Lista com Respostas Process Digital de Sinais-2023 1

(a) zeros ±j j poles \frac{1}{2} , \infty (b) poles at ±j j zeros at ±1 3.37. From the pole-zero diagram \displaystyle{X(z) = \frac{z}{(z^{2}-z+\frac{1}{2})(z+\frac{3}{4})}} \quad|z|>\frac{3}{4} \displaystyle{y[n] = x[-n+3] = x[-(n-3)]} \Rightarrow\displaystyle{Y(z) = z^{-3}X(z^{-1}) = \frac{z^{-3}z^{-1}}{(z^{-2}-z^{-1}+\frac{1}{2})(z^{-1}+\frac{3}{4})}} \displaystyle{\hspace{6em}= \frac{\frac{8}{3}}{z(2-2z+z^{2})(\frac{4}{3}+z)}} Poles at 0, -\frac{4}{3}, 1 \pm j, zeros at \infty x[n] causal\Rightarrow x[-n+3] is left-sided\Rightarrow ROC is 0 < |z| < 4/3. 3.35. (a) \displaystyle{X(z) = \log_{2}(\frac{1}{2}-z)} \quad \displaystyle{|z| < \frac{1}{2}} \displaystyle{X(z) = \log(1-2z) = -\sum^{\infty}_{i=1} \frac{(2z)^{i}}{i} = -\sum^{\infty}_{t=-\infty} \frac{1}{t} (2z)^{-t} = \sum^{\infty}_{t=-\infty} \frac{1}{t} \left( rac{1}{2}\right)^{t} z^{-t}} Therefore, \displaystyle{x[n] = \frac{1}{n}\left(\frac{1}{2}\right)^{n}u[-n-1]} (b) \displaystyle{nz[n]} \displaystyle{-(1-2z)}nz[n] \displaystyle{= \frac{1}{n}\left(\frac{1}{2}\right)^{n}u[-n-1]} \displaystyle{z[n] = \frac{1}{n}\left(\frac{1}{2}\right)^{n}u[-n-1]} 3.23. (a) y[n] = 0\ \ n < 0 y[n] = \sum_{k=0}^{n}x[k]h[n-k] = \sum_{k=0}^{n}a^{n-k} = a^{n}\frac{1-a^{-(n+1)}}{1-a^{-1}} = \frac{1-a^{n+1}}{1-a}\ \ 0 \leq n < N-1 y[n] = \sum_{k=0}^{N-1}x[k]h[n-k] = \sum_{k=0}^{N-1}a^{n-k} = a^{n}\frac{1-a^{-N}}{1-a^{-1}} = a^{n+1}\frac{1-a^{-N}}{a^{-1}} ,\ \ n \geq N (b) H(z) = \sum_{n=0}^{\infty} a^{n}z^{-n} = \frac{1}{1-az^{-1}} \quad |z| > |a| X(z) = \sum_{n=0}^{N-1} z^{-n} = \frac{1-z^{-N}}{1-z^{-1}} \quad |z|>0 Therefore, Y(z) = \frac{1-z^{-N}}{(1-az^{-1})(1-z^{-1})} = \frac{1}{(1-az^{-1})(1-z^{-1})} - \frac{z^{-N}}{(1-az^{-1})(1-z^{-1})} \quad |z| > |a| Now, \frac{1}{(1-az^{-1})(1-z^{-1})} = \frac{1}{1-az^{-1}}\cdot\frac{1}{1-z^{-1}} = \frac{1}{(1-a)}\left(\frac{1}{1-z^{-1}} - \frac{a}{1-az^{-1}}\right) So y[n] = \left(\frac{1}{1-a}\right)[u[n] - a^{n+1}u[n] - u[n-M] - a^{n-N+1}u[n-N]] = \frac{1-a^{n+1}}{1-a}u[n] - \frac{1-a^{n-N+1}}{1-a}u[n-N] = \begin{cases} 0, & n < 0 \\ \frac{1-a^{n+1}}{1-a},& 0 \leq n \leq N-1 \\ a^{n+1}\left(\frac{1-a^{-N}}{a^{-1}}\right),& n \geq N \end{cases} 3.16. (a) To determine H(z), we first find X(z) and Y(z): X(z) = \frac{1}{1-z^{-1}}.