# Linearity and time-shifting of $\mathcal{F}\{0.8^n\cos(0.1πn)u[n]\}$

To preface, this is not a homework related question but purely for self-study purposes.

Hi there, I try to calculate $$\mathcal{F}\{0.8^n\cos(0.1πn)u[n]\}$$ by using the properties of Discrete time Fourier transform.

Based on the linearity property, we have:

\begin{align} \mathcal{F}\{0.8^nu[n]\cos(0.1πn)\}&=\mathcal{F}\{0.8^nu[n]\}\cdot\mathcal{F}\{\cos(0.1πn)\} \end{align}

I got

\begin{align} \frac{1} {1-0.8e^{-jw}}\pi( \delta(w+0.1\pi)+\delta(w-0.1\pi))\tag1 \end{align}

But if I applied the time-shifting property:

\begin{align} \mathcal{F}\{0.8^n\cos(0.1πn)u[n]\} &= \mathcal{F}\left\{0.8^n\frac{e^{j0.1\pi n}+e^{j0.1\pi n}}{2} u[n]\right\}\\ &= 0.5\left(\frac{1}{1-0.8e^{-j(w+0.1\pi) }}+\frac{1}{1-0.8e^{-j(w-0.1\pi) }}\right) \tag2 \end{align} I am not quite sure if equation (1) and (2) are equivalent to each other. It looks like we have $$\frac{\pi} {1-0.8e^{-jw}}$$ at (1) when $$w$$= $$\pm0.1\pi$$, but I cannot get the same value at (2).

Why I get the different result here? Am I doing something wrong?

• Hi! Check again the linearity property.
– GKH
Commented Jan 22, 2020 at 8:39
• Thank you GKH. Yeah, linearity property cannot applied here Commented Jan 22, 2020 at 9:01

$$\displaystyle \hbox{ DTFT}(\alpha x_1 + \beta x_2) = \alpha\cdot \hbox{ DTFT}(x_1) + \beta \cdot\hbox{ DTFT}(x_2)$$