Signals

• Energy signal: a signal whose total energy is finite and non-zero, $0 < E < \infty$. Its average power is then zero. Example: a single pulse, $e^{-2t}u(t)$.
• Power signal: a signal whose average power is finite and non-zero, $0 < P < \infty$. Its total energy is then infinite. Example: $\cos\omega_0 t$, $u(t)$, any periodic signal.

For a continuous-time signal:

$$E = \lim_{T\to\infty}\int_{-T}^{T}|x(t)|^2\,dt, \qquad P = \lim_{T\to\infty}\frac{1}{2T}\int_{-T}^{T}|x(t)|^2\,dt$$

For a discrete-time signal:

$$E = \sum_{n=-\infty}^{\infty}|x[n]|^2, \qquad P = \lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^{N}|x[n]|^2$$

A signal that has neither finite energy nor finite non-zero power is neither energy nor power signal (e.g. $e^{t}$, $t\,u(t)$).

Checking $x(t) = e^{-3t}$ (defined for all $t$)

$|x(t)|^2 = e^{-6t}$.

Energy:

$$\begin{aligned} E &= \lim_{T\to\infty}\int_{-T}^{T} e^{-6t}\,dt = \lim_{T\to\infty}\frac{e^{6T}-e^{-6T}}{6} = \infty \end{aligned}$$

Power:

$$\begin{aligned} P &= \lim_{T\to\infty}\frac{1}{2T}\cdot\frac{e^{6T}-e^{-6T}}{6} = \lim_{T\to\infty}\frac{e^{6T}}{12T} = \infty \end{aligned}$$

(The exponential grows faster than $T$, so the limit is infinite.)

Since both $E$ and $P$ are infinite, $x(t) = e^{-3t}$ is neither an energy signal nor a power signal. The signal blows up as $t \to -\infty$.

Note: if the signal is taken as one-sided, $x(t) = e^{-3t}u(t)$, then

$$E = \int_0^{\infty} e^{-6t}\,dt = \frac{1}{6}\ \text{J}, \qquad P = 0$$

and it is an energy signal.

Answer: $e^{-3t}$ for all $t$: neither ($E = \infty$, $P = \infty$); $e^{-3t}u(t)$: energy signal with $E = 1/6$.

A discrete-time signal is periodic with period $N$ (a positive integer) if $x[n+N] = x[n]$ for all $n$.

For $x[n] = e^{j\omega_0 n}$:

$$\begin{aligned} x[n+N] &= e^{j\omega_0 (n+N)} = e^{j\omega_0 n}\,e^{j\omega_0 N} \end{aligned}$$

For $x[n+N] = x[n]$ we need

$$e^{j\omega_0 N} = 1 \;\Rightarrow\; \omega_0 N = 2\pi m, \quad m \text{ an integer}$$ $$\frac{\omega_0}{2\pi} = \frac{m}{N} = \text{rational number}$$

So $e^{j\omega_0 n}$ is periodic only if $\omega_0/2\pi$ is a rational number. The fundamental period is the smallest positive integer $N = 2\pi m/\omega_0$, taking $m$ as the smallest integer that makes $N$ an integer (with $m$ and $N$ having no common factor). The same condition holds for $\cos(\omega_0 n+\theta)$ and $\sin(\omega_0 n+\theta)$, because they are sums of such exponentials.

This differs from continuous time, where $e^{j\omega_0 t}$ is periodic for every $\omega_0 \neq 0$ (period $2\pi/|\omega_0|$). In discrete time $n$ takes only integer values, so $N$ must be an integer.

Example: $e^{j(3\pi/4)n}$: $\omega_0/2\pi = 3/8$ (rational), so periodic with $N = 8$. But $e^{j2n}$: $\omega_0/2\pi = 1/\pi$ (irrational), so it is not periodic.

A signal is a function of one or more independent variables (usually time) that carries information, e.g. speech, ECG, temperature readings.

