Signal and System Notes

  • 0 views

Preparation notes for upcomming Signal and System test.

Bulbs

  1. Get edges of image

    \[\begin{align} h'(t) &= \delta(t) * h'(t) \\ &= \delta'(t) * h(t) \end{align}\]

Examples

Question
All possible original functions in t-domain for given Laplace transform \(F(s) = \frac{3s^2-30s+74}{(s-4)(s-5)(s-6)}\).

Solution
\(F(s)\) could be reduced to

\[F(s) = \frac{1}{s-4} + \frac{1}{s-5} + \frac{1}{s-6} \,.\]

Therefore,

\[\begin{align} f(t) = \begin{cases} (e^{4t}+e^{5t}+e^{6t})u(t), &\mathrm{Re}[s] \gt 6 \\ (e^{4t}+e^{5t})u(t) - e^{6t}u(-t), &5 \lt \mathrm{Re}[s] \lt 6 \\ e^{4t}u(t) - (e^{5t}+e^{6t})u(-t), &4 \lt \mathrm{Re}[s] \lt 5 \\ -(e^{4t}+e^{5t}+e^{6t})u(-t), &\mathrm{Re}[s] \lt 4 \end{cases} \end{align}\]

Note
Denote the norm of the only singular point of \(F(s)\) as \(v\). Within convergence domain, i.e. \(\vert s \vert \gt v\), have

\[\mathcal{L}^{-1}\{F(s)\} = f(t)u(t) \,.\]

Out of convergence domain, i.e. \(\vert s \vert \lt v\), get corresponding left-sided sequence

\[\mathcal{L}^{-1}\{F(s)\} = -f(t)u(-t) \,.\]

Question
Given the Laplace transform of a causal system \(F(s)=\frac{s^2+3s+2}{s^2+4s+4}\), \(f(0^+)=\underline{\>-1\>}\), \(f(+\infty)=\underline{\>0\>}\).

Solution

\[\begin{align} F(s) &= \frac{s+1}{s+2} \\ &= 1 - \frac{1}{s+2} \,. \end{align}\]

Therefore,

\[f(t) = \delta(t) - e^{-2t}u(t) \,,\]

and get \(f(0^+) = -1\), \(f(\infty) = 0\).

Question
Given the differential equation of a discrete causal system as \(y(k) - \alpha y(k-1) = e(k)\), where \(\alpha \in \mathbb{R}\).

  1. \(H(z)\) and its convergence domain, unit impulse response \(h(k)\).
  2. Range of \(\alpha\) which allow system to be stable.
  3. For \(\alpha=\frac{1}{2}\), \(y(-1)=4\), \(e(k)=0\), solve system response \(y(k)\;(k\geq0)\).

Solution

  1. Apply Z-Transform on both sides, get

    \[\begin{align} &Y(z) - \alpha \frac{1}{z} Y(z) = E(z) \\ \Rightarrow\; &(1 - \alpha \frac{1}{z}) Y(z) = E(z) \\ \Rightarrow\; &H(z) = \frac{Y(z)}{E(z)} = \frac{1}{1 - \alpha z^{-1}} = \frac{z}{z - \alpha} \,, \end{align}\]

    where \(\vert \alpha \vert \lt \vert z \vert \lt 1\).

    Note \(\vert z \vert \lt 1\) is given by the conditon of convergence of the Z-Transform \(v^ku(t) \leftrightarrow \frac{z}{z-v}\).

    Unit impulse response is \(h(k) = \alpha^k u(t)\).

  2. Having \(\vert \alpha \vert \lt \vert z \vert \lt 1\), must have \(\vert \alpha \vert \lt 1\).
  3. Assume system response, i.e. zero-input response in this case, have the format \(y(k) = c \alpha^k u(k)\).
    Substitute with given constraints, get \(y(k) = (\frac{1}{2})^{k-1}u(t)\).

Question
Given the Fourier transform of signal \(f(t)\) to be \(F(j\omega)\), the Fourier transform of \(g(t)=e^{j4t}f(2t-4)\) should be \(\underline{\> \frac{1}{2} F(j\frac{1}{2}(\omega-4)) e^{-j2(\omega-4)} \>}\).

