Properties of Discrete Fourier Transform MCQ Quiz - Objective Question with Answer for Properties of Discrete Fourier Transform - Download Free PDF

Explanation:

Discrete Fourier Transform (DFT) and Symmetry Properties

Introduction: The Discrete Fourier Transform (DFT) is a mathematical technique used to analyze the frequency content of discrete signals. When a signal is real-valued, its DFT exhibits specific symmetry properties that simplify computations and provide insights into the relationship between various frequency components. In this problem, we are tasked with finding the value of X(6) of an 8-point DFT, given the values of X(0), X(1), X(2), and X(3). The key to solving this lies in exploiting the symmetry properties of the DFT.

Symmetry Properties of DFT for Real-Valued Sequences:

• Conjugate Symmetry: If the input sequence is real-valued, the DFT coefficients satisfy the following property:

  X(N-k) = conjugate(X(k)) for k = 1, 2, ..., N-1.

  Here, N is the number of points in the DFT (in this case, N = 8).

• Special Cases:
  • X(0) is always real because it represents the sum of all input values (DC component).
  • If N is even, X(N/2) (middle frequency) is also real.

In this problem, the input sequence is real-valued, so we can apply the conjugate symmetry property to find X(6).

Given:

• X(0) = 5
• X(1) = 1 + j
• X(2) = 3 + j
• X(3) = 2 + 3j

We need to determine the value of X(6). Using the conjugate symmetry property of the DFT:

X(N-k) = conjugate(X(k))

For N = 8, the relationship for X(6) becomes:

X(6) = conjugate(X(2))

From the given data:

X(2) = 3 + j

Taking the complex conjugate:

conjugate(X(2)) = 3 - j

Therefore:

X(6) = 3 - j

Correct Answer: Option 2

Analysis:

\(h\left( n \right) = \frac{{{e^{j\frac{\pi }{2}n}}\sin \left( {n\pi /4} \right)}}{{\pi n}}\)

\( = \frac{{{e^{j\frac{\pi }{2}n}}\sin \left( {n\pi /4} \right)}}{{\pi n}} = {e^{j\frac{\pi }{2}n}}{h_1}\left( n \right)\)

As the Fourier transform of a discrete and aperiodic signal is continuous and periodic. 

 ∴ H₁(ω) is periodic with a period of 2π.

Now, \({e^{j\frac{\pi }{2}n}}\) shifts the spectrum by \(\frac{\pi }{2}\)

∴ In the output,  10 will not pass.

At n = -1

\(50\cos \left( { - \pi + \frac{\pi }{2}} \right) = 50\cos \left( { - \frac{\pi }{2}} \right) = 0\)

\(\sin \left( {\frac{{ - 2\pi }}{{15}}} \right)\) will not pass.

Explanation:

D.T.F.T

Discrete-time Fourier transform is defined by

\(X\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right) = \mathop \sum \limits_{n = - \infty }^\infty x\left[ n \right]{e^{ - j\omega n}}\) 

It is periodic with 2π

Periodicity of DTFT

For any integer k,

\(X\left( {{e^{j\left( {\omega + 2\pi k} \right)}}} \right) = \mathop \sum \limits_{n = - \infty }^\infty x\left[ n \right]{e^{ - j\left( {\omega + 2\pi k} \right)n}}\) 

\(= \mathop \sum \limits_{n = - \infty }^\infty x\left[ n \right]{e^{ - j\omega n}}{e^{ - j2\pi kn}}\) 

\(X\left( {{e^{j\left( {\omega + 2\pi k} \right)}}} \right) = x\left[ n \right]{e^{ - j\omega n}}\)

\(X\left( {{e^{j\left( {\omega + 2\pi k} \right)}}} \right) = X\left( {{e^{j\omega }}} \right)\)

Discrete LPF

Discrete-time LPF is shown below

\(X\left( {\rm{\Omega }} \right) = \left\{ {\begin{array}{*{20}{c}} {1;\;\left| {\rm{\Omega }} \right| \le \omega }\\ {0;else\;\omega < \left| {\rm{\Omega }} \right| \le \pi } \end{array}} \right\}\)

From the response itself, it is clear that the LPF is periodic with 2π as a period.

