# Sampling Theorem and Dirac Comb

I am reading "The Scientist and Engineer's Guide to Digital Signal Processing" and trying to understand Figure 3.5 below which is about the sampling theorem and aliasing.

I do not understand the picture of the frequency domain. The signal can be represented as $$f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(x)e^{ixt}dx$$ where $F(x)$ is the Fourier transform of $f$. This represents $f$ with the "frequencies" $e^{ixt}$ and these can have $x$ negative. How then am I suppose to understand Figure 3-5 which shows the frequency domain for only greater than zero and moreover real and not complex.

Also why does the frequency domain of the impulse train have the original frequency spectrum appear flipped half the time? My current understanding is that the repetition comes from something like $$\mathcal{F^{-1}} \{f(t)\Delta(t)\} = F(x)*\Delta(x)$$ where $\Delta$ is the dirac comb and probably some kind of constant like $2\pi$ is needed in there somewhere. How does the flipping half the time in the frequency domain arise from $F(x)*\Delta(x)$. The image I think for convolution is that $\Delta$ is flipped and as it is moved to the right by $s$ the "area" of $F(x)\Delta(s-x)$ is the value of the convolution at $s$. This would seem to replicate the original spectrum multiple times but without the flipping half the time as shown in Figure 3.5. • Maybe you should ask the moderators to move this to dsp.SE Feb 10 '12 at 2:33
• I'm going to migrate this question to the dsp.SE site. There will be a link that appears below the question here that you can follow to the new location of your question. Feb 11 '12 at 5:33

Let $X(f)$ denote the Fourier transform of $x(t)$ where \begin{align*} X(f) &= \int_{-\infty}^{\infty} x(t) \exp(-j2\pi ft) \mathrm dt\\ x(t) &= \int_{-\infty}^{\infty} X(f) \exp(+j2\pi ft) \mathrm df \end{align*} which I will denote via $x(t) \leftrightarrow X(f)$. The following transform pairs will be needed in what follows. \begin{align*} \delta(t) &\leftrightarrow 1\\ \delta(t-t_0) &\leftrightarrow \exp(-j2\pi f \,t_0)\\ \sum_{n=-\infty}^{\infty}\delta(t-nT) &\leftrightarrow \frac{1}{T}\sum_{k=-\infty}^{\infty}\delta\left(f-\frac{k}{T}\right)\\ \end{align*}

Given a signal $x(t)$, its sampled pulse train (at intervals of $T$ seconds) is $$x(t) \sum_{n=-\infty}^{\infty}\delta(t-nT) = \sum_{n=-\infty}^{\infty} x(t)\delta(t-nT) = \sum_{n=-\infty}^{\infty} x(nT)\delta(t-nT).$$ Since multiplication in the time domain corresponds to convolution in the frequency domain, we have \begin{align*} x(t)\sum_{n=-\infty}^{\infty}\delta(t-nT) &\leftrightarrow X(f)\circledast\frac{1}{T}\sum_{k=-\infty}^{\infty}\delta\left(f-\frac{k}{T}\right)\\ &\leftrightarrow \frac{1}{T}\sum_{k=-\infty}^{\infty} X(f)\circledast\delta\left(f-\frac{k}{T}\right)\\ &\leftrightarrow \frac{1}{T}\sum_{k=-\infty}^{\infty} \int_{-\infty}^{\infty} X(f-w)\delta\left(w-\frac{k}{T}\right) \mathrm dw\\ x(t)\sum_{n=-\infty}^{\infty}\delta(t-nT) &\leftrightarrow \frac{1}{T}\sum_{k=-\infty}^{\infty} X\left(f-\frac{k}{T}\right) = \hat{X}(f) \end{align*} Thus, the Fourier transform of the impulse train formed by sampling $x(t)$ at $T$ second intervals is $\hat{X}(f)$ which is obtained by repeating $X(f)$ along the $f$ axis at intervals of $T^{-1}$ Hz and summing the result. Furthermore, $\hat{X}(f)$ is a periodic function of the frequency variable $f$ with period $T^{-1}$ Hz. That is, for all $f$, $$\hat{X}\left(f + \frac{1}{T}\right) = \hat{X}(f).$$ Note that all of this holds regardless of what $X(f)$ is: $X(f)$ could be nonzero for all $f$, and the result would still be valid.

