EAST WEST UNIVERSITYDEPARTMENT OF EEE

Course code: EEE 201Course name: Electrical circuit ΙΙ

Lab reportExperiment no: 07

Experiment name: Frequency response of passive filters.

Student name: B. M. ADNANId: 2011-1-80-020

Section: 02

Date of performance: 27-03-12Date of submission: 03-04-12

OBJECTIVE: The main aim of this experiment is to learn about different type of passive filters. As well as learn how to determine the cut-off frequencies of the passive filters.

CIRCUIT DIAGRAM:

Figure 01: Low pass and high pass filter.

Figure 02: Band stop and band pass filter.

ANSWER TO THE LAB-REPORT QUESTIONS:

ANS: 01

Low pass filter:

Cut-off frequency, fc = 1/ (2π×100×1×10^-6) = 1591.5494 Hz ω = 2π × fc = 1591.5494 × 2π = 10000 rad/sec Voltage across capacitor, Vc = [(-j100)/ (100-j100)] × 1 = 0.707<-45° V

High pass filter:

Cut-off frequency, fc = 1/ (2π×100×1×10^-6) = 1591.5494 Hzω = 2π × fc = 1591.5494 × 2π = 10000 rad/sec Voltage across resistor, Vr = 100/ (100-j100) × 1 = 0.707<45° V

Band pass filter:

Lower cut-off frequency, ω1 = - (R/ 2L) + √ ((R/2L) ² + (1/LC)) = - (2/2*10*10^-3) + √ (2/2*10*10^-3) + (1/10*10^- 3*250*10^-6) = 532.53 rad/secFrequency, f1 = (ω1/2π) = 532.53/ 2π = 84.75 HzNow, voltage across resistor, Vr = [2/ (2-j7.4032+j5.4031)]*20 = 14.14<45° V

Higher cut-off frequency, ω2 = (R/ 2L) + √ ((R/2L) ² + (1/LC)) = (2/2*10*10^-3) + √ (2/2*10*10^-3) + (1/10*10^- 3*250*10^-6) = 742.46 rad/sec

Frequency, f2 = (ω2/2π) = 742.46 / 2π

= 118.17 Hz

Now, voltage across resistor, Vr = [2/ (2-j7.4032+j5.4031)]*20 = 14.14<45° V

Band stop filter:

Lower cut-off frequency, ω1 = - (R/ 2L) + √ ((R/2L) ² + (1/LC)) = - (2/2*10*10^-3) + √ (2/2*10*10^-3) + (1/10*10^- 3*250*10^-6) = 532.53 rad/sec

Frequency, f1 = (ω1/2π) = 532.53/ 2π = 84.75 HzImpedance, Z = j (ω1L – 1/ω1C) = j (532.53*10*10^-3 – 1/532.53*250*10^-6) = 2<-90° ΩNow, voltage across Impedance, Vz = (2<-90) / (2+2<-90)*20 = 14.1421<-45° V

Higher cut-off frequency, ω2 = (R/ 2L) + √ ((R/2L) ² + (1/LC)) = (2/2*10*10^-3) + √ (2/2*10*10^-3) + (1/10*10^- 3*250*10^-6) = 742.46 rad/sec

Frequency, f2 = (ω2/2π) = 742.46 / 2π = 118.17 Hz

Impedance, Z = j (ω1L – 1/ω1C) = j (532.53*10*10^-3 – 1/532.53*250*10^-6) = 2<-90° Ω

Now, voltage across Impedance, Vz = (2<-90) / (2+2<-90)*20 = 14.1421<-45° V

COMPARISONComparison between theoretical and experimental values for low pass filter:

Elements Experimental values Theoretical valuesCut-off frequency 1588..40 Hz 1591.55 Hz

Voltage 0.708 V 0.707 V Comment: Here little bit difference occurs between experimental and theoretical values. It’s very small which is negligible.

Comparison between theoretical and experimental values for high pass filter:

Elements Experimental values Theoretical valuesCut-off frequency 1593.10 rad/sec 1591.55 Hz

Voltage 0.707 V 0.707 V

Comment: Here little bit difference occurs between experimental and theoretical values. It’s very small which is negligible.

Comparison between theoretical and experimental values for band pass filter:Elements Experimental values Theoretical values

Lower cut-off frequency 85.77 Hz 85.99 HzVoltage 14.032 V 14.14 V

Higher cut-off frequency 117.79 Hz 117.82 HzVoltage 14.16 V 14.14 V

Comment: Here little bit difference occurs between experimental and theoretical values. It’s very small which is negligible.

Comparison between theoretical and experimental values for band stop filter:Elements Experimental values Theoretical values

Lower cut-off frequency 85.77 Hz 85.99 HzVoltage 14.23 V 14.14 V

Higher cut-off frequency 117.91 Hz 117.82 HzVoltage 14.11 V 14.14 Hz

Comment: Here little bit difference occurs between experimental and theoretical values. It’s very small which is negligible.

ANS. 02

Band width, B = R / L = 2/ (10*10^-3) = 200 rad/sec Resonant frequency, ω = (1/√LC) = 1/ √ (10*10^-3*250*10^-6) = 632.46 rad/secSo, frequency, f = (ω/2π) = (632.46/2π) = 100.66 HzQuality factor, Q = ω/ B = 632.45/ 200 = 3.1632When the resistor value is doubled, then the value of resistance will be 4 Ω

So, band width, B = R / L = 4/ (10*10^-3) = 400 rad/sec When the resistor value is doubled, then the value of band width will double.

When the inductor value is doubled, then the value of the inductor wills 20 mH.

So, band width, B = R / L = 2/ (20*10^-3) = 100 rad/sec

When the inductor value is doubled, then the value of the band width will half.

DISCUSSION: In this experiment we learnt how to calculate different type of passive filters. By this way we cal easily recognize this type of circuit. We also revised the cut-off frequency calculation.

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