LTI Systems
LTI Systems
Linear, Time-Invariant => LTI systems are fundamental to signal analysis because it allows us to break down a signal into simpler pieces (superposition). Then we analyze the simpler signals.
For example,
Impulse Response
A signal can be considered as a sum of impulses times some coefficient per impulse.
Given an LTI system, we may be interested in see how it behaves given an impulse \(\delta (t)\). The corresponding output \(h(t)\) is the impulse response. If we have the impulse response, we can calculate the output signal without knowing the LTI system itself.
In discrete systems (LTID):
\[x[n] = \sum x[k] \delta [n-k]\] \[y[n] = \sum_{k=-\infty}^{\infty} x[k]\ h[n-k] = x[n] * h[n]\]
In continuous systems (LTIC):
\[x(t) = \int_{-\infty}^{\infty} x(\tau)\ \delta(t - \tau)\ d\tau\] \[y(t) = \int_{-\infty}^{\infty} x(\tau)\ h(t - \tau)\ d\tau = x(t) * h(t)\]
Where
- \(x[n],\ x(t)\) is input signal (linear combination of delayed impulses)
- \(y[n],\ y(t)\) is output signal
- \(h[n],\ h(t)\) is impulse response of the LTI system
- \(k,\ \tau\) is time delays
Given the system is linear and time-invariant, the impulse inputs and outputs behave linearly:
\[\delta (t-t_0) \Longrightarrow h(t-t_0)\] \[a\delta (t-t_0) \Longrightarrow ah(t-t_0)\]Example
Given:
\[y(t) = x(t-1) + 2x(t-3)\]Find:
- Impulse response
Given:
- Impulse response \(h(t) = e^{-3t}u(t)\)
- Input signal \(x(t) = \delta(t+1) + 3\delta(t-2) + 2\delta(t-6)\)
Find:
- Output \(y(t)\)
Convolution Properties
- \(x(t) * h(t) = h(t) * x(t)\) (Commutative)
- \(x(t) * (h_1(t) + h_2(t)) = x(t) * h_1(t) + x(t) * h_2(t)\) (Distributive)
- \((x(t) * h_1(t)) * h_2(t) = x(t) * (h_1(t) * h_2(t))\) (Associative)
- Given \(x(t) * h(t) = y(t)\), then \(x(t - T_1) * h(t - T_2) = y(t - T_1 - T_2)\) (Shifting)
- \(x(t) * \delta(t- t_0) = x(t - t_0)\) (Unit impulse shifting)
These apply for both continuous and discrete systems.
LTI System Properties
- Memoryless if \(h(t) = 0\) when \(t \neq 0\)
- Causal if \(h(t) = 0\) when \(t < 0\)
- Stable if \(\int_{-\infty}^{\infty} \mid h(t) \mid dt < \infty\)