### Unstable system
https://www.youtube.com/watch?v=XggxeuFDaDU
https://www.simscale.com/blog/2018/07/tacoma-narrows-bridge-collapse/
### Stable or not
https://www.youtube.com/watch?v=p9qrHdPEe28
### Laplace transform
https://www.youtube.com/watch?v=6MXMDrs6ZmA&list=RDCMUCJ0yBou72Lz9fqeMXh9mkog&index=4
### e to the pi i for dummies
https://www.youtube.com/watch?v=-dhHrg-KbJ0
Unstable system¶
不穩定的系統¶
https://www.youtube.com/watch?v=XggxeuFDaDU https://www.simscale.com/blog/2018/07/tacoma-narrows-bridge-collapse/
Stable or not¶ 穩定與否¶
https://www.youtube.com/watch?v=p9qrHdPEe28
Laplace transform¶
拉普拉斯變換¶
https://www.youtube.com/watch?v=6MXMDrs6ZmA&list=RDCMUCJ0yBou72Lz9fqeMXh9mkog&index=4
e to the pi i for dummies¶
e 到傻瓜的 pi i 中¶
For a signal x(t), after sampling by a impulse function $\delta (t)$ we have
\begin{equation}
x(t)\delta (t) = \sum_{n=-\infty}^{\infty} x[nT] \delta (t-nT). \tag{1}
\end{equation}
Then, let $s = \sigma + j \omega$ and $t>0$
\begin{equation}
\begin{split}
X(s) &= F\{ x(t)\}=\int_{0}^{\infty} x(t)e^{-st} dt \\
&= \int_{0}^{\infty} \sum_{n=0}^{\infty} x[nT] \delta (t-nT) e^{-st} dt
&= \sum_{n=0}^{\infty} x[nT] e^{-snT}
\end{split}. \tag{2}
\end{equation}
Let $z = e^{sT}$ and $T=1$, then we have
\begin{equation}
X(z)= \sum_{n=0}^{\infty}x[n]z^{-n}. \tag{3}
\end{equation}
For a signal x(t), after sampling by a impulse function
對於信號 x（t），在通過脈衝函數
Then, let
然後，let
Let
設
As we know, $ e^{snT} = e^{(\sigma + j\omega)nT}$, then we have
\begin{equation}
e^{snT} = e^{\sigma nT} \cdot e^{j\omega nT}. \tag{4}
\end{equation}
where $e^{\sigma nT}$ is a constant (A) when T is constant.
Hence,
\begin{equation}
e^{snT} = A \cdot [\cos (\omega nT) + j\sin(\omega nT)]. \tag{5}
\end{equation}
Try to plot the graph! <br>
Assume $A=1$, it means given a unit circle.
\begin{equation}
e^{\omega} = \cos (\omega) + j\sin(\omega). \tag{6}
\end{equation}
where $nT=1$. Try to plot it! <br>
As we know,
我們知道，
Hence, 因此
Try to plot the graph!
嘗試繪製圖表！
Assume
假設
where
其中
[3]:
import numpy as np
import numpy.random as random
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
plt.rcParams["figure.figsize"] = 10,10
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
N = 1000 # in points
t = 10 # in sec
n = np.linspace(0, t, N)
T = 0.1 #
sigma = 2 # [-2, 0, 2]
f = np.linspace(0, 10, N)
R = np.exp(sigma*n*T)*np.cos(2* np.pi *f * n * T)
I = np.exp(sigma*n*T)*np.sin(2* np.pi *f * n * T)
ax.set_title("Expoential jw")
ax.set_xlabel('Re axis')
ax.set_ylabel('Im axis')
ax.set_zlabel('Mag. frequency ')
ax.plot(R, I, f)
plt.show()
Definition: Given a discret seqence $x[n]$, then the z can be defined as follows.
\begin{equation}
X(z)=Z\{ x[n] \} = \sum_{n=-\infty}^{\infty}x[n]z^{-n}
\end{equation}
The Inverse z Transform can be defined as follows.
\begin{equation}
x[n]=Z^{-1}\{ X(z) \} = \frac{1}{2\pi j}\int_{c} X(z)z^{n-1}dz
\end{equation}
where, $C$ is an anti-clock close path in the region of convergence (ROC) including origins.
