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Then we can calculate the current over the resistor R .
1
V i
i = (5.305)
1
R 1
Also note that no current would flow into the op-amp, then the same current must pass
through the R and C combination (noting Kirchoff’s current law),
2
V (t)
i
i (t) = = i (t) + i (t) (5.306)
c
1
R2
R 1
V (t)
o
= + i (t) (5.307)
c
R
2
t
1
V (t) = C ∫ 0 i ( )d (5.308)
o
c
where we use the fact that the current flow over the R and C is determined by the output
2
voltage and the ground voltage at the negative input (inverting) terminal. We will add the
negative sign to the input–output voltage relation at the end of the derivation. If we take the
Laplace transform of the above equations, we can easily find the transfer function betwen
input voltage and output voltage,
V (s) V (s)
i
o
i (s) = = + Cs ⋅ V (s) (5.309)
1 o
R R
1 2
V (s) R 2 1
o
=− ⋅ (5.310)
V (s) R 1 1 + R Cs
2
i
where we added the negative sign to indicate that the sign relationship between input voltage
and output voltage is opposite.
V (s) R 2 1
o
=− ⋅ (5.311)
V (s) R R Cs + 1
i 1 2
For the case of R = R = R we have
1 2
V (s) =− 1 (5.312)
o
V (s) RCs + 1
i
The same reasoning applies for the derivation of the transfer function between input
voltage and output voltage for the high pass filter. The only difference is the definition of
the output voltage. In the above derivation, we would use
V (t) = R ⋅ i(t) (5.313)
o
and it can be shown, by following the same procedure for the low pass filter, that the transfer
function for the high pass filter is
V (s) RCs
o
=− (5.314)
V (s) RCs + 1
i
5.7 DIGITAL ELECTRONIC DEVICES
The logic ON state is represented by 1 and the logic OFF state is represented by 0. Two of the
most popular digital device types are transistor–transistor logic (TTL) and complementary