Properties of LTI systems

Two systems in a chain: flip Order and watch the output trace.

Does the order matter?

Chain two guitar pedals together, an overdrive into a reverb, and you get one sound. Swap their order, reverb into overdrive, and you get a different sound, at least for most pairs of pedals. But two LTI systems (linear, time-invariant, exactly the kind convolution builds) are not like that. I can wire them in either order and the output never changes at all, and the last page’s own algebra is why.

Convolution is associative: $(x*h_1)*h_2=x*(h_1*h_2)$. It’s also commutative: $h_1*h_2=h_2*h_1$. Put those together and a cascade of two systems, output of the first fed into the second, can be replaced by one system with impulse response $h_1*h_2$, and it doesn’t matter which of $h_1$, $h_2$ you convolve first.

Take $h_1=1,1$ and $h_2=1,-1$ and feed the chain $x=1,2,3$. Convolving $h_1$ then $h_2$ first gives $h_1*h_2=1,0,-1$, and $x*(h_1*h_2)=1,2,2,-2,-3$ . Going the other way, $x*h_1=1,3,5,3$ (5.2’s own worked example), then $(x*h_1)*h_2=1,2,2,-2,-3$ Identical.

Go back to the chain at the top of the page and read Difference between orders. It stays at 0.00 no matter which way you set it: the output trace doesn’t move a single sample.

Two paths, or one

Now split, instead of chain. Run the same input through two systems in parallel, one path into $h_1$, a separate path into $h_2$, and add the two outputs together at the end, the way a “dry” signal and a reverb signal get mixed back together. Convolution distributes over addition, $x*(h_1+h_2)=x*h_1+x*h_2$, so adding the two paths’ outputs gives exactly the same result as building one combined system with impulse response $h_1+h_2$ and running the input through that instead.

With the same $h_1=1,1$ and $h_2=1,-1$ from before, $h_1+h_2=2,0$ : two paths added, or one system with $h=2,0$, land on the same numbers either way.

Switch Route between “Two paths” and “One path” and read Difference between routes.

Notice it too stays at 0.00: whichever route you build, the output trace lands on the identical values, sample for sample.

Reading causality and stability off h

System-properties (4.2) tested causality and stability by trying inputs on a system and watching what came out. For an LTI system you don’t need to try anything: both properties are sitting right there in $h[n]$ itself.

A system is causal when its output can’t depend on an input sample that hasn’t happened yet. For an LTI system that’s exactly the statement $h[n]=0$ for every $n<0$: a nonzero $h$ before $n=0$ would mean the system’s response has already started before the spike that’s supposed to cause it arrives.

A system is BIBO stable when every bounded input produces a bounded output. For an LTI system, add up the size of every sample of $h$, ignoring sign, $\lvert h[0]\rvert+\lvert h[1]\rvert+\lvert h[2]\rvert+\dots$, and call that running total “the total of $h$‘s sizes.” If that total stays finite, however far out you add, the system is stable; if it keeps climbing without limit, it isn’t. Written with a summation sign, that total is $\sum_n\lvert h[n]\rvert$.

Take $h[n]=(0.5)^n$ for $n\ge0$ (a decaying echo settling toward zero). Its total of sizes is the geometric series $\sum_{n=0}^\infty 0.5^n=\tfrac{1}{1-0.5}=2$, finite, so it’s stable. Take $h[n]=(1.5)^n$ for $n\ge0$ instead (a growing feedback loop): that geometric series has ratio $1.5>1$, so it diverges, and the system is unstable.

Step Impulse response through Echo, Smoother, Settling and Growing, and read Causal? and Stable?.

Notice the Growing preset’s total never settles down, however far right you imagine looking, while the other three all reach a fixed number and stop climbing.

Does the response ever really stop?

Some impulse responses hit exactly zero after some point and stay there. Others only ever get smaller, forever, without ever landing on zero exactly. Both kinds show up constantly, and it’s worth having names for them.

