I have one signal with $N$ samples and I want to downsample it by factor of $M$. Essentially my downsampler is comb function which takes every M$^{th}$ sample while others are set to zero, i.e.,

$$ \check{Y}_k=\sum_{n=0}^{N-1}\sum_{m=0}^{\frac{N}{M}-1}x[n]\delta(n-mM)e^{-j\frac{2\pi nk}{N}}\tag{1} $$ where $x[n]$ is original signal, $\check{Y}_k$ is downsampled signal, and $n$ are time samples .

  • How is comb function $\delta(n-mM)$ define in terms of exponential function ? i.e.


  • Also what is the definition of comb function in frequency domain?

1 Answer 1


The frequency domain representation of the discrete time downsampling operation (i.e. $X_M(e^{j\omega})$ , the $DTFT\{x[Mn]\}$, in terms of $X(e^{j\omega})$, the $DTFT\{x[n]\}$ ) can be most easily observed by the following approach. We will begin from the end, $x[Mn]$ and proceed to its pre-compressed form by first deriving the frequency domain relation between the compressed signal $x[Mn]$ and its expanded form (pre-compressed) $x_s[n]$ and then we shall show the spectral relation between an input signal and its sampled form $x_s[n]$ (which is also the expanded signal at the input of the compressor)

First consider a periodic pulse train , $p_M[n] \triangleq \sum_{k=-\infty}^{\infty}{ \delta[n-kM]}$ whose period is $M$ and defined as :

$$p_M[n]= \begin{cases} 1 ~~~~, ~~~ \text{n = 0, M, 2M, ..., kM , ...} \\ 0 ~~~~, ~~~ \text{otherwise} \\ \end{cases} \tag{1}$$

Then the signal $x[n]$ will be first multiplied by this perodic pulse train $p_M[n]$, to produce $x_s[n]$ which will be selecting the required samples as follows:

$$x_s[n]=x[n]\cdot p_M[n] = \begin{cases} x[n] ~~~~, ~~~ \text{n = 0, M, 2M, ..., kM , ...} \\ 0 ~~~~, ~~~ \text{otherwise} \tag{2}\\ \end{cases}$$

Now, we shall take every $M^{th}$ sample of $x_s[n]$ and discard the rest, which is called as the compression operation as graphically represented by:

$$ x_s[n] \longrightarrow \boxed{\downarrow M} \longrightarrow x_s[Mn]=x[Mn]=x_M[n] \tag{3}$$

Note that eventhough the conversion between $x[n]$ and $x[Mn]$ is not invertable, the conversion between $x_s[n]$ and $x_s[Mn]$ is so, as should be clear:

$$ x_s[Mn]=x_{sM}[n] \longrightarrow \boxed{\uparrow M} \longrightarrow x_s[n] \tag{4}$$

i.e. the inverse is the expander operator and we shall exploit it for easily showing the spectral relation between $x_s[n]$ and $x_M[n]$ as it can be shown that:

$$X_{s}(e^{j\omega}) = X_{sM}(e^{j\omega M}) \Longrightarrow X_{sM}(e^{j\omega}) = X_{s}(e^{j\omega/ M}) \tag{5}$$

Based on this we can argue that (noting the equation (3)), $$ \boxed{ DTFT\{x[Mn]\} \triangleq X_M(e^{j\omega}) = X_{s}(e^{j\omega/ M}) } \tag{6}$$ where $X_{s}(e^{j\omega})$ is the DTFT of the sampled signal $x_s[n]$.

What comes next is the relation between the DTFTs of the signals $x_s[n]$ and $x[n]$ to complete the picture, which will be a straight forward observation if we consider representing the periodic impulse train by another form namelmy by its DFS (Discrete Fourier Series) representations (as is possible for any periodic discrete time signal with period M) :

$$ p_M[n] \triangleq \sum_{k=-\infty}^{\infty}{ \delta[n-kM]} = \frac{1}{M} \sum_{k=0}^{M-1}{e^{j \frac{2\pi}{M}kn} } \tag{7}$$

Then we shall express $x_s[n]$ as : $$ x_s[n] \triangleq x[n]\cdot p_M[n] = x[n] \big( \frac{1}{M} \sum_{k=0}^{M-1}{e^{j \frac{2\pi}{M}kn} }\big) = \frac{1}{M} \sum_{k=0}^{M-1}{x[n]e^{j \frac{2\pi}{M}kn} } \tag{8}$$

What comes next is using the linearity and modulation properties of DTFT operator and argue that: $$DTFT\{x_s[n]\} = DTFT\{\frac{1}{M} \sum_{k=0}^{M-1}{x[n]e^{j \frac{2\pi}{M}kn} } \} = \frac{1}{M} \sum_{k=0}^{M-1}{ DTFT \{ x[n]e^{j \frac{2\pi}{M}kn} \} } = $$ $$ \boxed{X_s(e^{j\omega})=\frac{1}{M} \sum_{k=0}^{M-1}{ X(e^{j \big(\omega -\frac{2\pi}{M}k \big)} ) } } \tag{9}$$

The final step is the mergal of the steps (6) and (9) to produce the result: $$ \boxed{ X_M(e^{j\omega}) = X_s(e^{j\omega/M})=\frac{1}{M} \sum_{k=0}^{M-1}{ X(e^{j \big(\frac{\omega}{M} -\frac{2\pi}{M}k \big)} ) } = \frac{1}{M} \sum_{k=0}^{M-1}{ X(e^{j \big(\frac{\omega-2\pi k}{M} \big)} ) } } \tag{10} $$

I hope it was a clear way of expressing the spectrum of downsampled signal $x_M[n]=x[Mn]$ to that of the original signal $x[n]$

Also note that a more complete understanding of the spectral relation would only be possible if you fully explore the relation by a graphical plot as well, which would clearly demonstrate the resulting frequency stretching towards the $\omega =\pi$ boundary, by a factor of $M$ after which a consideration of aliasing becomes clear...

  • $\begingroup$ Just one comment: shouldn't be in Eq. (7) $p_m[n]=..=\frac{1}{M}\sum_{k=0}^{M-1}e^{j\frac{2\pi nk}{M}}$? $\endgroup$
    – Cali
    Jul 1, 2016 at 19:47
  • $\begingroup$ @Cali yes you are right! It should be M, not N... let me correct. $\endgroup$
    – Fat32
    Jul 1, 2016 at 20:01
  • $\begingroup$ Then I think the same editing should be done in Eq. (8). However, a nice and detailed answer. Thank you! $\endgroup$
    – Cali
    Jul 1, 2016 at 20:16
  • 1
    $\begingroup$ yeees... latex itself is tiring to catch a typo... hope you understood it. I suggest you interpret the last equation (10) carefully. By adding M copies of scaled and shifted versions of $X(e^{j\omega})$ ... In particular observe the period of the scaled verison... which is not $2\pi$ but $M2\pi$ $\endgroup$
    – Fat32
    Jul 1, 2016 at 20:30
  • $\begingroup$ if there is a step you feel insecure, please don't hesitate to ask it as a new question... ;-) $\endgroup$
    – Fat32
    Jul 1, 2016 at 20:42

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