082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

23.9 Frequency response of a linear system 753
Eliminatingiyields
v i
= v o
jωCR+v o
= v o
(1+jωRC)
v o
v i
=
1
1 +jωRC
(23.10)
Equation (23.10) relates the output of the system to the input of the system.
Therefore,
G(jω) =
1
1 +jωRC
Itis convenient toconvertG(jω) into polar form:
1̸ 0
G(jω) = √
1 + (ωRC)
2̸
=
1
√ ̸
1 + (ωRC)
2
tan −1 ωRC
−tan −1 ωRC
Therefore,
|G(jω)| =
1
√
1+ (ωRC)
2
(23.11)
̸ G(jω) = −tan −1 ωRC (23.12)
The amplitude and phase characteristics for the circuit of Figure 23.27 are shown in
Figure 23.28. These show the variation of |G(jω)| and ̸ G(jω) with angular
frequency ω.
|G(jv)|
1
0.707
0
1/RC
v
R
G(jv)
1
v
y i
Figure23.27
Circuit forEngineering
application 23.3.
i C y o
– p –4
– p –2
Figure23.28
Amplitude andphase characteristics forthe circuit
ofFigure23.27.
Note that the circuit is a low-pass filter; it allows low frequencies to pass easily
and rejects high frequencies. The cut-off point of the filter, that is the point at
which significant frequency attenuation begins to occur, can be varied by changing ➔