Signals, Systems and Transforms

Periodicity of continuous-time signals

A continuous-time signal x(t) is periodic when x(t+T)=x(t) holds for every t and some T>0. The smallest such T is the fundamental period. A sinusoid \cos(\omega t) has fundamental period 2\pi/\omega only when its argument is linear in t. If the argument is a nonlinear function of t, such as a square root, a shift by T changes the argument by an amount that depends on t, so no constant period exists. If x(t) has fundamental period T, then x(at) has fundamental period T/|a|, because x(a(t+T/a))=x(at+T)=x(at). A sum of sinusoids is periodic only when the frequencies are rational multiples of a common \omega_0. Writing each frequency as an integer multiple of \omega_0 and taking the largest such \omega_0 gives the fundamental period 2\pi/\omega_0. For example, components at 6\pi and 4\pi rad/s share \omega_0=2\pi rad/s, so the fundamental period is 1 s. The sum of two complex exponentials satisfies e^{j\omega_1 t}+e^{j\omega_2 t}=2e^{j(\omega_1+\omega_2)t/2}\cos\!\left(\frac{(\omega_1-\omega_2)t}{2}\right), so its magnitude is 2\left|\cos\!\left(\frac{(\omega_1-\omega_2)t}{2}\right)\right|. Taking the absolute value folds the negative half-cycles onto the positive ones, so this magnitude can repeat faster than the signal itself.

Discrete-time convolution

For a discrete-time LTI system with impulse response h[n], the output to x[n] is the convolution y[n]=\sum_k x[k]\,h[n-k]. It is computed by reversing one sequence, shifting it, multiplying sample by sample and summing. If the lengths are L_x and L_h, the full output has L_x+L_h-1 samples.

• impulse response — the output produced by the unit impulse \delta[n]; it completely characterizes an LTI system. Example: if h[n] is nonzero only at n=0 and n=1, the response to \delta[n-3] is the same pattern starting at n=3.
• peak of the output — the largest sample of y[n], which must be found from the full convolution, because the largest input or impulse-response value need not be the largest output value.

A short FIR filter with taps -1, 2, -1 acts as a discrete second difference. It gives zero wherever the input is locally linear, such as on a constant plateau, and produces nonzero samples only near changes in the input. Each isolated step therefore contributes a fixed number of nonzero output samples, two on either side of the step. A causal system that starts at rest responds to a delayed impulse with its impulse response shifted by the same delay, so y[n]=h[n-p]. For a first-order recursion y[n]=c\,y[n-1]+x[n] at rest, the impulse response is h[n]=c^{n}u[n], which is nonzero for all n\ge 0. The output one sample after the impulse therefore equals c, independent of the delay.

Continuous-time convolution

For continuous-time LTI systems, y(t)=\int_{-\infty}^{\infty}x(\tau)\,h(t-\tau)\,d\tau. For causal signals the limits shrink to [0,t], and the integrand is nonzero only where both signals are nonzero. Convolving e^{-at}u(t) with u(t) gives \int_0^t e^{-a\tau}\,d\tau=\frac{1}{a}\left(1-e^{-at}\right) for t\ge 0, so the output approaches the steady value 1/a as t\to\infty. For an impulse response of the form h(t)=\delta(t)+k\,\delta(t-T), the output is y(t)=x(t)+k\,x(t-T). A sinusoid delayed by T acquires a phase shift \omega T. When \omega T is an odd multiple of \pi, the delayed copy equals minus the original, so the two terms combine to (1-k) times the input. The delay therefore changes the sign of the delayed term rather than its magnitude. Convolution in time becomes multiplication in the Laplace domain, Z(s)=X(s)Y(s), with the ROC containing the intersection of the individual ROCs. A rectangular pulse of width w starting at a is u(t-a)-u(t-a-w), with transform \frac{e^{-as}-e^{-(a+w)s}}{s}. Convolving such a pulse with an exponential therefore gives a product of simple factors.

Time-shift property of Laplace transform

For a causal signal, delaying by T multiplies the Laplace transform by e^{-sT}: if \mathcal{L}\{x(t)\}=X(s), then \mathcal{L}\{x(t-T)u(t-T)\}=e^{-sT}X(s). The factor must be e^{-sT}, not e^{-s/T} and not e^{+sT}, because a positive exponent corresponds to an advance, which is not causal.

• time delay — a shift to the right on the time axis, which appears as the factor e^{-sT} in s.
• time advance — a shift to the left, which is not allowed for a causal signal.

For a ramp starting at t=1, the signal (t-1)u(t-1) equals the shifted version of t\,u(t), whose transform is 1/s^2. Its transform is therefore e^{-s}/s^2, with the shift carried entirely by the exponential factor. The transform of e^{-at}u(t-T) is \frac{e^{-(s+a)T}}{s+a} with ROC \mathrm{Re}\{s\}>-a, because the shift and the exponential's pole interact in the exponent. When a system is described by a differential equation with zero initial conditions, the derivative \frac{dy}{dt} transforms to sY(s) and the delayed input x(t-T)u(t-T) transforms to e^{-sT}X(s). Collecting terms gives the transfer function H(s)=Y(s)/X(s) as the ratio of these coefficients, and its pole is the root of the coefficient of Y(s).

