3-Lesson_Notes_Lecture27_ee205 - EE 205 Dr. A. Zidouri...

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EE 205 Dr. A. Zidouri - 1 - Electric Circuits II Frequency Selective Circuits (Filters) Low and High Pass Filters Lecture #27
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EE 205 Dr. A. Zidouri - 2 - The material to be covered in this lecture is as follows: o Series RC Circuit as a Low Pass filter o Relating The Frequency Domain to the Time Domain o Series RC Circuit as High-Pass filter o Series RC Circuit- Qualitative Analysis o The Series RC Circuit- Quantitative Analysis After finishing this lecture you should be able to: ¾ Understand the Behavior of High-Pass Filters ¾ Distinguish between Low-Pass and High-Pass Filters ¾ Relate The Cutoff Frequency to the Time Constant ¾ Analyze Qualitatively And Quantitatively A High-Pass Filter ¾ Design A Simple High-Pass Filter
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EE 205 Dr. A. Zidouri - 3 - The Series RC Circuit Low-Pass Filter The series RC Circuit shown in Fig. 37-1 also behaves as a low pass filter. Note that the circuit’s output is defined as the output across the capacitor. As we did in the previous qualitative analysis in Lec36, we use three frequency regions to develop the behavior of the series RC circuit in Fig. 37-1: 1. Zero Frequency 0 ω = impedance of the capacitor C Z =∞ , the capacitor acts as an open circuit. The input and output voltages are the same as shown in Fig. 37-2a 0 =
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EE 205 Dr. A. Zidouri - 4 - 2. Frequencies increasing from zero. The impedance of the capacitor C Z R ± , The output voltage is smaller than the source voltage which divides between R and C. 3. Infinite frequency ω =∞ impedance of the capacitor 0 C Z = , the capacitor acts as a short circuit. The output voltage is zero as shown in Fig. 37-2b Based on this analysis we see that the series RC circuit functions as a low-pass filter. The following example explores the quantitative analysis of the circuit. Example 37-1 For the series RC circuit in Fig. 37-1 above: ± Find the transfer function () o i V Hj V = ± Determine the cutoff frequency in the circuit ± Choose values of R and C that will yield a low pass filter with 3 C kHz = . Take 1 CF µ = .
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EE 205 Dr. A. Zidouri - 5 - Solution: To derive the transfer function, consider the s-domain circuit in Fig. 37-3:
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3-Lesson_Notes_Lecture27_ee205 - EE 205 Dr. A. Zidouri...

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