Question

LJULJAL JL • 1.5(c-d) Using instead y[n] = x(C)[n] – x(d) [n] (In words, make a new signal y[n] by subtracting the signal inSketch and label the even and odd components of the signals shown in Fig. 1-24. Using Eqs. (1.5) and (1.6), the even and odd(pोtX.In) X[n] -5-4-3-2-1 -5-4-3-2-1 0 1 2 3 4 5 n 1 2 3 4 5 1 x In) -4-3-2-1 -4-3-2-1 0 1 2 3 4 1 2 3 4 (d) Fig. 1-25It was given new conditions to solve the problems for c-d.

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Answer #1

[2° Се и 4ete) и 1 м (Е) х(-n L-E) Molt) { th1 + (-nz} + {{n} - C-27 » С (+) + x-ti] > [~[+) - xt-t] от та 3 кг 7 - - - - - -

Even part of signal X(t)

X(t) - X(t) +X(st)

Odd part of signal X(t)

X.(t) - X(t) - X(-t

In que 2 the equation of the graph is given. Thus X(-t) is found by substituting -t in place of t .

If X(t)= X(-t) then it is an even signal thus it doesn't have any odd part.

If X(t) =X(-t) then it is an odd signal thus it doesn't have any even part.

If a signal doesn't satisfy any of the above condition then it is NEITHER ODD NEITHER EVEN signal . Such signals have both even and odd parts.

You can verify above points.

Please consider the labeling as X(t) on y-axis of continuous time signals and X[n] on y-axis of all descrete time signals and t on x-axis of all graphs.

I have answered all your questions.

Please rate a thumbs up if satisfied.

Have a nice time ahead :)

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