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I am taking a linear systems and signals course and we are currently looking at system transfer function realization. The textbook states the following,

A transfer function \$H(s)\$ can be realized by using integrators or differentiators along with adders and multipliers. We avoid use of differentiators for practical reasons discussed in Sections 2.1. Hence, in our implementation, we shall use integrators along with scalar multipliers and adders. We are already familiar with representation of all these elements except the integrator. The integrator can be represented by a box with integral sign (time domain representation) or by a box with a transfer function \$\frac{1}{s}\$ (frequency domain representation).

I'm not entirely sure i understand why \$\frac{1}{s}\$ is the frequency domain representation for an integrator. When i hear the word integrator i'm thinking 'integration, finding the area underneath the curve, summing of areas...'. Could anyone explain to me why \$\frac{1}{s}\$ represents integration? I am not seeing the link.

Thanks in advance.

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    \$\begingroup\$ the key for understanding this is understanding the Laplace transform (and its inverse) \$\endgroup\$
    – Curd
    Commented Oct 11, 2017 at 12:13
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    \$\begingroup\$ If you integrate a unit impulse, \$\delta(t)\$, which has LT \$\Delta (s)=1\$, you get a unit step with LT, \$H(s)=\frac{1}{s}\$. Hence the TF of the integrator is \$\frac {1}{s}\$. \$\endgroup\$
    – Chu
    Commented Oct 11, 2017 at 16:43
  • \$\begingroup\$ Can you see that a multiplication with \$s\$ is a derivation in time domain? \$\endgroup\$ Commented Oct 12, 2017 at 18:19

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From a mathematical view, the effect of differentiation in the Laplace Domain is just multiplication by s right? So the inverse operation of integration should have the inverse of s in the Laplace Domain, or 1/s.

Intuitively you could think of integration as having a low-pass or averaging effect which has a 1/s type frequency response.

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The integrator is part of the proportional-integral-derivative or PID block you can find in a control system. As highlighted in the above replies, the integrator "integrates" or accumulates the error over time. The error corresponds to the deviation between the target value and what the system delivers. In theory, as long as the error is present, the integrator output increases until the system corrects the deviation and cancels the static error. The below sketch shows the principle in action:

enter image description here

Practically speaking, an integrator can be realized around an op am with a capacitor and a resistance. The transfer function of such a block is \$H(s)=\frac{1}{s\tau}\$ in which \$\tau=R_1C_1\$ if you consider the below circuit:

enter image description here

The division by \$s\$ confirms the integration and represents a so-called pole at the origin. The \$\tau\$ offers a means to adjust the magnitude crossover point, the frequency at which the magnitude is 1 or 0 dB. As pointed out in some of the replies, in dc, the gain is theoretically infinite and there is no static error in the control system. Practically speaking, however, the gain is limited by the op amp open-loop gain \$A_{OL}\$ which sets a low-frequency pole as illustrated below and determined in the book I published here. The magnitude in the left side exhibits a -1-slope or a 20-dB decay per decade. The phase lag in the right side is a permanent -270° or 90°. You will get more information on PIDs and their applications in switching converters here.

enter image description here

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In the time domain 1/s (or integration) is finding the area under a curve or, by extension, providing a circuit that generates the product of the average input signal level and time period.

In the frequency domain, an integrator has the transfer function 1/s and relates to the fact that if you doubled the frequency of a sine input, the output amplitude would halve. At DC (s=0) the gain is infinite.

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The Laplace transform of an integrator is 1/s. Hence whenever one sees a 1/s in a transform, its evidence of an integrator.

$$ \mathcal{L} \left(\int_0^t f(\tau)\ d\tau\right) = \frac{F(s)}{s} $$

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