Fourier Analysis Questions and Answers – Fourier Half Range Series

«
»

This set of Fourier Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Fourier Half Range Series”.

1. In half range Fourier series expansion, we know the nature of the function in its full time period.
a) True
b) False
View Answer

Answer: b
Explanation: In half range Fourier series expansion, we know the nature of the function only in its half of the period. It can be either odd or even function in its full range. We assume it to be even function when we find half range cosine series and we assume it to be odd function when we find half range sine series.
advertisement

2. In half range cosine Fourier series, we assume the function to be _________
a) Odd function
b) Even function
c) Can’t be determined
d) Can be anything
View Answer

Answer: b
Explanation: In half range Fourier series expansion, the nature of the function in half of its period is only known. So when we find half range cosine series, there are only cosine terms which imply that the function is even function. f(x) = f(-x).

3. Find the half range sine series of the function f(x) = x, when 0<x<\(\frac{\pi}{2} \) and (π-x) when \(\frac{\pi}{2} \)<x< π.
a) \(\frac{8}{\pi}[\frac{sinx}{1^{2}} – sin\frac{(3x)}{3^{2}} + sin \frac{(5x)}{5^2} – sin \frac{(7x)}{7^{2}} +……] \)
b) \(\frac{4}{\pi}[\frac{sinx}{1^{2}} + sin\frac{(3x)}{3^{2}} + sin \frac{(5x)}{5^2} + sin \frac{(7x)}{7^{2}} +……] \)
c) \(\frac{8}{\pi}[\frac{sinx}{1^{2}} + sin\frac{(3x)}{3^{2}} + sin \frac{(5x)}{5^2} + sin \frac{(7x)}{7^{2}} +……] \)
d) \(\frac{4}{\pi}[\frac{sinx}{1^{2}} – sin\frac{(3x)}{3^{2}} + sin \frac{(5x)}{5^2} – sin \frac{(7x)}{7^{2}} +……] \)
View Answer

Answer: d
Explanation: bn = \(\frac{2}{\pi}[\int_{0}^{\pi⁄2}xsinnxdx+\int_{\pi⁄2}^{\pi}(\pi-x)sinnxdx\)]
= \(\frac{4}{\pi}\frac{sin\left(n\frac{\pi}{2}\right)}{n^2} \) [this term is zero whenever n is even and when odd, it gives 1 or -1]
= \(\frac{4}{\pi}[\frac{sinx}{1^{2}} – sin\frac{(3x)}{3^{2}} + sin \frac{(5x)}{5^2} – sin \frac{(7x)}{7^{2}} +……]. \)

4. Find bn when we have to find the half range sine series of the function x2 in the interval 0 to 3.
a) -18 \( \frac{cos(nπ)}{nπ} \)
b) 18 \( \frac{cos(nπ)}{nπ} \)
c) -18 \( \frac{cos(n \pi⁄2)}{nπ} \)
d) 18 \( \frac{cos(n \pi⁄2)}{nπ} \)
View Answer

Answer: a
Explanation: bn = \(\frac{2}{3} \int_0^3 x^{2} sin\left(n\pi\frac{x}{3}\right)dx \)
= \(\frac{2}{3} \left(-27\frac{cos(nπ)}{nπ}\right) \)
= \( -18\frac{cos(nπ)}{nπ}.\)

5. What is the formula for Parseval’s relation in Fourier series expansion?
a) \( \int_{-l}^l (f(x))^2 dx=l[\frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2+b_n^2 ) ] \)
b) \( \int_{-l}^l (f(x))^2 dx=l[\frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2 ) ] \)
c) \( \int_{-l}^l (f(x))^2 dx=l⁄2 [\frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2+b_n^2 ) ] \)
d) \( l\int_{-l}^l (f(x))^2 dx=[\frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2+b_n^2 ) ] \)
View Answer

Answer: a
Explanation: The real life significance of Parseval’s relation is to find the energy of the signal in its time period (1 time period). This can be found using the Fourier series coefficients. \(\int_{-l}^l (f(x))^2 dx=l[\frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2+b_n^2 ) ] \) is the realtion between the function and the Fourier series coefficients.

advertisement

6. In Parseval’s relation of Half range Fourier cosine series expansion, which of the following terms doesn’t appear?
a) a0
b) an
c) bn
d) all terms appear
View Answer

Answer: c
Explanation: In the expansion of a function in half range Fourier cosine series, only an and a0 appear. As the function is considered to be an even function, bn term becomes a null value as the integral becomes zero. So, only bn doesn’t appear in the parseval’s relation of half range Fourier cosine series.