• Continuous-time (CT) signal: defined for every value of time $t$ in an interval. The independent variable is continuous and written in round brackets, $x(t)$. Examples: speech voltage from a microphone, $x(t) = 5\sin(100\pi t)$, room temperature varying with time.
• Discrete-time (DT) signal: defined only at discrete instants of time, $t = nT_s$, where $n$ is an integer. It is a sequence of numbers written $x[n]$. Examples: daily closing share price, samples of speech stored in a computer, $x[n] = (0.5)^n u[n]$.

 CT signal x(t)              DT signal x[n]
   |   .--.                    |   o
   |  /    \                   |   |  o
   | /      \      .           | o |  |  o
   |/        \    /            | | |  |  |
 --+----------\--/---> t     --+-+-+--+--+----> n
                '--             -1 0  1  2

A DT signal is often obtained by sampling a CT signal: $x[n] = x(nT_s)$.

• Energy signal: a signal whose total energy is finite and non-zero, $0 < E < \infty$. Its average power is then zero. Example: a single pulse, $e^{-2t}u(t)$.
• Power signal: a signal whose average power is finite and non-zero, $0 < P < \infty$. Its total energy is then infinite. Example: $\cos\omega_0 t$, $u(t)$, any periodic signal.

For a continuous-time signal:

$$E = \lim_{T\to\infty}\int_{-T}^{T}|x(t)|^2\,dt, \qquad P = \lim_{T\to\infty}\frac{1}{2T}\int_{-T}^{T}|x(t)|^2\,dt$$

For a discrete-time signal:

$$E = \sum_{n=-\infty}^{\infty}|x[n]|^2, \qquad P = \lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^{N}|x[n]|^2$$

A signal that has neither finite energy nor finite non-zero power is neither energy nor power signal (e.g. $e^{t}$, $t\,u(t)$).

Examples

1. Energy signal: $x(t) = e^{-2t}u(t)$

$$E = \int_0^\infty e^{-4t}\,dt = \frac{1}{4}, \qquad P = 0$$

2. Power signal: $x(t) = A\cos\omega_0 t$

$$P = \frac{1}{T_0}\int_0^{T_0} A^2\cos^2\omega_0 t\,dt = \frac{A^2}{2}, \qquad E = \infty$$

3. DT energy signal: $x[n] = (0.5)^n u[n]$, $E = \sum_{n=0}^\infty 0.25^n = \frac{1}{1-0.25} = \frac{4}{3}$.

4. DT power signal: $x[n] = u[n]$, $P = \lim_{N\to\infty}\frac{N+1}{2N+1} = \frac{1}{2}$.

In general, time-limited (finite-duration) bounded signals are energy signals, and periodic signals are power signals.

These are basic operations on the independent variable $t$.

Time shifting

$y(t) = x(t - t_0)$.

• $t_0 > 0$: the signal is delayed (shifted right) by $t_0$.
• $t_0 < 0$: the signal is advanced (shifted left).

Example: if $x(t)$ is a pulse from 0 to 1, then $x(t-2)$ is the same pulse from 2 to 3. A radar echo is a delayed copy of the transmitted signal.

Time scaling

$y(t) = x(at)$, $a > 0$.

• $a > 1$: signal is compressed in time (plays faster).
• $0 < a < 1$: signal is expanded (plays slower).

Example: for the pulse $x(t)$ on $0 \le t \le 2$, $x(2t)$ lies on $0 \le t \le 1$ and $x(t/2)$ on $0 \le t \le 4$. Playing an audio tape at double speed gives $x(2t)$.

Time inversion (reflection/folding)

$y(t) = x(-t)$: the signal is mirrored about $t = 0$. A value at $t = 2$ in $x(t)$ appears at $t = -2$ in $x(-t)$. Example: playing a recording backwards.

 x(t)            x(t-2)          x(2t)          x(-t)
 1 +--+          1    +--+      1 +-+       1 +--+
   |  |               |  |        | |         |  |
 --+--+---t    --+----+--+-t  ----+-+---t  ---+--+-+--t
   0  1          0    2  3        0 .5       -1  0

Combined operation $x(at - b)$: first shift by $b$ to get $x(t-b)$, then scale by $a$ (replace $t$ by $at$). Example: $x(2t-2)$ for the pulse on $[0,1]$ lies on $1 \le t \le 1.5$.

