# Decimating a discrete time cosine function

Let $x[n] = \cos(wn)\ \$ be our signal. Let the other signal be $x_2[n] = \cos(2wn)\ \$ . The spectrum of $x[n]$ and $x_2[n]$ are well defined and consist of impulses at the relevant frequencies with weight $\pi$. However, if we interpret $x_2[n]$ as the decimated version of $x[n]$, the spectrum of $x_2[n]$ must have impulses with weight $\frac{\pi}{2}$. I have the explanation but I would like to hear the alternatives, what is the best way to resolve these seemingly contradicting facts?

• How do you determine that the impulses at $w$ have weight $\pi$? – Jim Clay Mar 23 '13 at 20:34
• @JimClay: Jim, the DTFT of a discrete-time cosine function with frequency w is a periodic train of impulses at frequencies -w+2\pi k and w+2\pi k, k=-\inf to \inf. All impulses have weight \pi. This comes from the fact that the DTFT of a complex exponential e^jw0t is 2\pi \delta(w-w0). 2\pi is there to cancel the 1 / 2\pi factor in the IDTFT. – Dost Mar 24 '13 at 19:03

The following correction of Jason R's answer completely resolves the problem: The frequency spectrum of $X_2(ω)$ consists of two components, which are scaled versions of $X(ω)$. Those are $4\pi$-periodic instead of being $2\pi$-periodic. When those $4\pi$-periodic versions are added up, the sum becomes $2π$-periodic again. THERE IS NO ALIASING or IMPULSES SUPERIMPOSED ON EACH OTHER. The derivation from $$X_2(\omega) = \frac{1}{2} \pi\left(\delta\left(\frac{\omega}{2}-\omega_0\right) + \delta\left(\frac{\omega}{2}+\omega_0\right) + \delta\left(\frac{\omega}{2}-\omega_0-\pi\right) + \delta\left(\frac{\omega}{2}+\omega_0-\pi\right)\right)$$ to $$X_2(\omega) = \frac{1}{2} \pi\left(\delta\left(\omega-2\omega_0\right) + \delta\left(\omega+2\omega_0\right) + \delta\left(\omega-2\omega_0-2\pi\right) + \delta\left(\omega+2\omega_0-2\pi\right)\right)$$ is not correct. Instead it must read $$X_2(ω)=π(δ(ω−2ω_0 )+δ(ω+2ω_0)+δ(ω−2ω_0−2π)+δ(ω+2ω_0−2π)) \qquad (∗)$$ This is due to the scaling property of impulses $$δ(aω)=\frac{δ(ω)}{|a|}$$ Then, observing that (*) is $2π$-periodic, two of the impulses are redundant and hence the DTFT of $x_2[n]=x[2n]$ becomes $$X_2(ω)=π(δ(ω−2ω_0)+δ(ω+2ω_0))$$

I'm not sure where you get the idea that $x_2[n]$ must have weight $\frac{\pi}{2}$. You can conclude that the weights are the same for both cases using a simple change of varible in the DTFT equation:

$$x[n] = \cos(\omega_0n)$$

\begin{align} X(\omega) &= \sum_{n=-\infty}^{\infty} x[n] e^{-j\omega n} \\ &= \sum_{n=-\infty}^{\infty} \cos(\omega_0 n) e^{-j\omega n} \\ &= \pi(\delta(\omega-\omega_0) + \delta(\omega+\omega_0)) \end{align}

Now consider the other signal:

$$x_2[n] = \cos(2\omega_0 n)$$

\begin{align} X_2(\omega) &= \sum_{n=-\infty}^{\infty} x_2[n] e^{-j\omega n} \\ &= \sum_{n=-\infty}^{\infty} \cos(2 \omega_0 n) e^{-j\omega n} \end{align}

Perform a change of variable:

$$n' = 2n$$

Since the limits of the summation over $n$ extend to infinity, we also sum from $-\infty$ to $\infty$ over $n'$.

$$X_2(\omega) = \sum_{n'=-\infty}^{\infty} \cos(\omega_0 n') e^{-j\omega \frac{n'}{2}}$$

