EE 205 Dr. A. Zidouri. Electric Circuits II. Frequency Selective Circuits (Filters) Low Pass Filter. Lecture #36

Transcription

1 EE 05 Dr. A. Zidouri Electric ircuits II Frequency Selective ircuits (Filters) ow Pass Filter ecture #36 - -

2 EE 05 Dr. A. Zidouri The material to be covered in this lecture is as follows: o Introduction to the ow Pass filter o The Series ircuit- Qualitative Analysis o Defining The utoff Frequency o The Series ircuit- Quantitative Analysis - -

3 EE 05 Dr. A. Zidouri After finishing this lecture you should be able to: Understand the Behavior of ow-pass Filters Understand the cutoff frequency. elate cutoff frequency to component values Analyze Qualitatively And Quantitatively A ow-pass Filter Design A Simple ow-pass Filter - 3 -

4 EE 05 Dr. A. Zidouri Introduction to ow Pass Filter Filters that will be considered here and in following lectures are passive filters. We will examine in this and next lecture two circuits that behave as low-pass filters Fig. 36- o Series ircuit o Series ircuit Discover what characteristics of these circuits determine the cutoff frequency. Note across which element the output signal is taken for the above two circuits - 4 -

5 EE 05 Dr. A. Zidouri The Series ircuit Qualitative Analysis et us start with a ow-pass Filter, series ircuit shown in Fig. 36-a Assume the circuit s input is a sinusoidal voltage source with varying frequency The circuit s output is the voltage across. Behavior of will not change with changing frequency Behavior of inductor will change with changing frequency ecall that the impedance of an inductor is jω at low frequencies the inductor s impedance is very small compared with the resistor s impedance, The inductor effectively functions as a short circuit. The term low frequencies thus refers to any frequencies for which ω The equivalent circuit for ω = 0 is shown in Fig. 36-b At this frequency Vi= Vo both in magnitude and phase angle V i V o Fig. 36-b A Series ow-pass filter at ω = 0-5 -

6 EE 05 Dr. A. Zidouri The Series ircuit Qualitative Analysis (cont) Increasing the frequency causes the inductor s impedance to increase. Increasing the inductor s impedance causes a corresponding increase in the magnitude of the voltage drop across Increasing the inductor s impedance causes a corresponding decrease in the magnitude of the output voltage drop across Increasing the inductor s impedance also introduces a shift in phase angle between V the voltage across the inductor and I the current in the inductor This results in a phase angle difference between the input and output voltage, V o lags frequency increases this phase lag approaches 90 The inductor effectively functions as an open circuit. The term high frequencies thus refers to any frequencies for which ω The equivalent circuit for ω = is shown in Fig. 36-c V o V i as the ω = - 6 -

7 EE 05 Dr. A. Zidouri The Series ircuit Qualitative Analysis (cont) V = at ω = this Based on the behavior of the output voltage magnitude at high frequencies o 0 series circuit selectively passes low-frequency inputs to the output. This circuit response is shown in Fig The ideal filter exhibits a discontinuity in magnitude at the H(j cutoff frequency ω Ideal Plot Not possible to use real components to construct a circuit that has an abrupt transition in magnitude. The change is gradually for a real circuit. So what do we mean by cutoff frequency? θ(j 0 0 o c Actual Plot -90 o Fig Frequency esponse for Series ircuit in Fig. 36-a - 7 -

8 EE 05 Dr. A. Zidouri Defining the cutoff frequency The transition between Passband and Stopband is not abrupt so there is not a single frequency ω but a band of transition. Therefore we need to define the cutoff frequency. The definition for cutoff frequency widely used by electrical engineers is the frequency for which the transfer function magnitude is decreased by the factor ( ) H max H jω = (36-) from its maximum value. ecall that the average power delivered by any circuit to a load is proportional to V where is the amplitude of the voltage drop across the load: V P = (36-) Define P max as the value of the average power delivered to a load when the magnitude of the load voltage is maximum: P max V max = (36-3) - 8 -

9 EE 05 Dr. A. Zidouri Defining the cutoff frequency (cont.) If we vary the frequency of the source voltage V i the load voltage is a maximum when the magnitude of the circuit s transfer function is also a maximum: V H V max = max i (36-4) When the frequency is equal to ω the magnitude of the load voltage is: Substituting equ. (36-5) into equ. (36-) ( ω ) = ( ω ) i V j H j V = Hmax V i = V max (36-5) V ( ) ( ) jω P jω = V = max - 9 -

10 EE 05 Dr. A. Zidouri Defining the cutoff frequency (cont.) = P V max = max (36-6) Therefore at the cutoff frequency ω, the average power delivered by the circuit is one half the maximum average power. Thus ω is also called the half-power frequency, or the corner frequency

11 EE 05 Dr. A. Zidouri The Series ircuit Quantitative Analysis onsider the circuit of Fig s V i (s) V o (s) Fig s-domain equivalent for the ircuit of Fig. 36-a The voltage transfer function for this circuit is: H ( s) = (36-7) s + To study the frequency response we replace s = jω in equation (36-7): H( jω ) = jω + (36-8) - -

12 EE 05 Dr. A. Zidouri The Series ircuit Quantitative Analysis (cont) Equation (36-8) expressed in polar form gives the magnitude and phase: ( ω) = H j ω + θ ( jω) tan ω (36-9) = (36-0) lose examination of these equations provide quantitative support for the plots seen in Fig We can now compute ω in terms of the circuit elements: ecall from (36-) that H( jω ) = Hmax, for low-pass filter, Hmax H( j0) H jω = = 36-3 thus: ( ) ω + (36-), solving we get: = as seen in Fig. ω = (36-). - -

13 EE 05 Dr. A. Zidouri Example 36. hoose values for and a commonly available value of =00 mh in the circuit in Fig. 36-a, such that the circuit could be used in an EG (electrograph) to filter noise above 0 Hz and pass the electric signals from the heart at or near Hz. ompute the magnitude of V o at Hz, 0 Hz, and 60 Hz to see how well the filter performs, take V i = V. Answer: The problem is to design a filter with a cutoff frequency of f c =0 Hz. ω = π ( 0) = 0πrad s Solving for from equ. (36-) ω 0π o ( ) V H j V = ω i we get: - 3 -

14 EE 05 Dr. A. Zidouri V o ( ω) = ω + V i = 0π ω + 400π V i For f c = Hz, V 0.995V For f c =0 Hz, V 0.707V For f c =60 Hz, V 0.64 o = almost equal to the input voltage as expected o = equal to = V i as 0 Hz is the cutoff frequency. o = V very small compared to the input so the filter is performing well

15 EE 05 Dr. A. Zidouri Self Test 36: A series low-pass filter with a cutoff frequency of a) ; H jω at 50 khz; and b) ( ) c) θ ( j ) ω at 50 khz ω = khz is needed. Using 5k = Ω, compute: Answer: a) 0.40 H b) 0.04; c) o - 5 -