LTI & convolution formulas

Must hold for all inputs $x_1,x_2$ and all constants $a,b$ (complex constants if the signals may be complex). It combines additivity ($a=b=1$) and homogeneity ($b=0$). Zero input giving zero output is necessary, not sufficient.

Discrete time: $x[n-n_0]\to y[n-n_0]$ for every integer $n_0$. Also called shift invariance.

Both must hold for every $t_0$. A memoryless (static) system is always causal. A causal system need not be memoryless.

Every bounded input must give a bounded output. One bounded input with an unbounded output proves the system unstable.

$\tau$ is the integration variable and $t$ is held fixed while integrating. In practice the limits shrink to the interval where both factors are nonzero.

Each output sample is a sum of products of samples whose indices add to $n$.

When $x$ and $h$ are both zero before the origin, $y$ is zero there too. If either signal has an impulse at the origin, take the lower limit as $0^{-}$ so that the impulse is included.

Then $y(t)={\int }_0^{\infty }h(\tau )x(t-\tau )d\tau$ uses only present and past input. Necessary and sufficient for an LTI system.

Necessary and sufficient for an LTI system. The integral of $|h|$ must be finite, not the integral of $h$. The output is then bounded by $|y|\le M_x\int |h|dt$. If $h(t)$ contains impulses, each term $c\delta (t-t_i)$ adds $|c|$ to the integral, so $h(t)=\delta (t)$ (a plain wire) is stable. A doublet term ${\delta }^{\prime }(t)$ (a differentiator) makes the system unstable.

Here $s(t)$ is the step response, not the Laplace variable. The step response is the running integral of $h$. For a causal $h$, $s(t)={\int }_0^{t}h(\tau )d\tau$. For a stable system the final value is $s(\infty )={\int }_{-\infty }^{\infty }h(t)dt=H(0)$, the DC gain. The ramp response is the running integral of the step response.

The step response is the running sum of $h$, and $h$ is the first difference of the step response.

In any transform domain a cascade multiplies, $H=H_1{H}_2$, and a parallel connection adds, $H=H_1+H_2$. The order of LTI blocks in a cascade does not change the overall response.

Examples: the delay $\delta (t-t_0)$ has inverse $\delta (t+t_0)$. The accumulator $h[n]=u[n]$ has inverse $\delta [n]-\delta [n-1]$, because $u[n]-u[n-1]=\delta [n]$. An inverse need not be causal or stable: for $t_0>0$ the delay is causal, but its inverse, an advance, is not.

Holds when the system is at rest before the input starts (initial rest), which makes it LTI and causal. $h$ is the inverse transform with the ROC to the right of the rightmost pole of $H(s)$, or outside the outermost pole of $H(z)$. Examples: $\frac{dy}{dt}+ay=x(t)$ gives $H(s)=\frac{1}{s+a}$ and $h(t)=e^{-at}u(t)$, stable only for $a>0$. $y[n]-ay[n-1]=x[n]$ gives $H(z)=\frac{1}{1-a{z}^{-1}}$ and $h[n]=a^{n}u[n]$, stable only for $|a|<1$. An $h[n]$ of finite length is called FIR and one of infinite length IIR. A difference equation with no past outputs ($N=0$) is always FIR, with $h[n]=b_n/a_0$.

Hold for continuous and discrete convolution. Associativity gives the cascade rule and distributivity gives the parallel rule.

Discrete time: $x[n]\ast \delta [n-n_0]=x[n-n_0]$. Convolving with an impulse train repeats the signal: $x(t)\ast {\sum }_k\delta (t-k{T}_0)={\sum }_{k}x(t-k{T}_0)$. Do not confuse convolution with multiplication by an impulse, which samples: $x(t)\delta (t-t_0)=x(t_0)\delta (t-t_0)$, for $x$ continuous at $t_0$.

Shifts add. The same holds in discrete time with integer shifts.

Convolving with $u$ is running integration (or running summation). Convolving with ${\delta }^{\prime }(t)$ differentiates: $x(t)\ast {\delta }^{\prime }(t)=x^{\prime }(t)$.

Useful for piecewise-constant signals, whose derivatives are impulses. A result found this way must be integrated back, using the fact that $y$ is zero before $t=a_1+a_2$.

Holds whenever both areas (sums) are finite, for example when $x$ and $h$ are absolutely integrable (summable). The quickest check on any such convolution. In transform terms it is $Y(0)=X(0)H(0)$.

