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

Correct Answer: a<sub>n</sub> are zero for all n and b<sub>n</sub> are zero for n even

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
$$x\left( t \right) = \sum\limits_{k = - \infty }^\infty {{a_k}{e^{jk{{2\pi } \over T}t}}} .$$
The same function x(t) can also be considered as a periodic function with period T' = 40. Let bk 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

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. . . . . . . .

An atomic state of hydrogen is represented by following wave function $$\psi \left( {r,\,\theta ,\,\phi } \right) = \frac{1}{{\sqrt 2 }}{\left( {\frac{1}{{{a_0}}}} \right)^{\frac{3}{2}}}\left( {1 - \frac{r}{{2{a_0}}}} \right){e^{\frac{{ - r}}{{2{a_0}}}}}\cos \theta $$         where, a0 is a constant. The quantum numbers of the state are

The Fourier transform of a signal
h(t) is $$H\left( {j\omega } \right) = {{\left( {2\cos \omega } \right)\left( {\sin \omega } \right)} \over \omega }$$
The value of h(0) is

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

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?

Two monochromatic waves having frequencies $$\omega $$ and $$\omega + \Delta \omega \left( {\Delta \omega \ll \omega } \right)$$    and corresponding wavelengths $$\lambda $$ and $$\lambda - \Delta \lambda \left( {\Delta \lambda \ll \lambda } \right)$$    of same polarization, travelling along X-axis are superimposed on each other. The phase velocity and group velocity of the resultant wave are respectively given 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 Fourier series expansion of a real periodic signal with fundamental frequency f0 is given by
$${g_p}\left( t \right) = \sum\limits_{n = - \infty }^\infty {{c_n}{e^{j2\pi {f_0}t}}} ;$$
it is given that c3 = 3 + j5. Then c3 is