Solution

\[\begin{align} g(t) &= e^{j4t}f(2(t-2)) \\ \end{align}\]

Break down step-by-step

\[\begin{align} f(2t) &\leftrightarrow \frac{1}{2} F(j\frac{1}{2}\omega) \\ f(2(t-2)) &\leftrightarrow \frac{1}{2} F(j\frac{1}{2}\omega) e^{-j2\omega} \\ e^{j4t}f(2(t-2)) = g(t) &\leftrightarrow \frac{1}{2} F(j\frac{1}{2}(\omega-4)) e^{-j2(\omega-4)} \\ \end{align}\]

Question

\[\underline{\> x(k) = \begin{cases} \begin{align} k, \hspace{1em} &k=1, 2 \\ 0, \hspace{1em} &otherwise \end{align} \end{cases} \>} \leftrightarrow X(z) = z^{-1} + 2z^{-2} \\\]

Solution

Knowing \(\mathcal{Z}\{\delta(k)\} = 1\) and \(\mathcal{Z}\{f(k-n)\} = z^{-n}F(z)\), combined have

\[x(k) = \delta(k-1)\cdot1 + \delta(k-2)\cdot2 \leftrightarrow X(z) = z^{-1}\cdot1 + z^{-2}\cdot2\]

Question

System \(H(s) = \frac{e^s}{s+1},\,\mathrm{Re}\{s\}\gt-1\) is  non-causal stable system .

Solution

Knowing \(\mathcal{L}\{e^{\alpha t}u(t)\} = \frac{1}{s-\alpha}\) and \(\mathcal{L}\{u(t-\tau)\} = \frac{1}{s}e^{-\tau s}\), combined have

\[h(t) = e^{-(t+1)}u(t+1) \leftrightarrow H(s) \,. \\\]

Obviously the system converges.
Noticing that non-zero value spans to \(t\lt0\), therefore the system is non-causal.

Question

Given \(h_1(t)=\delta(t-1)\), \(h_2(t)=e^{-2t}u(t-2)\), \(h_3(t)=e^{-t}u(t)\), and \(r(t) = e(t) * [h_1(t) + 1] * [h_2(t) + h_3(t)]\).

  1. Unit impulse response \(h(t)\).
  2. Causal? Stable? Reasons!

Solution

  1. Knowing \(e^{\alpha t}u(t) \leftrightarrow \frac{1}{s-\alpha}\) and \(f(t-t_c) \leftrightarrow F(s)e^{-st_c}\) (delay), have

    \[\begin{align} H_1(s) &= e^{-s} \,, \\ H_2(s) &= \mathcal{L}\{e^{-2(t-2)-4}u(t-2)\} \\ &= e^{-4}\frac{1}{s+2}e^{-2s} \\ &= e^{-2(s+2)}\frac{1}{s+2} \> (\mathrm{Re}\{s\}\gt-2) \,, \\ H_3(s) &= \frac{1}{s+1} \> (\mathrm{Re}\{s\}\gt-1) \,. \\ \end{align}\]

    Therefore,

    \[\begin{align} R(s) &= E(s) \cdot (1+e^{-s}) \cdot (e^{-2(s+2)}\frac{1}{s+2} + \frac{1}{s+1}) \\ &= E(s) \cdot (\frac{1}{s+2}e^{-2s-4} + \frac{1}{s+1} + \frac{1}{s+2}e^{-3s-4} + \frac{1}{s+1}e^{-s}) \,. \end{align}\]

    Back in t-domain,

    \[\begin{align} r(t) &= e(t) * \Big[ e^{-4}e^{-2(t-2)}u(t-2) + e^{-t}u(t) + e^{-4}e^{-2(t-3)}u(t-3) + e^{-(t-1)}u(t-1) \Big] \\ \Rightarrow h(t) &= e^{-t}u(t) + e^{-(t-1)}u(t-1) + e^{-2t}u(t-2) + e^{-2(t-1)}u(t-3) \,. \\ \end{align}\]

  2. \(h(t)=0,\, t\lt0\), therefore causal.
    System function \(H(s)\) converges for all \(j\omega\), therefore stable.

Transforms

Fourier Transform

Frequently Seen Pattern

\[\begin{align} \mathcal{F}\{\delta(t)\} &= 1 \\ \mathcal{F}\{1\} &= 2 \pi \delta(\omega) \\ \mathcal{F}\{u(t)\} &= \pi\delta(\omega) + \frac{1}{j\omega} \\ \mathcal{F}\{e^{-\alpha t} u(t)\} &= \frac{1}{\alpha + j\omega} \\ \mathcal{F}\{t e^{-\alpha t} u(t)\} &= \frac{1}{(\alpha + j\omega)^2} \\ \mathcal{F}\{\cos\omega_ct\} &= \pi [ \delta(\omega+\omega_c) + \delta(\omega-\omega_c) ] \\ \mathcal{F}\{\sin\omega_ct\} &= j\pi [ \delta(\omega+\omega_c) + \delta(\omega-\omega_c) ] \\ \mathcal{F}\{G_\tau(t)\} = \mathcal{F}\{u(t+\frac{\tau}{2})-u(t-\frac{\tau}{2})\} &= \tau\text{Sa}\Big(\tau\frac{\omega}{2}\Big) = \tau\frac{\sin(\tau\omega/2)}{\tau\omega/2} \\ \mathcal{F}\{\text{Sa}\Big(\frac{\Omega t}{2}\Big)\} &= \frac{2\pi}{\Omega}\Big[u(\omega+\frac{\Omega}{2})-u(\omega-\frac{\Omega}{2})\Big] \\ \mathcal{F}\{\delta_T(t)\} = \mathcal{F}\{\sum_{n=-\infty}^{+\infty}\delta(t-nT)\} &= \Omega\delta_{\Omega}(\omega) \hspace{1em} (\Omega = \frac{2\pi}{T}) \\ \end{align}\]

