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Let x(t) be a periodic function with period T = 10. The Fourier series coefficients for this series are denoted by $${a_k}$$ , that is<br>$$x\left( t \right) = \sum\limits_{k = - \infty }^\infty {{a_k}{e^{jk{{2\pi } \over T}t}}} .$$<br>The same function x(t) can also be considered as a periodic function with period T' = 40. Let b<sub>k</sub> be the Fourier series coefficients when period is taken as T'. If $$\sum\limits_{k = - \infty }^\infty {\left| {{a_k}} \right|} = 16,$$    then $$\sum\limits_{k = - \infty }^\infty {\left| {{b_k}} \right|} $$  is equal to

Correct Answer: 16

Let x(t) be a continuous time periodic signal with fundamental period T = 1 seconds. Let {ak} be the complex Fourier series coefficients of x(t), where k is integer valued. Consider the following statements about x(3t):
1. The complex Fourier series coefficients of x(3t) are {ak} where k is integer valued.
2. The complex Fourier series coefficients of x(3f) are {3ak} where k is integer valued.
3. The fundamental angular frequency of x(3t) is 6π rad/s.
For the three statements above, which one of the following is correct?

For a function g(t), it is given that
$$\int\limits_{ - \infty }^{ + \infty } {g\left( t \right){e^{ - j\omega t}}dt = \omega {e^{ - 2{\omega ^2}}}} $$    for any real value $$\omega $$ . If $$y\left( t \right) = \int\limits_{ - \infty }^t {g\left( \tau \right)} d\tau ,\,{\rm{then}}\,\int\limits_{ - \infty }^{ + \infty } {y\left( t \right)} dt$$       is. . . . . . . .

If a + b + c + d = 4, then find the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?

If a + b + c + d = 4, then the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?

A periodic signal x(t) has a trigonometric Fourier series expansion
$$x\left( t \right) = {a_0} + \sum\limits_{n = 1}^\infty {\left( {{a_n}\,\cos \,n{\omega _0}t + {b_n}\sin \,n{\omega _0}t} \right)} $$
If $$x\left( t \right) = - x\left( { - t} \right) = - x\left( {{{t - \pi } \over {{\omega _0}}}} \right),$$      we can conclude that

The function f(t) has the Fourier transform f(ω)
The Fourier transform of
$$g\left( t \right) = \left( {\int\limits_{ - \infty }^\infty {g\left( t \right){e^{ - j\omega }}dt} } \right)$$     is

Let the function
\[{\text{f}}\left( \theta \right) = \left| {\begin{array}{*{20}{c}} {\sin \theta }&{\cos \theta }&{\tan \theta } \\ {\sin \left( {\frac{\pi }{6}} \right)}&{\cos \left( {\frac{\pi }{6}} \right)}&{\tan \left( {\frac{\pi }{6}} \right)} \\ {\sin \left( {\frac{\pi }{3}} \right)}&{\cos \left( {\frac{\pi }{3}} \right)}&{\tan \left( {\frac{\pi }{3}} \right)} \end{array}} \right|\

The Fourier series representation of an impulse train denoted by
$$s\left( t \right) = \sum\limits_{n = - \infty }^\infty {\delta \left( {t - n{T_0}} \right)} \,{\rm{is}}\,{\rm{given}}\,{\rm{by}}$$

If {x} is a continuous, real valued random variable defined over the interval (-$$\infty $$, +$$\infty $$) and its occurrence is defined by the density function given as:
$${\text{f}}\left( {\text{x}} \right) = \frac{1}{{\sqrt {2\pi } * {\text{b}}}}{{\text{e}}^{\frac{{ - 1}}{2}{{\left( {\frac{{{\text{x}} - {\text{a}}}}{{\text{b}}}} \right)}^2}}}$$     where 'a' and 'b' are the statistical attributes of the random variable {x}. The value of the integral $$\int_{ - \infty }^{\text{a}} {\frac{1}{{\sqrt {2\pi } * {\text{b}}}}{{\text{e}}^{\frac{{ - 1}}{2}{{\left( {\frac{{{\text{x}} - {\text{a}}}}{{\text{b}}}} \right)}^2}}}{\text{dx}}} $$

The value of the expression $$\frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} + $$   $$\frac{{{{\left( {b - c} \right)}^2}}}{{\left( {a - b} \right)\left( {c - a} \right)}} + $$    $$\frac{{{{\left( {c - a} \right)}^2}}}{{\left( {a - b} \right)\left( {b - c} \right)}}$$   = ?