A DT sinusoid $\sin(\omega_0 n+\theta)$ is periodic if $\omega_0/2\pi$ is rational; its period is the smallest integer $N = 2\pi m/\omega_0$. A sum is periodic if each term is periodic, and its period is the LCM of the individual periods.

Term 1: $5\sin\left(\frac{3\pi}{8}n - \frac{\pi}{2}\right)$, $\omega_1 = \frac{3\pi}{8}$

$$\frac{\omega_1}{2\pi} = \frac{3}{16}\ (\text{rational}) \;\Rightarrow\; N_1 = \frac{2\pi m}{3\pi/8} = \frac{16m}{3} \;\Rightarrow\; N_1 = 16\ (m = 3)$$

Term 2: $2\cos\left(\frac{7\pi}{12}n\right)$, $\omega_2 = \frac{7\pi}{12}$

$$\frac{\omega_2}{2\pi} = \frac{7}{24}\ (\text{rational}) \;\Rightarrow\; N_2 = \frac{24m}{7} \;\Rightarrow\; N_2 = 24\ (m = 7)$$

Both terms are periodic, so $x[n]$ is periodic with

$$N = \text{LCM}(16, 24) = 48$$

The phase $-\pi/2$ does not affect the period.

Fundamental frequency:

$$\omega_0 = \frac{2\pi}{N} = \frac{2\pi}{48} = \frac{\pi}{24}\ \text{rad/sample}, \qquad f_0 = \frac{1}{48}\ \text{cycles/sample}$$

Answer: $x[n]$ is periodic; $N = 48$ samples, $\omega_0 = \pi/24$ rad/sample ($f_0 = 1/48$).

• Even signal: symmetric about the vertical axis, $x(-t) = x(t)$ (DT: $x[-n] = x[n]$). Examples: $\cos\omega t$, $t^2$, $e^{-|t|}$.
• Odd signal: antisymmetric about the origin, $x(-t) = -x(t)$. An odd signal is always zero at $t = 0$. Examples: $\sin\omega t$, $t$, $t^3$.

Even/odd decomposition

Let any signal be written as the sum of an even part and an odd part:

$$x(t) = x_e(t) + x_o(t) \qquad (1)$$

where $x_e(-t) = x_e(t)$ and $x_o(-t) = -x_o(t)$. Replace $t$ by $-t$:

$$x(-t) = x_e(-t) + x_o(-t) = x_e(t) - x_o(t) \qquad (2)$$

Adding (1) and (2), and subtracting (2) from (1):

$$\begin{aligned} x_e(t) &= \frac{1}{2}\left[x(t) + x(-t)\right] \\ x_o(t) &= \frac{1}{2}\left[x(t) - x(-t)\right] \end{aligned}$$

These parts always exist for any $x(t)$, so every signal can be decomposed into even and odd components. The same holds for DT signals: $x_e[n] = \frac{1}{2}(x[n]+x[-n])$, $x_o[n] = \frac{1}{2}(x[n]-x[-n])$.

Example

$x(t) = e^{t}$:

$$x_e(t) = \frac{e^{t}+e^{-t}}{2} = \cosh t, \qquad x_o(t) = \frac{e^{t}-e^{-t}}{2} = \sinh t$$

Check: $\cosh t + \sinh t = e^{t}$.

Example 2: $x(t) = u(t)$ gives $x_e(t) = \frac{1}{2}$ for all $t \neq 0$ and $x_o(t) = \frac{1}{2}\,\text{sgn}(t)$.

Useful properties: even × even = even, odd × odd = even, even × odd = odd; the integral of an odd signal over $[-T, T]$ is zero.

Continuous time

Unit step $u(t)$:

$$u(t) = \begin{cases} 1, & t > 0 \\ 0, & t < 0 \end{cases}$$

It is discontinuous at $t = 0$ (value there is undefined or taken as 1/2). It is used to switch signals on at $t = 0$.