Definition: Given a discret seqence
定義： 給定一個離散序列
The Inverse z Transform can be defined as follows.
Inverse z Transform 可以定義如下。
where,
其中，
<div>
<img src="attachment:image.png" width="400"/>
</div>
Compared with DTFT, <br>
\begin{equation}
X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} x[n] e^{-j\omega n}
\end{equation}
That is
\begin{equation}
z = e^{j\omega}
\end{equation}
Compared with DTFT, 與 DTFT 相比，
That is
#### 收斂區域(Region of Convergence, ROC)
是指使得z轉換收斂的複數平面上的z點集合，定義如下：
\begin{equation}
ROC = \sum_{n=-\infty}^{\infty}x[n]z^{-n}
\end{equation}
其中，C是包含原點的逆時針封閉路徑，落在收斂區域內。
http://homepage.ntu.edu.tw/~fengli/Teaching/SignalsSystems/PDF/Y15-7-1_2.pdf
收斂區域(Region of Convergence, ROC)¶
收斂區域（Region of Convergence， ROC）¶
是指使得z轉換收斂的複數平面上的z點集合，定義如下：
其中，C是包含原點的逆時針封閉路徑，落在收斂區域內。 http://homepage.ntu.edu.tw/~fengli/Teaching/SignalsSystems/PDF/Y15-7-1_2.pdf
#### ROC 為圓環 ring 或圓盤 disk，其中心為原點。
ROC 包含單位圓，則離散時間傅立葉轉換為收斂。 <br>
ROC 不可包含任何極點。<br>
有限長度訊號之 ROC 為整個 z 平面。<br>
右邊訊號的 ROC 從離原點最遠的有限極點延伸至 $\infty$。<br>
左邊訊號的 ROC 從離原點最近的有限極點延伸至原點。<br>
雙邊訊號的 ROC 只能是圓環<br>
ROC 為一連通區域。<br>
ROC 為圓環 ring 或圓盤 disk，其中心為原點。¶
ROC 包含單位圓，則離散時間傅立葉轉換為收斂。
ROC 不可包含任何極點。
有限長度訊號之 ROC 為整個 z 平面。
右邊訊號的 ROC 從離原點最遠的有限極點延伸至
左邊訊號的 ROC 從離原點最近的有限極點延伸至原點。
雙邊訊號的 ROC 只能是圓環
ROC 為一連通區域。
### Example 1.
Given a unit impule function $\delta[n]$, please find its z-transform and ROC. <br>
\begin{equation}
X(z)=Z\{ \delta[n] \} = \sum_{n=-\infty}^{\infty}\delta[n]z^{-n} = \delta[0] \cdot z^{-0}=1
\end{equation}
ROC is all z.
### Example 2.
Given a unit function $u[n]$, please find its z-transform and ROC. <br>
\begin{equation}
X(z)=Z\{ u[n] \} = \sum_{n=0}^{\infty}u[n]z^{-n} = \sum_{n=0}^{\infty}z^{-n} =\frac{1}{1-z^{-1}}
\end{equation}
ROC: $|z^{-1}|<1$ or $|z|>1$.
### Example 3.
Given a discrete sigal function $x[n]={1,2,4,3,2,1}, n=0,1,...,5$, please find its z-transform and ROC. <br>
\begin{equation}
X(z)=Z\{ x[n] \} = \sum_{n=-\infty}^{\infty}u[n]z^{-n} = 1+2z^{-1}+4z^{-2}+3z^{-3}+2z^{-4}+z^{-5}
\end{equation}
ROC: all z, $z \neq 0$
<div>
<img src="attachment:image.png" width="300"/>
</div>
### Example 4.
Given a discrete sigal function $x[n]=(0.5)^n u[n]$, please find its z-transform and ROC. <br>
\begin{equation}
X(z)=Z\{ x[n] \} = \sum_{n=-\infty}^{\infty}(0.5)^n u[n]z^{-n} = \sum_{n=0}^{\infty}(0.5 z)^{-n}=\frac{1}{1-0.5z^{-1}}
\end{equation}
ROC: $|0.5z^{-1}| <1$ or $|z|>0.5$
<div>
<img src="attachment:image-2.png" width="300"/>
</div>
Consider the system block diagram of a LTI system, as shown in the following.
\begin{equation}
H(z)= \frac{Y(z)}{X(z)}
\end{equation}
Consider the system block diagram of a LTI system, as shown in the following.
