331 lines
11 KiB
Markdown
331 lines
11 KiB
Markdown
# Chapter 2 - Systems
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A *system*, $S$, takes in an input signal, $x(t)$, and outputs a signal, $y(t)$.
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$$x(t) \rightarrow \boxed{S} \rightarrow y(t)$$
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$$y(t) = S[x(t)]$$
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## Simple System Properties
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#### Linear Systems
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$S$ is said to be linear if it satisfies both hoogeneity and superposition.
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#### Homogeneity
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Scaling an input by a factor, $a$, scales the output by the same factor.
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$$ay(t) = S[ax(t)]$$
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#### Superposition
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Adding two inputs lead to the corresponding outputs being added.
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If
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$$y_1(t) = S[x_1(t)]$$
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$$y_2(t) = S[x_2(t)]$$
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Then
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$$y_1(t) + y_2(t) = S[x_1(t) + x_2(t)]$$
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#### Time Invariance
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$S$ is said to be time-invariant if delaying or advancing the input gives the same delay or advance in the output.
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$$y(t - \alpha) = S[x(t - \alpha)]$$
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#### Causality
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$S$ is said to be causal if the output does not depend on any future inputs. Theoretically, noncausal systems cannot be realized.
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#### Signal Boundedness
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$x(t)$ is said to be bounded if $\lvert x(t) \rvert \le B_x \lt \infty$
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Bounded example:
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$$x(t) = e^{-t} u(t)$$
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Unbounded example:
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$$x(t) = e^t u(t)$$
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#### Bounded-Input, Bounded-Output Stability
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$S$ is said to be bounded-input, bounded-output (BIBO) stable if every bounded input gives rise to a bounded output.
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Fact: The system is BIBO unstable if just one bounded input gives rise to an unbounded output.
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Every bounded $x(t)$ gives rise to a bounded $y(t)$.
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### Example 2.1
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$$y(t) = S[x(t)] = x^2(t)$$
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##### Homogeneity:
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$$S[ax(t)] = [ax(t)]^2 = a^2x^2(t)$$
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$$ay(t) = ax^2(t)$$
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Since $S[ax(t)] \ne ay(t)$, $S$ is not homogeneous, and therefore not linear.
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##### Time invariance:
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$$S[x(t-\alpha)] = [x(t-\alpha)]^2 = x^2(t-\alpha) =y(t-\alpha)$$
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Since $S[x(t-\alpha)] = y(t-\alpha)$, $S$ is causal
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##### BIBO stability:
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$$\lvert x(t) \rvert \le B_x \lt \infty$$
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$$\lvert y(t) \rvert = \lvert x^2(t) \rvert \le B_x^2 \lt \infty$$
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### Example 2.2
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$$y(t) = S[x(t)] = \cos(x(t))$$
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##### Homogeneity:
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$$S[ax(t)] = \cos(ax(t))$$
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$$ay(t) = a\cos(x(t))$$
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Since $S[ax(t)] \ne ay(t)$, $S$ is not homogeneous and therefore not linear.
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##### Time invariance:
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$$S[x(t-\alpha)] = \cos(x(t-\alpha)) = y(t-\alpha)$$
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##### BIBO stability:
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$$\lvert x(t) \rvert \le B_x \lt \infty$$
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$$\lvert y(t) \rvert \le 1$$
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### Example 2.3
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$$y(t) = S[x(t)] = \lvert x(t) \rvert$$
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##### Homogeneity:
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$$S[ax(t)] = \lvert ax(t) \rvert$$
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$$ay(t) = a \lvert x(t) \rvert$$
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Since $\lvert a \rvert \ne a$, $S$ is not homogeneous and therefore not linear.
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##### Time invariance:
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$$S[x(t-\alpha)] = \lvert x(t-\alpha)\rvert$$
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$$y(t-\alpha) = \lvert x(t-\alpha)\rvert$$
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Since $S[x(t-\alpha)] = y(t-\alpha)$, $S$ is time invariant.