Now, regroup the exponent inside the sum to yield:

$$X_2(\omega) = \sum_{n=-\infty}^{\infty} \cos(\omega_0 n') e^{-j\frac{\omega}{2} n'}$$

Then via inspection we can observe that $X_2(\omega) = X\left(\frac{\omega}{2}\right)$. Then, in terms of the equation that we found previously, we conclude that:

\begin{align} X_2(\omega) &= X\left(\frac{\omega}{2}\right) \\ &= \pi\left(\delta\left(\frac{\omega}{2}-\omega_0\right) + \delta\left(\frac{\omega}{2}+\omega_0\right)\right) \\ &= \pi\left(\delta\left(\omega-2\omega_0\right) + \delta\left(\omega+2\omega_0\right)\right) \end{align}

Edit: After thinking about it a bit more and looking at the notation that you posted in the below comment, I assert that the two are equivalent, although that may not be readily apparent. While the equation given in your comment has a typo, you can find a derivation on slide 22 in this presentation. That structure makes explicit the aliasing that can happen when decimating a signal. The correct form is:

$$x[an] \Leftrightarrow \frac{1}{a} \sum_{k=0}^{a-1} X\left(\frac{\omega}{a} - \frac{2\pi k}{a}\right)$$

So, for the case you posed, we can write:

\begin{align} x_2[n] = x[2n] \Leftrightarrow X_2(\omega) &= \frac{1}{2} \sum_{k=0}^{1} X\left(\frac{\omega}{2} - \frac{2\pi k}{2}\right) \\ &= \frac{1}{2} \sum_{k=0}^{1} X\left(\frac{\omega - 2 \pi k}{2}\right) \\ &= \frac{1}{2} \left(X\left(\frac{\omega}{2}\right) + X\left(\frac{\omega}{2} - \pi\right)\right) \end{align}

We then substitute the known expression for $X(\omega)$ to yield:

$$X_2(\omega) = \frac{1}{2} \pi\left(\delta\left(\frac{\omega}{2}-\omega_0\right) + \delta\left(\frac{\omega}{2}+\omega_0\right) + \delta\left(\frac{\omega}{2}-\omega_0-\pi\right) + \delta\left(\frac{\omega}{2}+\omega_0-\pi\right)\right)$$

We then manipulate the arguments to the impulses, noting that they are nonzero when their argument is equal to zero:

$$X_2(\omega) = \frac{1}{2} \pi\left(\delta\left(\omega-2\omega_0\right) + \delta\left(\omega+2\omega_0\right) + \delta\left(\omega-2\omega_0-2\pi\right) + \delta\left(\omega+2\omega_0-2\pi\right)\right)$$

Which, noting the $2\pi$-periodicity of $X(\omega)$, is equivalent to:

\begin{align} X_2(\omega) &= \frac{1}{2} \pi\left(\delta\left(\omega-2\omega_0\right) + \delta\left(\omega+2\omega_0\right) + \delta\left(\omega-2\omega_0\right) + \delta\left(\omega+2\omega_0\right)\right) \\ &= \pi\left(\delta\left(\omega-2\omega_0\right) + \delta\left(\omega+2\omega_0\right)\right) \\ &= X\left(\frac{\omega}{2}\right) \end{align}

• Jason, what you say is absolutely correct. However, the contradiction is that; $x_2[n] = x[2n]$ which is the downsampled version of $x[n]$ by a factor of 2. And the downsampling operation by a factor of 2 brings a 1/2 multiplicative factor in the frequency domain. – Dost Mar 27 '13 at 19:07
• Do you have a reference for your claim? Details like extra multiplicative constants in the result are dependent upon the exact transform definition that is used. – Jason R Mar 27 '13 at 19:38
• The following is textbook knowledge: The frequency response of the decimated-by-M signal $x_d[n]$ of $x[n]$ is $$X_d(\omega)=\frac{1}{M}\sum_{i=0}^{M-1}X\left(\frac{\omega}{M}-\frac{2\pi j}{M}\right)$$ – Dost Mar 27 '13 at 22:09
• @JasonR I don't think your change of variable works. If $n' = 2n$, then the summation in $n'$ will have to be zero for $n'$ odd....which is a different summation. – Peter K. Apr 2 '13 at 16:22