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Hey there in the signal processing course I am studying there is an excercise that reads:

The sequence $x(n)$ is given $x(n)=\{-1\quad2\quad \underline{-3}\quad 2\quad -1\}$ and the fouriertransform $X(\omega)$. Without explicitly performing the fouriertransform solve the following: Sekvensen x(n) är given av x(n) =

a) $X(0)$

b) $argX(\omega)$

c)$\int_{-\pi}^{\pi}X(\omega)d\omega$

d)$X(\pi)$

e)$\int_{-\pi}^{\pi}|X(\omega)|^2d\omega$

I could not figure out how to solve any of this without doing the actual transform. I eventually solved all of it with the transform, but still I am none the wiser how to solve any of this without doing the transfom. Suggestions?

Please and thank you!

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  • $\begingroup$ What does the line under $\underline{-3}$ mean? $\endgroup$ Jun 29 at 19:35
  • $\begingroup$ @MarcusMüller That is the implied t= 0 $\endgroup$ Jun 30 at 3:37
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    $\begingroup$ @DanBoschen oh, that makes sense! Actually, quite beatiful! $\endgroup$ Jun 30 at 5:41
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Some key take-aways /properties to know about the Fourier Transform will help reveal the answers. There is an method to the madness in extracting some key take-aways and high level understanding of what the Fourier Transform represents that makes this exercise useful.

The line under the 3 indicates the assumed position of the vertical axis, meaning $t=0$ and thus we see we have a symmetric real waveform in time. A symmetric real waveform will always have a real transform in frequency (and vice versa). This is referred to as an even function. Similarly an odd function (where the positive is the same as the negative but sign reversed) will always have an imaginary transform. (This is the proof, once those points are proven, that a causal function in time (as the sum of an even and odd function) must always have a complex transform. Remember this relationship, it's useful.

So knowing that allows you to solve (B).

To solve (A), consider what X(0) represents. (hint, what is the Fourier Transform of a 9V battery??). Without actually solving the Fourier Transform, take a look at the formula when $\omega=0$ and see what it simply is for that case. Remember that one too!

To solve (D), do the same as above when $\omega$ is $\pi$ and look what happens to the formula as a simplification. Note how +1, -1, +1 , -1 ...comes into the picture and how easy it can make it to do that one in your head the next time around.

Note how (C) and (E) are integrated over the entire unique range of $\omega$. Go through a few cases to see what always occurs in that integration between the time domain signal and the frequency domain signal. Studying the formulas carefully will help you to do that again in your head in the future. Look into Parseval's Theorem and really understand what it is describing.

Hope this helps!

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  • $\begingroup$ Thank you, I understand most of it now. There is just this one hang up. For (B), the functin x(n) is symmetrical so the transform must be real. Meaning that $\measuredangle X(\omega)$ is either 0 or $\pi$ (or maybe both?) since $X(\omega)$ is on the real axis. 0 if $X(\omega)$ is positve and $\pi$ if $X(\omega)$ is negative. Can't seem to figure this last bit out though! $\endgroup$
    – Aedrha
    Jun 30 at 17:03
  • $\begingroup$ Real signals have a phase of 0 degrees (imaginary signals have a phase of 90 degrees). Omega is the independent variable (which happens to always be real but that isn’t our concern: our concern is with X(omega) — consider if that result is real or imaginary or complex. $\endgroup$ Jun 30 at 17:06

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