Analog filter desing
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Analog filter desing

Using ptototype lowpass filter to design highpass filter;

So, the transfer function of prototype lowpass filter: $H_{LP}(s)$

There is the Spectral transformation : $ s = F(\hat{s}) $

Then, we can get the Designed Filter : $HD (\hat{s}) = H{LP} (s)|_{s = F(\hat{s})}$

Highpass Filter Design

Theory

Define desired passband edge of highpass filter: $\hat\Omega_p$

and passband edge of prototype lowpass filter: $\Omega_p$ .

Here gives the Spectral Transformation:
$$
s = \frac{\Omega_p \hat\Omega_p } {\hat{s}}
$$

And on the imaginary axis
$$
\Omega = -\frac{\Omega_p \hat\Omega_p}{\hat\Omega}
$$

Here is the figure showing the relationship between $\Omega$ and $\hat\Omega$

Example

Suppose the passband ripple $\deltap (s) = 0.1$ and the $H{LP}(s) = \frac{1}{1+s}$

Then, Since $ |H_{LP}(\Omega_p) |= 1-\delta_p(s)$, we can get the $\Omega_p = 0.4843$

If we want to get the passband frequency of HighPass Filter $\hat\Omega_p = 5$ . Then the Spectral Transformation is:

$$
s = \frac{0.4843 \times 5}{\hat{s}}
$$

So the Designed HighPass Filter is

$$
HD(\hat{s}) = \frac{1}{1+s} | {s = F(\hat{s})} = \frac{1}{1+\frac{0.4843 \times 5}{\hat{s}}} = \frac{\hat{s}}{\hat{s} + 2.4215}
$$

Here is the plot of this HighPass Filter and it shows that the $\hat\Omega_p = 5$

Bandpass Filter Design

Theory

Desiered passband edges of bandpass filter: $\hat\Omega{p1} , \hat\Omega{p2}$

Passband edge of prototype lowpass filter: $\Omega_p$

Spectral Transformation

$$
s = \Omegap \frac{\hat{s}^2 + \hat\Omega{p1}\hat\Omega{p2}}{\hat{s}(\hat\Omega{p2} - \hat\Omega_{p1})}
$$

And on the imaginary axis

$$
\Omega = -\Omegap \frac{\hat\Omega{p1}\hat\Omega{p2} - \hat\Omega^2}{\hat\Omega(\hat\Omega{p2}-\hat\Omega_{p1})}
$$

Here is the figure showing the relationship between $\Omega$ and $\hat\Omega$

Both passband and stopband edge frequencies must exhibit symmetry with respect to $\hat\Omega_0$

$$
\hat\Omega0^2 = \hat\Omega{p1}\hat\Omega{p2} = \hat\Omega{s1}\hat\Omega_{s2}
$$

Example

Passband is [5,7]
Stopband: [0,3] and above 9
Passband ripple: $\delta_p = 0.01$
Stopband ripple: $\delta_s = 0.005$

The product of passband edge frequencies : $5 \times 7 = 35$
The product of stopband edge frequencies : $3 \times 9 = 27$
Adjust lower stopband edge frequency: $35 / 9 = 3.89$

Suppose the $\Omega_p = 1$, the Spectral Transformation:

$$
s = \Omegap \frac{\hat{s}^2 + \hat\Omega{p1}\hat\Omega{p2}}{\hat{s}(\hat\Omega{p2} - \hat\Omega_{p1})} = \frac{\hat{s}^2 + 35}{2\hat{s}}
$$

On imaginary axis

$$
\Omega = -\Omegap \frac{\hat\Omega{p1}\hat\Omega{p2} - \hat\Omega^2}{\hat\Omega(\hat\Omega{p2}-\hat\Omega_{p1})} = - \frac{35-\hat\Omega^2}{2\hat\Omega}
$$

Stopband edge of prototype lowpass filter

$$
\Omega_s = -\frac{35 -81}{2 \times 9} = 2.56
$$

Then design the specification for prototype lowpass filter

Passband:[0,1]
stopband: above 2.56
Passband ripple: $\delta_p = 0.01$
Stopband ripple: $\delta_s = 0.005$

Here shows the calculation of the prototype lowpass filter

8 order Butterworth filter with cutoff frequency 1.3178 satisfies specifications.

then from Spectral Transformation in imaginary axis function get the magnitude response

$$
|H_a(j\hat\Omega) |^2 = \frac{1}{1+(\Omega/\Omegac)^{2N}}|{\Omega == - \frac{35-\hat\Omega^2}{2\hat\Omega} }
$$

Then plot in MATLAB

Bandstop Filter Design

Desired stopband edges of bandstop filter: $\hat\Omega{s1},\hat\Omega{s2}$
Stopband edge of prototype lowpass filter:: $\Omega_s$

Spectral Transformation

$$
s = \Omegap \frac{\hat{s}(\hat\Omega{s2} - \hat\Omega{s1})}{\hat{s}^2 + \hat\Omega{s1}\hat\Omega_{s2}}
$$

On the imaginary axis

$$
\Omega = -\Omegap \frac{\hat\Omega(\hat\Omega{s2}-\hat\Omega{s1})}{\hat\Omega{s1}\hat\Omega_{s2} - \hat\Omega^2}
$$

Band edges exhibit symmetry with respect to centre frequency

$$
\hat\Omega0^2 = \hat\Omega{p1}\hat\Omega{p2} = \hat\Omega{s1}\hat\Omega_{s2}
$$