7. Find the value of \(\frac{1}{1^2} +\frac{1}{3^2} +\frac{1}{5^2} +\frac{1}{7^2} \) +….when finding the Half range Fourier sine series of the function f(x) = 1 in 0<x<π.
a) \(\frac{\pi^2}{4} \)
b) \(\frac{\pi^2}{8} \)
c) \(\frac{\pi^2}{2} \)
d) \(3\frac{\pi^2}{8} \)
View Answer

Answer: b
Explanation: Using parseval’s relation for half range Fourier sine series,
\(\int_{-l}^l(f(x))^2 dx=l[\frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2+b_n^2 ) ] \)
L.H.S. = \(\int_0^\pi(1)^2 dx = \pi \)
bn = \(\frac{2}{\pi} \int_0^π1.sin(nx)dx \)
= \(\frac{2}{\pi n} (1-(-1)^n ) \)
Using parseval’s formula,
\(\pi = \frac{\pi}{2} * \frac{4}{\pi^2} * 2 (\frac{1}{1^2} +\frac{1}{3^2} +\frac{1}{5^2} +\frac{1}{7^2} +….) \)
Therefore,
\(\frac{1}{1^2} +\frac{1}{3^2} +\frac{1}{5^2} +\frac{1}{7^2} +….= \frac{\pi^2}{8}.\)

8. Find the value of\( \frac{1}{1^4} +\frac{1}{3^4} +\frac{1}{5^4} +\frac{1}{7^4} \) +….by finding the half range Fourier cosine series of the function f(x) = x in the interval 0<x<l.
a) \(\frac{\pi^4}{12} \)
b) \(\frac{\pi^4}{48} \)
c) \(\frac{\pi^4}{24} \)
d) \(\frac{\pi^4}{96} \)
View Answer

Answer: d
Explanation: Using Parseval’s relation,
\(\int_{-l}^l(f(x))^2 dx=l[\frac{a_0^2}{2}+\sum_{n=1}^{∞}(a_n^2+b_n^2 ) ] \)
L.H.S. = \(\int_0^l x^2 dx= \frac{l^3}{3} \)
a0 = \(\frac{2}{l} \int_0^l xdx= l \)
an = \(\frac{2}{l} \int_0^lxcos(\frac{nπx}{l})dx \)
= \(\frac{-4l}{\pi^2} (\frac{1}{1^2} +\frac{1}{3^2} +\frac{1}{5^2} +…) \)
R.H.S. = \(\frac{l}{2} (\frac{l^2}{2}+\left(-4\frac{l}{\pi}\right)^2 (\frac{1}{1^4} +\frac{1}{3^4} +\frac{1}{5^4} +…..)) \)
Therefore,
\(\frac{1}{1^4} +\frac{1}{3^4} +\frac{1}{5^4} +\frac{1}{7^4} +….= (\frac{l^3}{3}-\frac{l^3}{4})\left(2\frac{\pi^4}{(16l^3 )}\right) \)
= \(\frac{\pi^4}{96}.\)

9. In Parseval’s formula for half range Fourier series, the formula contains l/2 multiplied with the square of individual coefficients.
a) True
b) False
View Answer

Answer: a
Explanation: The Parseval’s formula for half range Fourier cosine series is \(\int_0^l(f(x))^2 dx=\frac{l}{2} [\frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2)] \)
The Parseval’s formula for half range Fourier sine series is \(\int_0^l(f(x))^2 dx=\frac{l}{2} ∑_{n=1}^∞(b_n^2) .\)
advertisement

10. What is the value of a0 if the function is f(x) = x3 in the interval 0 to 5?
a) 25/4
b) 125/4
c) 625/4
d) 5/4
View Answer

Answer: c
Explanation: \(\int_0^5 x^3 dx= \int_{-l}^l(f(x))^2 dx=l[\frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2+b_n^2)] \) (from 0 to 5)
= \(\frac{5^4}{4}= \frac{625}{4}.\)

Sanfoundry Global Education & Learning Series – Fourier Analysis.

To practice all areas of Fourier Analysis, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
advertisement
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn