Напиши подробное решение номера 1, на английском, ...

Problem 1 is: design a Chebyshev analog low-pass filter with given specs and then obtain the digital filter via impulse invariance, then plot impulse and magnitude responses for both. (HW3_2025.pdf)

Below is a detailed solution in English plus ready-to-run MATLAB code at the end.

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1. Given specifications

• Passband edge frequency (analog):
  $f_p = 18\,\text{MHz}$
• Stopband edge frequency (analog):
  $f_s = 20\,\text{MHz}$
• Passband ripple:
  $A_p = 0.3\,\text{dB}$
• Stopband attenuation:
  $A_s = 60\,\text{dB}$
• Sampling frequency for the digital filter:
  $F_s = 50\,\text{MHz}$
  so sampling period
  $T = \dfrac{1}{F_s} = 20\,\text{ns}$

We first design the analog Chebyshev Type I low-pass filter, then convert it to a digital IIR filter by impulse invariance.

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2. Convert specs to angular frequencies

Use

$$\Omega = 2\pi f$$

• Passband edge (rad/s): $$\Omega_p = 2\pi f_p = 2\pi \cdot 18 \times 10^6$$
• Stopband edge (rad/s): $$\Omega_s = 2\pi f_s = 2\pi \cdot 20 \times 10^6$$

In MATLAB we’ll store these as:

matlab
Fp = 18e6;         % Hz
Fs1 = 20e6;        % Hz (stopband edge)
Wp = 2*pi*Fp;      % rad/s
Ws = 2*pi*Fs1;     % rad/s

Note: the ratio of stopband to passband edge (in rad/s or Hz) is the same:

$$\frac{\Omega_s}{\Omega_p} = \frac{f_s}{f_p} = \frac{20}{18} \approx 1.1111$$

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3. Chebyshev Type I filter parameters

A Chebyshev Type I lowpass filter has magnitude squared:

$$|H_a(j\Omega)|^2 = \frac{1}{1 + \varepsilon^2\,T_N^2\left(\frac{\Omega}{\Omega_c}\right)}$$

Where:

• $N$ is filter order,
• $T_N(\cdot)$ is the Chebyshev polynomial of degree $N$,
• $\varepsilon$ is the ripple factor,
• $\Omega_c$ is the cutoff parameter.

3.1 Ripple factor $\varepsilon$

Passband ripple in dB is related to $\varepsilon$ via:

$$A_p = 10\log_{10}\left(1+\varepsilon^2\right) \quad\Rightarrow\quad \varepsilon = \sqrt{10^{A_p/10} - 1}$$

For $A_p = 0.3\,\text{dB}$:

$$\varepsilon = \sqrt{10^{0.3/10} - 1} \approx 0.2674$$

(We’ll let MATLAB compute this numerically for us.)

3.2 Stopband condition and order $N$

The stopband attenuation is:

$$A_s = 10\log_{10}\left(1+\varepsilon^2 T_N^2\left(\frac{\Omega_s}{\Omega_c}\right)\right)$$

For a lowpass with passband edge at $\Omega_p$, we usually normalize such that $\Omega_p$ is at the passband edge and we compare $\Omega_s/\Omega_p$.

The formula for the minimum Chebyshev order is:

$$N \ge \frac{\cosh^{-1}\left(\sqrt{\frac{10^{A_s/10}-1}{10^{A_p/10}-1}}\right)} {\cosh^{-1}\left(\frac{\Omega_s}{\Omega_p}\right)}$$

Let

$$\begin{aligned} \varepsilon &= \sqrt{10^{A_p/10} - 1},\\ A &= \sqrt{10^{A_s/10} - 1}, \\ k &= \frac{\Omega_s}{\Omega_p} = \frac{f_s}{f_p}. \end{aligned}$$

Then

$$N \ge \frac{\cosh^{-1}(A/\varepsilon)}{\cosh^{-1}(k)}.$$

Numerically (approximate):

• $\varepsilon \approx 0.2674$
• $A \approx 1000$
• $k = 20/18 \approx 1.1111$
• $N \approx 19.09 \Rightarrow N = 20$ (must be an integer, so we round up).

