Question

The following signal's zero-point is at 2:

x[n] = {1, 0, −1, 2, 3}

The discrete-time Fourier Transform is expressed as X(e^jw) = X_r(e^jw)+jX_i(e^jw). Find the signal with Fourier Transform X_i(e^jw) + X_r(e^jw)e^j2w

Answer 1

The signal is x[n] is defined as

$x[n]=\left \{ 1,0,-1,2,3 \right \}$

The DTFT of any signal x[n] is defined as:

$X(e^{j\omega })=\sum_{n=-\infty }^{\infty}x[n]e^{-j\omega n}$

Since the sample at n=0 is 2, the DTFT can be obtained as

$X(e^{j\omega })=\sum_{n=-3 }^{1}x[n]e^{-j\omega n}=1e^{3j\omega}+0e^{2j\omega}-1e^{j\omega}+2+3e^{-j\omega}$

$X(e^{j\omega })=cos(3\omega)+jsin(3\omega)-cos(\omega)-jsin(\omega)+2+3cos(\omega)-j3sin(\omega)$

Hence the real part of the DTFT of x[n] is obtained as:

$X_{r}(e^{j\omega })=cos(3\omega)+2cos(\omega)+2$

And the imaginary part of the DTFT of x[n] is obtained as:

$X_{i}(e^{j\omega })=sin(3\omega)-4sin(\omega)$

From the real and imaginary parts of DTFT of x[n], the DTFT of the new sequence is obtained as:

$X_{i}(e^{j\omega })+X_{r}(e^{j\omega })e^{j2\omega }=\left (sin(3\omega)-4sin(\omega) \right )+\left (cos(3\omega)+2cos(\omega)+2 \right )e^{j2\omega }$

$X_{i}(e^{j\omega })+X_{r}(e^{j\omega })e^{j2\omega }=\left (\frac{e^{3j\omega}-e^{-3j\omega}}{2j}-4\frac{e^{j\omega}-e^{-j\omega}}{2j} \right )+\left (\frac{e^{3j\omega}+e^{-3j\omega}}{2}+2\frac{e^{j\omega}+e^{-j\omega}}{2}+2 \right )e^{j2\omega }$

$X_{i}(e^{j\omega })+X_{r}(e^{j\omega })e^{j2\omega }=\left (\frac{e^{3j\omega}-e^{-3j\omega}}{2j}-4\frac{e^{j\omega}-e^{-j\omega}}{2j} \right )+\left (\frac{e^{5j\omega}+e^{-j\omega}}{2}+2\frac{e^{3j\omega}+e^{j\omega}}{2}+2 \right )$

$X_{i}(e^{j\omega })+X_{r}(e^{j\omega })e^{j2\omega }=\sum_{n=-\infty }^{\infty}x_{1}[n]e^{-j\omega n}=\left (\frac{e^{3j\omega}-e^{-3j\omega}}{2j}-4\frac{e^{j\omega}-e^{-j\omega}}{2j} \right )+\left (\frac{e^{5j\omega}+e^{-j\omega}}{2}+2\frac{e^{3j\omega}+e^{j\omega}}{2}+2e^{j2\omega } \right )$

$X_{i}(e^{j\omega })+X_{r}(e^{j\omega })e^{j2\omega }=\sum_{n=-\infty }^{\infty}x_{1}[n]e^{-j\omega n}=\frac{e^{5j\omega}}{2}+e^{3j\omega}\left \{ 1+\frac{1}{2j} \right \}+2e^{2j\omega}+e^{j\omega}\left \{ 1-\frac{2}{j} \right \}+e^{-j\omega}\left \{ \frac{1}{2}+\frac{2}{j} \right \}$

From the above equation, by comparing coefficients, the desired new sequence x₁[n] can be obtained as follows:

$x_{1}[n]=\left \{ \frac{1}{2},0, 1+\frac{1}{2j},2,1-\frac{2}{j},0,\frac{1}{2}+\frac{2}{j} \right \}$

The second zero coming in the sequence x₁[n] is the sample that corresponds to n=0.