Continuous time: the widths add, $T_y=T_x+T_h$. Discrete time: $n_1$ and $n_2$ are the first indices of $x$ and $h$, and each length is counted from the first nonzero sample to the last. The first and last output samples are then single products of nonzero end samples, so they are nonzero; interior samples may be zero.

Continuous time only. Scaling both signals by $a$ compresses the output time axis by $a$ and scales its height by $1/|a|$.

The ROC of $Y$ contains the intersection of the ROCs of $X$ and $H$. Multiplication in time is convolution in frequency: $x(t)p(t)\leftrightarrow \frac{1}{2\pi }X(j\omega )\ast P(j\omega )$ in the $\omega$ convention, and $X(f)\ast P(f)$, without the $1/2\pi$, in the $f$ convention.

Holds for all $t$ when $s_0$ lies in the ROC of $H(s)$, the bilateral Laplace transform of $h$. The input exists for all time, $-\infty <t<\infty$.

$z_0$ must lie in the ROC of $H(z)$. With $z_0=e^{j{\Omega }_0}$ the eigenvalue is the frequency response $H(e^{j{\Omega }_0})$.

Exists as an ordinary function when $h$ is absolutely integrable (summable), that is, when the system is BIBO stable. For a real $h$, $|H|$ is even and $\angle H$ is odd in frequency.

Needs a real $h$ and a stable system. Discrete time: $A\cos ({\Omega }_{0}n+\varphi )\to A|H(e^{j{\Omega }_0})|\cos ({\Omega }_{0}n+\varphi +\angle H(e^{j{\Omega }_0}))$. For a sum of sinusoids, apply the rule to each term and add (superposition).

The $\omega =0$ (or $\Omega =0$) case of the sinusoid rule. For $x[n]=(-1{)}^n$ use $\Omega =\pi$: $y[n]=H(e^{j\pi })(-1{)}^n$, with $H(e^{j\pi })={\sum }_k(-1{)}^{k}h[k]$.

Correlation is convolution with the second signal reversed and conjugated. Because of the reversal it is not commutative: $R_{yx}(\tau )=R_{xy}^{\ast }(-\tau )$.

For a real, even signal, $x(-\tau )=x(\tau )$, so $R_{xx}=x\ast x$.

$m$ is the integer lag. The length rule of convolution applies: sequences of lengths $N_1$ and $N_2$ give $N_1+N_2-1$ lags, so the autocorrelation of an $N$-sample sequence has $2N-1$ lags.

For a real signal, $R_{xx}$ is real and even, and its largest value is at $\tau =0$. The last line is the area property of convolution; it needs $x$ absolutely integrable.

The inequality is the Cauchy-Schwarz inequality. Unlike $R_{xx}$, the cross-correlation need not peak at $\tau =0$.

$|X(j\omega ){|}^2$ is the energy spectral density. In the $f$ convention, $E_x=\int |X(f){|}^{2}df$ (Parseval). Discrete time: $R_{xx}[m]\leftrightarrow |X(e^{j\Omega }){|}^2$ and $E_x={\sum }_n|x[n]{|}^2=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }|X(e^{j\Omega }){|}^{2}d\Omega$. Because $|X{|}^2$ discards phase, $x(t)$ and every shifted copy $x(t-t_0)$ have the same autocorrelation.

For a periodic signal of period $T_0$, average over one period; $R_{xx}$ is then periodic with period $T_0$ and returns to its peak $R_{xx}(0)$ at every multiple of $T_0$. The cross-correlation of power signals uses the same average with $y^{\ast }(t)$ in place of $x^{\ast }(t)$. The power spectral density is the Fourier transform of the autocorrelation, $S_{xx}(\omega )=\mathcal{F}\{R_{xx}(\tau )\}$, and $P_x=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{S}_{xx}(\omega )d\omega$. Example: $A\cos ({\omega }_{0}t+\varphi )$ has $R_{xx}(\tau )=\frac{A^2}{2}\cos {\omega }_0\tau$, whatever $\varphi$ is, so $R_{xx}(0)=A^2/2$ is its average power and $S_{xx}(\omega )=\frac{\pi A^2}{2}\left[\delta (\omega -{\omega }_0)+\delta (\omega +{\omega }_0)\right]$.

For $y=h\ast x$, in the convention above. In frequency, $S_{yx}(\omega )=H(j\omega )S_{xx}(\omega )$ and $S_{yy}(\omega )=|H(j\omega ){|}^2{S}_{xx}(\omega )$, where $S$ is the energy (or power) spectral density.