Frequecy Response: Low Pass Filter And Cut-Off Frequency
Strings (SiPjAjk) = S7P5A51 Base Sequence = 12735 String Sequence = 12735 - 5 - 51
Figure 7.49 shows a circuit of a simple RC filter. Determine:
(a) The phasor form of the frequency response, H(jω) in terms of ω, R and C.
(b) The cutoff frequency of the RC filter.
S7P5A51 (Physical change).
Pj Problem of interest is of type change. Frequency problems are change problems. They are similar to velocity, acceleration and duration problems which are also change problems.
In general, frequency response is a measure of the variation in a load-related parameter as a function of the frequency of the excitation element. In electric circuits, the load-related parameter is usually the voltage across a load or the current through it and the excitation element is usually a sinusoidal signal. Consequently, any of the following is an acceptable definition of the frequency response of a circuit:
HV(jω) = VL(jω)/Vs(jω)
Where HV(jω) is frequency response of load; VL(jω) is voltage across load; Vs(jω) is frequency dependent voltage source.
HI(jω) = IL(jω)/Is(jω)
Where HI(jω) is frequency response of load; IL(jω) is current through load; Is(jω) is frequency dependent current source.
(a) H(jω) = Vo(jω)/Vi(jω)
By the voltage divider rule,:
Vo(jω) = Vi(jω)[(1/jωC)/(R + 1/jωC)] = Vi(jω)[1/(1 + jωRC)].
So, H(jω) = [1/(1 + jωRC)]----(1)
In phasor form:
1 = 1ej0
1 + jωRC = [1 + (jωRC)2)]1/2(ejarctan(ωCR)
So, in Phasor form:
H(jω) = [1/(1 + (jωRC)2)]1/2(e-jarctan(ωCR)----(2)
Where |H(jω)| = [1/(1 + (jωRC)2)]1/2
And the phase angle is:
<H(jω) = -arctan(ωCR).
Equations (1) and (2) reveals that at ω = 0,
Vo(jω = 0) = Vi(jω = 0).
That is, no filtering at ω = 0 (DC signal).
As the signal frequency increases, the magnitude of the frequency response decreases.
(b) Cutoff frequency, ω0 = 1/RC.
At low frequencies (When ω << 1/RC), filter is called a low-pass filter. A low-pass filter passes signals at low frequencies and filters out signals at high frequencies (ω >> 1/RC).
The value of |H(jω)| at the cufoff frequency, is 1/√2 = .707.
The cufoff frequency depends only on the values of R and C. As a result filtering characteristics can be selected for various values of R and C.
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