Step 1: Find the impedance of inductor and resistor pair.Zin = R + (jL) = 1 + (j)Step 2: Use voltage division to establish output to input ratio. R 1V outV in R+jL= 1+jH(j)==EE201 Lecture 35 P. 5Magnitude 10.50.01 0.1 1 10 100Phase(degrees) 0-20-40-60-80-100Freq (rad/sec)“Low Pass”Filter0.01 0.1 1 10 100H(j0) = 10º H(j) = 0 -90º|H(j)| 1 as 0 |H(j)| 0 as H(j) 0º as 0 H(j) -90º as EE201 Lecture 35 P. 61 Zin(j)+_Example: Frequency response of RLC circuits0.1 H 1 mFVout+_VinFind ratio of phasor output to inputZin = R+jL+(1/jC) = 1 + j0.1+ (1/j0.01)By voltage divisionV outV in=R R+jL+ (1/jC)= 11+ j0.01+ (1/j0.001)Multiply by and reduce j0.001j0.001EE201 Lecture 35 P. 7V outV in-2+ j10 +104= H(j)=j10|H(j)| 0 as |H(j)| 0 as 0H(j) -90 º as H(j) 90 º as 0Look at other frequencies|H(j)| 0.01 as 10 rad/sec|H(j)| 1 as 100 rad/sec|H(j)| 0.01 as 1000 rad/secEE201 Lecture 35 P. 81Plotting magnitude and phase“Band Pass” Response1 10 100 1000 10000 Magnitude1 10 100 1000
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