Attributes

Symmetry Attribute

For Fourier transform

\[f(t) \leftrightarrow f(j\omega) = R(\omega) \,, \\\]

have

\[R(t) \leftrightarrow 2 \pi f(\omega) \,, \\\]

or

\[\frac{1}{2 \pi} R(t) \leftrightarrow f(\omega) \,. \\\]
Derivative Attribute
\[\frac{d^nf(t)}{dt^n} \leftrightarrow (j\omega)^nF(j\omega) \\\]

Laplace Transform

Frequently Seen Pattern

\[\begin{align} \mathcal{L}\{\delta(t)\} &= 1 \\ \mathcal{L}\{u(t)\} &= \frac{1}{s} \\ \mathcal{L}\{t^n e^{\alpha t} u(t)\} &= \frac{n!}{(s-\alpha)^{n+1}} \\ \mathcal{L}\{u(t+\tau)\} &= \frac{1}{s}e^{\tau s} \\ \mathcal{L}\{(t+\tau)^n e^{\alpha(t+\tau)} u(t+\tau)\} &= \frac{n!}{(s-\alpha)^{n+1}}e^{\tau s} \\ \end{align}\]

Attributes

Derivative Attribute in t-domain
\[\mathcal{L}\{\frac{d^nf(t)}{dt^n}\} = s^nF(s) - \sum_{i=0}^{n-1} s^{n-i-1} f^{(i)}(0^-) \,, \\\]

in causal systems,

\[\mathcal{L}\{\frac{d^nf(t)}{dt^n}\} = s^nF(s) \,. \\\]

Z-Transform

Discrete System

Differential Equation and Characteristic Equation

For a differential equation, with constant coefficiency,

\[y(k+n)+ a_{n-1}y(k+n-1) + \cdots + a_0y(k) = 0 \,,\]

introducing the shifting operator \(S\), equation above could be denoted as

\[(S^n + a_{n-1}S^{n-1} + \cdots + a_0) = 0 \,,\]

from which you can get the corresponding characteristic equation

\[S^n + a_{n-1}S^{n-1} + \cdots + a_0 = 0 \,.\]

Assuming that \(v_1\), \(v_2\), \(\cdots\), \(v_n\) are characteristic roots, or eigenvalues, of the characteristic equation above, the equation could also be written in the following format

\[(S-v_1)^m (S-v_{m+1}) \cdots (S-v_n) = 0 \,,\]

where \(v_1\) to \(v_m\) are repeated roots.
Then the zero-input response is

\[y_{zi}(k) = (c_1 + c_2k + \cdots + c_mk^{m-1})v_1^k + c_{m+1}v_{m+1}^{k} + \cdots + c_nv^k \,.\]
Left-Sided and Right-Sided Sequence
\[\begin{cases}\begin{align} \text{Right-sided:}\> &f(k)u(k) \\ \text{Left-sided:}\> &-f(k)u(-k-1) \\ \end{align}\end{cases}\]

Be very careful to the \(1\) offset!

Frequently Used Pattern

\[\begin{align} \mathcal{Z}\{\delta(k)\} &= 1 \\ \mathcal{Z}\{v^ku(k)\} &= \frac{z}{z-v} \\ \mathcal{Z}\{ku(k)\} &= \frac{z}{(z-1)^2} \\ \mathcal{Z}\{kv^{k-1}u(k)\} &= \frac{z}{(z-v)^2} \\ \end{align}\]

Attributes

Time Shifting (Delaying) Attribute
\[\begin{align} \mathcal{Z}\{f(k+1)\} &= \mathcal{Z}\{Sf(k)\} \\ &= z[F(z) - f(0)] \,, \end{align}\]

or more general

\[f(k+n) \leftrightarrow z^n \Big[ F(z) - \sum_{i=0}^{n-1} f(i) z^{-1} \Big] \qquad n \gt 0 \,.\]

If all zero-states are \(0\), e.g. in a causal system, could be reduced to

\[f(k-n) \leftrightarrow z^{-n}F(z) \,.\]