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Given the transfer function $$T(s) = \frac{100}{1 + \frac{s}{10^{6}}}$$ and the input $$v_i(t) = 0.1 \sin(100t)$$ find the output, $v_o(t)$.

My approach was to use $v_o(t) = \mathcal{L^{-1}}\left\{T(s)\ V_i(s)\right\}$, where $V_i(s) = \mathcal{L\left\{v_i(t)\right\}}$. This gives

$$v_o(t) = \left(\frac{10^5}{10^8+1}\right) \mathrm{e}^{-10^6 \,t} - \left(\frac{10^5}{10^8+1}\right) \cos\left(100\,t\right) + \left(\frac{10^9}{10^8+1}\right) \sin\left(100\,t\right)$$

in MATLAB. However, my textbook does the following:

enter image description here

Which one is correct? The difference between the two functions is of the order $10^{-3}$, and the first function is not reducible to the second in MATLAB.


Edit:

This is the whole question — Example 1.5 from Sedra & Smith's Microelectronic Circuits (7th ed.):

enter image description here

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  • $\begingroup$ Note that you are misusing the conventional notation. The $s$ notation usually represents the Laplace transformation, Where the $j\omega$ represents the Fourier transform. Please review your question and this may help you understand. Also, more details like the full question may be useful. $\endgroup$ – havakok Apr 27 at 7:55
  • $\begingroup$ Go to the book's index and search for "Laplace". Which Laplace transform do they use? I used that book over a decade ago, but don't recall such details. $\endgroup$ – Rodrigo de Azevedo Apr 27 at 21:08
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    $\begingroup$ Yes, they don't talk about the Laplace transforms, but I was trying to use what I had learnt in earlier courses like advanced engineering mathematics and circuit analysis. In fact, I just realized that phasors can be used here since the input is sinusoidal. $\endgroup$ – Leponzo Apr 27 at 22:08
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You cannot solve this problem using the Laplace transform. The reason is that the Laplace transform of the input signal doesn't exist. You could use the Fourier transform, but in this case there's an even simpler way to determine the output signal. You need to know one important property of linear time-invariant (LTI) systems: their response to a sinusoidal input is a sinusoidal signal with the same frequency, but with its amplitude and phase altered according to the system's frequency response evaluated at the input frequency.

So for an input signal

$$x(t)=A\sin(\omega_0t+\phi)\tag{1}$$

the output is given by

$$y(t)=A\big|H(j\omega_0)\big|\sin\left(\omega_0t+\phi+\arg\big\{H(j\omega_0)\big\}\right)\tag{2}$$

where

$$H(j\omega)=\big|H(j\omega)\big|e^{j\arg\{H(j\omega_0)\}}\tag{3}$$

is the system's frequency response.

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  • $\begingroup$ Is the reason that the Laplace transform of the input does not exist that it is not $0$ for $t < 0$? $\endgroup$ – Leponzo Apr 27 at 11:51
  • $\begingroup$ Also, is there a way to get this output using MATLAB? I tried fourier and ifourier but those don't seem to give this result. $\endgroup$ – Leponzo Apr 27 at 12:56
  • $\begingroup$ @Leponzo: Yes, a sinusoid has a constant envelope for all values of $t$, so the Laplace integral doesn't converge. $\endgroup$ – Matt L. Apr 27 at 15:20
  • $\begingroup$ @Leponzo: I don't know how to obtain the same result with Matlab, but I think such exercises are best solved by hand, in order to learn and understand the basic properties of LTI systems. $\endgroup$ – Matt L. Apr 27 at 15:21
  • $\begingroup$ Without the adjective "bilateral", isn't your answer incorrect? $\endgroup$ – Rodrigo de Azevedo Apr 27 at 22:20

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