Unit impulse (Dirac delta) $\delta(t)$: zero everywhere except at $t = 0$, with unit area:

$$\delta(t) = 0 \text{ for } t \neq 0, \qquad \int_{-\infty}^{\infty}\delta(t)\,dt = 1$$

It is the limit of a rectangular pulse of width $\Delta$ and height $1/\Delta$ as $\Delta \to 0$. Sifting property: $\int x(t)\delta(t-t_0)\,dt = x(t_0)$.

Discrete time

$$u[n] = \begin{cases} 1, & n \ge 0 \\ 0, & n < 0 \end{cases} \qquad \delta[n] = \begin{cases} 1, & n = 0 \\ 0, & n \neq 0 \end{cases}$$

$\delta[n]$ is a simple sequence with value exactly 1 at $n = 0$ (no limiting process needed). Any sequence can be written as $x[n] = \sum_k x[k]\delta[n-k]$.

 u(t)              delta(t)
 1 +---------      ^ area = 1
   |               |
 --+--------> t  --+--------> t
   0               0

 u[n]
  1         o  o  o  o ...
            |  |  |  |
 ---o--o----+--+--+--+---> n
   -2 -1    0  1  2  3

 delta[n]
  1   o
      |
 --o--+--o--> n
  -1  0  1

Relationship

Continuous time:

$$u(t) = \int_{-\infty}^{t}\delta(\tau)\,d\tau, \qquad \delta(t) = \frac{du(t)}{dt}$$

The step is the running integral of the impulse; the impulse is the derivative of the step (the jump of height 1 at $t=0$).

Discrete time:

$$u[n] = \sum_{k=-\infty}^{n}\delta[k] = \sum_{k=0}^{\infty}\delta[n-k], \qquad \delta[n] = u[n] - u[n-1]$$

The step is the running sum of the impulse; the impulse is the first difference of the step.

Signals are classified by their energy and average power.

• Energy signal: total energy is finite, $0 < E < \infty$, so average power $P = 0$. Usually time-limited or decaying signals. Example: $e^{-at}u(t)$ ($a>0$), $E = 1/2a$.
• Power signal: average power is finite and non-zero, $0 < P < \infty$, so $E = \infty$. Usually periodic or everlasting signals. Example: $A\cos\omega t$, $P = A^2/2$; $u(t)$, $P = 1/2$.

$$E = \int_{-\infty}^{\infty}|x(t)|^2\,dt, \qquad P = \lim_{T\to\infty}\frac{1}{2T}\int_{-T}^{T}|x(t)|^2\,dt$$

(For DT signals the integrals become sums over $n$ and $2T$ becomes $2N+1$.)

A signal cannot be both. Some signals are neither, e.g. $e^{2t}$ or the ramp $t\,u(t)$, where both $E$ and $P$ are infinite.

A discrete-time complex exponential is

$$x[n] = C\alpha^{n}$$

where $C$ and $\alpha$ are in general complex numbers. (With $\alpha = e^{\beta}$ it is also written $Ce^{\beta n}$.) Its behaviour depends on $C$ and $\alpha$:

1. Real exponential ($C$, $\alpha$ real)

• $\alpha > 1$: grows exponentially.
• $0 < \alpha < 1$: decays exponentially.
• $-1 < \alpha < 0$: alternates in sign and decays.
• $\alpha < -1$: alternates in sign and grows.
• $\alpha = 1$: constant; $\alpha = -1$: alternates between $+C$ and $-C$.

2. Purely imaginary exponent ($|\alpha| = 1$), sinusoidal

With $\alpha = e^{j\omega_0}$: $x[n] = Ce^{j\omega_0 n} = C(\cos\omega_0 n + j\sin\omega_0 n)$. The magnitude is constant; real and imaginary parts are sampled sinusoids. It is periodic only if $\omega_0/2\pi$ is rational, and frequencies $\omega_0$ and $\omega_0 + 2\pi$ give identical signals.

