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If I apply a unit step input to a linear system and get the following as output:

$$(1-e^{-4t})u(t)$$

does that mean the transfer function is then:

$$T(t) = (1-e^{-4t})$$

My friend seems to think so but I disagreed. I proceeded otherwise to find the transfer function.

$$Laplace((1-e^{-4t})u(t))= \frac{1}{s} - \frac{1}{s+4} = (\frac{1}{s})\frac{4}{s+4}$$

I can then take out the unit step input (1/s) to get the transfer function so:

$$T(s) = \frac{4}{s+4}$$

$$T(t) = 4e^{-4t}$$

Which of us is correct?

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    \$\begingroup\$ Well, \$1-e^{-4t}\$ is not a function of s, so I don't see how it could be T(s). \$\endgroup\$ – The Photon Sep 5 '14 at 1:18
  • \$\begingroup\$ @ThePhoton I meant T(t) \$\endgroup\$ – codedude Sep 5 '14 at 1:31
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The impulse response \$h(t)\$ is the time derivative of the step response. Thus, for the step response given

$$h(t) = \frac{d}{dt}S(t) = 4e^{-4t}u(t) + (1 - e^{-4t})\delta (t) = 4e^{-4t}u(t)$$

Then, the transfer function \$H(s)\$ is the transform of the impulse response which is as you alternately derived,

$$H(s) = \frac{4}{s + 4}$$

Finally we have

$$y(t) = h(t) \star x(t) = \int_0^t h(\tau)x(t - \tau) d\tau = \int_0^t x(\tau)h(t - \tau) d\tau$$

and

$$Y(s) = H(s)X(s) $$

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  • \$\begingroup\$ Hey, the convolution integrals should go from 0 to t, not to inf. I tried editing your answer but I don't have enough reputation :) \$\endgroup\$ – raggot Sep 10 '14 at 13:08
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The important thing to note here is the following two statements:

$$Y(s) = H(s)X(s)$$

$$y(t) \neq h(t)x(t)$$

where y is the output, h is the transfer function and x is the input applied.

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You are right.

A way to confirm your intuition is that the step response exhibits a single time constant (i.e., one pole) \$\tau=-4\$ and a steady state of \$y(\infty)=1\$. Even without going through calculations, you can then argue

$$ T(s)=\frac{4}{s+4}. $$

Anyway your friend's answer is wrong not only because of the chosen parameters or form, but because of the use he makes of it. What defines an input response of a system in the time domain is the result of a convolution product, not of a scalar product:

$$ y(t) = T(t)\star u(t) = \int_0^tT(\tau)\cdot u(t-\tau)dt \neq T(t)\cdot u(t)$$

What \$u(t)\$ does to the system at a time \$t=t_x\$ has an impact on \$y(t)\$ (theoretically) forever. In a scalar multiplication you simply don't have that. Obviously the type of multiplication used is not the result of a design. It's derived mathematically. But this is a simple argument to show your friend he can't be right.

The main reason Laplace transform is so popular is exactly because the Laplace transform of a convolution product is a simple product:

$$ L[T(t)\star u(t)] = T(s)U(s).$$

It makes representations easier to deal with and more elegant.

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