Now suppose that $X(f)$ is zero if $f < a$ or $f > b$. Then, $\hat{X}(f)$, obtained by repeating $X(f)$ periodically along the $f$ axis, is nonzero only in the intervals $$\ldots, \left[a-\frac{2}{T}, b-\frac{2}{T}\right], \left[a-\frac{1}{T}, b-\frac{1}{T}\right], [a, b] \left[a+\frac{1}{T}, b+\frac{1}{T}\right], \left[a+\frac{2}{T}, b+\frac{2}{T}\right], \ldots$$ and so if $$b -\frac{1}{T} < a \Rightarrow b-a < \frac{1}{T},$$ that is, the support $b-a$ of $X(f)$ is smaller than the repetition interval $T^{-1}$ Hz, then the repetitions of $X(f)$ do not overlap. Indeed as the OP's figures show, for a real-valued signal whose spectrum extends from $a = -f_0$ to $b = f_0$ (support of $X(f)$ is of length $2f_0$), sampling $x(t)$ at intervals of $T = (3f_0)^{-1}$ (and thus repeating $X(f)$ at intervals of $3f_0$ Hz on the frequency axis) leads to no overlap while sampling $x(t)$ at intervals of $T = (1.5f_0)^{-1}$ leads to overlap of the repetitions of $X(f)$.

Finally, the OP asks about negative frequencies and the interpretation of the pictures which show positive frequencies only. For a real-valued signal, $X(f)$ has conjugate symmetry, meaning that $X(-f) = X^*(f)$, and so specifying $X(f)$ for positive values of $f$ suffices. In any case, the pictures he is looking at are of $|X(f)|$ and $|\hat{X}(f)|$ which are even functions of $f$, and so showing only the positive axis saves space, though it does make the pictures look a bit lopsided since only half the lobe is shown at low frequencies. For complex-valued signals, $X(f)$ does not have conjugate symmetry and $|X(f)|$ need not be an even function of $f$ and so the whole axis would need to be shown. But, the general development above is still applicable, and we still need to sample at rate exceeding $b-a$ Hz, and it helps to keep in mind that in this general case, each sample is actually two real numbers, not one, since we are sampling a complex-valued signal.

• Great answer. Hopefully we don't have anyone who wants to debate whether multiplication with a Dirac impulse train is a sensible model for sampling a signal discretely in time. Feb 13 '12 at 20:09
• Thanks. With regard to "sensible model for sampling a signal discretely in time", I suppose we could begin with a periodic function of $f$ as in $\hat{X}(f)$ above, and ask what is its complex exponential Fourier series which would be a bunch of points in time... Feb 13 '12 at 20:24

For real-valued signals $f(t)$, the plots of $|\mathcal{F}\{f\}(x)|$ you show are symmetric across the real axis, because $\mathcal{F}\{f\}(-x) = \overline{\mathcal{F}\{f\}(x)}$ (the overbar to the right of the equality indicates complex conjugation).

This also explains the "flipping" you observe, I think. Imagine the top right subplot in your figure reflected across the x-axis; it's that spectrum which is being duplicated due to finite sampling.

EDIT:

\begin{align*} \mathcal{F}\{f\}(-x) &= \int_{t=-\infty}^{t=\infty} f(t) e^{-2\pi i t(-x)} dt \\ &= \int_{t=-\infty}^{t=\infty} f(t) \cos(2\pi t(-x)) dt - i\int_{t=-\infty}^{t=\infty} f(t) \sin(2\pi t(-x)) dt \\ &= \overline{\int_{t=-\infty}^{t=\infty} f(t) \cos(2\pi t(-x)) dt + i\int_{t=-\infty}^{t=\infty} f(t) \sin(2\pi t(-x)) dt} \\ &= \overline{\int_{t=-\infty}^{t=\infty} f(t) e^{2\pi i t(-x)} dt} \\ &= \overline{\int_{t=-\infty}^{t=\infty} f(t) e^{-2\pi i tx} dt} \\ &= \overline{\mathcal{F}\{f\}(x)} \end{align*}

• Could you write out the proof of the duplication in rigorous mathematics? Feb 12 '12 at 11:47
• All the pictures about the Nyquist sampling show the mirror symmetry. Does the Nyquist sampling theorem apply when the graph in frequency domain isn't symmetric. More precisely does the Nyquist sampling theorem apply for complex signals $f(t)$ where $\mathcal{F}\{f\}(-x) = \overline{\mathcal{F}\{f\}(x)}$ is not true in general. And of course with the condition that $\mathcal{F}\{f\}(x)$ is zero outside a bounded set. Feb 13 '12 at 11:24