## Friday, November 7, 2014

### Discrete-time signals: time shift vs phase shift

First let's consider a continuous time sinusoidal signal: $$x(t)=\cos(\omega_0t)$$ Applying a time shift of $t_0$ gives $$x(t+t_0)=\cos(\omega_0(t+t_0))=\cos(\omega_0t+\omega_0t_0)=\cos(\omega_0t+\phi)$$ with $\phi=\omega_0t_0$. So for a continuous-time sinusoidal signal a time shift always corresponds to a phase shift, but, more importantly, the opposite is also always true: a phase shift always corresponds to a time-shift. The time shift is given by
$$t_0=\frac{\phi}{\omega_0}$$ With discrete-time signals, things are different. A time-shift always corresponds to a phase shift, but the opposite is generally not true. Let $$x[n]=\cos(n\theta_0)$$ Applying a time-shift of $n_0\in\mathbb{Z}$ gives $$x[n+n_0]=\cos((n+n_0)\theta_0)=\cos(n\theta_0+n_0\theta_0)$$ which corresponds to a phase shift of $\phi=n_0\theta_0$. However, assume another signal $$y[n]=\cos(n\theta_0+\phi),\quad\phi\in\mathbb{R}$$ If $n$ were a continuous variable, the signal $y[n]$ could be obtained from $x[n]$ by applying an appropriate time-shift. However, in discrete-time $y[n]$ cannot generally be obtained by time-shifting  $x[n]$. This is only possible if $$\phi=n_0\theta_0$$ is satisfied for some integer $n_0$. This can be seen as follows: $$y[n]=\cos(n\theta_0+\phi)=\cos((n+\phi/\theta_0)\theta_0)$$ which only equals $x[n+n_0]$ if $n_0=\phi/\theta_0$ is an integer.