We will find the magnitude spectrum \(\left \vert H\left ( e^{j\omega }\right ) \right \vert \) as the digital frequency \(\omega \) is changed from \(0\) radians to \(\pi \) radians. At each different value of \(\omega \), the magnitude of the frequency response is \(\left \vert H\left ( e^{j\omega }\right ) \right \vert =\frac{{\displaystyle \prod \limits _{i=1}^{M}} \left \vert \omega -z_{i}\right \vert }{{\displaystyle \prod \limits _{i=1}^{N}} \left \vert \omega -p_{i}\right \vert }\) where \(\left \vert \omega -z_{i}\right \vert \) is the length of the vector from the point \(\omega \) (which is the point on the unit circle) to the point where the \(i^{th}\) zero is located. And similarly, \(\left \vert \omega -p_{i}\right \vert \) is the length of the vector from the point \(\omega \) to the point where the \(i^{th}\) pole is located. So, by estimating these products, one can estimate a value for \(\left \vert H\left ( e^{j\omega }\right ) \right \vert \) as \(\omega \) is moved around the unit circle.
At \(\omega =0^{0}\), \(\left \vert H\left ( e^{j\omega }\right ) \right \vert \approx \frac{.25\times .25}{.2\times .2}\) \(\approx 1.5\)
At \(\omega =15^{0}\) where the zero is located, \(\left \vert H\left ( e^{j\omega }\right ) \right \vert =0\)
At \(\omega =90^{0}\), \(\left \vert H\left ( e^{j\omega }\right ) \right \vert \approx \frac{.7\times 1.2}{.65\times 1.1}\approx \allowbreak 1.1\)
At \(\omega =180^{0},\left \vert H\left ( e^{j\omega }\right ) \right \vert \approx \frac{1.9\times 1.9}{1.7\times 1.7}\approx \allowbreak 1.3\)
Hence this is a sketch
So this is a notch filter
At \(\omega =0^{0}\), \(\left \vert H\left ( e^{j\omega }\right ) \right \vert \approx \frac{1}{2^{5}}\) \(\approx 0.03\)
At \(\omega =90^{0}\), \(\left \vert H\left ( e^{j\omega }\right ) \right \vert \approx \frac{1}{.8\times 1\times 1.4\times 1.6\times 1.7}\approx \allowbreak 0.328\,26\)
At \(\omega =180^{0},\left \vert H\left ( e^{j\omega }\right ) \right \vert \approx \frac{1}{small\ values}\approx \) large
So this allows frequencies very close to \(\pi \) to pass. So high pass filter
At \(\omega =0^{0}\), \(\left \vert H\left ( e^{j\omega }\right ) \right \vert \approx \frac{.7\times 1\times 1.4\times 1.6\cdots \times 2\times 1.8\times 1.6\cdots }{1}\) \(\approx 20\)
At \(\omega =90^{0}\), \(\left \vert H\left ( e^{j\omega }\right ) \right \vert \approx \frac{\text{smaller values than above since vector is smaller now}}{1}\approx 10\)
At \(\omega =180^{0},\left \vert H\left ( e^{j\omega }\right ) \right \vert \approx \frac{\,\text{much smaller values than above since close to zeros }}{1}\approx \) \(0\)
So, this is low pass filter
At \(\omega =0^{0}\), \(\left \vert H\left ( e^{j\omega }\right ) \right \vert \approx \frac{1}{.3\times .5\times .7\times .3\times .5\times .7}\) \(\approx \) large value
At \(\omega =30^{0}\), \(\left \vert H\left ( e^{j\omega }\right ) \right \vert \approx \frac{1}{\text{very small values due to being close to poles}}\approx \allowbreak \)much larger value the above
At \(\omega =90^{0}\), \(\left \vert H\left ( e^{j\omega }\right ) \right \vert \approx \frac{1}{\text{larger values than the above due to vectors below x-axis being further away}}\approx \allowbreak \)smaller than where at \(\omega =0^{0}\)
At \(\omega =180^{0},\left \vert H\left ( e^{j\omega }\right ) \right \vert \approx \frac{\,1}{\text{much larger values than the above }}\approx \) \(0\)
