These notes DON’T contain the part about circuit as I know nothing about it.

EE120 Signals and Systems Lecture 1 #

Signals #

Signals are functions of one or more variables. A continuous-time signal, denoted $x(t)$, depends on a real-value time variable $t$, and a discrete-time signal, $x[n]$, depends on the integer-valued variable $n$ that indexes instants of time.

Two special discrete-time signals:

• Unit Impulse: $\delta[n] = \begin{cases}1, if\ n = 0\ 0, if\ n \not = 0\end{cases}$
• Unit Step: $u[n] = \begin{cases}1, if\ n \geq 0\ 0, if\ n < 0\end{cases}$

Note that $\delta[n] = u[n] - u[n - 1]$ and $u[n] = \sum_{k = 0}^{\infty}\delta[n - k]$ .

Systems #

For the purposes of this course, a system is defined as a process by which input signals are transformed to output signals. Inputs are typically denoted as $x$ and outputs as $y$.

Example: A moving average filter: $y[n] = \frac{1}{3}(x[n - 1] + x[n] + x[n + 1])$

Memory #

A system is called memoryless if its output at a given time depends on the input only at that time.

Example: The moving average filter is not a memoryless system since $y[n]$ depends on $x[n - 1]$ and $x[n + 1]$ , which means that the system would require memory registers to implement.

Causality #

A system is called causal if its output depends on the input at present and past times only, not on future times.

Example: The moving average filter is not a causal system since $y[n]$ depends on the future input $x[n + 1]$ .

Stability #

A system is called stable if all bounded inputs generate bounded outputs. It is unstable if there exists a bounded input for which the output grows unbounded.

Example: The moving average filter is stable, since the output is simply an average of input values and remains bounded when the input is bounded.

Example: The “accumulator” system, defined by $$ y[n] - y[n - 1] = x[n], y[-1] = 0 $$

when $x[n] = 0$ for $n < 0$ has the solution $$ y[n] = x[0] + x[1] + … + x[n], n \geq 0 \tag {*} $$ which can grow unbounded with bounded inputs, such as the unit step input which gives $y[n] = n + 1$ . The continuous-time analogue of the accumulator is the integrator: $$ \frac{dy(t)}{dt} = x(t), y(0) = 0 $$ which is likewise unstable.

Linearity #

A system is called linear if it satisfies these two conditions:

• Scaling: For any input-output pair $x(t) → y(t)$ and constant $a$, we have: $$ ax(t) → ay(t) $$
• Superposition: For any two input-output pairs $x_1(t) → y_1(t)$, $x_2(t) → y_2(t)$ , we have: $$ x_1(t) + x_2(t) → y_1(t) + y_2(t) $$

Corollary: If the input to a linear system is $0$, the output must be $0$ .

• Using the scaling property and choose $a = 0$ .

Example: The moving average filter is linear.

Example: The median filter: $$ y[n] = med{x[n - 1], x[n], x[n + 1]} $$ is nonlinear.

Time-Invariance #

A system is called time-invariant if a time shift in the input results is an identical time shift in the output: $$ x(t - T) \rightarrow y(t - T) $$

for any input-output pair and any amount of shift $T$. (The system’s parameters may depend on the time $t$ ). The moving average and median filters above are both time-invariant because the rule for generating the output (take the average or median of previous, current, and next input) does not change with time. On the other hand, the system $y[n]$ = $nx[n]$ is time-varying, as it applies a different rule at each time (multiply the input with time n).

Linear Time-Invariant (LTI) Systems and Convolution #

Systems that are both linear and time-invariant, referred to as LTI systems, are amenable to powerful analysis tools that we will study in this course. In particular they have the remarkable property that knowing the response to a unit impulse input is enough to predict the response to any other input. To see this in discrete-time, let $h[n]$ denote the response of a LTI system to the unit impulse $δ[n]$ . Rewrite the input $x[n]$ as $$ x[n] = … + x[-1]δ[n + 1] + x[0]δ[n] + x[1]δ[n - 1] + … $$

Since $δ[n]\rightarrow h[n]$ , by time-invariance: $δ[n - k]\rightarrow h[n - k], \forall k$ . Then, by linearity: $$ y[n] = … + x[-1]h[x + 1] + x[0]h[n] + x[1]h[n - 1] + … $$

Thus, for input $x[n]$ , the output is : $$ y[n] = \sum_{k=-\infty}^{\infty}x[k]h[n - k] $$

The operation on the right-hand side of the formula above is called the “convolution” of signals $x$ and $h$ , denoted $x * h$ : $$ (x * h)[n] := \sum_{k = -\infty}^{\infty}x[k]h[n - k] $$

Example: For the accumulator system the impulse response is the unit step, $u[n]$ . By using the convolution, we can acquire the solution (*).