I'm trying to understand the plots on this page. It's a book about the Discrete Fourier Transform and it's discussing how a a function $x(t)=\cos(w_0t)$ or $y(t)=\sin(w_0t)$ is composed of a positive and a negative frequency component. I get why the spectrum of $\cos(wt)$ has two real components and none imaginary. But i don't get why $\sin(wt)$ have two imaginary components, as in b) of the following image.

This is the link for the image, from the web mentioned page, that i don't understand

I think i get how $x(t)=\cos(wt)$ is the sum of two complex sinusoids of frequencies of opposite signs that results in an zero imaginary part:

$$x(t)=\cos(wt)=\frac{e^{jwt}+e^{-jwt}}{2}$$ $$ x(t)=\frac{\cos(wt)+j\sin(wt)+\cos(-wt)+j\sin(-wt)}{2} $$

Since $$\cos(-x)=\cos(x)$$ $$\sin(-x)=-\sin(x)$$


$$ x(t)=\frac{\cos(wt)+j\sin(wt)+\cos(wt)-j\sin(wt)}{2} $$


$$Re\{ \ x(t)\ \} = \frac{\cos(wt)+\cos(wt)}{2}=\cos(wt)$$


$$Im\{ \ x(t)\ \} = \frac{\sin(wt)-\sin(wt)}{2}=0$$

That explains why $\cos(wt)$ have two real parts on the graph, of same amplitude and "opposite" frequencies.

I will try to to the same with $\sin(wt)$:


Using $\cos(-x)=\cos(x)$

$$y(t)=\frac{ \cos(wt)+j\sin(wt) -(\cos(wt)+j\sin(-wt)) }{ 2j }$$ $$y(t)=\frac{ \cos(wt)+j\sin(wt) -\cos(wt)-j\sin(-wt) }{ 2j }$$

$$Re\{ \ y(t) \ \}=\frac{\cos(wt)-\cos(wt)}{2j}=0$$

$$Im\{ \ y(t) \ \}=\frac{\sin(wt)-\sin(-wt)}{2j}$$

I'm not sure how to follow from there. How come an imaginary part contains $j$? Or maybe $j$ should not be included? But in the case of

$$Im\{ \ y(t) \ \}=\frac{\sin(wt)-\sin(-wt)}{2}$$

Where that $j^-1$ went? This looks wrong to me because $$j \cdot Im\{ \ y(t) \ \}\neq\frac{\sin(wt)-\sin(-wt)}{2j}$$

What did I do wrong here? This looks so silly, I'm sorry.

  • $\begingroup$ You lose a minus sign early on, and then you pick the imaginary part (the coefft of $j$) when you are picking the real part: check out your sums. $\endgroup$ – ancientmathematician Apr 13 '17 at 7:22
  • $\begingroup$ And if $w$ is real, it's clear that the real part of $\sin wt$ is just, well, $\sin wt$. $\endgroup$ – ancientmathematician Apr 13 '17 at 7:24
  • $\begingroup$ Oh, i see. I've corrected the minus sign. It does reflect the plot where the sinusoid components have "opposite" amplitudes. But i'm not sure a get what you said about picking the real part. You're saying that actually what i've though was the $Re\{ \ y(t) \ \}$ and $Im\{ \ y(t) \ \}$ are both the real part of $y(t)$? That makes sense, since $\cos(x)=\sin(x+\frac{\pi}{2})$ Indeed it is strange to think that a difference in phase shift would make that change on the real and imaginary parts of the spectrum. If that is the case, then is the book's image incorrect? $\endgroup$ – Henri Augusto Apr 13 '17 at 11:07
  • $\begingroup$ The real part is the part without $j$, for goodness sake get the $j$ out of the denominator, then collect together all the terms without $j$ and all those with $j$ and then you'll see that the real and imaginary parts are $\sin wt$ and $0$. I really can't see what your problem is: the real part of a real function is itself, so why go all round the houses to see this? $\endgroup$ – ancientmathematician Apr 13 '17 at 11:49
  • $\begingroup$ The image on the book (linked on the question) shows that $Im\{ \ Y(\omega) \ \}$ have two components at frequencies $\omega$ and $-\omega$ and opposite signs amplitudes. I was trying to get to that conclusion. Is the image incorrect? Should the lines in b) be on the $Re\{ \ Y(\omega) \ \}$ axis? $\endgroup$ – Henri Augusto Apr 13 '17 at 13:29

I think that the answer is less complex than you are making it. Once you have: $$\cos(\omega t)=\frac{e^{j \omega t}+e^{j (-\omega) t}}{2} $$ that shows that there is a '$\frac{1}{2}$' magnitude at '$\omega$' and a '$\frac{1}{2}$' magnitude at '$-\omega$', both in the positive real direction.

Taking the same logic: $$\sin(\omega t)=\frac{e^{j \omega t}-e^{j (-\omega) t}}{2j} $$ Multiply both numerator and denominator of the fraction by $\frac{j}{2}$: $$\sin(\omega t)=\frac{j\frac{1}{2} e^{j \omega t} - j\frac{1}{2} e^{j (-\omega) t}}{\frac{1}{2}\times 2j^2}$$ Where $j^2=-1$. This eliminates the denominator and multiplies the numerator by $-1$: $$\sin(\omega t)=-j\frac{1}{2} e^{j \omega t} + j\frac{1}{2} e^{j (-\omega) t}$$ This shows that there is a '$-\frac{1}{2}j$' point at '$\omega$' and a '$(+)\frac{1}{2}j$' point at '$-\omega$'. In this case, both are in the imaginary plane. The one at positive $\omega$ has a negative sense, and the one at negative $\omega$ has a positive sense.

Does that help?


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