Concept:

N-point DFT of a sequence x(n) defined for n = 0, 1, 2 ... N-1 is given by:

\(X\left( k \right)=\sum_{n=0}^{N-1}x\left( n \right){{e}^{-jk\omega n}}\)

Also If \(x(n)\overset{DFT}{\longleftrightarrow} X(k)\)

The Conjugate symmetric property of DFT states:

X(k) = X*(N – k) 'or' X(N – k) = X*(k) 

Calculation:

Given, X(k) = {5, 1 - 3j, 0, 3 – 4j, 0, 3 + 4j, A, B]

Where A and B are the missing variable values.

Using the conjugate symmetric property of DFT: X(k) = X*(N – k)

We find X(6) = A = X*(8 – 6) = X*(2)

With X(2) = 0, X*(2) = 0

So, X(6) = A = 0

Similarly,

B = X(7) = X*(8 – 7) = X*(1)

With, X(1) = 1 – 3j

X(7) = X*(1) = 1+3j

So, A = 0 and B = 1 + 3j

Concept:

\({\alpha ^n}u\left( n \right)\mathop \to \limits^{D.T.F.T.} \frac{1}{{1 - \alpha {e^{ - j\omega }}}}\)

Calculation:

\(x\left( n \right) = {\left( {\frac{1}{2}} \right)^n}u\left( n \right)\)

\(y\left( n \right) = {x^2}\left( {\rm{n}} \right) = {\left( {\frac{1}{2}} \right)^{2n}}\)

\(u\left( n \right) = {\left( {\frac{1}{4}} \right)^n}u\left( n \right)\)

\({\rm{Y}}\left( {{e^{j\omega }}} \right) = \frac{1}{{1 - \frac{1}{4}{e^{ - j\omega }}}}\)

\({\rm{Y}}\left( {{e^{j0}}} \right) = \frac{1}{{1 - \frac{1}{4}{e^{ - j0}}}}\)

\({\rm{Y}}\left( {{e^{j0}}} \right) = \;\frac{4}{3}\)

Concept:

The Fourier series representation of discrete time periodic sequence in given by:

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

x(n) = … + a_–1 e^-jω_oⁿ + 1 + a₁ e^jω_oⁿ + …   ---(1) 

a_k = Fourier series coefficient periodic by N.

ω₀ = Fundamental frequency.

Calculation:

\(x\left( n \right) = \sin \left( {\frac{{\pi n}}{5}} \right)\)

The period of the given sequence will be:

\(N = \frac{{2\pi }}{{\pi /5}} = 10\)

So \({\omega _0} = \frac{{2\pi }}{{10}} = \frac{\pi }{5}\)

x(n) can be written as:

\(x\left( n \right) = \frac{1}{{2j}}\left( {{e^{j\frac{\pi }{5} \cdot n\;}} - {e^{ - j\frac{\pi }{5}n}}} \right)\) (Euler’s identity)

\(\Rightarrow \frac{1}{{2j}} \cdot {e^{j\frac{\pi }{5} \cdot n}} - \frac{1}{{2j}} \cdot {e^{ - j\frac{\pi }{5}n}} = \frac{1}{{2j}}{e^{j{\omega _0}n}} = \frac{{ - 1}}{{2j}}\;{e^{ - j{\omega _0}n}}\)

Comparing this with Equation-(1), we get:

\(\Rightarrow {a_{ - 1}} = \frac{{ - 1}}{{2j}}\;\;and\;\;{a_1} = \frac{1}{{2j}}\)

Since the fourier coefficient are periodic,

a_k = a_k+N

So, a_-1 = a_-1+10 = a_-1+10+10 = a_-1+10+10+10 … = a_-1+m(10)

For any integer m.

Similarly,

a₁ = a₁₊₁₀ = a_1+10+10 = a_1+10+10+10 ---- = a_1+m(10)

So, a_k will be non-zero for k = ± 1 + m(10)

For any other value of k, a_k will be zero.