Bilateral Laplace transform

The bilateral Laplace transform is X(s)=\int_{-\infty}^{\infty}x(t)\,e^{-st}\,dt, and its region of convergence (ROC) is part of its definition. The same algebraic expression can belong to different signals when the ROC differs, so the ROC must be stated. A right-sided signal has its ROC to the right of its rightmost pole, a left-sided signal has its ROC to the left of its leftmost pole, and a two-sided signal has a strip-shaped ROC between poles.

• left-sided signal — nonzero only for t<0. For example, e^{at}u(-t) has transform -\frac{1}{s-a} with \mathrm{Re}\{s\}<a. The minus sign is easy to forget.
• strip ROC — a region between two poles that identifies a two-sided signal.

Since \int_{-\infty}^{0}e^{(a-s)t}\,dt converges only when \mathrm{Re}\{a-s\}>0, the left-sided transform follows directly from the integral. Partial-fraction expansion of a rational X(s), followed by assigning each term to a right-sided or left-sided form according to the strip, gives the time signal. For instance, \frac{1}{(s+1)(s+2)}=\frac{1}{s+1}-\frac{1}{s+2} on -2<\mathrm{Re}\{s\}<-1 corresponds to a left-sided term from the pole at -1, since that pole lies to the right of the strip, and a right-sided term from the pole at -2, since that pole lies to the left.

System properties (linearity, time-invariance)

A system T is linear if it obeys superposition, T\{a\,x_1(t)+b\,x_2(t)\}=a\,T\{x_1(t)\}+b\,T\{x_2(t)\}. It is time-invariant if a shift of the input by T_0 produces the same shift of the output. Multiplying the input by a function of t is linear, but if that function depends explicitly on t the system is time-variant, because delaying the input does not delay the output in the same way. Several common operations can be classified as follows.

• squaring — nonlinear, since (x_1+x_2)^2\ne x_1^2+x_2^2, and time-invariant.
• time reversal — linear and time-varying, since reversing a shifted input gives x(-t+T_0) rather than the shifted output x(-t-T_0).
• absolute value — nonlinear, because |a\,x|\ne a\,|x| for negative a.
• shift — linear and time-invariant.

Classify an operation by checking superposition and the time-shift property separately, since a system can be linear and time-varying at the same time.

Fourier series symmetry properties

The Fourier series of a periodic function with period T uses the basis \cos(n\omega_0 t) and \sin(n\omega_0 t) with \omega_0=2\pi/T. Symmetry removes entire families of terms. An even function, with f(-x)=f(x), has only cosine terms, and an odd function, with f(-x)=-f(x), has only sine terms and zero DC term. The symmetry is checked by computing f(-x) over one full period. For a period-2 function the fundamental angular frequency is \omega_0=\pi, so an odd function of period 2 is expanded in \sin(n\pi x) for n=1,2,3,\dots only.

Time scaling and bandwidth of signals

If y(t)=x(at) with a>0, the spectrum becomes Y(f)=\frac{1}{a}X\!\left(\frac{f}{a}\right), so every frequency edge of x is multiplied by a. Compressing the time axis widens the band, and stretching it narrows the band. A time shift x(t-t_0) only multiplies the spectrum by a phase factor e^{-j2\pi f t_0} and leaves the band unchanged. Example: a signal band-limited to 1 kHz to 3 kHz, after y(t)=x(3t), is band-limited to 3 kHz to 9 kHz.

Bilinear transformation discretization

The bilinear transformation maps an analog transfer function H_a(s) into a discrete-time one by substituting s=\frac{2}{T}\cdot\frac{1-z^{-1}}{1+z^{-1}}, where T is the sampling period. It maps the left half of the s-plane inside the unit circle, so a stable analog filter stays stable. It maps s=0 to z=1 and s=\infty to z=-1. Each factor (s+a) becomes \frac{(2/T+a)-(2/T-a)z^{-1}}{1+z^{-1}}. For example, with T=0.5 s the factor s+2 becomes \frac{6-2z^{-1}}{1+z^{-1}}. The poles at infinity in the s-domain produce zeros at z=-1 in the discrete-time transfer function.

Sinusoidal steady-state response

For a stable LTI system with transfer function H(s), a sinusoidal input A\sin(\omega t+\phi) produces a steady-state output of the same frequency, with amplitude A\,|H(j\omega)| and phase shifted by \angle H(j\omega). Transient terms caused by starting the input at a finite time decay and do not affect this steady-state result. To find an input that produces a given steady-state output, divide the output phasor by H(j\omega). The required input amplitude is the output amplitude divided by |H(j\omega)|, and its phase is the output phase minus \angle H(j\omega). An all-pass system has |H(j\omega)|=1 at every frequency, so only the phase changes. For H(s)=\frac{s-a}{s+a} with a>0, H(ja)=j, which carries phase +90^\circ. A unit-step factor on the input affects only the transient.