Double integration problem for IDSFT

[Kurd]

Homework Statement

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The 2D Discrete Space Fourier transform (DSFT) X(w1,w2) of the sequence x(n1,n2) is given by,

$$X(w_1,w_2) = 5 + 2j sin(w_2) + cos(w_1) + 2e^{(-jw1-jw2)}$$

determine x(n1,n2)

Homework Equations

By definition inverse DSFT is,

$$x(n_1,n_2) = \dfrac{1}{(2π)^2} \int_{-π}^{π}\int_{-π}^{π} X(w_1,w_2) e^{(jw_1n_1+jw_2n_2)} dw_1dw_2$$

The Attempt at a Solution

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I got zero as a final answer.

Solving the double integral for each term I get zero when substituting pi, is it correct or did I made a mistake somewhere, what gets me confused is that when doing DSFT for some simple problem I can get cos(w1) for example but if I did the inverse DSFT I will get zero. can someone help.

Thanks in advance.

 

[haruspex]

I got zero as a final answer

Please post your working.

 

[Kurd]

Taking each term,

$$X_1(w_1,w_2) = 5$$
$$X_2(w_1,w_2) = 2jsin(w_2)$$
$$X_3(w_1,w_2) = cos(w_1)$$
$$X_4(w1,w2) = 2e^{(-jw_1-jw_2)}$$

then,

$$ x_1(n_1,n_2) = \dfrac{5}{(2π)^2} \int_{-π}^{π}\int_{-π}^{π} e^{jw_1n_1+jw_2n_2} dw_1dw_2$$
$$x_1(n_1,n_2) = \dfrac{5}{(2π)^2} \int_{-π}^{π}e^{jw_2n_2}[ \int_{-π}^{π} e^{jw_1n_1} dw_1]dw_2$$
$$x_1(n_1,n_2) = \dfrac{5}{(2π)^2} \int_{-π}^{π}e^{jw_2n_2} [\dfrac{e^{jw_1n_1}}{jn_1}]_{-π}^{π} dw_2$$
$$x_1(n_1,n_2) = \dfrac{5}{(2π)^2} \int_{-π}^{π}e^{jw_2n_2} [\dfrac{1}{jn_1}(e^{jπn_n1} - e^{-jπn_1})] dw_2$$

since n1 and n2 are discrete, then
$$(e^{jπn_1} - e^{-jπn_1})] = cos(jπn_1) + jsin(jπn_1) - cos(-jπn_1) + jsin(-jπn_1) = 0$$

I solved the remaining terms in a similar fashion, do you find anything wrong here.

 

[Kurd]

Ok I got it.

I didn't notice that

$$\dfrac{sin(πn_1)}{πn_1}$$

which is a sinc function and equals 1 at n1 = 0.

 

[haruspex]

Ok I got it.

I didn't notice that

$$\dfrac{sin(πn_1)}{πn_1}$$

which is a sinc function and equals 1 at n1 = 0.

Well done.