Signals and Systems

Discrete-time convolution

Discrete-time convolution gives the output of a linear time-invariant system from its input and impulse response, y[n]=\sum_{k}x[k]\,h[n-k]. To compute one output sample, pair each input sample x[k] with the shifted, time-reversed impulse response sample h[n-k], multiply the pairs, and add all the products. Convolution is commutative and associative, so the roles of x and h can be swapped without changing the result.

For finite-length sequences, the support of the output is the sum of the supports. If x[n] is nonzero for 0\le n\le N_x-1 and h[n] for 0\le n\le N_h-1, then y[n] is nonzero only for 0\le n\le N_x+N_h-2, so the output has length N_x+N_h-1. Whether a late output sample is zero must be checked against this finite support, not only against the first nonzero value. A useful global check is that the sum of all output samples equals the product of the sums of the two sequences, because Y(e^{j0})=X(e^{j0})H(e^{j0}).

When some impulse response samples are unknown, each output sample gives one linear equation in them. Two specified output samples determine two unknowns only if the coefficient matrix of that system is nonsingular, so the unknowns are found by solving a small linear system.

If h[n]=\delta[n]-\delta[n-1], then y[n]=x[n]-x[n-1], a first difference. For any finitely supported x, the samples of this difference telescope to zero when summed over all n. To find the peak of an output, evaluate y[n] sample by sample and compare the computed values; for example, x=[1,2] and h=[1,1,1] give y=[1,3,3,2].

Sampling and spectral aliasing

Ideal sampling with sampling frequency f_s produces a spectrum that repeats at all integer multiples of f_s, namely X_s(f)=f_s\sum_k X(f-kf_s). A frequency component at f therefore appears at every f-kf_s, and the aliased frequency that is observed is the copy lying in the band [-f_s/2,\,f_s/2]. A spectral point lying above f_s/2 folds back toward the band edge, so its apparent frequency is f_s-(f-kf_s) for the nearest copy.

The Nyquist criterion requires f_s>2B for a band-limited signal of bandwidth B. A sinusoid at 70 Hz sampled at 100 Hz is indistinguishable from a 30 Hz sinusoid, which is the standard example of aliasing. For a sum of sinusoids, the requirement is set by the highest frequency component, so each angular frequency \omega must be converted to f=\omega/2\pi before comparing with the sampling rate. Lower-frequency components do not raise the requirement.

For a band-limited spectrum, replicas do not overlap when f_s\ge 2B. When a spectral value at a given frequency is needed, evaluate the original spectrum at the shifted argument f-kf_s and multiply by the scaling factor, so the ratio of two replica values is found by shifting the arguments into the base band.

When a single ADC serves N channels in turn, each channel receives only every N-th conversion, so the ADC rate must be N times the per-channel Nyquist rate. For example, four channels each band-limited to 50 Hz require an ADC rate of 4\times 2\times 50=400 Hz.

Fourier transform properties

The inverse continuous-time Fourier transform is x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(j\omega)e^{j\omega t}\,d\omega. Setting t=0 shows that \frac{1}{2\pi}\int X(j\omega)\,d\omega equals the time-domain value x(0), which is a convenient shortcut when the spectrum is given in integral or closed form.

A standard pair is \mathcal{F}\{e^{-a|t|}\}=\dfrac{2a}{a^2+\omega^2} for a>0. Splitting the integral at t=0 gives \frac{1}{a-j\omega}+\frac{1}{a+j\omega}, and combining these yields the result. The transform is real and even, as required for a real even signal.

For a real signal, X(-j\omega)=X^*(j\omega). If x(t) is real and odd, its cosine integral vanishes, so the transform is purely imaginary and odd in \omega. The same symmetry rules hold for the DTFT X(e^{j\omega})=\sum_n x[n]e^{-j\omega n}: a real odd sequence has an imaginary odd transform. For instance, x[n]=\delta[n+1]-\delta[n-1] gives X(e^{j\omega})=2j\sin\omega.