3. General complex exponential

With $C = |C|e^{j\theta}$ and $\alpha = |\alpha|e^{j\omega_0}$:

$$x[n] = |C||\alpha|^n\cos(\omega_0 n+\theta) + j|C||\alpha|^n\sin(\omega_0 n+\theta)$$

• $|\alpha| = 1$: constant-amplitude sinusoid.
• $|\alpha| < 1$: sinusoid with decaying envelope.
• $|\alpha| > 1$: sinusoid with growing envelope.

 0<a<1 decay        -1<a<0 alternating decay
 o                    o
 | o                  |   o
 | | o o .            | . | . .
-+-+-+-+-+-> n     ---+-+-+-+-+-> n
                        o

• Periodic signal: repeats itself after a fixed interval. CT: $x(t+T) = x(t)$ for all $t$; DT: $x[n+N] = x[n]$ for all $n$, $N$ a positive integer. The smallest such $T$ or $N$ is the fundamental period. Examples: $\sin t$, $\cos(\pi n/4)$.
• Aperiodic signal: does not repeat for any finite period. Examples: $e^{-t}u(t)$, $\delta[n]$, $\cos(2n)$.

Condition for $x[n] = \cos(2\pi f_0 n + \theta)$ to be periodic

For period $N$:

$$\begin{aligned} x[n+N] &= \cos(2\pi f_0 n + 2\pi f_0 N + \theta) \end{aligned}$$

This equals $x[n]$ for all $n$ only if the extra angle is a multiple of $2\pi$:

$$2\pi f_0 N = 2\pi k \;\Rightarrow\; f_0 = \frac{k}{N}, \quad k, N \text{ integers}$$

So $x[n]$ is periodic only if $f_0$ is a rational number. If $f_0 = k/N$ in lowest terms, the fundamental period is $N$.

Example: $f_0 = 3/10$ gives $N = 10$. $f_0 = 1/(2\pi)$ (i.e. $\cos n$) is irrational, so the signal is aperiodic.

$x(t) = e^{-a|t|}$, $a > 0$, is a two-sided decaying exponential (even signal).

$|x(t)|^2 = e^{-2a|t|}$.

Energy: using even symmetry,

$$\begin{aligned} E &= \int_{-\infty}^{\infty} e^{-2a|t|}\,dt = 2\int_{0}^{\infty} e^{-2at}\,dt \\ &= 2\left[\frac{e^{-2at}}{-2a}\right]_0^{\infty} = 2\cdot\frac{1}{2a} = \frac{1}{a} \end{aligned}$$

This is finite and non-zero since $a > 0$.

Power:

$$P = \lim_{T\to\infty}\frac{1}{2T}\int_{-T}^{T}e^{-2a|t|}\,dt = \lim_{T\to\infty}\frac{1}{2T}\cdot\frac{1-e^{-2aT}}{a} = 0$$

Answer: $x(t) = e^{-a|t|}$ is an energy signal with $E = 1/a$ and $P = 0$.

Even and odd signals

• Even: $x(-t) = x(t)$, symmetric about the vertical axis. Example: $\cos t$.
• Odd: $x(-t) = -x(t)$, antisymmetric about the origin, $x(0) = 0$. Example: $\sin t$.

  Even: x(t) = |t|           Odd: x(t) = t
  \      |      /                    |     /
    \    |    /                      |   /
      \  |  /                        | /
 --------+--------> t       ---------+---------> t
         0                         / |
                                 /   |
  mirror image about        180 deg symmetry
  vertical axis             about origin

Any signal can be split as $x_e(t) = \frac{1}{2}[x(t)+x(-t)]$, $x_o(t) = \frac{1}{2}[x(t)-x(-t)]$.

(i) $x(t) = 5A\cos(\omega t + \phi)$

It is periodic with $T = 2\pi/\omega$, so compute power over one period:

$$\begin{aligned} P &= \frac{1}{T}\int_0^{T} 25A^2\cos^2(\omega t+\phi)\,dt \\ &= \frac{25A^2}{T}\int_0^{T}\frac{1+\cos(2\omega t+2\phi)}{2}\,dt = \frac{25A^2}{2} \end{aligned}$$

$E = \infty$ (infinite duration). Power signal, $P = 12.5A^2$.