Concept:

N-point DFT of a sequence x(n) defined for n = 0, 1, 2 ... N-1 is given by:

\(X\left( k \right)=\sum_{n=0}^{N-1}x\left( n \right){{e}^{-jk\omega n}}\)

Also If \(x(n)\overset{DFT}{\longleftrightarrow} X(k)\)

The Conjugate symmetric property of DFT states:

X(k) = X*(N – k) 'or' X(N – k) = X*(k) 

Calculation:

Given, X(k) = {5, 1 - 3j, 0, 3 – 4j, 0, 3 + 4j, A, B]

Where A and B are the missing variable values.

Using the conjugate symmetric property of DFT: X(k) = X*(N – k)

We find X(6) = A = X*(8 – 6) = X*(2)

With X(2) = 0, X*(2) = 0

So, X(6) = A = 0

Similarly,

B = X(7) = X*(8 – 7) = X*(1)

With, X(1) = 1 – 3j

X(7) = X*(1) = 1+3j

So, A = 0 and B = 1 + 3j

\(\frac{1}{{\rm{\pi }}}\mathop \smallint \limits_{ - {\rm{\pi }}}^{\rm{\pi }} {\rm{X}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right){\sin ^2}\left( {2{\rm{\omega }}} \right){\rm{d\omega }}\)

We have

\({\sin ^2}\left( {2{\rm{\omega }}} \right) = \frac{{1 - \cos 4{\rm{\omega }}}}{2}\)

Thus,

\(\begin{array}{l} \frac{1}{{\rm{\pi }}}\mathop \smallint \limits_{ - {\rm{\pi }}}^{\rm{\pi }} {\rm{X}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right)\frac{{\left( {1 - {\rm{cos}}4{\rm{\omega }}} \right)}}{2}{\rm{\;d\omega }}\\ = \frac{1}{{2{\rm{\pi }}}}\mathop \smallint \limits_{ - {\rm{\pi }}}^{\rm{\pi }} {\rm{X}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right){\rm{d\omega }} - \frac{1}{{2{\rm{\pi }}}}\mathop \smallint \limits_{ - {\rm{\pi }}}^{\rm{\pi }} {\rm{X}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right){\rm{cos}}4{\rm{\omega d\omega }} \end{array}\)

We have

\(\frac{1}{{2{\rm{\pi }}}}\mathop \smallint \limits_{ - {\rm{\pi }}}^{\rm{\pi }} {\rm{X}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right){\rm{d\omega }} = {\rm{x}}\left[ 0 \right] = 8\)

And from the \(\cos 4{\rm{\omega }}\) term

\(\begin{array}{l} \frac{1}{{2{\rm{\pi }}}}\mathop \smallint \limits_{ - {\rm{\pi }}}^{\rm{\pi }} {\rm{X}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right) \times \left( {\frac{{{{\rm{e}}^{{\rm{j}}4{\rm{\omega }}}} + {{\rm{e}}^{ - {\rm{j}}4{\rm{\omega }}}}}}{2}} \right){\rm{d\omega }}\\ = \frac{1}{2}\left[ {\frac{1}{{2{\rm{\pi }}}}\mathop \smallint \limits_{ - {\rm{\pi \;}}}^{\rm{\pi }} {\rm{X}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right){{\rm{e}}^{{\rm{j}}4{\rm{\omega }}}}{\rm{d\omega }} + \frac{1}{{2{\rm{\pi }}}}\mathop \smallint \limits_{ - {\rm{\pi \;}}}^{\rm{\pi }} {\rm{X}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right){{\rm{e}}^{ - {\rm{j}}4{\rm{\omega }}}}{\rm{d\omega }}} \right]\\ = \frac{1}{2}\left[ {{\rm{x}}\left[ 4 \right] + {\rm{x}}\left[ { - 4} \right]} \right] = \frac{1}{2}\left[ {0 + 0} \right] = 0 \end{array}\)

Explanation:

D.T.F.T

Discrete-time Fourier transform is defined by

\(X\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right) = \mathop \sum \limits_{n = - \infty }^\infty x\left[ n \right]{e^{ - j\omega n}}\) 

