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I made a DC-DC converter its ideal transfer function is: $$ V_0(D) = DV_i $$

I took measurements of its response and then made a linear regression, I got this result:

$$ V_0(D) = 0.98DV_i -0.42 $$

Quite similar to the expected value. Now, can I say that the system has a linear response?

strictly speaking, the system response should be zero when the input is zero, but in this case, if D = 0:

$$ V_0 = -0.42 $$

What would be the correct way to refer at the response of this system?

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  • \$\begingroup\$ Right up until it clips, yes. \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 5 '16 at 0:39
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    \$\begingroup\$ Linear is exactly correct. However, it is not proportional, which would only occur if c is zero. That is, the output is not proportional to the input (if you double the input you do not double the output. Of course, you can say that changes in the output are proportional to changes in the input, but this is not (exactly) the same thing. \$\endgroup\$ – WhatRoughBeast Jun 5 '16 at 1:04
  • \$\begingroup\$ You did a regression with \$D\$ as the independent variable or \$V_i\$? \$\endgroup\$ – The Photon Jun 5 '16 at 2:10
  • \$\begingroup\$ Possibly related: affine transformation. \$\endgroup\$ – The Photon Jun 5 '16 at 2:12
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    \$\begingroup\$ Strictly speaking that is not a linear relationship between D and \$V_o\$, it is an affine relationship (see previously commented link). However its very rare that the distinction matters in engineering and you're likely to hear people call it "linear" anyway. \$\endgroup\$ – The Photon Jun 5 '16 at 2:28
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Strictly speaking that is not a linear relationship between \$D\$ and \$V_o\$, it is an affine relationship. However its very rare that the distinction matters in engineering and you're likely to hear people call it "linear" anyway.

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  • \$\begingroup\$ Great, that was exactly what I was looking for. So could I say that Vo(D) = 0.98DVi - 0.42 is an affine model of the system? \$\endgroup\$ – Luis Ramon Ramirez Rodriguez Jun 5 '16 at 2:34
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    \$\begingroup\$ Honestly I think it would be more common to just call it a linear model. \$\endgroup\$ – The Photon Jun 5 '16 at 2:41
  • \$\begingroup\$ I would say that it is misleading to call it linear. When someone says linear I understand T(a + b) = T(a) + T(b) \$\endgroup\$ – user110971 Jun 5 '16 at 7:39
  • \$\begingroup\$ I understand that it is affine, but I really have trouble understanding why it's not just linear, since linearity means y=mx+b. Could you write something about the distinction? \$\endgroup\$ – sweber Jun 5 '16 at 7:50
  • \$\begingroup\$ @sweber, see user110971's comment. That's the definition of a linear system, and it doesn't apply to OP's system. \$\endgroup\$ – The Photon Jun 5 '16 at 14:57
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Can a system with a response Y(x) = ax + c be called linear?

If two systems, A and B, are linear then for a cascade of the two systems, the order does not matter. That is, AB = BA:

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For example, let system A be an ideal gain of 10 stage while system B is an ideal 1st order low-pass filter with unity DC gain.

Since both stages are linear, the cascade of the two systems is a low-pass filter with a DC gain of 10 regardless of whether B follows A or A follows B in the cascade.

Now, see that a system with gain and offset is not a linear system. For example, let system A be as before but system B is now unity gain with a constant offset of 1.

For the cascade AB, the output is the input scaled by 10 plus an offset of 1.

However, for the cascade BA, the output is the input scaled by 10 plus an offset of 10 and so system B is not a linear system.


Another definition of linearity is the following: if \$y_1\$ is the output of a system given input \$x_1\$ and \$y_2\$ is the output of the same system given input \$x_2\$, then given the input \$x_3 = a_1x_1 + a_2x_2\$, the output is \$y_3 = a_1y_1 + a_2y_2\$ if and only if the system is linear.

For the case of system B with unity gain and offset 1, we have

$$y_1 = x_1 + 1$$ $$y_2 = x_2 + 1$$ $$y_3 = a_1x_1 + a_2x_2 + 1 \ne a_1y_1 + a_2y_2 = a_1x_1 + a_2x_2 + a_1 + a_2$$

Thus, system B is therefore not a linear system and so the answer to the quoted question is no.

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  • \$\begingroup\$ The equation doesn't satisfy the superposition theorem so isn't linear. There is a very widespread misuse of the word linear in engineering I think \$\endgroup\$ – Luis Ramon Ramirez Rodriguez Jun 5 '16 at 3:31

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