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##### BIBO stability
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$$\lvert x(t) \rvert \le B_x \lt \infty$$
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$$\lvert y(t) \rvert = \lvert x(t) \rvert \le B_x \lt \infty$$
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### Example 2.4
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$$y(t) = S[x(t)] = mx(t) + b$$
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##### Homogeneity:
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$$S[ax(t)] = max(t) + b$$
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$$ay(t) = amx(t) + ab$$
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##### Superposition:
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$$y_1(t) = mx_1(t) + b$$
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$$y_2(t) = mx_2(t) + b$$
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$$y_1(t) + y_2(t) = m(x_1(t) + x_2(t)) + 2b$$
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For $b=0$, $S$ is linear, otherwise, neither homogeneity nor superposition are satisfied.
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##### Time invariance:
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$$S[x(t-\alpha)] = mx(t-\alpha) + b$$
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$$y(t-\alpha) = mx(t-\alpha) + b$$
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Since $y(t-\alpha) = x(t-\alpha)$, $S$ is time invariant.
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##### BIBO stability
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$$\lvert y(t) \rvert = \lvert mx(t) + b \rvert \le \lvert m \rvert \lvert x(t) \rvert + \lvert b \rvert \le \lvert m \rvert B_x \lt \infty$$
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### Example 2.5
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$$y(t) = S[x(t)] = tx(t+3)$$
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##### Homogeneity:
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$$S[ax(t)] = atx(t+3)$$
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$$ay(t) = atx(t+3)$$
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##### Superposition:
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$$y_1(t) = tx_1(t+3)$$
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$$y_2(t) = tx_2(t+3)$$
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$$S[x_1(t) + x_2(t)] = t[x_1(t+3) + x_2(t+3)]$$
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$$y_1 + y_2 = tx_1(t+3) + tx_2(t+3)$$
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Since both homogeneity and superposition are satisfied, $S$ is a linear system.
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##### Time invariance:
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$$S[x(t-\alpha)] = tx(t+3-\alpha)$$
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$$y(t-\alpha) = (t-\alpha)\cdot x (t+3-\alpha)$$
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Not time invariant.
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##### Causality:
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No
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##### BIBO stability:
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$$\lvert y(t) \rvert = \lvert tx(t+3) \rvert = \lvert t \rvert \lvert x(t+3)\rvert$$
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Since $\lvert t \rvert$ is not bounded, ther is no $B_y$ such that:
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$$\lvert y(t) \rvert \le B_y \lt \infty$$
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### Example 2.6 - Expansion
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$$y(t) = S[x(t)] = x({t \over 10})$$
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##### Homogeneity
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$$S[ax(t)] = ax({t \over 10}) = ay(t)$$
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##### Superposition
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$$y_1(t) = x_1({t \over 10})$$
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$$y_2(t) = x_2({t \over 10})$$
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$$S[x_1(t) + x_2(t)] = x_1({t \over 10}) + x_2({t \over 10})$$
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##### Time Invariance
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$$S[x(t - \alpha)] = x({t \over 10} - \alpha)$$
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$$y(t - \alpha) = x({t - \alpha \over 10})$$
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##### Causal
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$$y(-10) = x(-1)$$
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##### BIBO Stability
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Expansion affects input space, not output space
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### Example 2.7
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$$y(t) = S[x(t)] = {1 \over T} \int\limits_{t-T}^t x(\tau)d\tau + B$$
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##### Homogeneity
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$$S[ax(t)] = {1 \over T} \int\limits_{t-T}^t ax(\tau)d\tau + B$$
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$$ay(t) = {a \over T} \int\limits_{t-T}^t x(\tau)d\tau + B$$
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Not homogeneous unless $B = 0$.
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##### Superposition
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$$y_1(t) = {1 \over T} \int\limits_{t-T}^t x_1(\tau)d\tau + B$$
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$$y_2(t) = {1 \over T} \int\limits_{t-T}^t x_2(\tau)d\tau + B$$
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$$S[x_1(t) + x_2(t)] = {1 \over T} \int\limits_{t-T}^t [x_1(\tau) + x_2(\tau)]d\tau + B$$
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$$y_1(t) + y_2(t) = {1 \over T} \int\limits_{t-T}^t x_1(\tau)d\tau + {1 \over T} \int\limits_{t-T}^t x_2(\tau)d\tau + 2B$$
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##### Time Invariance
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$$S[x(t - \alpha)] = {1 \over T} \int\limits_{t-T}^t x(\tau - \alpha)d\tau + B$$
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Substitute $v = \tau - \alpha$:
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$$S[x(t - \alpha)] = {1 \over T} \int\limits_{t-T-\alpha}^{t-\alpha} x(v)dv + B$$
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$$y(t - \alpha) = {1 \over T} \int\limits_{t-T-\alpha}^{t-\alpha} x(\tau)d\tau + B$$
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##### Causality
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$y(t)$ depends on $x(t)$ which uses inputs from $t-T$ to $t$, and no future inputs. Therefore, the system is causal.
