Introduction to
Signals & Systems

Learning outcomes

By the end of this lecture, you will be able to:

• Classify signals as periodic or aperiodic, causal or non-causal, energy or power, and even or odd.
• Test whether a system is linear and/or shift-invariant.
• Explain why the impulse response completely describes an LTI or LSI system.
• Use convolution to calculate a system output and simplify systems connected in cascade or in parallel.

iNotation used in this lecture

Three classic DSP problems

Case study · Hubble's spherical aberration

$$ \frac{U_{\text{in}} - U_{\text{out}}}{R} = \frac{1}{L}\int_{-\infty}^{t} U_{\text{out}}(\tau)\,d\tau + C\,\frac{dU_{\text{out}}}{dt}. $$

(1.1)

$$ \frac{U_{\text{in}}'-U_{\text{out}}'}{R}=\frac{1}{L}U_{\text{out}}+C\,U_{\text{out}}''. $$

$$ C\,U_{\text{out}}''+\frac{1}{R}U_{\text{out}}'+\frac{1}{L}U_{\text{out}}=\frac{1}{R}U_{\text{in}}' \quad\Big|\ \div\,C $$

$$ U_{\text{out}}'' + \frac{1}{RC}U_{\text{out}}' + \frac{1}{LC}U_{\text{out}} = \frac{1}{RC}U_{\text{in}}'. $$

(1.2)

Why an LCCDE is the target

$$\sum_{k=0}^{N}a_k\frac{d^{k}y}{dt^{k}}=\sum_{m=0}^{M}b_m\frac{d^{m}x}{dt^{m}}.$$

Analog circuits are not limited by a sampling rate and introduce no data-conversion delay. Passive networks may need no power. They are also essential around every converter: an anti-alias filter is needed before an ADC, and a reconstruction filter is needed after a DAC.

Digital filters are reproducible and easy to retune. They can provide exact linear phase, adapt while operating, work at several sample rates, and process stored data forwards or backwards. The main cost of a more complex filter is additional computation.

Where this module is heading

Deterministic vs. random

A resistor produces a small random voltage because its charge carriers move due to thermal energy. This is called Johnson–Nyquist noise. We cannot predict the next sample, but we can describe its statistics. The amplitudes have a zero-mean Gaussian distribution with variance \(\sigma^{2}=4kTRB\), where \(k\) is Boltzmann's constant, \(T\) is temperature, \(R\) is resistance and \(B\) is measurement bandwidth. Its power spectral density is approximately flat, so the noise is called white. For a \(1\,\text{k}\Omega\) resistor at room temperature over a \(10\,\text{kHz}\) bandwidth, \(\sigma\approx0.4\,\mu\text{V}\). This is small, but it contributes to the noise floor of every measurement.

Gaussian and white describe different properties. Gaussian describes the distribution of amplitude values. White describes how power is distributed over frequency. A signal can have one property without the other. Passing white Gaussian noise through the resonator in Figure 1.1.d keeps the amplitude distribution Gaussian, but the output spectrum is shaped by \(|H|^{2}\) and is no longer white.

Why this matters. The first part of the module mainly uses deterministic signals, because their values are known and can be transformed exactly. Real measurements also contain random noise, and signals such as speech, biosignals and radar returns are naturally variable. Statistical signal processing describes these signals using correlation, power spectral density, detection and estimation. The deterministic tools developed here remain essential because they describe how a known system acts on both the desired signal and the noise.

Periodic vs. aperiodic

Why this matters. Periodic and aperiodic signals are usually measured differently:

A pulse or decaying exponential can have finite total energy. A nonzero periodic signal continues forever, so its total energy is infinite. However, its average power can be finite. To see this, divide a periodic signal into identical periods. Each period contains the same finite energy \(E_k\), but there are infinitely many periods:

Therefore, a nonzero periodic signal is treated as a power signal rather than an energy signal. This distinction also determines the frequency-domain tool used in Lecture 2. Periodic signals are represented by a Fourier series with discrete harmonics, while aperiodic energy signals use a continuous Fourier transform. Parseval's theorem then connects energy or power in time with energy or power in frequency.

Causal vs. non-causal

Why this matters. A real-time filter, control system or live audio effect must be causal because future samples do not yet exist. Offline processing can use samples on both sides of the current point because the whole data record is already available. For example, an ideal low-pass filter has a two-sided, non-causal impulse response. A practical FIR design delays and truncates that response so it can run in real time.

Causality is also linked to stability. A rational transfer function may represent different time-domain signals depending on its region of convergence. For a causal continuous-time system, the region of convergence lies to the right of the rightmost pole. For the same system to be BIBO stable, this region must include the \(j\Omega\)-axis. Therefore, all poles of a causal stable system must lie in the open left-half-plane.

