Signals and Systems Midterm Review

Fourier Transform and Series

CFS

$$a_k=\frac{1}{T}{\int }_0^{T}x(t)\exp (-jk{\omega }_{0}t)\mathrm{d}t,{\omega }_0=\frac{2\pi }{T}$$ $$x(t)=\sum _{k=-\infty }^{\infty }{a}_k\exp (jk{\omega }_{0}t)$$

DFS

$$a_k=\frac{1}{N}\sum _{n=0}^{N-1}x[n]\exp (-jk{\omega }_{0}n),{\omega }_0=\frac{2\pi }{N}$$ $$x[n]=\sum _{k=-\infty }^{\infty }{a}_k\exp (jk{\omega }_{0}n)$$

CTFT(aka CFT)

$$X(\omega )={\int }_{-\infty }^{\infty }x(t)\exp (-j\omega t)\mathrm{d}t$$ $$x(t)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }X(\omega )\exp (j\omega t)\mathrm{d}\omega$$

DTFT

$$\begin{array}{rl}X(\omega )& =\sum _{n=-\infty }^{\infty }x[n]\exp (-j\omega n)\\ x[n]& =\frac{1}{2\pi }{\int }_{-\pi }^{\pi }X(\omega )\exp (j\omega n)\mathrm{d}\omega \end{array}$$

System Properties

• Memoryless or Algebraic or Non-Dynamic: The outputs at any instant do not depend on values of the inputs at any other instant: $y[n_0]=T\left\{x[n_0]\right\}$ for all $n_0$
• Linear: The response to an arbitrary linear combination (or "superposition") of inputs signals is always the same linear combination of the individual responses to these signals: $T\left\{a{x}_A[n]+b{x}_B[n]\right\}=aT\left\{x_A[n]\right\}+bT\left\{x_B[n]\right\}$
• Time-Invariant: The response to an arbitrarily translated set of inputs is always the response to the original set, but translated by the same amount: If $x[n]\to y[n]$ then $x[n-n_0]\to y[n-n_0]$ for all $x$ and $x_0$
• Linear and Time-Invariant (LTI): The system, model or mapping is both linear and time-invariant. See LTI systems
• Causal: The output at any instant does not depend on future inputs: for all $n_0$, $y[n_0]$ does not depend on $x[n]$ for $n>n_0$. For a LTI system, this means for all $t<0$, its unit response $h(t)=0$
• BIBO Stable: The response to a bounded input is always bounded: $\left|x_n\right|\le M_x<\infty$ for all $n$ implies that $\left|y[n]\right|\le M_y<\infty$ for all $n$; For a LTI system, this means that its unit pulse response $h[t]$ or $h(t)$ is absolutely summable or absolutely integrable over $(-\infty ,\infty )$
• Invertible: A system is called invertible if it produces distinct output signals for distinct input signals.
• Initial Rest: The system's output is zero before any input is applied.

Fundamental Signals

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

Unit Step

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

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

Square Wave

$$x(t+NT)=\{\begin{array}{ll}1,& \left|t\right|<T_1\\ 0,& T_1<\left|t\right|<\frac{T}{2}\end{array},N=0,1,2,\dots$$

whose Fourier series is

$$a_k=\frac{\sin \left(k{\omega }_0{T}_1\right)}{k\pi }=\frac{2T_1}{T}\mathrm{sinc}\left(k{\omega }_0{T}_1\right)=\frac{2T_1}{T}\mathrm{sinc}\left(2k\pi \frac{T_1}{T}\right)$$

If we let $\lambda =\frac{2T_1}{T}$ be valid length of the square signal, we can just have $a_k=\lambda \mathrm{sinc}\left(\lambda k\pi \right)$

例题

T1

判断如下信号是否是线性的、时不变的和稳定的

1. $y(t)=x(t-2)+x(2-t)$
2. $y(t)=tx(t)$
3. $y(t)=x^{\prime \prime }(t)+3x^{\prime }(t)+2x(t)$
4. $y[n]=\{\begin{array}{ll}x[n],& n\ge 0\\ -x[n],& n<0\end{array}$
5. $y[n]=nx[n-1]+x[n+1]$