An impulse response that’s exactly zero past some finite point is a finite impulse response, usually shortened to FIR. A three-sample moving average, $h=\tfrac13,\tfrac13,\tfrac13$, is one: it has exactly three nonzero samples and is zero everywhere else, so its total of sizes is automatically finite too, an FIR is always stable. An impulse response that goes on forever, even while shrinking toward zero, is an infinite impulse response, or IIR: the settling loop from the last instrument, $h[n]=(0.5)^n$, never actually reaches zero, no matter how far out you look.

Switch Response between “Finite (FIR)” and “Infinite (IIR)” and read Last nonzero sample.

Notice the finite trace finds an exact last nonzero sample and stops there, while the infinite trace’s readout admits there isn’t one, it only keeps getting smaller.

What comes out when a spinning signal goes in?

Back in 3.4, a phasor was a point spinning around the origin at a constant rate. Measured once per sample, that rate is $\Omega$ radians per sample, its across value tracing a cosine and its up value tracing a sine. Feed that spinning point into an LTI system, and something clean happens: it comes back out still spinning at exactly the same rate, only scaled in size and turned to a new starting angle. One fixed complex number does both jobs at once, and I’ll call it $H(e^{j\Omega})$. A signal that a system hands back with nothing changed but its size and angle is called an eigenfunction of that system (from the German eigen, “its own”), and spinning points are eigenfunctions of every stable LTI system (for a growing $h$ like the one above, the sum that gives $H$ never settles, so there is no single number to scale by).

Here’s why. Convolving $x[n]=e^{j\Omega n}$ with $h$ gives $y[n]=\sum_k h[k]\,e^{j\Omega(n-k)}=e^{j\Omega n}\sum_k h[k]\,e^{-j\Omega k}$. The sum $\sum_k h[k]\,e^{-j\Omega k}$ doesn’t depend on $n$ at all, it’s one fixed complex number, so $y[n]=H(e^{j\Omega})\,e^{j\Omega n}$ with $H(e^{j\Omega})=\sum_k h[k]\,e^{-j\Omega k}$. The input’s own spin rate, $\Omega$, survives untouched; only its size and angle change.

Take $h=1,0.5$ (a spike now, half a spike one sample later) and a phasor spinning a quarter turn per sample, $\Omega=\pi/2$. Directly, $y[n]=x[n]+0.5\,x[n-1]=e^{j\pi n/2}\big(1+0.5\,e^{-j\pi/2}\big)=e^{j\pi n/2}(1-0.5j)$. That matches $H(e^{j\pi/2})=h[0]+h[1]e^{-j\pi/2}=1-0.5j$, a single complex number with magnitude $\sqrt{1^2+0.5^2}=1.11803$ and angle about $-26.565°$ (a little below the horizontal axis): the spinning point comes out $1.118\times$ its original size, turned back by $26.565°$, spinning at the exact same rate it went in at.

Not every system does this. A squarer, $y[n]=x[n]^2$, is not LTI (4.2 already showed it fails linearity): squaring a phasor spinning at $\Omega$ doubles its spin rate instead, $\big(e^{j\Omega n}\big)^2=e^{j2\Omega n}$, something no single scaling number could ever produce.

Switch System between Echo and Squarer and read What happened to it.

Notice the echo’s output always spins at the input’s own rate, just resized and turned, while the squarer’s output visibly spins twice as fast, a doubled rate that no single complex number could explain.

Where you’ll meet this

Reordering a cascade of guitar pedals or camera filters, or folding two parallel effects into one, both lean on the associativity and distributivity you just used. Reading stability straight off $h$ is exactly how a feedback design gets checked before it’s ever built, instead of by listening for it to run away. FIR and IIR are the two families every filter you’ll design later falls into. And the eigenfunction property, one fixed complex number turning a spinning input into a scaled, rotated version of itself, is the exact idea the frequency response builds on starting in Chapter 12.

Reference card

That closes out the foundations, Chapters 1 through 5, every idea from here on builds on this set.