Difference between revisions of "Integrators"
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{{NumBlk|::|<math> | {{NumBlk|::|<math> | ||
\begin{align} | \begin{align} | ||
− | \overline{v_{o,\text{T}}^2} & = \int_0^\infty \left|H_1\left(s\right)\right|^2\cdot \frac{\overline{v_{n1}^2}}{\Delta f} df + \int_0^\infty \left|H_2\left(s\right)\right|^2\cdot \frac{\overline{v_{n2}^2}}{\Delta f} df \\ | + | \overline{v_{o,\text{T}}^2} & = \int_0^\infty \left|H_1\left(s\right)\right|^2\cdot \frac{\overline{v_{n1}^2}}{\Delta f}\cdot df + \int_0^\infty \left|H_2\left(s\right)\right|^2\cdot \frac{\overline{v_{n2}^2}}{\Delta f}\cdot df \\ |
& = \int_0^\infty \frac{1}{1 + \omega^2 R_1^2 C^2}\cdot 4kTR_1\cdot df + \int_0^\infty \frac{1}{1 + \omega^2 R_2^2 C^2}\cdot 4kTR_2\cdot df \\ | & = \int_0^\infty \frac{1}{1 + \omega^2 R_1^2 C^2}\cdot 4kTR_1\cdot df + \int_0^\infty \frac{1}{1 + \omega^2 R_2^2 C^2}\cdot 4kTR_2\cdot df \\ | ||
& = \frac{kT}{C} + \frac{kT}{C} = \frac{2kT}{C} \\ | & = \frac{kT}{C} + \frac{kT}{C} = \frac{2kT}{C} \\ |
Revision as of 20:19, 1 April 2021
Contents
The Ideal Integrator
The ideal integrator, shown in Fig. 1, with symbol shown in Fig. 2, makes use of an ideal operational amplifier with , , and . The current through the resistor, , can be expressed as:
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(1)
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Thus, we can write the integrator output voltage, , as:
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(2)
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In the Laplace domain:
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(3)
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Or equivalently:
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(4)
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The magnitude and phase response of an ideal integrator is shown in Figs. 3 and 4.
Fig. 5 shows a multiple-input integrator, with output voltage:
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(5)
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Integrator Noise
Fig. 6 shows an integrator where the output is fed back to one of its inputs, giving us:
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(5)
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Ignoring the noise from the amplifier, the output noise of the integrator in Fig. 6 can be expressed as:
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(7)
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The total integrated noise is then:
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(8)
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