1. 考虑$x_3(t)=\alpha x_1(t)+\beta x_2(t)$, 则有

$$\begin{array}{rl}{y}_3(t)& =x_3(t-2)+x_3(2-t)=\alpha x_1(t-2)+\beta x_2(t-2)+\alpha x_1(2-t)+\beta x_2(2-t)\\ & =\alpha x_1(t-1)+\alpha x_1(2-t)+\beta x_2(t-2)+\beta x_2(2-t)=\alpha y_1(t)+\beta y_2(t)\end{array}$$

故该系统是线性的, 再考虑$x_2(t)=x_1(t+T)$, 则

$$y_2(t)=x_2(t-2)+x_2(2-t)=x_1(t-2+T)+x_1(2-t+T)\ne y_1(t+T)$$

所以该系统不是时不变的. 另外, 该系统显然是稳定的. 2. 考虑$x_3(t)=\alpha x_1(t)+\beta x_2(t)$, 则有

$$y_3(t)=t{x}_3(t)=\alpha t{x}_1(t)+\beta t{x}_2(t)=\alpha y_1(t)+\beta y_2(t)$$

故该系统是线性的, 再考虑$x_2(t)=x_1(t+T)$, 则

$$y_2(t)=t{x}_2(t)=t{x}_1(t+T)\ne y_1(t+T)$$

3. 考虑$x_3(t)=\alpha x_1(t)+\beta x_2(t)$, 则有

$$\begin{array}{rl}{y}_3(t)& =x_3^{\prime \prime }(t)+3x_3^{\prime }(t)+2x_3(t)\\ & =\alpha x_1^{\prime \prime }(t)+\beta x_2^{\prime \prime }(t)+3\alpha x_1^{\prime }(t)+3\beta x_2^{\prime }(t)+2\alpha x_1(t)+2\beta x_2(t)\\ & =y_1(t)+y_2(t)\end{array}$$

故该系统是线性的, 再考虑$x_2(t)=x_1(t+T)$, 则

$$y_2(t)=x_1^{\prime \prime }(t+T)+3x_1^{\prime }(t+T)+2x_1(t+T)=y_1(t+T)$$

所以该系统不是时不变的. 另外, 该系统对导数不存在的输入是不稳定的. 4. 考虑$x_3[n]=\alpha x_1[n]+\beta x_2[n]$, 则有

$$y_3[n]=\{\begin{array}{ll}\alpha x_1[n]+\beta x_2[n],& n\ge 0\\ -\alpha x_1[n]-\beta x_2[n],& n<0\end{array}=\alpha y_1[n]+\beta y_2[n]$$

故该系统是线性的, 再考虑$x_2[n]=x_1[n+N]$, 则

$$y_2[n]=\{\begin{array}{ll}{x}_1[n+N],& n\ge 0\\ -x_1[n+N],& n<0\end{array}\ne y_1[n+N]$$

所以该系统不是时不变的. 另外, 该系统显然是稳定的. 5. 考虑$x_3[n]=\alpha x_1[n]+\beta x_2[n]$, 则有

$$\begin{array}{rl}{y}_3[n]& =n\left(\alpha x_1[n-1]+\beta x_2[n-1]\right)+\alpha x_1[n+1]+\beta x_2[n+1]\\ & =\alpha n{x}_1[n-1]+\alpha x_1[n+1]+\beta n{x}_2[n-1]+\beta x_2[n+1]\\ & =\alpha y_1[n]+\beta y_2[n]\end{array}$$

故该系统是线性的, 再考虑$x_2[n]=x_1[n+N]$, 则

$$y_2[n]=n{x}_1[n-1+N]+x_1[n+1+N]\ne y_1[n+N]$$

所以该系统不是时不变的. 另外, 该系统在$n\to \infty$时是不稳定的

Note

• 判断系统的线性时可以忽略时间项$t,n$, 因为它们肯定时对应的
• 如果系统表达式中在输入信号之外也出现的时间项, 那么该系统大概率不是时不变的. 而如果没有出现, 也不能说明该系统是时不变的