[amp_mcq option1=”x(t + 4)” option2=”x(t – 4)” option3=”x(t + 2)” option4=”x(t – 2)” correct=”option1″]<\/p>\n
The correct answer is A. $x(t+4)$.<\/p>\n
The frequency response of the system is a delayed version of the unit impulse response, with a delay of $4$ samples. This means that the output of the system is the input signal delayed by $4$ samples.<\/p>\n
To see this, consider the following example. Let the input signal be $x(t) = u(t)$, where $u(t)$ is the unit step function. The frequency response of the system is then<\/p>\n
$$H(f) = e^{-j4\\pi f}$$<\/p>\n
The output of the system is then<\/p>\n
$$y(t) = \\int_{-\\infty}^{\\infty} x(t-\\tau) H(\\tau) d\\tau = \\int_{-\\infty}^{\\infty} u(t-\\tau) e^{-j4\\pi \\tau} d\\tau = e^{-j4\\pi t} u(t) = x(t+4)$$<\/p>\n
Therefore, the output of the system is the input signal delayed by $4$ samples.<\/p>\n
The other options are incorrect because they do not take into account the delay of the system.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”x(t + 4)” option2=”x(t – 4)” option3=”x(t + 2)” option4=”x(t – 2)” 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-50197","post","type-post","status-publish","format-standard","hentry","category-signal-processing","no-featured-image-padding"],"yoast_head":"\n