[amp_mcq option1=”$$\\frac{1}{2},\\frac{1}{2}x\\left( t \\right)$$” option2=”$$ – \\frac{1}{2},\\frac{1}{2}x\\left( t \\right)$$” option3=”$$\\frac{1}{2}, – \\frac{1}{2}x\\left( t \\right)$$” option4=”$$ – \\frac{1}{2}, – \\frac{1}{2}x\\left( t \\right)$$” correct=”option1″]<\/p>\n
The correct answer is $\\boxed{\\frac{1}{2}, -\\frac{1}{2}x(t)}$.<\/p>\n
The even part of a function is the part that is symmetric about the $y$-axis. The odd part of a function is the part that is antisymmetric about the $y$-axis.<\/p>\n
The unit step function $u(t)$ is defined as follows:<\/p>\n
$$u(t) = \\begin{cases} 1 & \\text{if } t \\ge 0 \\\\ 0 & \\text{if } t < 0 \\end{cases}$$<\/p>\n
The even part of the unit step function is $u(t) + u(-t)$, which is equal to $\\frac{1}{2} + \\frac{1}{2}u(t)$. The odd part of the unit step function is $u(t) – u(-t)$, which is equal to $\\frac{1}{2} – \\frac{1}{2}u(t)$.<\/p>\n
The function $x(t)$ is shown in the figure below.<\/p>\n
[asy]
\nunitsize(1 cm);<\/p>\n
draw((-2,0)–(2,0));
\ndraw((0,-1)–(0,1));<\/p>\n
label(“$t$”, (2,0), S);
\nlabel(“$y$”, (0,1), E);<\/p>\n
draw(graph(x,-2,2),red);<\/p>\n
draw((0,0)–(0,1.2));
\ndraw((0,0)–(1.2,0));<\/p>\n
label(“$x(t)$”, (0.5,1), S);
\n[\/asy]<\/p>\n
The even part of $x(t)$ is the part that is symmetric about the $y$-axis. This is the part of the graph that lies above the line $y=0$. The odd part of $x(t)$ is the part that is antisymmetric about the $y$-axis. This is the part of the graph that lies below the line $y=0$.<\/p>\n
The even part of $x(t)$ is equal to $\\frac{1}{2}x(t)$. The odd part of $x(t)$ is equal to $-\\frac{1}{2}x(t)$.<\/p>\n
Therefore, the even and odd parts of the unit step function $u(t)$ are respectively, $\\boxed{\\frac{1}{2}, -\\frac{1}{2}x(t)}$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”$$\\frac{1}{2},\\frac{1}{2}x\\left( t \\right)$$” option2=”$$ – \\frac{1}{2},\\frac{1}{2}x\\left( t \\right)$$” option3=”$$\\frac{1}{2}, – \\frac{1}{2}x\\left( t \\right)$$” option4=”$$ – \\frac{1}{2}, – \\frac{1}{2}x\\left( t \\right)$$” correct=”option1″]<\/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-57173","post","type-post","status-publish","format-standard","hentry","category-signal-processing","no-featured-image-padding"],"yoast_head":"\n