So the analog Chebyshev filter will be order 20.

In practice, MATLAB can calculate $N$ and the cutoff frequency $W_c$ for us with cheb1ord.

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4. Analog Chebyshev filter design (in MATLAB)

We now design the analog lowpass with these specs:

matlab
Rp = 0.3;         % passband ripple in dB
Rs = 60;          % stopband attenuation in dB

[N, Wc] = cheb1ord(Wp, Ws, Rp, Rs, 's');    % analog (s-plane) design
% N  = filter order  (should come out as 20)
% Wc = cutoff rad/s  (Chebyshev cutoff)

Then use cheby1 to get numerator and denominator of $H_a(s)$:

matlab
[ba, aa] = cheby1(N, Rp, Wc, 'low', 's');  % analog Chebyshev Type I, lowpass

So the analog transfer function is:

$$H_a(s) = \frac{B_a(s)}{A_a(s)} = \frac{b_0 s^{N} + b_1 s^{N-1} + \cdots + b_N} {s^{N} + a_1 s^{N-1} + \cdots + a_N}$$

where ba and aa store the coefficients of the numerator and denominator.

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5. Impulse invariance method

5.1 Concept

Let:

• $h_a(t)$ be the impulse response of the analog filter $H_a(s)$.
• Sampling period $T = 1/F_s$.

Impulse invariance defines the digital impulse response as:

$$h[n] = T \, h_a(nT), \quad n = 0,1,2,\ldots$$

This ensures that each analog pole $s_k$ maps to a digital pole

$$z_k = e^{s_k T}.$$

The corresponding digital transfer function can be found by a partial fraction expansion or, more conveniently, with MATLAB’s impinvar function:

matlab
[bz, az] = impinvar(ba, aa, Fs);

This directly gives you the discrete-time filter:

$$H(z) = \frac{B(z)}{A(z)} = \frac{b_0 + b_1 z^{-1} + \cdots + b_M z^{-M}} {1 + a_1 z^{-1} + \cdots + a_M z^{-M}}$$

Note: Impulse invariance preserves the shape of the impulse response (sampled) but causes aliasing in the frequency domain; here we assume that the analog filter is sufficiently attenuated above Nyquist (25 MHz) so that aliasing is not critical.

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6. Impulse responses (analog and digital)

6.1 Analog impulse response

We can obtain $h_a(t)$ for a selected time interval using Control System Toolbox:

matlab
t_max = 10/Fp;                 % e.g. several periods of passband edge
t = linspace(0, t_max, 2000);  % time vector
Ha_sys = tf(ba, aa);           % continuous-time transfer function
ha = impulse(Ha_sys, t);       % analog impulse response

6.2 Digital impulse response

For the digital filter we use impz:

matlab
N_imp = 200;                   % number of samples to show
[hd, n] = impz(bz, az, N_imp); % digital impulse response

We then plot and compare:

• Analog: $h_a(t)$ vs. $t$
• Digital: $h[n]$ vs. $n$

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7. Magnitude responses (analog and digital)

7.1 Analog magnitude response

Use freqs for continuous-time filters. To compare with the digital case we’ll plot Analog frequency in Hz rather than rad/s:

matlab
w_max = 2*pi*25e6;                 % up to 25 MHz (Nyquist of digital system)
w = linspace(0, w_max, 4096);      % rad/s
Ha_w = freqs(ba, aa, w);           % analog frequency response
fa = w / (2*pi);                   % Hz
MagHa_dB = 20*log10(abs(Ha_w));    % magnitude in dB

7.2 Digital magnitude response

Use freqz with sampling frequency Fs so the frequency axis is in Hz:

matlab
[Hd_w, fd] = freqz(bz, az, 4096, Fs);   % fd in Hz
MagHd_dB = 20*log10(abs(Hd_w));

We can then plot both on the same figure (or separate figures) to compare magnitude responses as functions of frequency up to 25 MHz.

