fix _default_dt for DelayLtiSystem to actually adapt to delay size

The previous heuristic always returned `0.07` because the appended `0` in
`minimum([abs.(real.(ps));0])` and `minimum([sys.Tau;0])` forced `r = 0`,
which then fell back to `r = 1.0`. As a result `step(delay(0.01))`
(issue #544) sampled at `dt = 0.07` and could not resolve the delay.

Rewritten to mirror `_default_dt(::LTISystem)`: take the fastest pole of
`sys.P.P` and the smallest positive delay, and pick the more demanding
of `1/(12·ω_max)` and `τ_min/12`. Add a regression test for #544.

Co-Authored-By: Claude Opus 4.7 (1M context) <noreply@anthropic.com>

Author: baggepinnen

lib/ControlSystemsBase/src/delay_systems.jl

@@ -80,16 +80,15 @@ end
 # Again we have to do something for default vectors, more or less a copy from timeresp.jl
 function _default_dt(sys::DelayLtiSystem)
     if !isstable(sys.P.P)
-        return 0.05   # Something small
-    else
-        ps = poles(sys.P.P)
-        r = minimum([abs.(real.(ps));0]) # Find the fastest pole of sys.P.P
-        r = min(r, minimum([sys.Tau;0])) # Find the fastest delay
-        if r == 0.0
-            r = 1.0
-        end
-        return 0.07/r
+        return 0.05
     end
+    ps = poles(sys.P.P)
+    ω_max = isempty(ps) ? 0.0 : maximum(abs, ps)
+    τ_min = minimum((τ for τ in sys.Tau if τ > 0); init = Inf)
+    dt_pole  = ω_max > 0      ? round(1/(12*ω_max), sigdigits=2) : Inf
+    dt_delay = isfinite(τ_min) ? round(τ_min/12,    sigdigits=2) : Inf
+    dt = min(dt_pole, dt_delay)
+    return isfinite(dt) ? dt : 0.05
 end
 
 

test/test_delay_timeresp.jl

@@ -133,3 +133,14 @@ P = 1 / (0.85*s + 1)*exp(-0.14*s)
 res = step(P, 5)
 @test res.t[end] > 4.5
 @test length(res.y) > 30
+
+# Issue #544: step(delay(τ)) with small τ used to throw BoundsError after
+# `dt <= dtmin` from the DDE solver; default dt now adapts to τ.
+let sys = delay(0.01)
+    res = step(sys)
+    @test res.t[2] - res.t[1] ≤ 0.01/5
+    @test res.t[end] > 0.1
+    @test all(isfinite, res.y)
+    @test res.y[1] ≈ 0 atol = 1e-8
+    @test res.y[end] ≈ 1 atol = 1e-3
+end