[Fourier transform and its application study notes] 24. Cascade, Pulse response

We studied in the last class.

1. In a discrete finite dimensional space, any linear system is obtained by multiplying the matrix.
2. In a continuous infinite dimensional space, any linear system is obtained by integrating the integral of the kernel function.

Impulse response (Impulse response)

cascaded Linear Systems (cascading linear system)

If both $l$ and $m$ are linear, there is

$w =mlv$

In continuous infinite dimensional space

$\begin{align*}
MLv
&=m\left (\int_{-\infty}^{\infty}k (x, y) v (y) dy \right) \ \
&\approx M\left (\sum_{i=-\infty}^{\infty}k (x,y_i) v (y_i) \delta y_i \right) \ \
&=\sum_{-\infty}^{\infty}m\left (k (x,y_i) v (y_i) \delta y_i \right) \ \
&=\sum_{-\infty}^{\infty}m_x \left (k (x,y_i) v (y_i) \delta y_i \right) \qquad m\ deal\ with\ the\ function\ that\ Lv\ o uput\ based\ on\ x \ \
&\approx \int_{-\infty}^{\infty}m_x (k (x, Y)) v (y) dy
\end{align*}$

Definition of impulse response

The above discussion on cascading is to elicit the following conclusion:

Any linear system is obtained by integrating the nucleus (function), which is the response of the linear system to the impulse function. (Any linear system was given by integration against a kernel (impulse response).)

The derivation process is as follows:

$\begin{align*}
V (x)
&= (\delta * v) (x) \ \
&=\int_{-\infty}^{\infty}\delta (x-y) v (y) dy \qquad (\delta\ shift\ property)
\end{align*}$

So the linear system has the following representation

$\begin{align*}
Lv (x)
&=l\left (\int_{-\infty}^{\infty}\delta (x-y) v (y) dy \right) \ \
&=\int_{-\infty}^{\infty}l_x\delta (x-y) v (y) dy
\end{align*}$

Make $h (x, y) = L_x\delta (x-y) $, then there is,

$\displaystyle{Lv (x) = \int_{-\infty}^{\infty}h (x, y) v (y) dy}$

where $k (x, y) $ is the kernel function of the linear system, it is obtained by $l_x\delta (X-y) $.

The impulse response is defined as follows

• $\delta (x-y) $ is the position of the pulse on the $y$, $L _x\delta (x-y) $ represents the pulse input to the linear system, at which time the system responds and outputs the impulse response $h (x, y) $.

Schwartz Kernel function theorem

If $l$ is a linear operator of the generalized function (distribution), that is, if $l$ transforms a generalized function into another generalized function on the basis of the principle of superposition, then there will be a unique kernel $k$, which makes $LV = <k,v>$

The impulse response of Fourier transform

When the input pulse function $\delta (x-y) $, the Fourier transform outputs

$h (x, y) = \mathcal{f} (\delta (x-y)) = E^{-2\pi ixy}$

In addition, the Fourier transform formula is as follows

$\mathcal{f}f (x) = \displaystyle{\int_{-\infty}^{\infty}e^{-2\pi ixy}f (y) dy}$

Its kernel function is

$k (x, y) = E^{-2\pi ixy}$

We note that the kernel function is the same as the impulse response, which has the following relationship:

• If a linear RP is represented as an integral form of the product of $k (x, y) $ and the input function, then $k (x, y) $ is the impulse response.
• Conversely, if we can get the impulse response of the linear system, we can express the linear operator of the linear system by integrating the product of the impulse response and the input function.

Impulse response of discrete finite-dimensional linear systems

In a continuous infinite-dimensional linear system, the impulse response is the response of the linear system to the input pulse $\delta (x-y) $. In discrete finite-dimensional linear systems, the response to the input pulse sequence is also the same, and the expression of matrix multiplication is as follows:

$A \cdot \left[\underline{\delta}_0,\underline{\delta}_1,\underline{\delta}_2,..., \underline{\delta}_{n-1} \right ]
=a\cdot\begin{bmatrix}
1 &0 &0 &, ... &0 \
0 &1 &0 &, ... &0 \
0 &0 &1 &, ... &0 \
\vdots &\vdots &\vdots & ... &\vdots \
0 &0 &0 &, ... &1
\end{bmatrix}=a$

Examples of impulse response

Switch

$Lv = \pi v$

Its impulse response is

$h (x, y) = L\delta (XY) = \PI (x) \delta (XY) = \pi (y) \delta (x-y) \qquad (\delta\ sampling\ property) $

The integral of the product of the impulse response and the input function gets the linear operator of the switch

$\begin{align*}
\int_{-\infty}^{\infty}h (x, y) v (y) dy
&=\int_{-\infty}^{\infty}\pi (y) \delta (x-y) v (y) dy\\
&=\int_{-\infty}^{\infty}\delta (x-y) \left (\pi (y) v (y) \right) dy\\
&=\left (\delta * (\pi v) \right) (x) \qquad (\delta\ shift\ property) \ \
&=\PI (x) v (x)
\end{align*}$

The result proves that the previous conclusion is correct.

Convolution and linear time invariant systems

Introducing time-shifted symbol $\tau$

$\tau_a V (x) = V (x-a) $.

Assume that the linear system of the useful convolution expression is as follows

$Lv = h*v$

If we delay the input $v$ $a$

$\begin{align*}
L\tau_a V (x)
&= (H*\TAU_AV) (x) \ \
&= \int_{-\infty}^{\infty}h (x-y) v (y-a) dy \ \
&= \int_{-\infty}^{\infty}h (x-z-a) v (z) DZ \qquad (letting\ z=y-a) \ \
&= \int_{-\infty}^{\infty}\left (H (x-z) *\delta (x-a) \right) v (z) DZ \qquad (\delta\ shift\ property) \ \
&= \left (\int_{-\infty}^{\infty}h (x-z) v (z) DZ \right) *\delta (x-a) \ \
&= (H*V) (x-a) \ \
&= \tau_a (H*V) (x) \ \
&= \tau_alv (x)
\end{align*}$

The results show that the delay of the output of the linear system is $a$ with the input delay, which is called linear time-invariant system.

The conclusion is:

• If a linear system is given by convolution, then he is the same.

The reverse is also true:

• If a linear system is constant, then it must be given by convolution.

The certification process is as follows:

Any continuous infinite dimensional linear system has the following representation

$Lv = \displaystyle{\int_{-\infty}^{\infty}l_x\delta (x-y) v (y) dy}$

We make $h (x) = L_x\delta (x) $, which is the impulse response of the linear system to the $\delta_0$. Then there are

$L _x\delta (XY) = l_x (\tau_y\delta (x)) $

If the $l$ is constant, then the output will have the same delay as the input

$L _x\delta (XY) = l_x (\tau_y\delta (x)) = \tau_y (L_x\delta (x)) = \tau_yh (x) = h (XY) $

That

$Lv = \displaystyle{\int_{-\infty}^{\infty}h (x-y) v (y) dy}$

[Fourier transform and its application study notes] 24. Cascade, Pulse response