Pràctica 4 (2013)Trabajo Inglés
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<CSL; Linear Circuits and Systems (Practice 4)>
Authors: Castán, Anna and Sánchez, Andrea
The purpose of this practice is to
measure experimentally the frequency
response curves (gain and phase angle)
of a band filter and a filter of order 2
side eliminated, using the equivalent
model of a real inductor.
If you have an inductance L and a real loss òhmiques parasitic capacity, the equivalent model is ideal coil (L) with a small series resistor (RP) and a capacitor in parallel (Cp).
Fig 1. Equivalent circuit of an inductance and capacity parasitic losses òhmiques So when we have a coil in a circuit, actually what we are three elements, and must be taken into account when designing or analyzing filter.
2. PRELIMINARY STUDY The following two circuits are two passive filters of order 2, a band-pass and band removed, considering that the coil feel real.
2.1 Check that the transfer function H (s) = Vo (s) / Vi (s) of previous filters are: 2.2 Using the Scilab software, get the gain and phase curves, the two filters above, in the following cases: <CSL; Linear Circuits and Systems (Practice 4)> 3. MINIMUM MATERIAL TO PRACTICAL REALIZATION 3.2 Make connections Fig. 4 • A plate connections (type board Protocol).
• Resistance value R 2 = 10 kΩ and 100kΩ • 1 coil value L = 33 mH • 1 C = 330pF capacitor value • two coaxial cables with BNC connectors, bananas.
• one coaxial cable with BNC-BNC connectors • 1 connector "T" instruments available • Function Generators • Oscilloscope.
3.1 Assemble the circuit of Fig.4 plate proto-board, with R = 10kΩ and L = 33 mH.
3.3 What kind of filter is it? Measurement bandwidth (BW) and the frequency of resonance (f0) of this filter.
It’s a high-pass filter.
Vmax= 11.9V Vo(f2-f1)= 8.41V BW= 30.6Hz <CSL; Linear Circuits and Systems (Practice 4)> 3.4 With measurements in the previous section, it determines how much capacity parasitic Cp, the coil circuit.
To find the value of the parasitic capacity, we use two equations since we have two unknowns; Cp and Rp.
3.6 Change the resistance of 10kΩ to 100kΩ one. Measure the new value of the BW filter. Repeat the Experience section 3.5. What will now exit? Justify your answer.
3.5 If the filter now excited before a square signal frequency equal to the resonant frequency of the filter, which signal to get out? Justify your answer.
Harmonics are more attentive so will output more similar to the breast.
3.7 Mount the circuit board in fig.6 proto-board, with R = 100kΩ, C = 330pF and L = 33 mH.
As we enter a square signal (sum of signals with multiple frequency of wo) to pass through the filter gain frequencies are different and the signal remains square.
It is a filter band-eliminated since w = 0 on the gain is 1 and w = 1 the gain is <CSL; Linear Circuits and Systems (Practice 4)> infinite. Additionally there frequency which is attenuated.
is a 3.8 Measure the circuit as you did in section 3.2. What kind of filter is it? fo= 9.5kHz Vo(f2-f1)= 8.62V BW= 0.33kHz It’s a band rejected.
3.9 Explain the response of this circuit because theoretically, it coincides with the measured.
The response of this circuit does not match the result as there is a parasitic resistance of the coil that makes it not ideal.
4.CONCLUSIONS This practical passive filters have observed the behavior of a low-pass filter is removed and passes. So if watching their behavior and measuring success with square signal bandwidth ... We can conclude by saying that the practice of performing calculations and was successful, although the first part of practice we used a coil 42mH.