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I'm trying to calculate the Fourier transform of an impulse-train sampled signal in two differnt ways but I end up with different results.

Impulse-train sampling of a continous signal $x(t)$ with sampling period $T_{s}$ is defined as

$$x_{p}(t) = x(t)p(t) \textrm{ where } p(t) = \sum_{n=-\infty}^{\infty} \delta(t - nT_{s})$$ Now consider the input signal $ x(t) = e^{-t}u(t)$ where the Fourier transform is $X(\omega) = \frac{1}{1 + j \omega}$. According to common procedure I calculate the Fourier transform of $x_{p}(t)$, using convolution in the frequency domain, as: $$X_{p}(\omega) = \frac{1}{T_{s}}\sum_{k=-\infty}^{\infty}X_{}(\omega- \frac{2\pi k}{T_{s}})$$ $$ \textrm{Inserting $X(\omega) $ I get } X_{p}(\omega)= \frac{1}{T_{s}}\sum_{k=-\infty}^{\infty}\frac{1}{1 + j (\omega - \frac{2\pi k}{T_{s}})} \tag{1}$$ The other way is to calculate $X_{p}$ directly from $x_{p}(t) = x(t)p(t) = \sum_{n=-\infty}^{\infty} x(nT_{s}) \delta(t - nT_{s})$. Now I get the result $$X_{p}(\omega) = \int_{-\infty}^{\infty} e^{-nT_{s}}\sum_{n=0}^{\infty} \delta(t - nT_{s}) e^{-j \omega t }dt = \sum_{n=0}^{\infty} e^{-nT_{s}(1+j \omega)} \tag{2}$$

I can't see that 1 and 2 are equal? Or are they? If not, what misstake did I do?

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  • $\begingroup$ Direct reduction of sum-2 into sum-1 may not be possible. Instead the equality between them is established by properties of Fourier transform and series, and then sum-2 equals sum-1 is concluded. Such dead ends happen in DSP (or math) when you need to evaluate certain integrals or sums whose direct analytic evaluation is not known yet. In such cases, either certain tricks r applied to convert the integral into a known form, or alternate methods of reaching the same result is utilised. Indeed for sampling theorem, equality of sum-1 to sum-2 is a theoretical concern rather than a practical one. $\endgroup$
    – Fat32
    Oct 10 at 4:04
  • $\begingroup$ This question confirms my 4-decade long belief and preference of the definition of the Fourier Transform using "ordinary frequency". $$ \mathscr{F} \Big\{ x(t) \Big\} \triangleq X(f) \triangleq \int\limits_{-\infty}^{\infty} x(t) \, e^{-j2 \pi f t} \ \mathrm{d}t $$ $$ \mathscr{F}^{-1} \Big\{ X(f) \Big\} \triangleq x(t) = \int\limits_{-\infty}^{\infty} X(f) \, e^{j2 \pi f t} \ \mathrm{d}f $$ It also confirms my belief that the sampling impulse train should always be multiplied by the sampling period $T$. Life is so much simpler and sometimes more correct. $\endgroup$ Oct 10 at 21:14
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    $\begingroup$ this and this $\endgroup$ Oct 10 at 21:21
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Both results are correct. Hence, the equality

$$\frac{1}{T}\sum_{k=-\infty}^{\infty}\frac{1}{1+ j \left(\omega - k\frac{2\pi }{T}\right)}=\sum_{n=0}^{\infty}e^{-nT(1+j\omega)}\tag{1}$$

holds true.

One way to prove this is to realize that the term on the left-hand side of Eq. $(1)$ is periodic with period $2\pi/T$. Consequently, we can express this term by its Fourier series

$$\frac{1}{T}\sum_{k=-\infty}^{\infty}\frac{1}{1+ j \left(\omega - k\frac{2\pi }{T}\right)}=\sum_{n=-\infty}^{\infty}c_ne^{jnT\omega}\tag{2}$$

It can be shown (try it!) that the Fourier coefficients $c_n$ are just (scaled) samples of the time domain function $x(t)$, so you obtain the right-hand side of Eq. $(1)$.

Eq. $(1)$ is an example of a more general result called Poisson's sum formula. The general form of Poisson's sum formula that specializes to Eq. $(1)$ is

$$\frac{1}{T}\sum_{k=-\infty}^{\infty}X\left(\omega-k\frac{2\pi}{T}\right)=\sum_{n=-\infty}^{\infty}x(nT)e^{-jnT\omega}\tag{3}$$

where $X(\omega)$ is the Fourier transform of $x(t)$.

Note that the right-hand side of Eq. $(3)$ is the discrete-time Fourier transform (DTFT) of the sequence $x(nT)$, which, according to Eq. $(3)$, equals the periodized spectrum $X(\omega)$ of the continuous-time function $x(t)$.

The dual form of Poisson's sum formula is

$$\sum_{k=-\infty}^{\infty}x(t-kT)=\frac{1}{T}\sum_{n=-\infty}^{\infty}X\left(n\frac{2\pi}{T}\right)e^{jn\frac{2\pi}{T}t}\tag{4}$$

i.e., the Fourier coefficients of a periodized time domain function $x(t)$ are samples of its Fourier transform $X(\omega)$.

Note that choosing $\omega=0$ in $(3)$, or, equivalently, $t=0$ in $(4)$ leads to the remarkable result

$$\sum_{k=-\infty}^{\infty}x(kT)=\frac{1}{T}\sum_{n=-\infty}^{\infty}X\left(n\frac{2\pi}{T}\right)\tag{5}$$

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  • $\begingroup$ Thanks' for this insightful and well written answer. Unfortunately I don't have the mathematical skills to work out the fourier coefficients in eq 1. Poisson's sum formula was a great help. Now I can verfiy that 1 and 2 from my post are equal. Thanks' $\endgroup$
    – Fredrik R
    Oct 11 at 21:10

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