(ii) $x[n] = \sin(n/4)$

Here $\omega_0 = 1/4$, $\omega_0/2\pi = 1/8\pi$ is irrational, so the sequence is not periodic, but it is bounded and lasts forever.

$$\begin{aligned} P &= \lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^{N}\sin^2\frac{n}{4} \\ &= \lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^{N}\frac{1-\cos(n/2)}{2} = \frac{1}{2} \end{aligned}$$

because the sum of $\cos(n/2)$ stays bounded while $2N+1 \to \infty$. Energy $E = \sum\sin^2(n/4) = \infty$.

Answer: (i) power signal, $P = 25A^2/2$; (ii) power signal (aperiodic), $P = 1/2$.

A discrete-time signal is periodic with period $N$ (a positive integer) if $x[n+N] = x[n]$ for all $n$.

For $x[n] = e^{j\omega_0 n}$:

$$\begin{aligned} x[n+N] &= e^{j\omega_0 (n+N)} = e^{j\omega_0 n}\,e^{j\omega_0 N} \end{aligned}$$

For $x[n+N] = x[n]$ we need

$$e^{j\omega_0 N} = 1 \;\Rightarrow\; \omega_0 N = 2\pi m, \quad m \text{ an integer}$$ $$\frac{\omega_0}{2\pi} = \frac{m}{N} = \text{rational number}$$

So $e^{j\omega_0 n}$ is periodic only if $\omega_0/2\pi$ is a rational number. The fundamental period is the smallest positive integer $N = 2\pi m/\omega_0$, taking $m$ as the smallest integer that makes $N$ an integer (with $m$ and $N$ having no common factor). The same condition holds for $\cos(\omega_0 n+\theta)$ and $\sin(\omega_0 n+\theta)$, because they are sums of such exponentials.

This differs from continuous time, where $e^{j\omega_0 t}$ is periodic for every $\omega_0 \neq 0$ (period $2\pi/|\omega_0|$). In discrete time $n$ takes only integer values, so $N$ must be an integer.

Conversely, if $\omega_0/2\pi = m/N$ is rational, then $e^{j\omega_0 (n+N)} = e^{j\omega_0 n}e^{j2\pi m} = e^{j\omega_0 n}$, so the signal is periodic. Hence the DT complex exponential is periodic if and only if its frequency (in cycles/sample, $f_0 = \omega_0/2\pi$) is rational. Hence proved.

Example: $e^{j(2\pi/5)n}$: $f_0 = 1/5$, periodic with $N = 5$. $e^{jn}$: $f_0 = 1/2\pi$, irrational, not periodic.

CT unit step

$$u(t) = \begin{cases} 1, & t \ge 0 \\ 0, & t < 0 \end{cases}$$

It switches from 0 to 1 at $t = 0$; $u(t) = \int_{-\infty}^{t}\delta(\tau)d\tau$.

Periodicity of $x(t) = e^{j\omega t}$

$x(t)$ is periodic with period $T$ if $x(t+T) = x(t)$ for all $t$:

$$e^{j\omega(t+T)} = e^{j\omega t}e^{j\omega T} = e^{j\omega t} \;\Rightarrow\; e^{j\omega T} = 1$$ $$\cos\omega T + j\sin\omega T = 1 \;\Rightarrow\; \omega T = 2\pi m, \quad m = \pm1, \pm2, \ldots$$

So the necessary condition is $\omega T = 2\pi m$. For $\omega \ne 0$ the smallest positive $T$ (with $m = 1$) is

$$T_0 = \frac{2\pi}{|\omega|}$$

Since $T$ can be any real number, this is satisfied for every non-zero $\omega$: a CT complex exponential is always periodic (for $\omega = 0$ it is a constant, periodic with any $T$). This differs from DT, where $\omega/2\pi$ must also be rational.

Energy signal: $0 < E < \infty$, $P = 0$, where $E = \sum_{n}|x[n]|^2$. Power signal: $0 < P < \infty$, $E = \infty$, where $P = \lim_{N\to\infty}\frac{1}{2N+1}\sum_{-N}^{N}|x[n]|^2$. Periodic sequences are power signals.