Impulse response of LTI systems

The impulse response h[n] (or h(t)) is the output produced by a unit impulse input, and it completely characterizes an LTI system because y=x*h. A system is causal when h[n]=0 for n<0, and it is bounded-input bounded-output (BIBO) stable when \sum_n|h[n]|<\infty, or \int|h(t)|\,dt<\infty in continuous time.

Recursive systems with feedback usually have infinitely long impulse responses. For the system y[n]=a\,y[n-1]+x[n] at rest, h[n]=a^n u[n]. This response is nonzero for every n\ge 0, so it has infinitely many nonzero samples. Its absolute sum is a geometric series, which converges when |a|<1, so the system is stable exactly in that case.

In continuous time, an LTI system described by a differential equation with zero initial conditions has transfer function H(s)=Y(s)/X(s), and the Laplace transform of its impulse response equals H(s). Differentiation maps to multiplication by s, since \mathcal{L}\{dy/dt\}=sY(s). For example, y'+2y=x gives H(s)=1/(s+2) and h(t)=e^{-2t}u(t).

Impulse responses of parallel branches add, and a delayed branch contributes a shifted copy of h. When two responses are compared at a chosen instant, equate their sampled values at that instant. This reduces to matching the coefficients of the delayed exponential terms that appear in each response.

Initial value theorem

For a causal function with Laplace transform F(s), the initial value theorem states f(0^+)=\lim_{s\to\infty}sF(s), provided the limit exists. It gives the value just after t=0 directly from the transform, without inverting it.

For a unit step input, Y(s)=H(s)/s, so y(0^+)=\lim_{s\to\infty}H(s). If H(s) is strictly proper, meaning it has more poles than zeros, this limit is zero. If the numerator and denominator have the same degree, the limit is the ratio of their leading coefficients.

For a derivative g(t)=df/dt, the transform is G(s)=sF(s)-f(0^+), so g(0^+)=\lim_{s\to\infty}s\,[sF(s)-f(0^+)]. The factor s must be kept inside the limit, because the constant term cancels and the next-order coefficient of sF(s) determines the result. For example, F(s)=\dfrac{2s+1}{s^2+3s+1} gives f(0^+)=\lim_{s\to\infty}sF(s)=2.

Digital filter structures

An FIR filter computes y[n]=\sum_{k=0}^{M}b_k\,x[n-k], so H(z)=\sum_k b_k z^{-k} is a polynomial in z^{-1} with all its poles at the origin. In the direct form, the input passes through a tapped delay line, and each tap is multiplied by its coefficient before the products are summed.

A chain of delays whose tap gains form a geometric sequence realizes a first-order recursive pole. Summing the series gives \sum_{k\ge 0}a^k z^{-k}=\dfrac{1}{1-az^{-1}} for |az^{-1}|<1. Matching the coefficient in a given transfer function to this form gives the parameter a, and the pole of the system lies at z=a. For instance, a chain with gains 0.3^k corresponds to H(z)=1/(1-0.3z^{-1}).

For a two-tap FIR filter H(z)=b_0+b_1z^{-1}, the magnitude response is |H(e^{j\omega})|=|b_0+b_1e^{-j\omega}|. It equals |b_0+b_1| at \omega=0 and |b_0-b_1| at \omega=\pi. A high-pass filter has its maximum at \omega=\pi, and the ratio of the maximum to the minimum magnitude then constrains the two coefficients. For example, b_0=0.6 and b_1=-0.2 give |H(0)|=0.4 and |H(\pi)|=0.8, so the ratio is 2.

Frequency response of LTI systems

The frequency response H(j\omega) is the transfer function evaluated on the imaginary axis, H(j\omega)=Y(j\omega)/X(j\omega). Its magnitude gives the gain applied to each sinusoid, and its phase gives the shift of that sinusoid.