# Why does a simple delay system with $h[n]=\delta[n-n_0]$ have a linear phase?

Let's consider a simple delay system with impulse response $$h[n]=\delta[n-n_0]$$ We also know that it has linear phase with group delay of $n_0$ .

In general, a FIR filter with symmetric (or anti-symmetric) coefficients with filter order $M$ always has linear phase and its (group) delay is $M/2$. Now my question is: 1) What is the filter order of the simple delay system? My guess is that the order is $n_0$ (since it is number of unit-delays) 2) Does it have symmetric filter coefficients? Apparently it is not, but not quite sure. 3) Why does it have a linear phase?

In response to "1) What is the filter order of the simple delay system?"

The order of a filter is the power of the highest nonzero coefficient of the Z-transform. For the simple delay system given, the Z-transform is given as $H(z) = z^{-n_0}$ so it has order $n_0$.

In response to "2) Does it have symmetric filter coefficients?"

The coefficients are not symmetric if you stop at $n_0$, but adding additional zero coefficients to the signal will not alter it and by doing so, one may describe an equivalent filter that has coefficient symmetry. For example, if the delay is 2 and our original filter is [0, 0, 1], then we can add two more zeros to obtain [0, 0, 1, 0, 0]. These two filters have identical behavior, but the second one can be said to have symmetric coefficients. I would argue that a simple delay does have symmetric coefficients since it can always be put into an equivalent form that has symmetric coefficients.

In response to "3) Why does it have a linear phase?"

The simplest way to understand why a delay results in a linear phase is to examine the output for a sinusoidal input that is delayed by a constant amount. If we have some sinusoidal signal such as $$x[n] = \cos\left(2 \pi f_0 n\right)$$ and we consider what happens after it is passed through our system $$y[n] = x[n] * h[n]$$ which in our case is simply $$y[n] = x[n-n_0]$$ or if we substitute in $x[n]$ we obtain $$y[n] = \cos\left(2 \pi f_0 n - 2 \pi f_0 n_0\right).$$ Notice that the resulting output of the system is a sinusoidal signal with the same frequency as the input, but with an additional phase shift. In this case, the phase shift is $2 \pi f_0 n_0$. However, if we are to consider the phase as a function of the input frequency of the sine wave (that is instead of fixing the frequency as $f_0$, we allow it to vary and rename it $f$), then we obtain a phase function $$\phi(f) = 2 \pi n_0 f$$ and from this, you can see that the phase response varies linearly in $f$ with a slope of $2 \pi n_0$ radians. We can use this phase function to predict the resulting phase offset at various frequencies. And this prediction is related to the frequency through a simple linear relationship. That is why we say that it has linear phase.

1- The filter order depends on the length of $h[n]$. you can determine the length, whether they are zero or nonzero filter taps (each coefficient is called a tap". 2-It is obvious that this filter is not symmetric; because it has just a nonzero coefficient in the positive part of $n$ axis (discrete time axis).

3-You know that if $h_1 [n]=\delta [n]$, then it's Fourier transform is "1", and if $h [n]=\delta [n-n_0]$ then it's Fourier transform is $e^{-j*w*n_0}$. We define phase as coefficient of $j$, and you see that $phase=-w*n_0$ which is a linear phase (something like $y = a*x$).