考慮 LTI 系統的系統框圖，如下所示。
<div>
<img src="attachment:image.png" width="300"/>
</div>
If a LTI can be denoted as follows.
\begin{equation}
H(z)= \frac{b_0+b_1 z^{-1} +...+b_M z^{-M}}{a_0+a_1 z^{-1} +...+a_N z^{-N}}
\end{equation}
Then, we can represent it as
\begin{equation}
H(z)= \frac{b_0}{a_0} z ^{N-M} \frac{\Pi_{l=1}^{M}(z-z_l)}{\Pi_{l=1}^{N}(z-p_l)}
\end{equation}
where $z=z_l, l=1,2,...,M$ are called zeros, and $z=p_l, l=1,2,...,N$ are called poles, $b_0/a_0$ is the system gain.
If a LTI can be denoted as follows.
如果 LTI 可以表示如下。
### Example 5.
If the transfer function of a LTI system is defined as
\begin{equation}
H(z)=\frac{1}{1-z^{-1}}
\end{equation}
Try to find its ROC, zero, and pole.
<div>
<img src="attachment:image.png" width="300"/>
</div>
### Stability example
ex1: https://www.youtube.com/watch?v=taaisxUnE1E <br>
ex2: https://www.youtube.com/watch?v=1MeunVzResU
[6]:
import numpy as np
import scipy.signal as signal
z = np.array( [ -0.8, 1 ] )
p = np.array( [ 0.6 + 0.8j, 0.6 - 0.8j, -1 ] )
k = 0.8
b, a = signal.zpk2tf( z, p, k )
print( "Numerator Polynomial Coefficients =", b )
print( "Denominator Polynomial Coefficients =", a )
Numerator Polynomial Coefficients = [ 0.8 -0.16 -0.64] Denominator Polynomial Coefficients = [ 1. -0.2 -0.2 1. ]
[4]:
import numpy as np
import scipy.signal as signal
import matplotlib.pyplot as plt
from matplotlib import patches
from matplotlib.markers import MarkerStyle
def zplane(z, p):
fig = plt.figure( )
ax = plt.subplot( 1, 1, 1 )
unit_circle = patches.Circle( ( 0,0 ), radius = 1, fill = False, color = 'black', ls = 'dashed' )
ax.add_patch( unit_circle )
plt.axvline( 0, color = 'black' )
plt.axhline( 0, color = 'black' )
plt.xlim( ( -2, 2 ) )
plt.ylim( ( -1.5, 1.5 ) )
plt.grid( )
plt.plot( z.real, z.imag, 'ko', fillstyle = 'none', ms = 12 )
plt.plot( p.real, p.imag, 'kx', fillstyle = 'none', ms = 12 )
return fig
def main( ):
b = np.array( [ 0.8, -0.16, -0.64 ] )
a = np.array( [ 1, -0.2, -0.2, 1 ] )
z, p, k = signal.tf2zpk( b, a )
print( "Zeros =", z )
print( "Poles =", p )
print( "Gain =", k )
zplane( z, p )
plt.show( )
main( )
Zeros = [ 1. -0.8] Poles = [ 0.6+0.8j 0.6-0.8j -1. +0.j ] Gain = 0.8
The Inverse z Transform can be defined as follows.
\begin{equation}
x[n]=Z^{-1}\{ X(z) \} = \frac{1}{2\pi j}\int_{c} X(z)z^{n-1}dz
\end{equation}
實際反z轉換過程中，通常採用下列方式： <br>
(1) 長除法(Long Division Method)。<br>
(2) 部份分式展開法(Partial Fraction Expansion Method)。<br>
(3) 餘數法(Residue Method)。<br>
The Inverse z Transform can be defined as follows.