It is periodic with 2π

Periodicity of DTFT

For any integer k,

\(X\left( {{e^{j\left( {\omega + 2\pi k} \right)}}} \right) = \mathop \sum \limits_{n = - \infty }^\infty x\left[ n \right]{e^{ - j\left( {\omega + 2\pi k} \right)n}}\) 

\(= \mathop \sum \limits_{n = - \infty }^\infty x\left[ n \right]{e^{ - j\omega n}}{e^{ - j2\pi kn}}\) 

\(X\left( {{e^{j\left( {\omega + 2\pi k} \right)}}} \right) = x\left[ n \right]{e^{ - j\omega n}}\)

\(X\left( {{e^{j\left( {\omega + 2\pi k} \right)}}} \right) = X\left( {{e^{j\omega }}} \right)\)

Discrete LPF

Discrete-time LPF is shown below

\(X\left( {\rm{\Omega }} \right) = \left\{ {\begin{array}{*{20}{c}} {1;\;\left| {\rm{\Omega }} \right| \le \omega }\\ {0;else\;\omega < \left| {\rm{\Omega }} \right| \le \pi } \end{array}} \right\}\)

From the response itself, it is clear that the LPF is periodic with 2π as a period.

Concept:

The Fourier series representation of discrete time periodic sequence in given by:

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

x(n) = … + a_–1 e^-jω_oⁿ + 1 + a₁ e^jω_oⁿ + …   ---(1) 

a_k = Fourier series coefficient periodic by N.

ω₀ = Fundamental frequency.

Calculation:

\(x\left( n \right) = \sin \left( {\frac{{\pi n}}{5}} \right)\)

The period of the given sequence will be:

\(N = \frac{{2\pi }}{{\pi /5}} = 10\)

So \({\omega _0} = \frac{{2\pi }}{{10}} = \frac{\pi }{5}\)

x(n) can be written as:

\(x\left( n \right) = \frac{1}{{2j}}\left( {{e^{j\frac{\pi }{5} \cdot n\;}} - {e^{ - j\frac{\pi }{5}n}}} \right)\) (Euler’s identity)

\(\Rightarrow \frac{1}{{2j}} \cdot {e^{j\frac{\pi }{5} \cdot n}} - \frac{1}{{2j}} \cdot {e^{ - j\frac{\pi }{5}n}} = \frac{1}{{2j}}{e^{j{\omega _0}n}} = \frac{{ - 1}}{{2j}}\;{e^{ - j{\omega _0}n}}\)

Comparing this with Equation-(1), we get:

\(\Rightarrow {a_{ - 1}} = \frac{{ - 1}}{{2j}}\;\;and\;\;{a_1} = \frac{1}{{2j}}\)

Since the fourier coefficient are periodic,

a_k = a_k+N

So, a_-1 = a_-1+10 = a_-1+10+10 = a_-1+10+10+10 … = a_-1+m(10)

For any integer m.

Similarly,

a₁ = a₁₊₁₀ = a_1+10+10 = a_1+10+10+10 ---- = a_1+m(10)

So, a_k will be non-zero for k = ± 1 + m(10)

For any other value of k, a_k will be zero.

Given:

X(0) = 4, X(1) = 1 – j1

And X(3) = 2 + j2

From property of conjugate symmetry,

X(k) = X* (N – k)

For k = 2 and N = 5,

X(2) = X* (5 – 2) = X* (3)

X(2) = 2 – j2

For k = 4 and N = 5,

X(4) = X* (5 – 4) = X*(1)

X(4) = 1 + j

Inverse discrete Fourier transform of x(n) is given by,

\(x\left( n \right) = \frac{1}{N}\mathop \sum \nolimits_{k = 0}^{N - 1} X\left( k \right){e^{j2\pi nk/N}}\)

For \(n = 0,\;x\left( 0 \right) = \frac{1}{5}\mathop \sum \nolimits_{k = 0}^4 X\left( k \right){e^0}\) 