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If instead, inputs ranged from $t-T$ to $t+T$, the system would rely on future inputs and would not be causal.
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##### BIBO Stability
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$$\lvert x(t) \rvert \le B_x < \infty$$
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$$|y(t)| = \left| {1 \over T} \int\limits_{t-T}^t x(\tau)d\tau + B \right|$$
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$$|y(t)| \le \left| {1 \over T} \int\limits_{t-T}^t x(\tau)d\tau \right| + |B|$$
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$$|y(t)| \le {1 \over T} \int\limits_{t-T}^t |x(\tau)|d\tau + |B|$$
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$$|y(t)| \le {1 \over T} \int\limits_{t-T}^t B_x d\tau + |B|$$
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$$|y(t)| \le {1 \over T} \left[\tau B_x\Big|_{t-T}^{t}\right] + |B|$$
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$$|y(t)| \le {1 \over T} T B_x + |B|$$
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$$|y(t)| \le B_x + |B|$$
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### Example 2.8 - AM
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$$y(t) = S[m(t)] = m(t)\cos(\omega_c t)$$
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##### Homogeneity
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$$S[am(t)] = am(t)\cos(\omega_c t) ay(t)$$
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##### Superposition
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$$y_1(t) = m_1(t)\cos(\omega_c t)$$
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$$y_2(t) = m_2(t)\cos(\omega_c t)$$
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$$S[m_1(t) + m_2(t)] = [m_1(t) + m_2(t)]\cos(\omega_c t)$$
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$$y_1(t) + y_2(t) = [m_1(t) + m_2(t)]\cos(\omega_c t)$$
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##### Time Invariance
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$$S[m(t-\alpha)] = m(t-\alpha)\cos(\omega_c t)$$
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$$y(t - \alpha) = m(t-\alpha)\cos(\omega_c (t -\alpha))$$
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##### Causality
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Yes
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##### BIBO Stability
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$$|m(t)| \le B_x$$
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$$|y(t)| = |m(t)\cos(\omega_c t)| \le |m(t)| \le B_x$$
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### Example 2.9 - FM
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$$y(t) = S[m(t)] = \cos\left(\omega_c t + \int\limits_{-\infty}^{t} m(\tau) d\tau \right)$$
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##### Homogeneity
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$$S[am(t)] = \cos\left(\omega_c t + \int\limits_{-\infty}^{t} am(\tau) d\tau \right)$$
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$$ay(t) = a\cos\left(\omega_c t + \int\limits_{-\infty}^{t} m(\tau) d\tau \right)$$
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##### Time Invariance
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$$S[m(t-\alpha)] = \cos\left(\omega_c t + \int\limits_{-\infty}^{t} m(\tau - \alpha) d\tau \right)$$
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$$y(t-\alpha) = \cos\left(\omega_c (t - \alpha) + \int\limits_{-\infty}^{t} m(\tau) d\tau \right)$$
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##### BIBO Stability
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$$|y(t)| \le 1$$
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## Linear, Time-Invariant Systems
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The implulse response $h(t)$ of an LTI system is the output of the system when
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$$\delta(t) \rightarrow \boxed{\text{LTI}} \rightarrow h(t)$$
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Causal if $h(t) = 0$ for $t<0$.
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BIBO stable if:
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$$\int |h(t)|dt < \infty$$
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If $h(t)$ has finite supports, it is *always* BIBO stable. If $h(t)$ has infinite support, BIBO stability must be checked.
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A linear, constant coefficient ODE with input $x(t)$ and output $y(t)$ is LTI under zero initial conditions and when $x(t)$ is a causal input.
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### Circuit Application
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An RLC circuit with no initial voltage across the capacitor and no initial current through the inductor is LTI.
