Frequecy Response: High Pass Filter And Cut-Off Frequency
Strings (SiPjAjk) = S7P5A51 Base Sequence = 12735 String Sequence = 12735 - 5 - 51
Figure 7.50 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)Frequency response, H(jω) = Vo/Vi(jω)
By the voltage divider rule:
Vo(jω) = Vi(jω)R/(R + 1/jωC) = Vi(jω)[jωRC/(1 + jωRC)]
So, H(jω) = Vo/Vi(jω) = [jωRC/(1 + jωRC)]----(1)
jωRC = (ωRC)ejπ/2
1 + jωRC = (1 + (jωRC)2)1/2(ejarctanωCR
So, phasor form:
H(jω) = [ωRC/(1 + (jωRC)2))1/2](ej[π/2-arctan(ωCR)]----(2)
where |H(jω)| = ωRC/(1 + (jωRC)2)1/2
And the phase angle:
<H(jω) = π/2-arctan(ωCR)
Equations (1) and (2) reveal that at ω = 0,
Vo(jω = 0) = Vi(jω = 0).
That is, no filtering at ω = 0 (DC signal).
As the signal frequency approaches infinity, the magnitude of the frequency response asymptotically approach 1.
(b) Cutoff frequency, ω0 = 1/RC.
When ω >> 1/RC, filter is called a high pass filter. A high-pass filter passes signals at high frequencies and filters out signals at low frequencies (ω << 1/RC).
The value of |H(jω)| at the cufoff frequency, is 1/√2 = ).707.
The cufoff frequency depends entirely on the values of R and C. As a result filtering characteristics can be selected for various values of R and C.
In essence the filter effect results in the scaling and phase angle shifting of the input signal. In mathematical terms, for the following input signal in phasor form:
Vi = |Vi|ejφi
The scaling of the output signal becomes:
Vo = |H|(Vi).
And the phase angle of the output signal becomes:
φo = <H + φi.
The point "." is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
Ordinary Differential Equations (ODEs)