Inverse z Transform 可以定義如下。
(1) 長除法(Long Division Method)。
（1） 長除法（Long Division Method）。
(2) 部份分式展開法(Partial Fraction Expansion Method)。
（2） 部份分式展開法（Partial Fraction Expansion Method）。
(3) 餘數法(Residue Method)。
（3） 餘數法（Residue Method）。
### Example 6. (Long Division Method)
If a transfer function is
\begin{equation}
X(z)=\frac{1+z^{-1}+2z^{-2}-z^{-3}+3z^{-4}}{1-z^{-1}+z^{-2}}
\end{equation}
Try to find its inverse Z transform.
Sol:
\begin{equation}
X(z)=\frac{1+z^{-1}+2z^{-2}-z^{-3}+3z^{-4}}{1-z^{-1}+z^{-2}} = 1+2z^{-1}+3z^{-2}
\end{equation}
Therefore, <br>
$x[n]=Z^{-1}\{ X(z) \} =Z^{-1}\{ 1+2z^{-1}+3z^{-2} \}=\delta[n]+2\delta[n-1]+3\delta[n-2]$ <br>
or, <br>
$x[n]={1,2,3}, n=0, 1, 2$ <br>
Sol:
溶膠：
or, 或
Hence, if a function is defined as<br>
\begin{equation}
X(z)= \frac{b_0+b_1 z^{-1} +...+b_M z^{-M}}{a_0+a_1 z^{-1} +...+a_N z^{-N}}
\end{equation}
and we can represent it as <br>
\begin{equation}
x[n]= \frac { b_n - \sum_{i=1}^{n} x[n-i]a_i} {a_0}, n=1,2,...
\end{equation}
where $x[0]=b_0/a_0$. <br>
Hence, if a function is defined as
因此，如果函數定義為
[7]:
import numpy as np
b = np.array( [ 1, 1, 2, -1, 3 ] )
a = np.array( [ 1, -1, 1, 0, 0 ] )
z, p, k = signal.tf2zpk( b, a )
print( "Zeros =", z )
print( "Poles =", p )
print( "Gain =", k )
zplane( z, p )
plt.show( )
Zeros = [-1. +1.41421356j -1. -1.41421356j 0.5+0.8660254j 0.5-0.8660254j ] Poles = [0.5+0.8660254j 0.5-0.8660254j 0. +0.j 0. +0.j ] Gain = 1.0
[8]:
M = b.size
N = a.size
x = np.zeros( M )
x[0] = b[0] / a[0]
for n in range( 1, M ):
sum = 0
k = n
if n > N:
k = N
for i in range( 1, k + 1 ):
sum = sum + x[n-i] * a[i]
x[n] = ( b[n] - sum ) / a[0]
print( x )
[1. 2. 3. 0. 0.]
### Example 7. (Partial Fraction Expansion Method)
If a transfer function is
\begin{equation}
X(z)=\frac{1}{1-3z^{-1}+2z^{-2}}=\frac{1}{(1-z^{-1})(1-2z^{-1})}
\end{equation}
Then, we have <br>
\begin{equation}
X(z)=\frac{1}{(1-z^{-1})(1-2z^{-1})} = \frac{-1}{(1-z^{-1})} + \frac{2}{(1-2z^{-1})}
\end{equation}
so, the discrete time sequence is <br>
\begin{equation}
x[n] =Z^{-1} \{ X(z) \} = Z^{-1} \{ \frac{-1}{(1-z^{-1})} + \frac{2}{(1-2z^{-1})} \}
=-u[n]+2\cdot (2)^n u[n]
\end{equation}
[9]:
import numpy as np
b = np.array( [ 1, 0, 0 ] )
a = np.array( [ 1, -3, 2] )
z, p, k = signal.tf2zpk( b, a )
print( "Zeros =", z )
print( "Poles =", p )
print( "Gain =", k )
zplane( z, p )
plt.show( )
Zeros = [0. 0.] Poles = [2. 1.] Gain = 1.0
## <font color='red'> Final Problem:</font>
### Reference
https://www.youtube.com/watch?v=AU3iQSAIiPo