\(x\left( 0 \right) = \frac{1}{5}\left[ {X\left( 0 \right) + X\left( 1 \right) + X\left( 2 \right) + X\left( 3 \right) + X\left( 4 \right)} \right]\)

\(x\left( 0 \right) = \frac{1}{5}\left[ {4 + 1 - j + 2 - j2 + 2 + j2 + 1 + j} \right]\)

x(0) = 2

Concept:

N-point DFT of a sequence x(n) defined for n = 0, 1, 2 ... N-1 is given by:

\(X\left( k \right)=\sum_{n=0}^{N-1}x\left( n \right){{e}^{-jk\omega n}}\)

Also If \(x(n)\overset{DFT}{\longleftrightarrow} X(k)\)

The Conjugate symmetric property of DFT states:

X(k) = X*(N – k) 'or' X(N – k) = X*(k) 

Calculation:

Given, X(k) = {5, 1 - 3j, 0, 3 – 4j, 0, 3 + 4j, A, B]

Where A and B are the missing variable values.

Using the conjugate symmetric property of DFT: X(k) = X*(N – k)

We find X(6) = A = X*(8 – 6) = X*(2)

With X(2) = 0, X*(2) = 0

So, X(6) = A = 0

Similarly,

B = X(7) = X*(8 – 7) = X*(1)

With, X(1) = 1 – 3j

X(7) = X*(1) = 1+3j

So, A = 0 and B = 1 + 3j

\(\frac{1}{{\rm{\pi }}}\mathop \smallint \limits_{ - {\rm{\pi }}}^{\rm{\pi }} {\rm{X}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right){\sin ^2}\left( {2{\rm{\omega }}} \right){\rm{d\omega }}\)

We have

\({\sin ^2}\left( {2{\rm{\omega }}} \right) = \frac{{1 - \cos 4{\rm{\omega }}}}{2}\)

Thus,

\(\begin{array}{l} \frac{1}{{\rm{\pi }}}\mathop \smallint \limits_{ - {\rm{\pi }}}^{\rm{\pi }} {\rm{X}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right)\frac{{\left( {1 - {\rm{cos}}4{\rm{\omega }}} \right)}}{2}{\rm{\;d\omega }}\\ = \frac{1}{{2{\rm{\pi }}}}\mathop \smallint \limits_{ - {\rm{\pi }}}^{\rm{\pi }} {\rm{X}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right){\rm{d\omega }} - \frac{1}{{2{\rm{\pi }}}}\mathop \smallint \limits_{ - {\rm{\pi }}}^{\rm{\pi }} {\rm{X}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right){\rm{cos}}4{\rm{\omega d\omega }} \end{array}\)

We have

\(\frac{1}{{2{\rm{\pi }}}}\mathop \smallint \limits_{ - {\rm{\pi }}}^{\rm{\pi }} {\rm{X}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right){\rm{d\omega }} = {\rm{x}}\left[ 0 \right] = 8\)

And from the \(\cos 4{\rm{\omega }}\) term

\(\begin{array}{l} \frac{1}{{2{\rm{\pi }}}}\mathop \smallint \limits_{ - {\rm{\pi }}}^{\rm{\pi }} {\rm{X}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right) \times \left( {\frac{{{{\rm{e}}^{{\rm{j}}4{\rm{\omega }}}} + {{\rm{e}}^{ - {\rm{j}}4{\rm{\omega }}}}}}{2}} \right){\rm{d\omega }}\\ = \frac{1}{2}\left[ {\frac{1}{{2{\rm{\pi }}}}\mathop \smallint \limits_{ - {\rm{\pi \;}}}^{\rm{\pi }} {\rm{X}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right){{\rm{e}}^{{\rm{j}}4{\rm{\omega }}}}{\rm{d\omega }} + \frac{1}{{2{\rm{\pi }}}}\mathop \smallint \limits_{ - {\rm{\pi \;}}}^{\rm{\pi }} {\rm{X}}\left( {{{\rm{e}}^{{\rm{j\omega }}}}} \right){{\rm{e}}^{ - {\rm{j}}4{\rm{\omega }}}}{\rm{d\omega }}} \right]\\ = \frac{1}{2}\left[ {{\rm{x}}\left[ 4 \right] + {\rm{x}}\left[ { - 4} \right]} \right] = \frac{1}{2}\left[ {0 + 0} \right] = 0 \end{array}\)

