![\"\"](\"/userfiles/30478-5_1_1.gif\")
(5.1.1)
\Step 1:
\Noninverter amplifier :
\
The input voltage Vin is applied directly to the non-inverting (+) input terminal, which means that the output gain of the amplifier becomes positive.
\Therefore the output signal is in-phase with the input signal. So it acts as a non-inverting amplifier.
\The voltage gain of the non-inverting amplifier is ![\"\"](\"/test/fckeditor/editor/plugins/fckeditor_wiris/integration/showimage.php?formula=9c627e56eafc28cfd14085832b3a4ec1.png\")
.
\
(5.1.2)
\Step 1:
\Integrator amplifier :
\![\"\"](\"/userfiles/30478-5_1_2.gif\")
If 0 volts is applied to the input voltage, then there will be no current flows through the resistor, therefore no charging of the capacitor. Then the output voltage will not change. Hence the output voltage will be constant.
\Apply positive voltage to the input, then the output of the op-amp will fall negative at a linear rate.
\Now apply negative voltage to the input, then the output will rising at a linear rate.
\The output voltage rate-of-change will be proportional to the value of the input voltage.
\The ouput voltage of the integrator amplifier is ![\"\"](\"/test/fckeditor/editor/plugins/fckeditor_wiris/integration/showimage.php?formula=fea304cc2460a99769a6a02b9f628c14.png\")
.
\
(5.1.3)
\Step 1:
\Inverter amplifier :
\![\"\"](\"/userfiles/30478-5_1_3.gif\")
\
The input voltage Vin is applied directly to the inverting (-) input terminal, which means that the output gain of the amplifier becomes negative.
\Therefore the output signal is out-phase with the input signal. So it acts as a inverting amplifier.
\The voltage gain of the inverting amplifier is ![\"\"](\"/test/fckeditor/editor/plugins/fckeditor_wiris/integration/showimage.php?formula=221988f21aea3ec761082878354de9e2.png\")
.
(5.1.4)
\Step 1:
\Differentiator amplifier :
\![\"\"](\"/userfiles/30478-5_1_4.gif\")
Apply linear positive voltage to the input, then the output of the op-amp will constant negative voltage.
\Now apply linear negative voltage to the input, then the output of the op-amp will constant positive voltage.
\The input voltage rate-of-change will be proportional to the value of the output voltage.
\The ouput voltage of the differentiator amplifier is ![\"\"](\"/test/fckeditor/editor/plugins/fckeditor_wiris/integration/showimage.php?formula=306a8427aea59d7b9d729f693a2e600d.png\")
.
\
(5.2)
\Step 1:
\The capacitance of the integrator amplifier is ![\"\"](\"/test/fckeditor/editor/plugins/fckeditor_wiris/integration/showimage.php?formula=d3d492d8a5465eba415ac9cfe5602a3f.png\")
.
Input voltage is ![\"\"](\"/test/fckeditor/editor/plugins/fckeditor_wiris/integration/showimage.php?formula=79468f0809d8f68719e207eb46283cbe.png\")
.
If positive voltage is applied to the input, then the output of the op-amp will fall negative at a linear rate.
\The rate of change of the outpuit voltage is ![\"\"](\"/test/fckeditor/editor/plugins/fckeditor_wiris/integration/showimage.php?formula=a9077a3eab9402319238c754cbbc8d6f.png\")
.
The ouput voltage of the integrator amplifier is ![\"\"](\"/test/fckeditor/editor/plugins/fckeditor_wiris/integration/showimage.php?formula=4f7e0d41dbaa5daa4472a3f9496e7794.png\")
.
![\"\"](\"/test/fckeditor/editor/plugins/fckeditor_wiris/integration/showimage.php?formula=98b7622fa13a11bd5e1e136d0b0a9211.png\")
The input resistance of the integrator amplifier is ![\"\"](\"/test/fckeditor/editor/plugins/fckeditor_wiris/integration/showimage.php?formula=ea1ba417952965340e6e1e1c4c5c08a8.png\")
.
Solution :
\The input resistance of the integrator amplifier is ![\"\"](\"http://www.mathskey.com/test/fckeditor/editor/plugins/fckeditor_wiris/integration/showimage.php?formula=ea1ba417952965340e6e1e1c4c5c08a8.png\")
.