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Let h[n] be the impulse response of a discrete-time linear time invariant (LTI) filter. The impulse response is given by

\(h\left[ 0 \right] = \frac{1}{3};\;h\left[ 1 \right] = \frac{1}{3};h\left[ 2 \right] = \frac{1}{3};\)  h[n] = 0 for n < 0 and n > 2 

Let H(ω) be the discrete-time Fourier transform (DTFT) of h[n]. where ω is the normalized angular frequency in radians. Given that H(ω0) = 0 and 0 < ω0 < π, the value of ω0 (in radians) is equal to ________.

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Concept:

The DTFT of an aperiodic signal x(n) is defined as:

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

Application:

\(h\left( n \right) = \left\{ {\frac{1}{3},\frac{1}{3},\frac{1}{3}} \right\}\)

\(H\left( {{e^{j\omega }}} \right) = \frac{1}{3} + \frac{1}{3}{e^{ - j\omega }} + \frac{1}{3}{e^{ - j2\omega }}\)

\(= \frac{1}{3}{e^{ - j\omega }}\left[ {{e^{j\omega }} + {e^{ - j\omega }}} \right] + \frac{1}{3}{e^{ - j\omega }}\)

\(= \frac{2}{3}{e^{ - j\omega }}\cos \omega + \frac{1}{3}{e^{ - j\omega }}\)

\(= \frac{1}{3}{e^{ - j\omega }}\left[ {1 + 2\cos \omega } \right]\)

At ω = ω0

\(H\left( {{e^{j{\omega _0}}}} \right) = 0\)

We can, therefore, write:

1 + 2 cos ω0 = 0

cos ω0 = -1/2

\(\cos {\omega _0} = \cos \frac{{2\pi }}{3}\)

\(\omega_o = \frac{{2\pi }}{3} = 2.09\;radians\)

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