{"id":56949,"date":"2024-04-16T00:53:35","date_gmt":"2024-04-16T00:53:35","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=56949"},"modified":"2024-04-16T00:53:35","modified_gmt":"2024-04-16T00:53:35","slug":"it-is-desired-to-find-three-tap-causal-filter-which-gives-zero-signal-as-an-output-to-and-input-of-the-form-xleft-n-right-c_1exp-left-fracjpi-n2-right-c_2exp","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/it-is-desired-to-find-three-tap-causal-filter-which-gives-zero-signal-as-an-output-to-and-input-of-the-form-xleft-n-right-c_1exp-left-fracjpi-n2-right-c_2exp\/","title":{"rendered":"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}}\\]"},"content":{"rendered":"

[amp_mcq option1=”a = -1, b = 1″ option2=”a = 0, b = 1″ option3=”a = 1, b = 1″ option4=”a = 0, b = -1″ correct=”option3″]<\/p>\n

The correct answer is $\\boxed{\\text{A}}$.<\/p>\n

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<\/p>\n

\\begin{align}
\ny[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.
\n\\end{align<\/em>}<\/p>\n

We are given that the output is zero for all $n$, so we must have<\/p>\n

\\begin{align}
\ny[0] &= \\left( {c_1} + {c_2} \\right) + \\left( {c_1} – {c_2} \\right) a + \\left( {c_1} + {c_2} \\right) b = 0 \\\\
\ny[1] &= \\left( {c_1} – {c_2} \\right) e^{-j\\pi} a + \\left( {c_1} + {c_2} \\right) e^{j\\pi} b = 0 \\\\
\ny[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.
\n\\end{align<\/em>}<\/p>\n

Solving these equations, we find that $a = -1$ and $b = 1$.<\/p>\n","protected":false},"excerpt":{"rendered":"

[amp_mcq option1=”a = -1, b = 1″ option2=”a = 0, b = 1″ option3=”a = 1, b = 1″ option4=”a = 0, b = -1″ correct=”option3″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[959],"tags":[],"class_list":["post-56949","post","type-post","status-publish","format-standard","hentry","category-signal-processing","no-featured-image-padding"],"yoast_head":"\nIt 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. 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