Periodicity of $x[n] = \cos\left(\frac{\pi n}{5}\right)\sin\left(\frac{\pi n}{3}\right)$

Use $\cos A\sin B = \frac{1}{2}[\sin(A+B) - \sin(A-B)]$:

$$\begin{aligned} x[n] &= \frac{1}{2}\left[\sin\left(\frac{\pi}{5}+\frac{\pi}{3}\right)n - \sin\left(\frac{\pi}{5}-\frac{\pi}{3}\right)n\right] \\ &= \frac{1}{2}\sin\left(\frac{8\pi}{15}n\right) + \frac{1}{2}\sin\left(\frac{2\pi}{15}n\right) \end{aligned}$$

• $\omega_1 = 8\pi/15$: $\omega_1/2\pi = 4/15$ (rational), $N_1 = 15$.
• $\omega_2 = 2\pi/15$: $\omega_2/2\pi = 1/15$ (rational), $N_2 = 15$.

$$N = \text{LCM}(15, 15) = 15$$

Check: $x[n+15] = \cos(\frac{\pi n}{5}+3\pi)\sin(\frac{\pi n}{3}+5\pi) = (-\cos\frac{\pi n}{5})(-\sin\frac{\pi n}{3}) = x[n]$.

(Taking the LCM of the periods of the two factors, 10 and 6, gives 30, which is a period but not the fundamental one.)

Since it is periodic, it is a power signal: $P = \frac{1}{4}\left(\frac{1}{2}+\frac{1}{2}\right) = \frac{1}{4}$.

Answer: periodic, fundamental period $N = 15$ samples ($\omega_0 = 2\pi/15$ rad/sample).

Evaluate each term range by range:

• $u[n+2] - u[n-3]$ = 1 for $-2 \le n \le 2$, else 0.
• $n u[n-3] - n u[n-6] = n\,(u[n-3]-u[n-6])$ = $n$ for $3 \le n \le 5$, else 0.

So $x[n] = \{1, 1, \underline{1}, 1, 1, 3, 4, 5\}$ for $n = -2$ to $5$ (underline marks $n = 0$).

 x[n]
  5 |                          o
  4 |                       o  |
  3 |                    o  |  |
  2 |                    |  |  |
  1 |     o  o  o  o  o  |  |  |
    |     |  |  |  |  |  |  |  |
 ---+--+--+--+--+--+--+--+--+--+--+--> n
     -3 -2 -1  0  1  2  3  4  5  6

$$x[n] = \sum_{k=-\infty}^{\infty}(-1)^k\delta[n-k]$$

Each impulse $\delta[n-k]$ is non-zero only at $n = k$, so at each $n$ exactly one term survives, with value $(-1)^n$:

$$x[n] = (-1)^n = \cos(\pi n) = \{\ldots, 1, -1, \underline{1}, -1, 1, \ldots\}$$

$x[n+2] = (-1)^{n+2} = (-1)^n = x[n]$, and $x[n+1] = -x[n] \neq x[n]$.

(Also $\omega_0 = \pi$, $N = 2\pi m/\pi = 2m$, smallest $N = 2$.)

Answer: fundamental period $N = 2$.

Given $x[n] = \{1, 2, \underline{0}, -2, 1\}$:

Let $y[n] = x[2n-3]$. For each integer $n$, the argument $m = 2n-3$ is always odd, so only the odd-index samples $x[-1]$ and $x[1]$ can appear (in downsampling, $x[-2], x[0], x[2]$ are lost).

For all other $n$, $2n-3$ is outside $-2 \ldots 2$, so $y[n] = 0$.

Answer:

$$y[n] = x[2n-3] = \{\underline{0}, 2, -2\} = 2\delta[n-1] - 2\delta[n-2]$$

(Method: first shift, $x[n-3]$ moves the sequence 3 steps right; then scale by 2, keeping only samples at even positions of the shifted sequence and halving their indices.)