\frac{1}{1-2z^{-1}}=\frac{1}{(1-z^{-1})(1-2z^{-1})} = \frac{a^{-1}}{(1-\frac{1}{3}z^{-1})(1-2z^{-1})} \quad\frac{1}{3} < |z| < 2 Y(z) = \frac{5}{1-\frac{1}{5}z^{-1}}=\frac{5}{\frac{1}{(1-\frac{1}{5}z^{-1})}} =(1-\frac{1}{3}z^{-1})^{-1}/(1-\frac{1}{5}z^{-1})^{-1} \quad |z|> \frac{2}{3} Now, H(z) = \frac{Y(z)}{X(z)}=\frac{1-2z^{-1}}{1-\frac{1}{3}z^{-1}} \quad |z| > \frac{2}{3} The pole-zero plot of H(z) is plotted below. (b) Taking the inverse z-transform of H(z), we get h[n] = \left(\frac{2}{3}\right)^nu[n]-2\left(\frac{1}{3}\right)^nu[n-1] = \left(\frac{2}{3}\right)\left(u[n] - 3u[n-1]\right) (c) Since H(z) = \frac{Y(z)}{X(z)} = \frac{1 -2z^{-1}}{1-\frac{1}{3}z^{-1}} we can write Y(z)(1-\frac{1}{3}z^{-1}) = X(z)(1-2z^{-1}), whose inverse z-transform leads to y[n] = -\frac{1}{3}y[n-1] = x[n] - 2x[n-1] (d) The system is stable because the ROC includes the unit circle. It is also causal since the impulse response h[n] = 0 for n < 0. 3.9. H(z) = \frac{1+x^{-1}}{(1-\frac{1}{2}z^{-1})(1+\frac{1}{2}z^{-1})} = \frac{2}{(1-\frac{1}{2}z^{-1})} - \frac{1}{(1+\frac{1}{2}z^{-1})} (a) h[n] causal \Rightarrow ROC outside |z| = \frac{1}{2} \Rightarrow |z| > \frac{1}{2}. (b) ROC includes |z| = 1 \Rightarrow stable. (c) y[n] = \frac{1}{3}\left(-\frac{1}{4}\right)^n u[n] - \frac{3}{4}(2)^nu[-n-1] Y(z) = \frac{-\frac{3}{4}z^{-1} + \frac{3}{4}}{1+x^{-1} + 2z^{-1}} = \frac{1}{1+\frac{1}{2}z^{-1}(1-2z^{-1})} \quad \frac{1}{4}|z| < 2 X(z) = \frac{Y(z)}{H(z)} = \left(\frac{1-\frac{1}{2}z^{-1}}{1-2z^{-1}} \right) & |z| < 2 x[n] = -(2)^n u[-n-1]+\frac{1}{2}(2)^n u[-n] (d) h[n] = 2\left(\frac{1}{2}\right)^n u[n] - \left(-\frac{1}{4}\right)^n u[n] (c) X(z) = \frac{1 - \frac{1}{2}z^{-1}}{1 + \frac{3}{4}z^{-1} + \frac{1}{8}z^{-2}} \quad |z| > \frac{1}{2} Partial Fractions: X(z) = \frac{-3}{1 + \frac{1}{4}z^{-1}} + \frac{4}{1 + \frac{1}{2}z^{-1}} \quad |z| > \frac{1}{2} x[n] = \left[-3 \left( -\frac{1}{4} \right)^n + 4 \left( -\frac{1}{2} \right)^n \right] u[n] Long division: \frac{1}{1 + \frac{3}{4}z^{-1} + \frac{1}{8}z^{-2}} 1 + \left(-\frac{3}{4}\right)z^{-1} + \left(\frac{-3}{16} + 1\right)z^{-2} + \ldots \overline{\smash[b]{\phantom{1}\frac{1}{1}\right) \frac{3}{4}z^{-1} + \frac{1}{8}z^{-2}}}} \\ \left(-\frac{3}{4}\right)z^{-1} \quad + \quad \phantom{-}\frac{3}{4}(\frac{3}{4})z^{-2} \quad + \phantom{-}\frac{3}{4}(\frac{-3}{4})z^{-3} \underbar{[}{-\frac{3}{8} + \frac{3}{4}(\frac{3}{4})]z^{-2} - \frac{3}{8}(\frac{3}{4})z^{-3}} \Rightarrow x[n] = \left[-3 \left( -\frac{1}{4} \right)^n + \left( -\frac{1}{2} \right)^{n-2} \right] u[n] (d) X(z) = \frac{1 - \frac{1}{2}z^{-1}}{1 - \frac{1}{2}z^{-2}} \quad |z| > \frac{1}{2} Partial Fractions: X(z) = \frac{1 - \frac{1}{2}z^{-1}}{1 + \frac{1}{2}z^{-1}} \quad |z| > \frac{1}{2} x[n] = \left( -\frac{1}{2} \right)^n u[n] Long division: see part (i) above. (e) X(z) = \frac{1 - az^{-1}}{z^{-1} - a} \quad |z| > |a^{-1}| Partial Fractions: X(z) = -a \cdot \frac{a^{-1}(1-a^2)}{1-a^{-1}z^{-1}} \quad |z| > |a^{-1}| x[n] = -a\delta[n] - (1-a^2)a^{-(n+1)}u[n] Long division: \underline{\phantom{-a + z^{-1}}}\right] + \dots x[n] = -a\delta[n] - (1-a^2)a^{-(n+1)}u[n] 3.5. X(z) = (1 + 2z)(1 + 3z^{-1})(1 - z^{-1}) = 2z + 5 - 4z^{-1} - 3z^{-2} = \sum_{n=-\infty}^{\infty} x[n]z^{-n} Therefore, x[n] = 2\delta[n + 1] + 5\delta[n] - 4\delta[n - 1] - 3\delta[n - 2] 3.4. The pole-zero plot of X(z) appears below. (a) For the Fourier transform of x[n] to exist, the z-transform of x[n] must have an ROC which includes the unit circle, therefore, \frac{1}{3} < |z| < |2|. Since this ROC lies outside \frac{1}{3}, this pole contributes a right-sided sequence. Since the ROC lies inside 2 and 3, these poles contribute left-sided sequences. The overall x[n] is therefore two-sided. (b) Two-sided sequences have ROC's which look like washers. There are two possibilities. The ROC's corresponding to these are: \frac{1}{3} < |z| < |2| and |2| < |z| < |3|. (c) The ROC must be a connected region. For stability, the ROC must contain the unit circle. For causality the ROC must be outside the outermost pole. These conditions cannot be met by any of the possible ROC's of this pole-zero plot. 3.2. { n , 0 ≤ n ≤ N - 1 x[n] = N , N ≤ n = n u[n] - (n - N)u[n - N] d d 1 n x[n] ⇔ -z⁻¹—— X(z) ⇒ n u[n] ⇔ -z⁻¹—— ———— |z| > 1 dz dz (1 - z⁻¹)² z⁻¹ n u[n] ⇔ ——— |z| > 1 (1 - z⁻¹)² X(z)· z⁻ⁿ⁰ ⇒ (n - N)u[n - N] ⇔ z⁻ⁿ⁻¹ x[n - n₀] ⇔ ————————— |z| > 1 (1 - z⁻¹)² therefore z⁻¹ - z⁻ⁿ⁻¹ z⁻¹(1 - z⁻ⁿ) X(z) = ————————— = (1 - z⁻¹)² (1 - z⁻¹)²