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8. Complete MATLAB code

Below is a single self-contained MATLAB script that:

1. Designs the analog Chebyshev filter,
2. Converts it to digital using impulse invariance,
3. Plots impulse responses of analog and digital filters,
4. Plots magnitude responses of analog and digital filters.

matlab
%% Problem 1: Chebyshev LPF + Impulse Invariance

clear; clc; close all;

%% Given specifications
Fp  = 18e6;        % passband edge frequency (Hz)
Fs1 = 20e6;        % stopband edge frequency (Hz)
Rp  = 0.3;         % passband ripple (dB)
Rs  = 60;          % stopband attenuation (dB)
Fs  = 50e6;        % sampling frequency for digital filter (Hz)
T   = 1/Fs;        % sampling period (s)

% Analog radian frequencies
Wp = 2*pi*Fp;      % passband edge (rad/s)
Ws = 2*pi*Fs1;     % stopband edge (rad/s)

%% 1) Analog Chebyshev Type I design
% Determine filter order and cutoff
[N, Wc] = cheb1ord(Wp, Ws, Rp, Rs, 's');
fprintf('Analog Chebyshev filter order N = %d\n', N);

% Design analog Chebyshev Type I lowpass (s-domain)
[ba, aa] = cheby1(N, Rp, Wc, 'low', 's');

% Analog transfer function H_a(s)
Ha_sys = tf(ba, aa);   % requires Control System Toolbox

%% 2) Impulse invariance: analog -> digital
% Obtain digital IIR filter using impulse invariance
[bz, az] = impinvar(ba, aa, Fs);

%% 3) Impulse responses

% --- Analog impulse response ---
t_max = 10/Fp;                     % choose a suitable time span
t = linspace(0, t_max, 2000);      % time vector
ha = impulse(Ha_sys, t);

% --- Digital impulse response ---
N_imp = 200;                       % number of samples for impulse response
[hd, n] = impz(bz, az, N_imp);     % digital impulse response

% Plot impulse responses
figure;
subplot(2,1,1);
plot(t, ha, 'LineWidth', 1.2);
grid on;
xlabel('t (s)');
ylabel('h_a(t)');
title('Analog Impulse Response');

subplot(2,1,2);
stem(n, hd, 'filled');
grid on;
xlabel('n');
ylabel('h[n]');
title('Digital Impulse Response (Impulse Invariance)');

%% 4) Magnitude responses

% --- Analog magnitude response ---
w_max = 2*pi*25e6;             % plot up to 25 MHz (Nyquist of digital system)
w = linspace(0, w_max, 4096);  % rad/s
Ha_w = freqs(ba, aa, w);       % analog freq response
fa = w / (2*pi);               % frequency axis in Hz
MagHa_dB = 20*log10(abs(Ha_w));

% --- Digital magnitude response ---
[Hd_w, fd] = freqz(bz, az, 4096, Fs);   % fd in Hz
MagHd_dB = 20*log10(abs(Hd_w));

% Plot magnitude responses
figure;
subplot(2,1,1);
plot(fa/1e6, MagHa_dB, 'LineWidth', 1.2);
grid on;
xlabel('Frequency (MHz)');
ylabel('|H_a(j\Omega)| (dB)');
title('Analog Chebyshev LPF Magnitude Response');
xlim([0 25]);    % show 0–25 MHz

subplot(2,1,2);
plot(fd/1e6, MagHd_dB, 'LineWidth', 1.2);
grid on;
xlabel('Frequency (MHz)');
ylabel('|H_d(e^{j\omega})| (dB)');
title('Digital IIR (Impulse Invariance) Magnitude Response');
xlim([0 25]);    % 0–25 MHz (Nyquist)

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If you’d like, I can also add comments showing how to extract and write down the difference equation of the resulting digital filter $H(z)$ from the bz and az coefficients.