Frequency Response Of Output voltage At High And Low Frequency
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Frequency Response Of Output voltage At High And Low Frequency

Figure 6.11 shows a circuit of the given elements. Determine the behavior of the frequency response of the voltage V0 at extremely high and low frequencies.

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

Pj Problem of interest is of type change. The 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) = ILs(jω)
Where HI(jω) is frequency response of load; IL(jω) is current through load; Vs(jω) is frequency dependent voltage source.

Z1 = R1 + Xc
Where Xc is capacitive reactance.
So, Z1 = R1 + (1/jωC).
Z2 = R2 || XL
Where XL is inductive reactance.
So, Z2 = [(R2)(jωL)]/[R2 + jωL].
So, by the voltage divide rule:
V0 = (Vi)[(R2)(jωL)/(R2 + jωL)]/[(R2)(jωL))/(R2 + jωL) + (R1 + (1/jωC)]----(1)
Extremely low frequency implies ω tends to zero.
When ω ----> 0; numerator of eqn (1) ----> 0
So, at extremely low frequency, V0 ----> 0.
Extremely high frequency implies ω tends to infinity.
When ω ----> ∞ numerator of eqn (1) ----> R2Vi since (jωL)/(R2 + jωL)----> 1.
Also, denominator of eqn (1) ----> R1 + R2 since 1/jωC ----> 0.
So, at extremely high frequency, V0 ----> (R2Vi)/(R1 + R2).

Blessed are they that have not seen, and yet have believed. John 20:29