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While solving Example 4.1 of Signals and Systems by Alan Oppenheim. Example 4.1 is: $$ x(t)=e^{-at}u(t), a>0$$ and the transform I get is: $$ X(j\omega)\frac{1}{a+j\omega}, a>0$$

The problem is understanding the sketch: The sketch is drawn for both the values $+a$ and $-a$> i dont understand why it $-a$ when $a>0$ is defined. enter image description here

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    $\begingroup$ $a>0$, no question about it. It is the frequency variable $\omega$ which can take on negative values, e.g. $\omega=-a$ (which is of course negative if $a$ is positive). $\endgroup$ – Matt L. Nov 19 '15 at 11:47
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    $\begingroup$ "The sketch is drawn for both the values $+a $ and $−a$. i dont understand why it $-a$ when $a>0$ is defined." The variable on the horizontal axis is $\omega$, the radian frequency which can take on all real number values. The $a$ and $-a$ are labels on the axes. Surely we can agree that if $a =1$, say, then there should be a point on the horizontal axis that bears the label $1$ and another with label $-1$? Ditto for labels $2a=2$ and $-2a=-2$ etc? The reason for not putting in these additional labels is that $a$ and $-a$ are of particular importance as the 3dB down points. $\endgroup$ – Dilip Sarwate Nov 19 '15 at 19:41
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The plot is of $$\mid X\left(i\omega\right) \mid = \sqrt{\left(\frac{1}{a+j\omega}\right)\left(\frac{1}{a-j\omega}\right)} = \frac{1}{\sqrt{a^2 + \omega^2}}$$

against $\omega$

In particular $\omega$ can be equal to $-a$.

This checks out with Wolfram alpha

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