Fundamental Signals

Exponential Signals

Comparison of the CT exponential signal and DT exponential signal

$\exp (j{\omega }_{0}t)$	$\exp (j{\omega }_{0}n)$
Distinct signals for distinct values of ${\omega }_0$	Identical signals for values of ${\omega }_0$ separated by multiplies of $2\pi$
Periodic for any choice of ${\omega }_0$	Periodic only if ${\omega }_0=\frac{2\pi m}{N}$ for some integers $N>0$ and $m$
Fundamental frequency ${\omega }_0$	Fundamental frequency $\frac{{\omega }_0}{m}$
Fundamental period $\{\begin{array}{ll}\text{undefined},& {\omega }_0=0\\ \frac{2\pi }{{\omega }_0},& {\omega }_0\ne 0\end{array}$	Fundamental period $\{\begin{array}{ll}\text{undefined},& {\omega }_0=0\\ m\frac{2\pi }{{\omega }_0},& {\omega }_0\ne 0\end{array}$

Unit Impulse Signal and Unit Step Signal

Unit Impulse

$$\delta [n]=\{\begin{array}{ll}0,& n\ne 0\\ 1,& n=0\end{array},\delta (t)=\{\begin{array}{ll}0,& t\ne 0\\ +\infty ,& t=0\end{array},\text{where}{\int }_{-\infty }^{\infty }\delta (t)\mathrm{d}t=1$$

Properties

$$f(t)\delta (t)=f(0)\delta (t)$$

since $\delta (t)=0$ for $t\ne 0$ 2.

$${\int }_{-\infty }^{\infty }f(t)\delta (t)\mathrm{d}t=f(0)$$

since ${\int }_{-\infty }^{\infty }f(t)\delta (t)\mathrm{d}t={\int }_{-\infty }^{\infty }f(0)\delta (t)\mathrm{d}t$ 3.

$$\delta (\alpha t)=\frac{1}{\left|\alpha \right|}\delta (t)$$

since ${\int }_{-\infty }^{\infty }f(t)\delta (\alpha t)\mathrm{d}t=\frac{1}{\alpha }{\int }_{-\infty }^{\infty }f\left(\frac{\alpha t}{\alpha }\right)\delta (\alpha t)\mathrm{d}(\alpha t)=\frac{1}{\left|\alpha \right|}f(0)$, where we take the absolute value $\left|\alpha \right|$ because the Dirac delta function must remain positive regardless of the sign of $\alpha$. 4.

$$\left[\frac{\mathrm{d}}{\mathrm{d}x}\delta (x-a)\right]f(x)=-\delta (x-a)f^{\prime }(x)$$

This is a symbolical result from

$$\int \left[\frac{\mathrm{d}}{\mathrm{d}x}{\delta }_i(x)\right]f(x)\mathrm{d}x=-\int {\delta }_i(x)\left[\frac{\mathrm{d}}{\mathrm{d}x}f(x)\right]\mathrm{d}x$$

where ${\delta }_i$ is a family of functions so that ${\lim }_{i\to \infty }\int {\delta }_i(x)f(x)\mathrm{d}x=f(a)$ and we can take derivatives at both sides 5.

$$f(t)\ast \delta (t-t_0)=f(t-t_0)$$

where $\ast$ stands for convolution. This is because

$$f(t)\ast \delta (t-t_0)={\int }_{-\infty }^{\infty }f(t-\tau )\delta (\tau -t_0)\mathrm{d}\tau =f(t-t_0)$$

$$f(t)\ast {\delta }^{\prime }(t)=f^{\prime }(t)$$

Unit Step

$$u[n]=\{\begin{array}{ll}0,& n<0\\ 1,& n\ge 0\end{array},u(t)=\{\begin{array}{ll}0,& t<0\\ 1,& t\ge 0\end{array}$$

The convolution of a signal with a unit step is the integration or summation of the signal to the current time

$$\begin{array}{rl}x[n]\ast u[n]& =u[n]\ast x[n]=\sum _{k=-\infty }^{n}x[k]\\ x(t)\ast u(t)& =u(t)\ast x(t)={\int }_{-\infty }^{t}x(\tau )\mathrm{d}\tau \end{array}$$

Similarly, if we define

$$u_{-2}(t)=u(t)\ast u(t)$$

which is a slope function, and we can have

$$u_{-2}(t)\ast x(t)={\int }_{-\infty }^t{\int }_{-\infty }^{\tau }x(\sigma )\mathrm{d}\sigma \mathrm{d}\tau$$

Relationship

$$\delta [n]=u[n]-u[n-1],u[n]=\sum _{m=-\infty }^n\delta [m]$$

and

$$u(t)={\int }_{-\infty }^t\delta (\tau )\mathrm{d}\tau$$

or equivalently

$$u(t)={\int }_{-\infty }^t\delta (\tau )\mathrm{d}\tau \stackrel{\sigma =t-\tau }{=}{\int }_{+\infty }^0\delta (t-\sigma )\mathrm{d}(-\sigma )={\int }_0^{+\infty }\delta (t-\sigma )\mathrm{d}\sigma$$

The interpretation of this equality can be related to convolution. We can rewrite this as

$$u(t)={\int }_0^{\infty }\delta (t-\tau )\mathrm{d}\tau ={\int }_{-\infty }^{\infty }\delta (t-\tau )u(\tau )\mathrm{d}\tau =u(t)\ast \delta (t)$$

which is a natural since $\delta (t)$ serves as a unit element in convolution algebra