Electronic – effects of feedback on noise and non-linearities

circuit analysiscircuit-designcontrol systemfeedbacknon-linear

This video lecture (start at 32:01), the professor showed a demo about the effects of feedback on noise and non-linearities.

The graph below is a system with input signal which is applied directly to an operational amplifier(the amplifier models the linear portion of the forward path gain), followed by a nonlinear element (has dead zone and two different degrees of compression of linear region). A potentiometer is put between the output and input of nonlinear element to moderate the effects on the non-linearity.

What I am confused is why the feedback path (potentiometer) is connected between the input and output of the non-linearity element as the figure 1 instead of the one as I modified in figure 2.

Can I use the feedback path with the voltage divider as figure 2?

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Figure 1: system from the lecture

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Figure 2: modified system

Best Answer

The professor in that video is trying to demonstrate the effect of feed back on noise and other non-linearities.

Lets start by assuming the amplifier is ideal: infinite gain, infinite bandwidth and capable of driving any load. I know these do not exist but it makes it easier for us to do a thought experiment.

Taking the circuit of Figure 1 then the system always has unity gain but we have a source of noise and non-linearity. The amplifier is forcing the voltage at the wiper of the potentiometer to match \$ v_i \$. If we imagine the wiper fully on the right hand side as drawn then the amplifier sees the effects of noise and non-linearity and corrects for it totally, With the wiper fully on the left hand side the amplifier does not see any of these effects so there is no correction and we see the full effects of non-linearity and noise. With the wiper somewhere in-between there is partial correction.

To clarify the point regarding always having unity gain I mean unity gain with some superimposed distortion resulting from noise and non-linearity. The position of the potentiometer controls the amount of distortion seen.

With your circuit of figure two the amplifier sees all the effects of non-linearity and noise so these are totally corrected for. However this circuit also introduces gain. The larger \$ R_1 \$ or the smaller \$ R_2 \$ the more gain and.

$$v_o = v_i \left(1+\frac{R_1}{R_2}\right)$$