Page 317 - Mechatronics with Experiments

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JWST499-Cetinkunt
JWST499-c05
ELECTRONIC COMPONENTS FOR MECHATRONIC SYSTEMS 303
op-amp is the same. Since current into the op-amp is zero, voltage at the (−) terminal is the
same as the output voltage. That is,
+
−
i = i = 0.0 (5.274)
−
V = V + (5.275)
= V (5.276)
o
+
Notice that the output voltage, V = V is the voltage across the capacitor C. The voltage
o
across the capacitor is the output voltage and is related to the current and capacitance value
as (assuming zero initial voltage at t = 0, V (0) = 0.0),
o
t
1
V (t) = C ∫ 0 i( )d (5.277)
o
1
V (s) = i(s) (5.278)
o
Cs
The current in the circuit is
V (t) − V (t)
i
o
i(t) = (5.279)
R
V (s) − V (s)
o
i
i(s) = (5.280)
R
1
R ⋅ i(s) = V (s) − i(s) (5.281)
i
Cs
( )
1
R + ⋅ i(s) = V (s) (5.282)
i
Cs
i(s) Cs
= (5.283)
V (s) RCs + 1
i
Hence the transfer function between the output voltage and input voltage is
V (s) 1
o
= (5.284)
V (s) RCs + 1
i
V (jw)
o 1
= (5.285)
V (jw) 1 + jRCw
i
1
Notice that when w = rad∕s, the magnitude ratio of the output voltage to input
RC
voltage is
|V (jw)| 1
o
| | = √ = 0.707 (5.286)
| |
i
| V (jw) | 2
1 1
where the value w = RC rad∕sor f = 2 RC Hz is called the cutoff frequency of the filter,
c
c
that is the frequency at which the output signal magnitude is 0.707 times the input sig-
nal magnitude in steady-state. In other words, the output signal is attenuated by 3 dB in
comparison to the input signal.
High Pass Filter Op-Amp The high pass filter suppresses the low frequency content
of the input signal, and passes the high frequency content. An op-amp implementation of