Подробно реши первый номер, ответ на английском, к...

Here is a detailed solution for Problem 1 (in English) with MATLAB code at the end. (HW3_2025.pdf)

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1. Problem restatement

We want a digital Chebyshev type I lowpass filter obtained via impulse invariance from an analog prototype.

Given:

• Passband edge frequency: $F_p = 18\,\text{MHz}$
• Stopband edge frequency: $F_s = 20\,\text{MHz}$
• Passband ripple: $R_p = 0.3\,\text{dB}$
• Stopband attenuation: $R_s = 60\,\text{dB}$
• Sampling frequency: $F_\text{samp} = 50\,\text{MHz}$

Tasks:

1. Design the analog Chebyshev I lowpass filter.
2. Convert it to a digital filter using the impulse invariance method.
3. Plot:
   • (a) Impulse responses of both analog and digital systems.
   • (b) Magnitude responses of both analog and digital systems.

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2. Design of the analog Chebyshev I lowpass filter

2.1 Convert to angular frequencies

We first convert the passband and stopband edges from Hz to rad/s:

$$\Omega_p = 2\pi F_p, \quad \Omega_s = 2\pi F_s$$

In MATLAB:

matlab
Fp    = 18e6;   % passband edge in Hz
Fstop = 20e6;   % stopband edge in Hz
Fs    = 50e6;   % sampling frequency in Hz

Wp = 2*pi*Fp;      % rad/s
Ws = 2*pi*Fstop;   % rad/s
Rp = 0.3;          % passband ripple (dB)
Rs = 60;           % stopband attenuation (dB)

2.2 Chebyshev I order formula (theory)

For a type I Chebyshev analog lowpass, the minimum order $N$ that meets the specs is

$$\varepsilon = \sqrt{10^{R_p/10} - 1}$$ $$A = \sqrt{10^{R_s/10} - 1}$$ $$N \ge \frac{\operatorname{arccosh}\left(\dfrac{A}{\varepsilon}\right)}{\operatorname{arccosh}\left(\dfrac{\Omega_s}{\Omega_p}\right)}$$

Numerically (already pre-computed):

• $\varepsilon \approx \sqrt{10^{0.3/10} - 1}$
• $A \approx \sqrt{10^{60/10} - 1}$
• $\Omega_s / \Omega_p = F_s/F_p = 20/18$

This gives approximately:

$$N \approx 19.09$$

The filter order must be an integer, so we choose:

$$\boxed{N = 20}$$

In MATLAB we let cheb1ord do that for us and also return the necessary edge frequency:

matlab
[N, Wc] = cheb1ord(Wp, Ws, Rp, Rs, 's');  % 's' → analog design
% N  will come out as 20
% Wc is the resulting passband edge (cutoff) in rad/s

2.3 Analog lowpass prototype

We now design the analog Chebyshev type I lowpass filter:

$$H_a(s) = \frac{B_a(s)}{A_a(s)}$$

In MATLAB:

matlab
[Ba, Aa] = cheby1(N, Rp, Wc, 's');   % analog Chebyshev I lowpass

Ba and Aa are the numerator and denominator polynomials in $s$.

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3. Impulse invariance: analog → digital

3.1 Theory

Let $h_a(t)$ be the impulse response of the analog filter $H_a(s)$.

Impulse invariance constructs a discrete-time filter whose impulse response is

$$h_d[n] = T \, h_a(nT), \quad T = \frac{1}{F_\text{samp}}$$

So the discrete-time impulse is simply the sampled analog impulse, scaled by $T$. The mapping in pole–residue form:

If

$$H_a(s) = \sum_k \frac{r_k}{s - p_k} \Rightarrow h_a(t) = \sum_k r_k e^{p_k t} u(t)$$

then after impulse invariance,

$$h_d[n] = T \sum_k r_k e^{p_k nT}$$

and the discrete-time transfer function becomes

$$H_d(z) = \sum_k \frac{T r_k}{1 - e^{p_k T} z^{-1}}$$

In MATLAB we don’t have to do this by hand; we use impinvar:

matlab
T = 1/Fs;
[Bd, Ad] = impinvar(Ba, Aa, Fs);   % digital filter coefficients

So the digital filter is

$$H_d(z) = \frac{B_d(z)}{A_d(z)} = \frac{b_0 + b_1 z^{-1} + \dots + b_M z^{-M}} {1 + a_1 z^{-1} + \dots + a_M z^{-M}}$$

with Bd containing $\{b_k\}$ and Ad containing $\{1, a_1,\dots,a_M\}$.

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4. Impulse responses (part a)

4.1 Analog impulse response

Using partial fraction expansion:

$$H_a(s) = \sum_{k=1}^N \frac{r_k}{s - p_k}$$

(For our proper filter, there is no polynomial $k(s)$ term.)

Then

$$h_a(t) = \sum_{k=1}^N r_k e^{p_k t} u(t)$$

We can compute this with residue and evaluate at a chosen time grid $t$.

