Calculate continuous unit hydrograph with given n and k (in the framework of the linear storage cascade)

1 | ```
unitHydrograph(n, k, t, force = FALSE)
``` |

`n` |
Numeric. Number of storages in cascade. |

`k` |
Numeric. Storage coefficient [1/s] (resistance to let water run out). High damping = slowly reacting landscape = high soil water absorbtion = high k. |

`t` |
Numeric, possibly a vector. Time [s]. |

`force` |
Logical: Force the integral of the hydrograph to be 1? DEFAULT: FALSE |

Vector with the unit hydrograph along t

The sum under the UH should always be 1 (if t is long enough). This needs yet to be checked...

Berry Boessenkool, berry-b@gmx.de, July 2013

`lsc`

on how to estimate n and k for a given discharge dataset. `deconvolution.uh`

in the package hydromad, http://hydromad.catchment.org

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 | ```
Time <- 0:100
plot(Time, unitHydrograph(n=2, k=3, t=Time), type="l", las=1,
main="Unit Hydrograph - linear storage cascade")
lines(Time, unitHydrograph(n=2, k=8, t=Time), col=2)
lines(Time, unitHydrograph(n=5.5,k=8, t=Time), col=4)
text(c(12, 20, 50), c(0.1, 0.04, 0.025), c("n=2, k=3","n=2, k=8","n=5.5, k=8"),
col=c(1,2,4), adj=0)
# try several parameters (e.g. in Monte Carlo Simulation to estimate
# sensitivity of model towards slight differences/uncertainty in parameters):
nreps <- 1e3 # 5e4 eg on faster computers
n <- rnorm(nreps, mean=2, sd=0.8); n <- n[n>0]
k <- rnorm(nreps, mean=8, sd=1.1); k <- k[k>0]
UH <- sapply(1:nreps, function(i) unitHydrograph(n=n[i], k=k[i], t=Time))
UHquant <- apply(UH, 1, quantile, probs=0:10/10, na.rm=TRUE)
if(interactive()) View(UHquant)
plot(Time, unitHydrograph(n=2, k=8, t=Time), type="l", ylim=c(0, 0.06), las=1)
# uncertainty intervals as semi-transparent bands:
for(i in 1:5)
polygon(x=c(Time, rev(Time)), y=c(UHquant[i,], rev(UHquant[12-i,])),
col=rgb(0,0,1, alpha=0.3), lty=0)
lines(Time, UHquant[6,], col=4)
lines(Time, unitHydrograph(n=2, k=8, t=Time))
# Label a few bands for clarity:
points(rep(24,3), UHquant[c(2,5,9),25], pch="+")
for(i in 1:3) text(25, UHquant[c(2,5,9)[i],25],
paste("Q", c(10,40,80)[i], sep=""), adj=-0.1, cex=0.7)
# And explain what they mean:
Explain <- "Q80: 80% of the 50000 simulations are smaller than this value"
legend("topright", bty="n", legend=Explain)
# Some n and k values are cut off at the left, that explains the shift from the
# median of simulations relative to the n2k8 line.
``` |

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