It is desired to find three-tap causal filter which gives zero signal as an output to and input of the form \[x\left[ n \right] = {c_1}\exp \left( { – \frac{{j\pi n}}{2}} \right) + {c_2}\exp \left( {\frac{{j\pi n}}{2}} \right),\] Where c1 and c2 are arbitrary real numbers. The desired three-tap filter is given by h[0] = 1, h[1] = a, h[2] = b and h[n] = 0 for n < 0 or n > 2. What are the values of the filter taps a and b if the output is y[n] = 0 for all n, when x[n] is as given above? \[\xrightarrow{{x\left[ n \right]}}\boxed{\begin{array}{*{20}{c}} {n = 0} \\ \downarrow \\ {h\left[ n \right] = \left\{ {1,a,b} \right\}} \end{array}}\xrightarrow{{y\left[ n \right] = 0}}\]

The correct answer is $\boxed{\text{A}}$.

The output of a linear time-invariant (LTI) system is given by the convolution of the input signal and the impulse response of the system. In this case, the input signal is $x[n] = {c_1}\exp \left( { – \frac{{j\pi n}}{2}} \right) + {c_2}\exp \left( {\frac{{j\pi n}}{2}} \right)$ and the impulse response is $h[n] = \left{ {1,a,b} \right}$. The output is therefore given by

\begin{align}
y[n] &= \sum_{k=-\infty}^{\infty} x[k] h[n-k] \\ &= {c_1}\exp \left( { – \frac{{j\pi n}}{2}} \right) \sum_{k=-\infty}^{\infty} h[n-k] + {c_2}\exp \left( {\frac{{j\pi n}}{2}} \right) \sum_{k=-\infty}^{\infty} h[n-k] \\ &= {c_1}\exp \left( { – \frac{{j\pi n}}{2}} \right) \left( h[0] + h[n-1] + h[n-2] \right) + {c_2}\exp \left( {\frac{{j\pi n}}{2}} \right) \left( h[0] + h[n-1] + h[n-2] \right) \\ &= \left( {c_1} + {c_2} \right) h[0] + \left( {c_1} – {c_2} \right) e^{-j\pi n/2} h[1] + \left( {c_1} + {c_2} \right) e^{j\pi n/2} h[2] \\ &= \left( {c_1} + {c_2} \right) + \left( {c_1} – {c_2} \right) e^{-j\pi n/2} a + \left( {c_1} + {c_2} \right) e^{j\pi n/2} b.
\end{align}

We are given that the output is zero for all $n$, so we must have

\begin{align}
y[0] &= \left( {c_1} + {c_2} \right) + \left( {c_1} – {c_2} \right) a + \left( {c_1} + {c_2} \right) b = 0 \\
y[1] &= \left( {c_1} – {c_2} \right) e^{-j\pi} a + \left( {c_1} + {c_2} \right) e^{j\pi} b = 0 \\
y[2] &= \left( {c_1} + {c_2} \right) + \left( {c_1} – {c_2} \right) e^{-j2\pi} a + \left( {c_1} + {c_2} \right) e^{j2\pi} b = 0.
\end{align}

Solving these equations, we find that $a = -1$ and $b = 1$.