Concept:

N-point DFT of a sequence x(n) defined for n = 0, 1, 2 ---- N-1 is given by:

\(X\left( k \right)=\sum_{n=0}^{N-1} x\left( n \right){{e}^{-jk\omega n}}\)

Also, If \(x(n) \overset{DFT}{\longleftrightarrow} X (k)\)

Then, X(k) = X*(N – k) or X(N – k) = X*(k) (Conjugate symmetric property of DFT)

The energy of a discrete-time sequence is defined as:

\(E=\sum_{N=0}^{N-1}{{x}^{2}}\left( n \right)=\frac{1}{N}\sum_{k=0}^{N-1}{{\left| X\left( k \right) \right|}^{2}}\)

Calculation:

Given, X(k) = {5, 1 - 3j, 0, 3 – 4j, 0, 3 + 4j, A, B]

Where A and B are the missing variable values.

Using the conjugate symmetric property of DFT: X(k) = X*(N – k)

We find X(6) = A = X*(8 – 6) = X*(2)

With X(2) = 0, X*(2) = 0

So, X(6) = A = 0

Similarly,

B = x(7) = X*(8 – 7) = X*(1)

With, X(1) = 1 – 3j

X(7) = X*(1) = 1+3j

So, A = 0 and B = 1 + 3j

X(k) is, therefore:

⇒ X(k) = {5, 1 – 3j, 0, 3 – 4j, 0, 3 + 4j, 0, 1+3j}

Now, the Energy of a discrete-time sequence is defined as:

\(E=\sum_{N=0}^{N-1}{{x}^{2}}\left( n \right)=\frac{1}{N}\sum_{k=0}^{N-1}{{\left| X\left( k \right) \right|}^{2}}\)

The Energy of the given sequence will be;

 \(=\frac{1}{8}\sum_{k=0}^{7},{{\left| X\left( k \right) \right|}^{2}}\)

\(=\frac{1}{8}[{{\left| X\left( 0 \right) \right|}^{2}}+{{\left| X\left( 1 \right) \right|}^{2}}+---+{{\left| X\left( 7 \right) \right|}^{2}}]\)

\(=\frac{1}{8}[{{\left( 5 \right)}^{2}}+{{\left| 1-3j \right|}^{2}}+0+{{\left| 3-4j \right|}^{2}}+0+{{\left| 3+4j \right|}^{2}}+0+{{\left| 1+3j \right|}^{2}}]\)

\(=\frac{1}{8}[25+{{\left( \sqrt{10} \right)}^{2}}+{{\left( \sqrt{25} \right)}^{2}}+{{\left( \sqrt{25} \right)}^{2}}+{{\left( \sqrt{10} \right)}^{2}}\)

 \(=\frac{1}{8}\left[ 25+10+25+25+10 \right]\)

\(=\frac{95}{8}=11.875\)

Analysis:

\(h\left( n \right) = \frac{{{e^{j\frac{\pi }{2}n}}\sin \left( {n\pi /4} \right)}}{{\pi n}}\)

\( = \frac{{{e^{j\frac{\pi }{2}n}}\sin \left( {n\pi /4} \right)}}{{\pi n}} = {e^{j\frac{\pi }{2}n}}{h_1}\left( n \right)\)

As the Fourier transform of a discrete and aperiodic signal is continuous and periodic. 

 ∴ H₁(ω) is periodic with a period of 2π.

Now, \({e^{j\frac{\pi }{2}n}}\) shifts the spectrum by \(\frac{\pi }{2}\)

∴ In the output,  10 will not pass.

At n = -1

\(50\cos \left( { - \pi + \frac{\pi }{2}} \right) = 50\cos \left( { - \frac{\pi }{2}} \right) = 0\)

\(\sin \left( {\frac{{ - 2\pi }}{{15}}} \right)\) will not pass.