Sinais e Sistemas Formulário completo

Prof. Dr. Marcelo de Oliveira Rosa

2 de outubro de 2014

1 Matemática

∑_{k=N_{1}}^{N_{2}} α^{k} = \frac{α^{N_{1}} − α^{N_{2}+1}}{1 − α} , N_{2} ≥ N_{1}  (1)

2 Domínio do tempo

E_{x} = ∑_{k=−∞}^{+∞} |x[k]|^{2} dt  (2)

P_{x} = lim_{N→+∞} \frac{1}{N} ∫_{k_{0}}^{k_{0}+N} |x[k]|^{2} dt  (3)

y[n] = ∑_{k=−∞}^{+∞} x[k]h[n − k]  (4)

\begin{aligned}y[n] & = x[n] ∗ h[n] ⇔ \\ y[an] & = |a| x[at] ∗ h[at] \end{aligned}  (5)

2.1 Equação à diferenças

Seja p o maior atraso de y[n] e q o maior atraso de x[n] de uma equação a diferenças linear com coeficientes constantes.

y_{p}[n] = \begin{cases}0 & p > q \\ Kδ[n] & p = q \\ \begin{matrix}K_{q−p}δ[n − (q − p)]+ \\ K_{q−p−1}δ[n − (q − p − 1)]+ \\ ⋯ + \\ K_{0}δ[n] \end{matrix} & p < q \end{cases}

(6)

3 Domínio de Fourier

e^{±ȷx} = cos(x) ± ȷ sen(x)  (7)

\begin{aligned}cos(x) & = \frac{e^{+ȷx} + e^{−ȷx}}{2} \\ sen(x) & = \frac{e^{+ȷx} − e^{−ȷx}}{2ȷ} \end{aligned}  (8)

cos(a ± b) = cos(a) cos(b) ∓ sen(a) sen(b)
sen(a ± b) = sen(a) cos(b) ± sen(b) cos(a)

(9)

X(e^{ȷω}) = ∑_{k=−∞}^{+∞} x[k]e^{−ȷωk}  (10)

x[n] = \frac{1}{2π} ∫_{−π}^{π} X(e^{ȷω})e^{+ȷωn}dω  (11)

\begin{aligned}X(e^{ȷω}) & = X_{p}(e^{ȷω}) + X_{i}(e^{ȷω}) \\ X_{p}(e^{ȷω}) & = \frac{1}{2} [X(e^{ȷω} + X^{∗}(e^{−ȷω}] \\ X_{i}(e^{ȷω}) & = \frac{1}{2} [X(e^{ȷω} − X^{∗}(e^{−ȷω}] \\ X_{p}(e^{ȷω}) & = X_{p}^{∗}(e^{−ȷω}) \\ X_{i}(e^{ȷω}) & = −X_{i}^{∗}(e^{−ȷω}) \end{aligned}  (12)

FT_{DT} {x[−n]} = X(e^{−ȷω})  (13)

FT_{DT} {x^{∗}[n]} = X^{∗}(e^{−ȷω})  (14)

FT_{DT} \{nx[n]\} = ȷ\frac{dX(e^{ȷω})}{dω}  (15)

FT_{DT} {1} = ∑_{k=−∞}^{+∞} 2πδ(ω + 2πk)  (16)

FT_{DT} \{u[n]\} = \frac{1}{1 − e^{−ȷω}} + ∑_{k=−∞}^{+∞} πδ(ω + 2πk)

(17)

FT_{DT} \{(n + 1)a^{n}u[n]\} = \frac{1}{(1 − ae^{−ȷω})^{2}} (18)

FT_{DT} \{\begin{cases}1, & 0 ≤ n ≤ M \\ 0, & c.c. \end{cases} \} =\frac{sen(ω(M + 1)/2)}{sen(ω/2)}

× e^{−jωM/2} (19)

FT_{DT} \{\frac{sen(ω_{c}n)}{πn}\} = \begin{cases}1, & |ω| < ω_{c} \\ 0, & ω_{c} < ω ≤ π \end{cases} (20)

FT_{DT} {e^{ȷω_{0}n}} = ∑_{k=−∞}^{+∞} 2πδ(ω − ω_{0} + 2πk) (21)

4 Domínio Z

X(z) = ∑_{n=−∞}^{+∞} x[n]z^{−n}  (22)
x[n] = \frac{1}{2πȷ} ∮_{C} X(z)z^{n−1}dz  (23)
𝒵 {z_{0}^{n}x[n]} = X(z/z_{0})  (24)
𝒵 {x^{∗}[n]} = X^{∗}(z^{∗})  (25)
𝒵 \{nx[n]\} = −z \frac{d}{dz} X(z), ROC = R_{x} (26)
𝒵 \{a^{n}u[n]\} = \frac{1}{1 − az^{−1}} , |z| > |a|  (27)
𝒵 \{−a^{n}u(−n − 1)\} = \frac{1}{1 − az^{−1}} , |z| < |a| (28)
𝒵 {δ[n]} = 1  (29)
𝒵 \{u[n]\} = \frac{1}{1 − z^{−1}} , |z| > 1  (30)
𝒵 \{−u[−n − 1]\} = \frac{1}{s}, |z| < 1  (31)
𝒵 \{na^{n}u[n]\} = \frac{az^{−1}}{(1 − az^{−1})^{2}} , |z| > |a| (32)

𝒵 \{−na^{n}u[−n − 1]\} = \frac{az^{−1}}{(1 − az^{−1})^{2}} , |z| < |a|

(33)

𝒵 \{r^{n} cos(ω_{0}t)u(t)\} = \frac{1 − (r cos(ω_{0}))z^{−1}}{1 − (2r cos(ω_{0}))z^{−1} + r^{2}z^{−2}} ,

|z| > |r| (34)

𝒵 \{r^{n} sen(ω_{0}t)u(t)\} = \frac{(r cos(ω_{0}))z^{−1}}{1 − (2r cos(ω_{0}))z^{−1} + r^{2}z^{−2}} ,

|z| > |r| (35)

\begin{aligned}Res [X(z)z^{n−1} em z = z_{0}] & = \frac{1}{(s − 1)!} [\frac{d^{s−1}Ψ(z)}{dz^{s−1}}]_{z=z_{0}} \\ Ψ(z) & = [X(z)z^{n−1}] (z − z_{0})^{s} \end{aligned}

(36)