Feedforward FIR filters are always stable when their coefficients are finite. Recursive IIR filters use past outputs and can achieve a sharp response with fewer coefficients, but their pole positions must be checked. This is an important design trade-off developed in Lecture 6.

Even vs. odd

Why this matters. Every signal can be written uniquely as the sum of an even part and an odd part:

This decomposition often simplifies analysis. It also explains several patterns that appear later. For a real system, complex poles occur in conjugate pairs \(p,p^*=-\alpha\pm j\beta\). Such a pair produces sine and cosine terms under the same decaying envelope.

The same idea appears in the Fourier transform. For a real-valued signal, \(X(\Omega)\) has conjugate symmetry: its real part is even and its imaginary part is odd.

Therefore, even and odd symmetry is useful in three connected places: decomposing a signal, describing damped sinusoids, and interpreting the real and imaginary parts of a spectrum.

The Dirac delta (unit impulse)

Figure 1.2.a · \(\delta(t)\), \(\delta[n]\)

Why this matters. The unit impulse is a standard test input. Any signal can be represented as a collection of shifted impulses with different weights. If we know how an LTI or LSI system responds to one impulse, we can add the shifted and scaled responses to find the output for any input. The response to the unit impulse is called \(h(t)\) or \(h[n]\). In discrete time, the impulse can also be written as the change in the unit step: \(\delta[n]=u[n]-u[n-1]\).

The unit step

Figure 1.2.b · \(u(t)\), \(u[n]\)

Why this matters. The unit step switches a signal on at \(t=0\). For example, \(e^{-at}u(t)\) is a decaying exponential that exists only for \(t\ge0\). The step response is also a useful way to test a real system because a sudden change is easier to generate than an ideal impulse. For an LTI system, the step response is the running integral of the impulse response.

The rectangular pulse

Figure 1.2.c · \(p(t)\), \(p[n]\)

Why this matters. A rectangular pulse can be built from two steps: \(p(t)=u(t-a)-u(t-b)\). The first step turns the pulse on at \(t=a\), and the second turns it off at \(t=b\). This gives a simple way to select a finite part of a signal. The MATLAB convolution example uses such a pulse, and Lecture 7 uses related window functions to design FIR filters.

Linearity

Why this matters in practice. Superposition lets us study one component at a time. For example, we can calculate the response to a useful signal and to noise separately, then add the two outputs. Transfer functions, frequency responses and pole–zero methods all depend on this property. Physical devices are linear only over a limited operating range. If an amplifier clips or saturates, it creates new frequency components and the linear model is no longer accurate.

Time invariance

Why this matters in practice. If a filter is time invariant, a response measured today can be used to predict its response tomorrow. Real components change slightly with temperature, age and supply voltage, so perfect time invariance is an ideal model. Calibration, regulated power supplies and temperature compensation help real hardware remain close enough to this model for fixed-coefficient filter theory to work.

Impulse response

Why this matters in practice. Once \(h\) is known, convolution predicts the output for any input. Audio engineers measure a room impulse response to reproduce its reverberation. Communication engineers estimate a channel response before equalization. Control engineers apply short test inputs to identify a plant. This single-response description works only because the system is both linear and time invariant.

The sifting property

$$ \int_{-\infty}^{\infty} x(\tau)\,\delta(\tau - t_0)\,d\tau = x(t_0), \qquad \sum_{k=-\infty}^{\infty} x[k]\,\delta[n-k] = x[n]. $$

$$ x[n] = 2\delta[n] - \delta[n-1] + 3\delta[n-2]. $$

$$ \delta[n] \Rightarrow \mathcal{T}\{\cdot\} \Rightarrow h[n], \qquad \delta[n-k] \Rightarrow \mathcal{T}\{\cdot\} \Rightarrow h[n-k]. $$

(1.11)

$$ x[k]\,\delta[n-k] \;\Rightarrow\; \mathcal{T}\{\cdot\} \;\Rightarrow\; x[k]\,h[n-k]. $$

(1.12)

$$\underbrace{\sum_{k=-\infty}^{\infty} x[k]\,\delta[n-k]}_{=\,x[n]} \;\Rightarrow\; \mathcal{T}\{\cdot\} \;\Rightarrow\; \underbrace{\sum_{k=-\infty}^{\infty} x[k]\,h[n-k]}_{\text{must equal }y[n]}.$$

(1.13)

$$\boxed{\,y[n] = \sum_{k=-\infty}^{\infty} x[k]\,h[n-k] \;=\; (x*h)[n]\,.}$$

(1.14)

$$\boxed{\,y(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau \;=\; x(t)*h(t) \;=\; h(t)*x(t)\,.}$$

(1.15)

Swapping the two functions does not change the result. Either one may be reversed and shifted during a graphical calculation.

