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High Pass Filter
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High Pass Filter

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.

The string:
S₇P₅A₅₁ (Physical change).
The math:

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:
H_V(jω) = V_L(jω)/V_s(jω)
Where H_V(jω) is frequency response of load; V_L(jω) is voltage across load; V_s(jω) is frequency dependent voltage source.

H_I(jω) = I_L(jω)/I_s(jω)
Where H_I(jω) is frequency response of load; I_L(jω) is current through load; I_s(jω) is frequency dependent current source.

(a)Frequency response, H(jω) = V_o/V_i(jω)
By the voltage divider rule:
V_o(jω) = V_i(jω)R/(R + 1/jωC) = V_i(jω)[jωRC/(1 + jωRC)]
So, H(jω) = V_o/V_i(jω) = [jωRC/(1 + jωRC)]----(1)
Phasor form:
jωRC = (ωRC)e^jπ/2
Phasor form:
1 + jωRC = (1 + (jωRC)²)^1/2(e^jarctanωCR
So, phasor form:
H(jω) = [ωRC/(1 + (jωRC)²))^1/2](e^{j[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,
V_o(jω = 0) = V_i(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:
V_i = |V_i|e^jφ_i
The scaling of the output signal becomes:
V_o = |H|(V_i).
And the phase angle of the output signal becomes:
φ_o = <H + φ_i.

Math

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