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

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:

 

 

 

 

(1)

Thus, we can write the integrator output voltage, , as:

 

 

 

 

(2)

In the Laplace domain:

 

 

 

 

(3)

Or equivalently:

 

 

 

 

(4)

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:

 

 

 

 

(5)

Integrator Noise

Fig. 6 shows an integrator where the output is fed back to one of its inputs, giving us:

 

 

 

 

(5)

Ignoring the noise from the amplifier, the output noise of the integrator in Fig. 6 can be expressed as:

 

 

 

 

(7)

The total integrated noise is then:

 

 

 

 

(8)

Integrator Non-Idealities

Finite Gain

Non-Dominant Poles

Capacitor Non-Idealities