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#### Without Initial Condition
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For a capacitor with no initial voltage:
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$$V(t) = S[I(t)] = {1\over C} \int\limits_0^t I(\tau) d\tau$$
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##### Homogeneity
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$$S[aI(t)] = {1\over C}\int\limits_0^t aI(\tau) d\tau = {a\over C}\int\limits_0^t I(\tau) d\tau = aV(t)$$
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##### Superposition
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$$V_1(t) = {1\over C}\int\limits_0^t I_1(\tau) d\tau$$
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$$V_2(t) = {1\over C}\int\limits_0^t I_2(\tau) d\tau$$
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$$S[I_1(t) + I_2(t)] = {1\over C} \int\limits_0^t [I_1(\tau) + I_2(\tau)]d\tau = V_1(t) + V_2(t)$$
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##### Time Invariance
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$$S[I(t - \alpha)] = {1\over C}\int\limits_{-\alpha}^{t-\alpha} I(\tau - \alpha) d\tau$$
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$$S[I(t - \alpha)] = {1\over C}\int\limits_{-\alpha}^{t-\alpha} I(p) dp$$
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If $I(t)$ is causal:
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$$S[I(t - \alpha)] = {1\over C}\int\limits_{0}^{t-\alpha} I(p) dp$$
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$$V(t - \alpha) = {1\over C}\int\limits_0^{t-\alpha} I(\tau) d\tau$$
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#### With Initial Condition
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For a capacitor with initial voltage:
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$$V(t) = {1\over C} \int\limits_0^t I(\tau) d\tau + V_0$$
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##### Homogeneity
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$$S[aI(t)] = {1\over C}\int\limits_0^t aI(\tau) d\tau + V_0$$
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$$aV(t) = {a\over C}\int\limits_0^t I(\tau) d\tau + aV_0$$
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#### Impulse Response
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$$i(t) = \delta (t), h(t) = V(t) = ?$$
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$$h(t) = {1 \over C} \int\limits_0^t \delta (\tau) d\tau =
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\begin{cases}
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{1\over C} & t \ge 0 \\
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0 & t < 0
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\end{cases}$$
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$$h(t) = {1 \over C} u(t)$$
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### Averager
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$$y(t) = S[x(t)] = {1 \over T} \int\limits_{t-T}^t x(\tau) d\tau$$
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#### Impulse response
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$$h(t) = {1 \over T} \int\limits_{t-T}^t \delta(\tau) d\tau$$
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3 possibilities for impulse response:
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1. $0 < T - t$ means that the impulse is to the left of the limits of integration, and $h(t)$ is 0.
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1. $T - t \le 0 \le t$ means that the impulse is within the bounds of integration, and $h(t)$ takes on a non-zero value.
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1. $t > 0$ means that the impulse is to the right of the bounds of integration, nad $h(t)$ is 0.
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For the second case, $T - t \le 0 \le t$, gives the response:
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$$h(t) = \begin{cases}
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{1 \over T} & 0 \le t \le T \\
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0 & \text{elsewhere}
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\end{cases} = {1\over T} \left[u(t) - u(t - T)\right]$$
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Since $h(t)$ is 0 for $t < 0$, it is a causal signal.
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## Convolution
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$$y(t) = x(t)*h(t) = \int x(\tau) h(t - \tau)d\tau$$
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##### Commutative:
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$$x(t) * h(t) = \int h(\tau) x(t-\tau)d\tau = h(t) * x(t)$$
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#### Case 1:
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Two finite support wich have the same region of support (domain).
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$$x(t) = h(t)$$
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##### Step 1:
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Decide to manipulate $x$ or $h$.
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$$h(t - \tau) = h(-\tau + t)$$
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Since we are looking at $\tau$ as the variable, there is a reflection and an advance by $t$.
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As $h(t-\tau)$ slides along the real number line, first there will be no overlap between $x(\tau)$ and $h(t - \tau)$. They multiply to 0, so $y(t) = 0$. Then, there will be some overlap, so $y(t)$ becomes:
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$$y(t) = \int\limits_0^t d\tau = t$$
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$h(t-\tau)$ continues to slide and soon there is full overlap, so $y(t) = 1$.
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