Frequency Response: Band Pass Filter And Resonant (Natural) Frequency
Strings (SiPjAjk) = S7P5A51 Base Sequence = 12735 String Sequence = 12735 - 5 - 51
Figure 7.56 shows a circuit of a simple RLC filter. Determine:
(a) The frequency response of the RLC filter in terms of the natural or resonant frequency.
(b) The bandwidth in terms of the natural or resonant frequency and the quality factor.
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)[R/(R + (1/jωC) + jωL)] = Vi(jω)[jωCR/(1 + jωCR + (jω)2LC)]
So, H(jω) = [jωCR/(1 + jωCR + (jω)2LC)]-------(1)
In phasor form:
H(jω) = ωCR/√[(1 -ω2LC)2 + (ωCR)2]<[π/2 -arctan(ωCR/(1 - ω2LC))] -------(2)
(b)The natural or resonant frequency:
ωn = √1/LC.
The quality factor:
Q = 1/(ωnCR) = (1/R)(√(L/C)).
The damping ratio:
1/(2Q) = (R/2)(√(C/L))
Bandwidth, B = ωn/Q.
So, H(jω) in equation (1) is expressed in terms of resonant frequency and quality factor as follows:
H(jω) = [(1/(Qωn))jω]/[(jω/ωn)2 + (1/(Qωn))jω + 1]
The bandwith (passband) is a frequency range. The band pass filter allows input signal frequencies within the range to pass through it.
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