In MATLAB:

matlab
% Time axis for viewing the analog impulse response
tmax = 4e-6;                 % 4 microseconds (adjust as needed)
Nt   = 2000;
t    = linspace(0, tmax, Nt);

% Partial fraction expansion for analog filter
[r, p, k] = residue(Ba, Aa); % Ba/Aa in s-domain

% Analog impulse response h_a(t) = sum r_k e^{p_k t}, k is empty for proper LPF
ha = zeros(size(t));
for i = 1:length(r)
    ha = ha + r(i)*exp(p(i)*t);
end

4.2 Digital impulse response

We use impz:

matlab
L  = 200;                   % number of samples to show
[hd, n] = impz(Bd, Ad, L);  % digital impulse response
t_samples = n / Fs;         % map discrete-time index to seconds

If you want to directly check impulse invariance:

matlab
% Sample the analog impulse at the same instants n*T
t_n = n / Fs;
ha_samples = zeros(size(t_n));
for i = 1:length(r)
    ha_samples = ha_samples + r(i)*exp(p(i)*t_n);
end

hd_theoretical = T * ha_samples;  % should match hd up to numerical error

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5. Magnitude responses (part b)

5.1 Analog magnitude response

We use freqs to evaluate $H_a(j\Omega)$ over a frequency range 0 to $F_\text{Nyquist} = F_\text{samp}/2$.

matlab
Nfreq = 2048;
w_max = 2*pi*(Fs/2);                    % up to Nyquist frequency in rad/s
w     = linspace(0, w_max, Nfreq);      % analog rad/s
[Ha, ~] = freqs(Ba, Aa, w);

fa    = w/(2*pi);                       % Hz
Ha_dB = 20*log10(abs(Ha));

5.2 Digital magnitude response

We use freqz with sampling frequency Fs (this makes the frequency vector in Hz):

matlab
[Hd, fd] = freqz(Bd, Ad, Nfreq, Fs);    % fd in Hz
Hd_dB    = 20*log10(abs(Hd));

Now we can plot both analog and digital magnitude responses on the same frequency axis (in MHz).

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6. Complete MATLAB code for Problem 1

matlab
%% Problem 1: Chebyshev I LPF via Impulse Invariance

clear; clc; close all;

%% Given specifications
Fp    = 18e6;    % passband edge in Hz
Fstop = 20e6;    % stopband edge in Hz
Fs    = 50e6;    % sampling frequency in Hz

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

T = 1/Fs;        % sampling period

%% 1) Analog Chebyshev I lowpass design
% Convert to rad/s
Wp = 2*pi*Fp;
Ws = 2*pi*Fstop;

% Find minimum order and cutoff
[N, Wc] = cheb1ord(Wp, Ws, Rp, Rs, 's');
fprintf('Analog Chebyshev-I order N = %d\n', N);

% Design analog filter
[Ba, Aa] = cheby1(N, Rp, Wc, 's');  % analog lowpass H_a(s) = Ba(s)/Aa(s)

%% 2) Impulse invariance to get digital filter
[Bd, Ad] = impinvar(Ba, Aa, Fs);    % digital H_d(z) via impulse invariance

%% 3) Impulse responses

% --- Analog impulse response h_a(t) using partial fractions ---
tmax = 4e-6;                        % end time (seconds) to view analog impulse
Nt   = 2000;
t    = linspace(0, tmax, Nt);

% Partial fraction expansion in s-domain
[r, p, k] = residue(Ba, Aa);        %#ok<ASGLU>  k should be empty for proper LPF

% h_a(t) = sum r_k * exp(p_k * t)
ha = zeros(size(t));
for i = 1:length(r)
    ha = ha + r(i).*exp(p(i).*t);
end

% --- Digital impulse response h_d[n] ---
L = 200;                            % number of samples to plot
[hd, n] = impz(Bd, Ad, L);          % discrete-time impulse response
t_samples = n/Fs;                   % map discrete index to seconds

% Optional: check impulse invariance (T * h_a(nT) vs hd)
ha_nT = zeros(size(n));
for i = 1:length(r)
    ha_nT = ha_nT + r(i).*exp(p(i).*t_samples);
end
hd_theoretical = T * ha_nT;

%% 4) Plot impulse responses (analog and digital)

figure;
subplot(2,1,1);
plot(t*1e6, ha, 'LineWidth', 1.2);
grid on;
xlabel('Time \mu s');
ylabel('h_a(t)');
title('Analog impulse response');

subplot(2,1,2);
stem(t_samples*1e6, hd, 'filled');
grid on;
xlabel('Time \mu s');
ylabel('h_d[n]');
title('Digital impulse response (impulse invariance)');

% (Optional overlay to visually compare T*h_a(nT) and h_d[n])
figure;
stem(t_samples*1e6, hd, 'filled'); hold on;
plot(t_samples*1e6, hd_theoretical, 'r--', 'LineWidth', 1.2);
grid on;
xlabel('Time \mu s');
ylabel('Amplitude');
legend('h_d[n]', 'T \cdot h_a(nT)', 'Location', 'best');
title('Check of impulse invariance: discrete vs sampled analog');

