निम्नलिखित अनुक्रम के लिए असतत फूरियर श्रेणी निरूपण है:

\(x\left( n \right) = \cos \frac{\pi }{4}n\)

This question was previously asked in
ESE Electrical 2019 Official Paper
View all UPSC IES Papers >
  1. \(\frac{1}{2}{e^{j{{\rm{\Omega }}_0}n}} + \frac{1}{2}{e^{ - j{{\rm{\Omega }}_0}n}}\) और \({{\rm{\Omega }}_0} = \frac{\pi }{8}\)
  2. \(\frac{1}{2}{e^{ - j{{\rm{\Omega }}_0}n}} + \frac{1}{2}{e^{ - j2{{\rm{\Omega }}_0}n}}\) और \({{\rm{\Omega }}_0} = \frac{\pi }{4}\)
  3. \(\frac{1}{2}{e^{ - j{{\rm{\Omega }}_0}n}} + \frac{1}{2}{e^{ - j{{\rm{\Omega }}_0}n}}\) और \({{\rm{\Omega }}_0} = \frac{\pi }{6}\)
  4. \(\frac{1}{2}{e^{j{{\rm{\Omega }}_0}n}} + \frac{1}{2}{e^{j7{{\rm{\Omega }}_0}n}}\) और \({{\rm{\Omega }}_0} = \frac{\pi }{4}\)

Answer (Detailed Solution Below)

Option 4 : \(\frac{1}{2}{e^{j{{\rm{\Omega }}_0}n}} + \frac{1}{2}{e^{j7{{\rm{\Omega }}_0}n}}\) और \({{\rm{\Omega }}_0} = \frac{\pi }{4}\)
Free
ST 1: UPSC ESE (IES) Civil - Building Materials
6.2 K Users
20 Questions 40 Marks 24 Mins

Detailed Solution

Download Solution PDF

संप्रत्यय:

असतत-समय आवर्ती अनुक्रम का फूरियर श्रेणी निरूपण इस प्रकार दिया गया है:

\(x\left( n \right) = \mathop \sum \limits_{k = 0}^{N - 1} {a_k}{e^{jk{\omega _0}n}}\)

\(x\left( n \right) = \ldots + {a_{ - 1}}{e^{ - j{\omega _0}n}} + {a_1}{e^{j{\omega _0}n}} + \ldots \)

ak = फूरियर श्रेणी गुणांक N द्वारा आवधिक।

ω0 = मूल आवृत्ति।

अनुप्रयोग:

दिया गया अनुक्रम है: \(x\left( n \right) = \cos \frac{\pi }{4}n\)

अनुक्रम \({\omega _0} = \frac{\pi }{4}\) की मूल आवृत्ति

हम जानते हैं कि, \(\cos \theta = \frac{{{e^{j\theta }} + {e^{ - j\theta }}}}{2}\)

अब, हम दिए गए अनुक्रम को इस प्रकार पुनर्लेखित कर सकते हैं

\(x\left( n \right) = \frac{{{e^{\frac{{j\pi }}{4}n}} + {e^{ - \frac{{j\pi }}{4}n}}}}{2}\)

\( = \frac{1}{2}{e^{\frac{{j\pi }}{4}n}} + \frac{1}{2}{e^{\frac{{ - j\pi }}{4}n}}\)

हम लिख सकते हैं \({e^{\frac{{ - j\pi }}{4}n}} = {e^{\frac{{j7\pi }}{4}n}}\)

अब, x(n) बन जाता है

\(x\left( n \right) = \frac{1}{2}{e^{\frac{{j\pi }}{4}n}} + \frac{1}{2}{e^{\frac{{j7\pi }}{4}n}}\)

\( = \frac{1}{2}{e^{j{{\rm{\Omega }}_0}n}} + \frac{1}{2}{e^{j7{{\rm{\Omega }}_0}n}}\) और \({{\rm{\Omega }}_0} = \frac{\pi }{4}\)
Latest UPSC IES Updates

Last updated on May 28, 2025

->  UPSC ESE admit card 2025 for the prelims exam has been released. 

-> The UPSC IES Prelims 2025 will be held on 8th June 2025.

-> The selection process includes a Prelims and a Mains Examination, followed by a Personality Test/Interview.

-> Candidates should attempt the UPSC IES mock tests to increase their efficiency. The UPSC IES previous year papers can be downloaded here.

More Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Questions

Get Free Access Now
Hot Links: rummy teen patti dhani teen patti teen patti game online lucky teen patti teen patti joy