Several systems in cascade may be grouped in any order. Their overall impulse response is \(h_1*h_2\).

For systems in parallel, add the branch impulse responses before or after convolving with the input.

Convolution with an impulse leaves a signal unchanged. Shifting the impulse shifts the output by the same amount.

Why not just stop at convolution?

save_convolution_animations.m · continuous-time part

%% ---------------- Continuous-Time Convolution Animation (Square * Right-Triangle)
% Parameters and signals (as in your code)
dt   = 0.0025;
tau  = -1:dt:4;
tmin = -1; tmax = 5;
tvec = -1:0.02:5;
y    = zeros(size(tvec));
 
x_of = @(t) double((t>=0) & (t<=2));
h_of = @(t) double((t>=0) & (t<=1)).*(1 - t);
x_tau = x_of(tau);
 
% Create figure
figCT = figure('Color','w','Position',[80 80 1200 880]);
 
ax1 = subplot(3,1,1);
plot(tau, x_tau, 'LineWidth', 2); hold on; grid on;
ph  = plot(tau, zeros(size(tau)), 'r--', 'LineWidth', 2);
xlabel('\tau'); ylabel('Amplitude');
title('x(\tau) and h(t-\tau)'); xlim([tmin tmax]); ylim([-0.2 1.2]);
legend('x(\tau)','h(t-\tau)','Location','northeast');
 
ax2 = subplot(3,1,2);
pprod = plot(tau, zeros(size(tau)), 'm', 'LineWidth', 2); hold on; grid on;
pfill = area(tau, zeros(size(tau)));
pfill.FaceAlpha = 0.15; pfill.FaceColor = [1 0 1];
xlabel('\tau'); ylabel('Product');
title('Product x(\tau)\cdot h(t-\tau) (shaded area = y(t))');
xlim([tmin tmax]); ylim([-0.2 1.2]);
 
ax3 = subplot(3,1,3);
py  = plot(tvec, zeros(size(tvec)), 'k', 'LineWidth', 2); hold on; grid on;
pc  = plot(NaN, NaN, 'ro', 'MarkerSize', 6, 'MarkerFaceColor', 'r');
xlabel('t'); ylabel('y(t)');
title('Convolution Output y(t) = \int x(\tau)h(t-\tau)\,d\tau','Interpreter','latex');
xlim([tmin tmax]); ylim([-0.05 1.1]);
 
sgtitle('Convolution Animation: Square x(t) * Right-Triangle h(t)','FontWeight','bold');
 
% Prepare video and gif writers for continuous animation
vidCT = VideoWriter('conv_continuous.mp4','MPEG-4');
vidCT.FrameRate = 30;
open(vidCT);
gifCT = 'conv_continuous.gif';
first_frame_ct = true;
 
% Animation loop (capture frames)
for i = 1:numel(tvec)
    t0 = tvec(i);
    h_shift = h_of(t0 - tau);
    prod_tau = x_tau .* h_shift;
    y(i)     = trapz(tau, prod_tau);
 
    set(ph,    'YData', h_shift);
    set(pprod, 'YData', prod_tau);
    set(pfill, 'YData', prod_tau);
    set(py, 'YData', y);
    set(pc, 'XData', t0, 'YData', y(i));
 
    title(ax1, sprintf('x(\\tau) and h(t-\\tau),   t = %.2f', t0));
    title(ax2, sprintf('Overlap area = y(%.2f) = %.4f', t0, y(i)));
 
    drawnow;
 
    % write video frame
    frame = getframe(figCT);
    writeVideo(vidCT, frame);
 
    % write gif frame
    [img, cmap] = frame2im(frame);
    [idx, cm] = rgb2ind(img, 256);
    if first_frame_ct
        imwrite(idx, cm, gifCT, 'gif', 'LoopCount', Inf, 'DelayTime', 1/30);
        first_frame_ct = false;
    else
        imwrite(idx, cm, gifCT, 'gif', 'WriteMode', 'append', 'DelayTime', 1/30);
    end
end
 
close(vidCT);
disp('Continuous-time animation saved: conv_continuous.mp4 and conv_continuous.gif');
 
pause(0.5); % small pause to ensure file flush

save_convolution_animations.m · discrete-time part

%% ---------------- Discrete-Time Convolution Animation (Square * Right-Triangle)
m = -20:1:20;
x_m = double(m>=0 & m<=10);
h_of_d = @(n) double(n>=0 & n<=5) .* (1 - n/5);
n_out = -20:20;
y_d = zeros(size(n_out));
 
figDT = figure('Color','w','Position',[80 80 1200 900]);
 