%% 5) Magnitude responses for analog and digital filters

Nfreq = 2048;
w_max = 2*pi*(Fs/2);                 % 0 to Nyquist (rad/s)

w = linspace(0, w_max, Nfreq);       % analog rad/s frequencies
[Ha, ~] = freqs(Ba, Aa, w);          % analog frequency response
fa = w/(2*pi);                       % Hz
Ha_dB = 20*log10(abs(Ha));

[Hd, fd] = freqz(Bd, Ad, Nfreq, Fs); % digital frequency response, fd in Hz
Hd_dB = 20*log10(abs(Hd));

% Plot magnitude responses (in dB) vs frequency in MHz
figure;
plot(fa/1e6, Ha_dB, 'LineWidth', 1.2); hold on;
plot(fd/1e6, Hd_dB, '--', 'LineWidth', 1.2);
grid on;
xlabel('Frequency (MHz)');
ylabel('Magnitude (dB)');
title('Magnitude responses: analog vs digital');
legend('Analog H_a(j\Omega)', 'Digital H_d(e^{j\omega})', 'Location', 'best');

% Optionally highlight passband/stopband edges
xline(Fp/1e6, 'g--', '18 MHz');
xline(Fstop/1e6, 'r--', '20 MHz');

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If you want, next we can also derive the difference equation for the digital filter from Bd and Ad, but for Problem 1 the main requirements are the filter design and the impulse/frequency plots.

Вот готовый MATLAB-код, который строит импульсные и частотные характеристики для Задачи 2 (аналоговый фильтр $H(s) = \dfrac{s+2}{s^2 + s + 1}$ и его цифровой прототип по билинейному преобразованию, $F_s = 20$ Гц). (HW3_2025.pdf)

matlab
%% Problem 2: Plots for analog & digital filters (bilinear transform)
clear; clc; close all;

%% Given data
Fs = 20;              % sampling frequency in Hz
T  = 1/Fs;            % sampling period

% Analog transfer function:
% H(s) = (s + 2) / (s^2 + s + 1)
Ba = [1 2];           % numerator in s
Aa = [1 1 1];         % denominator in s

%% Digital filter via bilinear transform
% H(z) is the digital prototype of H(s)
[Bd, Ad] = bilinear(Ba, Aa, Fs);  % requires Signal Processing Toolbox

%% ================== IMPULSE RESPONSES ==================

% ---- Analog impulse response h_a(t) via partial fractions ----
t_max = 3;            % seconds, adjust as needed
Nt    = 2000;         % number of time points
t     = linspace(0, t_max, Nt);

[r, p, k] = residue(Ba, Aa);   % partial fraction expansion in s
ha = zeros(size(t));
for i = 1:length(r)
    ha = ha + r(i) .* exp(p(i).*t);
end
if ~isempty(k)
    % if there were a direct term (here there shouldn't be, but just in case)
    ha = ha + k(1)*ones(size(t));
end

% ---- Digital impulse response h_d[n] ----
L = 80;                           % number of samples to show
[hd, n] = impz(Bd, Ad, L);        % discrete-time impulse response
t_d = n*T;                        % map sample index n to time nT (seconds)

% ---- 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(t_d, hd, 'filled');
grid on;
xlabel('t = nT (s)');
ylabel('h_d[n]');
title('Digital impulse response (bilinear transform)');

%% ================== FREQUENCY RESPONSES ==================

Nf = 1024;                         % number of frequency points

% ---- Analog magnitude response ----
% For comparison with the digital filter we usually look up to pi/T rad/s
w_max = pi/T;                      % max analog rad/s
w     = linspace(0, w_max, Nf);    % rad/s
[Ha, ~] = freqs(Ba, Aa, w);       % analog freq response H(jΩ)
fa = w/(2*pi);                     % convert rad/s to Hz

% ---- Digital magnitude response ----
[Hd, fd] = freqz(Bd, Ad, Nf, Fs);  % digital freq response, fd in Hz

% ---- Plot magnitude responses (linear scale) ----
figure;
plot(fa, abs(Ha), 'LineWidth', 1.2); hold on;
plot(fd, abs(Hd), '--', 'LineWidth', 1.2);
grid on;
xlabel('Frequency (Hz)');
ylabel('|H(j\Omega)|, |H(e^{j\omega})|');
title('Magnitude responses: analog vs digital');
legend('Analog', 'Digital', 'Location', 'best');

% ---- Plot magnitude responses (dB) ----
figure;
plot(fa, 20*log10(abs(Ha)+eps), 'LineWidth', 1.2); hold on;
plot(fd, 20*log10(abs(Hd)+eps), '--', 'LineWidth', 1.2);
grid on;
xlabel('Frequency (Hz)');
ylabel('Magnitude (dB)');
title('Magnitude responses (dB): analog vs digital');
legend('Analog', 'Digital', 'Location', 'best');

Если нужно, могу отдельно выписать разностное уравнение для цифрового фильтра из коэффициентов Bd и Ad.