6

I'm obtaining the following series as an analytical solution to a problem using a differential time $dt$

$$h(t) = \lim_{n\rightarrow\infty}_{m\rightarrow-\infty}_{\Delta t\rightarrow 0}\sum_{i=m}^n f(t-i\ \Delta t)\ g(i\ \Delta t)\ \Delta t$$

$f(t)$ is a function defined completely in $\mathbb{R}$.

I'm deriving expressions from scratch, and got stuck at this step. Does this series seem familiar to you in any way? Is there a way to lose the limit and summation expressions? I'd appreciate small hints.

Some background: water resources

It is based on the method of constructing arbitrary hydrographs from unit hyetographs in water resources. Basically, you have a function $i(t)$ that defines the amount of rainfall over a basin, and $u(t)$ that defines the water discharge through an exit point. Note that both are functions of time. $u(t)$ is also a functional of $i(t)$. A unit hyetograph is a function of $t$ defined through: $$i_u(t) = \left\{ \begin{array}{l l} 0 & \quad t<0 \ \ {\rm or}\ \ t>\Delta t\\ 1 & \quad t>0\ \ {\rm and}\ \ t<\Delta t \end{array} \right. $$

The corresponding hydrograph is obtained empirically. Just suppose that there is a function $u_u(t)$ that corresponds to this unit hyetograph. Note that the only characteristic of $i_u(t)$ is $\Delta t$.

Now that we have defined a unit hyetograph $i_u(t)$, we want to construct $u(t)$ from arbitrary $i(t)$ by superposition, which explains why $u(t)$ is defined as a functional of $i(t)$ and $u_u(t)$.

Below is a visual example (Implemented in MATLAB):

An implementation of the series above in MATLAB

After Vincent Tjeng's answer, I've considered taking the convolution of both functions. I've used the conv function from MATLAB. The result can be seen in the following figure. I've taken $\Delta t=0.25$ to demonstrate how the series expression converges to the convolution:

Series versus the convolution

EDIT: Non-calculus example on the subject:

example

  1. The unit hyetograph and the corresponding unit hydrograph are given.
  2. An arbitrary hyetograph is given. To determine the corresponding hydrograph, the unit hydrograph is multiplied by 2, offset by $t_r$. This is equal to $u(t)=u_u(t)+2\ u_u(t-t_r)$. The series above is obtained in a similar way.
  3. The two steps are summed together.
  4. The resulting hydrograph.

Some remarks: This example (which follows the convention used by all engineers), is defined over a discrete timestep $\Delta t = t_r$. However, I now realize that while I am trying to obtain a general mathematical solution, I am ending up with a system where the $t_r$ is differential. Furthermore, I now realize that the expression I derived above assumues $i_u$, the unit hydrograph to be equal to the Dirac Delta function, which is not the case in practice. Nevertheless, I find the following paragraph in Wikipedia:

An instantaneous unit hydrograph is a further refinement of the concept; for an IUH, the input rainfall is assumed to all take place at a discrete point in time (obviously, this isn't the case for actual rainstorms). Making this assumption can greatly simplify the analysis involved in constructing a unit hydrograph, and it is necessary for the creation of a geomorphologic instantaneous unit hydrograph.

It seems that I have successfully reinvented the wheel.

osolmaz
  • 331
  • You sum a constant? $f$ does not depend on $i$? –  Mar 18 '13 at 13:15
  • Apologies I made a typo. Correcting it right now – osolmaz Mar 18 '13 at 13:23
  • You mean $f(t+idt)$? –  Mar 18 '13 at 13:25
  • God, yes thanks :D – osolmaz Mar 18 '13 at 13:25
  • Maybe, you are considering the expression $$ \lim_{n\rightarrow\infty}{{\rm d}t\rightarrow 0}\frac{1}{n}\sum{i=0}^nf(t+i \Delta t ),$$ where $\Delta t=\frac{1}{n}.$ – Mhenni Benghorbal Mar 18 '13 at 13:29
  • What Mhenni said. There is perhaps a relation between $n$ and $dt$. – GEdgar Mar 18 '13 at 13:35
  • @GEdgarL It should be and I believe it is what I already mentioned in my previous comment. – Mhenni Benghorbal Mar 18 '13 at 13:35
  • I suppose not. There is no relationship; even $i$'s initial value is arbitrary, but I've simplified down to 0. – osolmaz Mar 18 '13 at 13:44
  • @nrs: How did you derive this? – Mhenni Benghorbal Mar 18 '13 at 13:49
  • @MhenniBenghorbal I added some background to the question body. I can add more details when necessary – osolmaz Mar 18 '13 at 14:27
  • To reinforce what Mhenni and GEdgar have said: without knowing of some relationship between $\Delta t$ and $m$ or $n$, the problem is absolutely impossible to solve because we don't have enough data to even determine the domain over which $f$ and $g$ are evaluated! To see this, look at the limit with $n\equiv \dfrac{C}{\Delta t}$; then $g$ is never evaluated at an argument larger than $C$ - but since we've set $C$ arbitrarily, then we could have a limit where the sum depends on all values of $g$, or only on those values less than 5, or only those values less than a million... – Steven Stadnicki Mar 26 '13 at 02:40
  • Also, you can't really take the limits 'sequentially' because if you take the limit over $\Delta t$ first then the sum goes to 0, whereas if you take the limit over $n$ (and/or $m$) first then that limit will be infinite unless $f$ and $g$ have finite support (which is certainly possible). There must be some relation between $\Delta t$ and $n$ for your expression to make sense. – Steven Stadnicki Mar 26 '13 at 02:44