ax1 = subplot(3,1,1);
stem(m, x_m, 'filled','LineWidth',1.4); hold on; grid on;
sh = stem(m, zeros(size(m)), 'r','filled','LineWidth',1.2);
xlabel('m'); ylabel('Amplitude');
title('x[m] and h[n_0 - m]');
xlim([min(m) max(m)]); ylim([-0.1 1.2]);
legend('x[m]','h[n_0-m]','Location','northeast');
 
ax2 = subplot(3,1,2);
sp = stem(m, zeros(size(m)), 'm','filled','LineWidth',1.4); hold on; grid on;
xlabel('m'); ylabel('Product');
title('x[m]\cdot h[n_0 - m]  (sum of stems = y[n_0])');
xlim([min(m) max(m)]); ylim([-0.1 1.2]);
 
ax3 = subplot(3,1,3);
syplot = stem(n_out, zeros(size(n_out)), 'k','filled','LineWidth',1.6); hold on; grid on;
scurr  = stem(nan, nan, 'r','filled','LineWidth',1.8);
xlabel('n'); ylabel('y[n]');
title('Convolution Output y[n] = \Sigma_m x[m]h[n-m]');
xlim([min(n_out)-1 max(n_out)+1]);
 
sgtitle('Discrete-Time Convolution Animation: Full Range (-\infty to +\infty)','FontWeight','bold');
 
% Prepare video and gif writers for discrete animation
vidDT = VideoWriter('conv_discrete.mp4','MPEG-4');
vidDT.FrameRate = 10; % coarser frame rate to match pauses
open(vidDT);
gifDT = 'conv_discrete.gif';
first_frame_dt = true;
 
% Animation loop (capture frames)
for ii = 1:numel(n_out)
    n0 = n_out(ii);
    h_shift = h_of_d(n0 - m);
    prod_m = x_m .* h_shift;
    y_d(ii) = sum(prod_m);
 
    set(sh, 'YData', h_shift);
    set(sp, 'YData', prod_m);
    set(syplot, 'YData', y_d);
    set(scurr, 'XData', n0, 'YData', y_d(ii));
 
    title(ax1, sprintf('x[m] and h[n_0 - m],  n_0 = %d', n0));
    title(ax2, sprintf('Sum of magenta stems = y[%d] = %.4f', n0, y_d(ii)));
 
    drawnow;
    pause(0.05); % small pause for smooth capture
 
    % write video frame
    frame = getframe(figDT);
    writeVideo(vidDT, frame);
 
    % write gif frame
    [img, cmap] = frame2im(frame);
    [idx, cm] = rgb2ind(img, 256);
    if first_frame_dt
        imwrite(idx, cm, gifDT, 'gif', 'LoopCount', Inf, 'DelayTime', 0.3);
        first_frame_dt = false;
    else
        imwrite(idx, cm, gifDT, 'gif', 'WriteMode', 'append', 'DelayTime', 0.3);
    end
end
 
close(vidDT);
disp('Discrete-time animation saved: conv_discrete.mp4 and conv_discrete.gif');

MATLAB corner · verifying the properties of convolution

PROBLEM 1.1

For \(x(t)=e^{-2t}u(t)\), decide whether the signal is periodic or aperiodic, causal or non-causal, and an energy or power signal. Then calculate its energy.

PROBLEM 1.2

An LTI system has impulse response \(h(t)=e^{-3t}u(t)\). Use convolution to find its response to the unit-step input \(x(t)=u(t)\).

PROBLEM 1.3

Determine whether \(x[n] = \cos\!\big(\tfrac{3\pi}{7}n - \tfrac{\pi}{8}\big)\) is periodic, and if so, find its fundamental period \(N\).

PROBLEM 1.4

Let \(x[n]=a^n u[n]\) and \(h[n]=b^n u[n]\), with \(0<a,b<1\) and \(a\ne b\). Find \(y[n]=x[n]*h[n]\).

PROBLEM 1.5

For each input–output relationship, determine whether the system is linear and/or shift-invariant: (a) \(y[n] = 2x[n] + 3\). (b) \(y[n] = x[n]\sin(\omega_0 n)\). (c) \(y[n] = \big(x[n]\big)^2\). (d) \(y[n] = \displaystyle\sum_{m=-\infty}^{n} x[m]\).

PROBLEM 1.6

Two LSI systems are cascaded: \(h_1[n]=\delta[n]-\delta[n-3]\) then \(h_2[n]=(0.8)^n u[n]\). For \(x[n]=u[n]\), find \(y[n]\) two ways — (a) \((x*h_1)*h_2\), (b) \(x*(h_1*h_2)\) — and confirm they agree.

PROBLEM 1.7*

(a) Show that \(x[n]=z^n\), \(z\) a complex constant, is an eigenfunction of any LSI system. (b) By a counterexample, show that \(x[n]=z^n u[n]\) is not an eigenfunction. (* optional)