Latest revision as of 10:25, 11 November 2011

Practice Question on Computing the Fourier Series discrete-time signal

Obtain the Fourier series the DT signal

$ x[n] = \left\{ \begin{array}{ll} 1, & \text{ for } -5\leq n \leq 5,\\ 0, & \text{ for } 5< |n| \leq 10. \end{array} \right. \ $

x[n] periodic with period 20.

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Answer 1

Solution 1:

1) $ w_o = \frac{2\pi}{T} = 2\frac{\pi}{20}=\frac{\pi}{10} $

2) a_o is the DC value of the AC signal and is therefore 1/2

3) $ a_k= 1/20 * \int_{-5}^{5} \! x(t)*e^{-j*k*(\pi/10)*t}\,dx $ = $ e^{j*k*\pi/2}-e^{-j*k*\pi/2}/(2*\pi*j*k) = \frac{sin(k\pi/2)}{(k\pi)} $(Clarkjv 18:25, 8 February 2011 (UTC))

Solution 2:

1)$ w_o=\frac{\pi}{10} $ (see solution 1)

From example 3.5 (sec. 3.3 pg 193 Signals and Systems 2nd edition Oppenheim) where T_1 is half the pulse width, $ a_k=\frac{sin(k * w_o * T_1)}{(k*\pi)} $,

2)a_o is still the DC value of the AC signal and therefore, a_o = 1/2

From $ a_k=\frac{sin(k * w_o * T_1)}{(k*\pi)} $

3)$ a_k=\frac{sin(k\pi/2)}{(k\pi)} $ (Clarkjv 18:25, 8 February 2011 (UTC))

TA's comment: The given signal is a DT signal and not a CT signal.

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