So, this is band pass filter
At \(\omega =0^{0}\), \(\left \vert H\left ( e^{j\omega }\right ) \right \vert \approx \frac{1}{\text{very small values due to being close to poles}}\) \(\approx \) large value
At \(\omega =90^{0}\), \(\left \vert H\left ( e^{j\omega }\right ) \right \vert \approx \frac{1}{1.3\times 1.4\times 1.5\times 1.6}\approx \allowbreak .2\)
At \(\omega =180^{0},\left \vert H\left ( e^{j\omega }\right ) \right \vert \approx \frac{\,1}{1.8\times 1.8\times 1.8\times 1.8}\approx \) smaller values than above
So, low pass filter
\(H\left ( s\right ) =\frac{s+2}{\left ( s+2\right ) ^{2}+4}\)
Using impulse invariance, \(H\left ( z\right ) ={\displaystyle \sum \limits _{i=1}^{N}} \frac{T\ A_{i}}{1-e^{p_{i}T}z^{-1}}\) where \(p_{i}\) are the poles of \(H\left ( s\right ) \) and \(A_{i}\) is the partial fraction result of expressing \(H\left ( s\right ) \) as \({\displaystyle \sum \limits _{i=1}^{N}} \frac{A_{i}}{s-p_{i}}\). Notice that this method works only for distinct poles in \(H\left ( s\right ) \). So the first step is to express \(H\left ( s\right ) \) is partial fraction form to determine \(A_{i}\). The poles of \(H\left ( s\right ) \) are roots of the denominator \(\left ( s+2\right ) ^{2}+4\) hence poles are roots of \(s^{2}+4s+8\) or \(-\frac{b}{2}\pm \frac{1}{2}\sqrt{b^{2}-4ac}=-1\pm \frac{1}{2}\sqrt{16-4\times 8}=-1\pm 2j\), hence \begin{align*} p_{1} & =-1+2j\\ p_{2} & =-1-2j \end{align*}
Then \(H\left ( s\right ) =\frac{s+2}{\left ( s-p_{1}\right ) \left ( s-p_{2}\right ) }=\frac{A_{1}}{s-p_{1}}+\frac{A_{2}}{s-p_{2}}\), then\[ A_{1}=\lim _{s\rightarrow p_{1}}\frac{s+2}{\left ( s-p_{2}\right ) }=\lim _{s\rightarrow p_{1}}\frac{-1+2j+2}{\left ( \left ( -1+2j\right ) -\left ( -1-2j\right ) \right ) }=\frac{1+2j}{4j}\] and\[ A_{2}=\lim _{s\rightarrow p_{2}}\frac{s+2}{\left ( s-p_{1}\right ) }=\lim _{s\rightarrow p_{2}}\frac{-1-2j+2}{\left ( \left ( -1-2j\right ) -\left ( -1+2j\right ) \right ) }=\frac{1-2j}{-4j}\] Hence\[ H\left ( s\right ) =\frac{\frac{1+2j}{4j}}{s-\left ( -1+2j\right ) }+\frac{\frac{1-2j}{-4j}}{s-\left ( -1-2j\right ) }\] And \[ H\left ( z\right ) =\frac{T\frac{1+2j}{4j}}{1-z^{-1}\exp \left ( -1+2j\right ) T}+\frac{T\frac{1-2j}{-4j}}{1-z^{-1}\exp \left ( -1-2j\right ) T}\] We can take \(T=1\) and the above becomes\[ H\left ( z\right ) =\frac{\frac{1+2j}{4j}}{1-z^{-}1\exp \left ( -1+2j\right ) }+\frac{\frac{1-2j}{-4j}}{1-z^{-}1\exp \left ( -1-2j\right ) }\] This can be simplified to\[ H\left ( z\right ) =\frac{z^{2}+0.32035z}{z^{2}+0.30618z+0.13535}\] The poles are \begin{align*} z_{1} & =-0.153-0.3345\,j\\ z_{2} & =-0.153+0.3345\,i \end{align*}
So, they are both inside the unit circle.
Using bilinear transformation, \(H\left ( z\right ) =\left . H\left ( s\right ) \right \vert _{s=\frac{2}{T}\frac{1-z^{-1}}{1+z^{-1}}}\) Since \(H\left ( s\right ) =\frac{s+2}{\left ( s+2\right ) ^{2}+4}\), then\begin{align*} H\left ( z\right ) & =\frac{\frac{2}{T}\frac{1-z^{-1}}{1+z^{-1}}+2}{\left ( \frac{2}{T}\frac{1-z^{-1}}{1+z^{-1}}+2\right ) ^{2}+4}\\ & =\frac{T\left ( 1+z\right ) \left ( z-1+T+Tz\right ) }{2\left ( \left ( z-1\right ) ^{2}+2T^{2}\left ( 1+z\right ) ^{2}+2T\left ( z^{2}-1\right ) \right ) } \end{align*}
For \(T=1\), the above simplifies to\[ H\left ( z\right ) =\frac{z+z^{2}}{1+2z+10z^{2}}\] The poles are located at roots of \(1+2z+10z^{2}\), which are\begin{align*} z_{1} & =-0.1-0.3i\\ z_{1} & =-0.1+0.3i \end{align*}
So, they are both inside the unit circle.
