# How to show that this filter is a HP filter

i got a system with transfer function given by:

$$H(\omega)=1-e^{-j\omega}$$ I already plot it, and that seems to be a periodic function with $H(0)=0$, $H(\pi)=2$, , is that enough to show that this is a FIR bandpass filter. thank you very much

• That is a highpass filter. How did you calculate $H(\infty)$? Since a digital filter has a response that is periodic, it doesn't really make sense to try to calculate its response at infinite frequency. – Jason R Apr 3 '13 at 15:20
• You need to consider what the definition is for a LP/BP/HP filter in the case of a discrete time system. Consider that the useful/meaningful range of frequencies maps to w being in the range 0-pi. – user2718 Apr 3 '13 at 15:45
• @BZ For complex filters (which this is) the range is $0$ to $2\pi$ or, equivalently, $-\pi$ to $\pi$. – Jim Clay Apr 3 '13 at 16:13
• @JimClay - Actually this is not a complex filter. All the coefficients are real. The term $$e^{-j\omega}$$ represents a real delay element. Certainly the frequency response is defined by a complex function, but most filter defintions are based only on the non-negative portion of the response. Not a big deal though, I think the proper point has been passed on to the OP. – user2718 Apr 3 '13 at 17:04

As was pointed out in the comments above, this is a highpass filter. There is no strict definition for what constitutes a lowpass, bandpass, or highpass filter. Instead, they possess certain qualities:

• Lowpass filters attenuate high frequencies while passing low frequencies.

• Bandpass filters attenuate low and high frequencies, while passing some band in the middle.

• Highpass filters attenuate low frequencies while passing high frequencies.

Note the lack of precise terms in the descriptions above. In order to classify a filter into a category like those above, one evaluates the character of its magnitude transfer function. That requires a little manipulation:

$$H(\omega) = 1 - e^{-j\omega}$$

\begin{align} |H(\omega)| &= |1 - e^{-j\omega}| \\ &= \left|e^{-j\frac{\omega}{2}} \left(e^{j\frac{\omega}{2}} - e^{j\frac{\omega}{2}}\right)\right| \end{align}

Using Euler's identity, we can simplify the expression inside the parentheses:

$$e^{j\frac{\omega}{2}} - e^{j\frac{\omega}{2}} = j2\sin\left(\frac{\omega}{2}\right)$$

$$|H(\omega)| = \left|e^{-j\frac{\omega}{2}} j2\sin\left(\frac{\omega}{2}\right)\right|$$

We can drop the $j$ and $e^{-j\frac{\omega}{2}}$ terms here, because each of them have unity magnitude and therefore don't affect $|H(\omega)|$. Therefore:

$$|H(\omega)| = \left|2\sin\left(\frac{\omega}{2}\right)\right|$$

We can check this result by comparing it with a plot from MATLAB's Filter Visualization Tool (fvtool):

Which looks about right. This is plotted over the region $\omega \in [0, \pi]$, and shows that the filter has zero response at DC (attenuating low frequencies) and a magnitude response of $2$ at $\omega = \pi$ (passing high frequencies). That looks like a highpass filter.