Question

Consider the signals x[n] = 2^n-1 u[-n + 2] and y[n] = 2^-n+2 u[n + 1], where u[n] is the unit step sequence. Let X(e^jω) and Y(e^jω) be the discrete-time Fourier transform of x[n] and y[n], respectively. The value of the integral

\(\dfrac{1}{2\pi}\displaystyle\int_0^{2\pi} X(e^{j\omega}) Y(e^{-j\omega})d\omega\)

(rounded off to one decimal place) is _____  

Answer 1

Concept:

\(x\left[ n \right]*y\left[ n \right]\;\begin{array}{*{20}{c}} {\mathop \leftrightarrow \limits^{Fourier} }\\^{Transform} \end{array}X\left( {{e^{j\omega }}} \right)Y\left( {{e^{j\omega }}} \right)\)

Let z(n) = x(n) * y(n)

Then Z(e^jω) = X(e^jω) Y(e^jω)

Also, using central ordinate theorem:

\(z\left[ 0 \right] = \frac{1}{{2\pi }}\mathop \smallint \nolimits_0^{2\pi } z\left( {{e^{j\omega }}} \right)d\omega \)

\( = \frac{1}{{2\pi }}\mathop \smallint \nolimits_0^{2\pi } X\left( {{e^{j\omega }}} \right)Y\left( {{e^{j\omega }}} \right)d\omega \)

Application:

x[n] = 2^n-1 u[-n + 2]

y[n] = 2^-n+2  u[n + 1]

Given:

\(\frac{1}{{2\pi }}\mathop \smallint \nolimits_0^{2\pi } X\left( {{e^{j\omega }}} \right)\;y\left( {{e^{ - j\omega }}} \right)d\omega \)

If y(n) = y(e^jω)

Then, using time reversal property:

y[-n] = y(e^-jω)

Hence z[n] = x[n] * y[-n]

z[n] = [2^n-1u[-n + 2]] * [2ⁿ⁺²u[-n + 1]]

\(z\left[ n \right] = \mathop \sum \nolimits_{k = - \infty }^\infty {2^{k - 1}}\;u\left[ { - k + 2} \right] \cdot {2^{n - k + 2}}\;u\left[ { - n + k + 1} \right]\)

\(z\left[ n \right] = \mathop \sum \nolimits_{k = - \infty }^2 {2^{k - 1}} \cdot {2^{n - k + 2}}\;u\left[ { - n + k + 1} \right]\)

\(z\left[ n \right] = \mathop \sum \nolimits_{k = - \infty }^2 {2^{k - 1 + n - k + 2}}\;u\left[ { - n + k + 1} \right]\)

\(z\left[ n \right] = \mathop \sum \nolimits_{k = - \infty }^2 {2^{n + 1}}\;u\left[ { - n + k + 1} \right]\;\)

\(\frac{1}{{2\pi }}\mathop \smallint \nolimits_0^{2\pi } X\left( {{e^{j\omega }}} \right)\;Y\left( {{e^{ - j\omega }}} \right) = z\left[ 0 \right]\)

Putting n = 0, we get:= 2

\(z\left[ 0 \right] = \mathop \sum \nolimits_{k = - \infty }^2 2 \cdot u\left[ {k + 1} \right]\)

\( = \mathop \sum \nolimits_{k = - 1}^2 2 \cdot \left( 1 \right)\)

= 2 × 4

= 8