Consider some \(H\left ( s\right ) =\frac{N\left ( s\right ) }{D\left ( s\right ) }\). Let \(D\left ( s\right ) \) be written in factored form \({\displaystyle \prod \limits _{i}^{N}} \left ( s-p_{i}\right ) \) where \(N\) is number of \(H\left ( s\right ) \) poles and \(p_{i}\) is the pole. For the purpose of this solution, we can assume there is one pole only. The same idea applied for all others. Hence, we have\begin{equation} H\left ( s\right ) =\frac{N\left ( s\right ) }{s-p}\tag{1} \end{equation} And now we want to show that if \(p<0\), then the transformation results in \(H\left ( z\right ) \) with a pole inside the unit circle. Let \[ s=\frac{1-z^{-1}}{T}\] then (1) becomes\[ H\left ( z\right ) =\frac{N\left ( z\right ) }{\frac{1-z^{-1}}{T}-p}=\frac{TN\left ( z\right ) }{1-z^{-1}-Tp}=\frac{zTN\left ( z\right ) }{z-1-zTp}=\frac{zTN\left ( z\right ) }{z\left ( 1-Tp\right ) -1}=\frac{\frac{zTN\left ( z\right ) }{1-Tp}}{z-\frac{1}{1-Tp}}\] Hence pole of the \(H\left ( z\right ) \) is \[ q=\frac{1}{1-Tp}\] Since \(p<0\) then the numerator of \(q\) is larger than one. Hence \(q<1\), hence a stable pole of \(H\left ( z\right ) \). Therefore, a stable pole of \(H\left ( s\right ) \) maps to a stable pole of \(H\left ( z\right ) \) Now we need to show that a stable pole of \(H\left ( z\right ) \) will not map to a stable pole of \(H\left ( s\right ) \). First we need to find the inverse transformation. Since \(s=\frac{1-z^{-1}}{T}\) then\begin{align*} sT & =1-z^{-1}\\ sT & =\frac{z-1}{z}\\ zsT & =z-1\\ zsT-z & =-1\\ z-zsT & =1\\ z\left ( 1-sT\right ) & =1 \end{align*}
Hence\[ z=\frac{1}{1-sT}\] Now, given \(H\left ( z\right ) =\frac{N\left ( z\right ) }{D\left ( z\right ) }=\frac{N\left ( z\right ) }{z-q}\) where \(q\) is a pole of \(H\left ( z\right ) \) where \(q\) is stable. Hence \(\left \vert q\right \vert <1\), i.e. pole is inside the unit circle. Now apply the above transformation\begin{align*} H\left ( s\right ) & =\frac{N\left ( z\right ) }{z-q}\\ & =\frac{N\left ( s\right ) }{\frac{1}{1-sT}-q}=\frac{N\left ( s\right ) \left ( 1-sT\right ) }{1-q\left ( 1-sT\right ) }=\frac{N\left ( s\right ) \left ( 1-sT\right ) }{1-q+qsT}=\frac{N\left ( s\right ) \frac{\left ( 1-sT\right ) }{qT}}{s+\frac{1-q}{qT}}=\frac{N\left ( s\right ) \frac{\left ( 1-sT\right ) }{qT}}{s-\left ( \frac{q-1}{qT}\right ) } \end{align*}
Hence \(H\left ( s\right ) \) pole is at \[ \frac{q-1}{qT}\] this pole will be stable only if the real part of it is less than zero. Let \(q=\frac{j}{2}\) a stable pole in the z plane. Then the above pole size becomes \(\frac{\frac{j}{2}-1}{\frac{j}{2}T}=\frac{-j\left ( \frac{j}{2}-1\right ) }{\frac{1}{2}T}=\frac{\left ( \frac{-j^{2}}{2}-j\right ) }{\frac{1}{2}T}=\frac{\left ( \frac{1}{2}-j\right ) }{\frac{1}{2}T}\). Hence the real part of this pole is \(\frac{1}{T}\), which is \(>0\) since \(T\) is positive. Hence \(H\left ( s\right ) \) is unstable. Hence, starting with stable \(H\left ( z\right ) \), using this transformation, the resulting \(H\left ( s\right ) \) is not always stable. (it depends on the location of the z pole), sometimes we get stable \(H\left ( s\right ) \) and sometimes unstable \(H\left ( s\right ) \). For example, if we have used \(q=\frac{1}{2}\), then doing the above results in \(\frac{\frac{1}{2}-1}{\frac{1}{2}T}=-\frac{1}{T}\) which is \(<0\) since \(T\) is positive. Hence we see that depending on the z pole, the resulting \(H\left ( s\right ) \) can be stable or not.