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Frequency Response of Netwroks (Electronic Engineering) - Lab Report Example

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Frequency Response of Netwroks (Electronic Engineering)

Current was determined by monitoring the voltage across 100 ? resistor. The CRO was used to record the current and voltage waveforms. The above procedure was repeated for the series connection of a resistor and inductor. Voltage across the inductor was measured at 100 Hz. The frequency response of the RC low-pass filter was measured over the frequency range 100 Hz to 100 kHz. At R = 1 k?, C = 0.01 µF, the attenuation at 15 to 20 frequencies were logarithmically recorded over this range. This procedure was repeated with R= 10 k?. The LF oscillator was connected to the RLC series circuit and with R=100 ? the voltage across the capacitor and inductor, and current through the circuit at frequencies between 1 kHz and 100 kHz determined by measuring the voltage across the series resistor. The band-pass filter circuit was constructed and voltages Vo and Vi measured over the range of frequencies 1kHz to 100 kHz. The band-stop filter circuit was also constructed and Vo and Vi again measured over the range of frequencies 1kHz to 100kHz. A Twin-T filter was then constructed with R1 = R2 = 100 ? and C1=C2=0.01µF. ...
1000 4.673 0.301 0.00301 1552.492 0.000644 2000 4.627 0.561 0.00561 824.7772 0.001212 3000 4.551 0.83 0.0083 548.3133 0.001824 4000 4.47 1.093 0.01093 408.9661 0.002445 5000 4.37 1.32 0.0132 331.0606 0.003021 6000 4.253 1.596 0.01596 266.4787 0.003753 7000 4.11 1.802 0.01802 228.0799 0.004384 8000 3.962 1.989 0.01989 199.1956 0.00502 9000 3.84 2.123 0.02123 180.8761 0.005529 10000 3.701 2.268 0.02268 163.1834 0.006128 Figure1 The capacitance is the gradient of the line which is 6?10-4 F By calculation C= 1/2?f Xc and at f =10000 Hz and Xc =163.1834 ? then, C = 1 / 2*?*10000*163.1834 = 0.4126 F which reasonably agrees with the experimental values. The small difference between the calculated value and the measured value may be due to inaccurate readings or as a result of rounded figures. 2) Measuring VL and VR in figure6 with R=100 ? and C=0.01uf Table.2 ? (Hz) VL (v) VR (V) I (A) XL (?) 1/XL 100 1.443 4.379 0.04379 32.95273 0.030347 1000 0.675 4.36 0.0436 15.48165 0.064593 2000 1.299 4.23 0.0423 30.70922 0.032564 3000 1.867 4.028 0.04028 46.35055 0.021575 4000 2.317 3.791 0.03791 61.11844 0.016362 5000 2.709 3.53 0.0353 76.74221 0.013031 6000 3.033 3.273 0.03273 92.66728 0.010791 7000 3.273 2.907 0.02907 112.5903 0.008882 8000 3.328 2.705 0.02705 123.0314 0.008128 9000 3.488 2.496 0.02496 139.7436 0.007156 10000 3.592 2.32 0.0232 154.8276 0.006459 Figure 2 From the gradient, the inductance obtained to be 15.6 mH The inductive reactance XL at 100Hz is XL=2fL=2?3.14?100?15.6?10-3= 9.8 ? At f = 100 Hz from the table above XL = 32.95273 ? The difference is due to experimental errors and errors in reading of results. 3) Measuring Vo , Vi , the gain and calculating the frequency response Table.3 ? (HZ) Vi (V) VO (V) Gain Gain in dBs 100 4.634 4.638 1.000863 0.007494 160 ...Show more


Frequency Response Networks Tutor The main objective of this experiment was to investigate how capacitive and inductive reactance’s vary with frequency, plot the frequency response of a simple RC filter, plot the variation of current with frequency in a series RLC circuit and determine the resonant frequency and Q factor and to evaluate simple band- pass and band-stop filters…
Author : ibreitenberg
Frequency Response of Netwroks (Electronic Engineering) essay example
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