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# 电子代写|数字信号处理代写Digital Signal Processing代考|ELEC9721 LTI Systems

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## 电子代写|数字信号处理代写Digital Signal Processing代考|LTI Systems

Recall that a DT system is LTI iff it maps the input $x[n]$ to the output $y[n]$ via (linear) convolution:
$$y[n]=h[n] * x[n] .$$
The sifting property of $\delta[n]$ tells us that the input $x[n]=\delta[n]$ produces the output
$$y[n]=h[n] * \delta[n]=h[n] .$$
For this reason, $h[n]$ is called the impulse response of the system. By the convolution property of the $\mathrm{ZT}$,
$$Y(z)=H(z) X(z) .$$

$H(z)$ is the transfer function of the system.
Since DT convolution is a linear operation, impulse responses and transfer functions of systems in parallel add:
\begin{aligned} h[n] & =h_1[n]+h_2[n], \ H(z) & =H_1(z)+H_2(z) \end{aligned}

## 电子代写|数字信号处理代写Digital Signal Processing代考|Difference Equations

An important class of digital filters is implemented through the use of difference equations:
$$a_N y[n+N]+\ldots+a_0 y[n]=b_M x[n+M]+\ldots+b_0 x[n] .$$
There is no harm in assuming $a_N \neq 0$ and $b_M \neq 0$, since otherwise $M$ and $N$ can be redefined. In fact, we may divide through the equation by $a_N$ and redefine coefficients accordingly. This makes $a_N=1$. Now suppose
$$a_0=\ldots=a_{K-1}=0 .$$
In other words, $K$ is the smallest index such that $a_K \neq 0$. The difference equation becomes
$$y[n+N]+a_{N-1} y[n+N-1]+\ldots+a_K y[n+K]=b_M x[n+M]+\ldots+b_0 x[n] .$$
The number $N-K$ is the order of the equation. For now, we will restrict ourselves to equations with order $N-K>0$.

A difference equation is very similar to a differential equation in that it expresses a relationship between shifts of the input $x[n]$ and the output $y[n]$, rather than derivatives of $x(t)$ and $y(t)$. Like a differential equation, a difference equation has infinitely many solutions corresponding to a given input signal $x[n]$. A single solution is determined uniquely by specifying $N-K$ initial conditions, typically adjacent values such as $y[-1], \ldots, y[K-N]$. For a given $x[n]$ and set of initial conditions, a difference equation can be solved using the same analytic methods as for differential equations: Find a particular solution, add the general homogeneous solution with $N-K$ free parameters, and apply the $N-K$ initial conditions.
Example $8.1$ Solve
$$y[n+2]-\frac{5}{2} y[n+1]+y[n]=x[n]$$
for the input
$$x[n]=n$$
and initial conditions
$$y[-1]=1, \quad y[-2]=0 .$$

## 电子代写|数字信号处理代写Digital Signal Processing代考|LTI Systems

$$y[n]=h[n] * x[n] .$$

$$y[n]=h[n] * \delta[n]=h[n] .$$

$$Y(z)=H(z) X(z) .$$
$H(z)$ 是系统的传递函数。

$$h[n]=h_1[n]+h_2[n], H(z) \quad=H_1(z)+H_2(z)$$

## 电子代写|数字信号处理代写Digital Signal Processing代考|Difference Equations

$$a_N y[n+N]+\ldots+a_0 y[n]=b_M x[n+M]+\ldots+b_0 x[n] .$$

$$a_0=\ldots=a_{K-1}=0 .$$

$$y[n+N]+a_{N-1} y[n+N-1]+\ldots+a_K y[n+K]=b_M x[n+M]+\ldots+b_0 x[n] .$$

$$y[n+2]-\frac{5}{2} y[n+1]+y[n]=x[n]$$

$$x[n]=n$$

$$y[-1]=1, \quad y[-2]=0 .$$

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