I have the following question in my textbook (since it is an electrical engineering textbook it uses the notation $j$ to denote the imaginary number $i$). Note that $u(t)$ is a step function:

If $i(t) = 5e^{3t}u(-t)$, find (a) $A_i(2)$ (b) $B_i(3)$ (c) $|F_i(j3)|$

For a function, $f(t)$:

$$F_f(\omega)=\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt$$

$F_f(\omega)=\int_{-\infty}^{\infty}f(t) \cos(\omega t)dt - i \int_{-\infty}^{\infty}f(t) \sin(\omega t)dt \tag{1}$

And $A_f(\omega)$ and $B_f$ are of the form:

$$F_f(\omega)=A_f(\omega)+iB_f(\omega) \tag{2}$$

Therefore, matching the form of (2) to (1):

$$ A_f(\omega)=\int_{-\infty}^{\infty}f(t) \cos(\omega t)dt$$

$$B_f(\omega)=-\int_{-\infty}^{\infty}f(t) \sin(\omega t)dt$$

The answers to this question are given in an arbitrary order. They are 0.833; 1.154; 1.179

I seem to have found that $A_i(2)=1.154$ and $B_i(3)=0.833$ (see below for the expressions I obtained through integration by parts), however, whichever way I evaluate the integral (or simply attempt to find the magnitude) of $F(j3)$, I always seem to come out with it being zero.

I have so far tried integrating by parts to separately find $A(\omega)$ and $B(\omega$ and then tried using the formula:


I obtain:



But this yields


I have also directly found the Fourier transform:


Again yielding the same answer according to the magnitude of an imaginary number (which is expected).

Am I doing something wrong, or misinterpreting the question?

  • $\begingroup$ What are $A_i$, etc.? $\endgroup$ – copper.hat Apr 7 '17 at 0:38
  • $\begingroup$ Ah sorry if that wasn't clear, $A_i$ just denotes the function $A$ for the Fourier transform the function $i(t)$ $\endgroup$ – Resquiens Apr 7 '17 at 0:46
  • $\begingroup$ It is far from clear. You have a function $i$ defined in terms of $A$ and then you have $A$ defined in terms of $f$. The function $f$ is defined nowhere. $\endgroup$ – copper.hat Apr 7 '17 at 0:55
  • $\begingroup$ The function $f$ is the function the Fourier transform ($F$) is being taken of, in this case $i$. The function $i$ is not in any way dependent on A, rather A would be defined as the real part of the Fourier Transform integral of $i$, as in the equation above. $\endgroup$ – Resquiens Apr 7 '17 at 1:00
  • 1
    $\begingroup$ Your definition of $i$ has an $A$ on the right hand side. Is this intentional? This was the content of my second comment above. $\endgroup$ – copper.hat Apr 7 '17 at 14:47

This really is just a matter of grinding carefully through the computation.

Let $\hat{i}$ be the Fourier transform of $i$, that is ${\hat{i}}(\omega) = \int_{-\infty}^\infty e^{-j \omega t} i(t) dt$.

Then $A_i(\omega) = \operatorname{re} \hat{i}(\omega)$ and $B_i(\omega) = \operatorname{im} \hat{i}(\omega)$.

Evaluating: \begin{eqnarray} {\hat{i}}(\omega) &=& \int_{-\infty}^\infty e^{-j \omega t} 5 e^{3t} u(-t) dt \\ &=& 5 \int_{-\infty}^0 e^{(3-j \omega )t} dt \\ &=& 5 \int_0^\infty e^{-(3-j \omega )t} dt \\ &=& \left( 5 {1 \over 3-j \omega }e^{-(3-j \omega )t} \right) \mid_{t=0}^\infty \\ &=& {5 \over 3-j\omega} \end{eqnarray} (a) We have ${\hat{i}}(2) = {5 \over 3-2j} = {1 \over 13} (15+10j)$, so $A_i(2) = {15\over 13} \approx 1.154 $.

(b) We have ${\hat{i}}(3) = {5 \over 3-3j} = {5 \over 6} (1+j)$, so $B_i(2) = {5\over 6} \approx 0.8333 $.

(c) This is a bit peculiar, the Fourier transform is only defined on the real line, so this part does not make any sense as stated. It is not clear whether you have made another mistake in transcribing the question, your teacher is trying to catch you out or the teacher wants you to blindly plug the numbers in.

If we look at the derivation of $\hat{i}$ above, we see that we can repeat the analysis above to get $\hat{i}(j \sigma) = {5 \over 3+\sigma}$, and the integration is valid as long as $\sigma > -3$.

This gives ${\hat{i}}(3j) = {5 \over 6}$ and so $|F_i(j3)| = {5 \over 6} \approx 0.833$.

However: I think you have most likely have made a mistake in the question and meant to write $|F_i(3)|$, and from above we can see that $|F_i(3)| = |\hat{i}(3)| = {5 \over 6} |1+j| = \sqrt{2} {5 \over 6} \approx 1.179$.

  • $\begingroup$ Thanks a lot! I suppose the textbook has misplaced a j :) $\endgroup$ – Resquiens Apr 8 '17 at 14:46

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