 Solution

For differencing, the only nonzero filter coefficients are $a_0=1$ and $a_1=-1$. Therefore $a(\lambda )=1-e^{i\lambda }$ andThe left sketch shows the filter gain increasing from zero at frequency zero to two at $\pi$. This differencing filter removes a constant component and attenuates slowly varying components while relatively emphasizing higher frequencies. It is a high-pass filter, rather than an averaging or smoothing filter.