2 Answers2

3

This looks to me like a convolution of the two functions $f(t), g(t)$.

$$h(t) = \lim_{n\rightarrow\infty}_{m\rightarrow-\infty}_{\Delta t\rightarrow 0}\sum_{i=m}^n f(t-i\ \Delta t)\ g(i\ \Delta t)\ \Delta t=\int^\infty_{-\infty}f(t-\tau)g(\tau)\, d\tau$$

Using the Hyetograph as $f(t)$, and defining $g(t)$ as $$g(t)=\begin{cases} 0 & t<0, t>1 \\ 1 & 0 \le t \le 1 \end{cases}$$

Convolving your Unit Hyetograph gives me a graph similar to yours:

enter image description here

However, convolving your Arbitrary Hyetograph gives me this constructed Hydrograph instead, and I'm wondering whether I understood your question wrongly.

enter image description here

My arbitrary Hyetograph as I define it is as follows:

$$f(t)=\begin{cases} 0 & t<0, t>6 \\ 2 & 0 \le t<2 \\ 3 & 2 \le t<4 \\ 1 & 4\le t<6 \end{cases}$$

Edit: As requested, here is my code in Mathematica. I use version 9.0.

hyetograph = 
  Function[t, 
   Piecewise[{{0, t < 0}, {2, 0 <= t < 2}, {3, 2 <= t < 4}, {1, 
      4 <= t < 6}, {0, t >= 6}}]];
Plot[{hyetograph[t], g[t]}, {t, 0, 10}, AxesLabel -> {"Time", ""}, 
 PlotLegends -> {"hyetograph[t]", "g[t]"}]
g = Function[t, HeavisidePi[t - 1/2]];
hydrograph = Convolve[hyetograph[t], g[t], t, y];
Plot[hydrograph, {y, 0, 10}, AxesLabel -> {"Time", ""}, 
 PlotLegends -> {"hydrograph[t]"}]
osolmaz
  • 331
Vincent Tjeng
  • 3,304
  • 2
  • 21
  • 34
  • this is exactly what I've been looking for. I also wonder why the numerical solution and the convolution are different. The result I've been trying to obtain is the one in your solution. How did you take the convolution? – osolmaz Mar 24 '13 at 18:20
  • I've updated the question with my calculation of the convolution. I can't figure out how our results ended up being different. – osolmaz Mar 24 '13 at 20:58
  • @nrs sorry for the late reply. I took my convolution using Mathematica, but I think that that shouldn't make a difference. I'd like to check something regarding my understanding of the question: firstly, would it be right to think of the hydrograph as the amount of runoff through a river system given rainfall over a period of time as defined by the hyetograph? In that case, for your unit hyetograph, is it meant to be i=1 till t=1 instead? Otherwise, the hydrograph doesn't really make sense to me in the context. – Vincent Tjeng Mar 25 '13 at 02:01
  • Also, if my understanding regarding the background of the question is correct, the area under the curve for any corresponding pair of hydrographs and hyetographs should be the same, given that they are tracking the same amount of rainwater flow. That's not the case for your unit example. – Vincent Tjeng Mar 25 '13 at 02:05
  • I'm also a little bit confused. i=1 till t=1 should be correct, but then it conflicts with the expression I derived (or does it?). As for your second comment, the area under the arbitrary hydrograph can be different for arbitrary hyetographs. Either way, your solution seems correct to me (if you assumed i=1 till t=1), and that is what I'm trying to achieve. Maybe I could have a peek at your Mathematica file? – osolmaz Mar 25 '13 at 13:27
  • @nrs I'll add the code to my answer tomorrow when I get access to my other computer. I did assume i=1 till t=1 as that made more sense to me. With regards to my second comment, I'm wondering whether the area under a given hyetograph should be the same as that for the hydrograph it produces (I assume each hyetograph should correspond to a given hydrograph - is this correct?) and not whether all hydrographs should have the same area under the curve. – Vincent Tjeng Mar 25 '13 at 13:36
  • There is no correspondence between those areas. Just to clarify the subject, I'm going to speak in terms of units. Hyetographs are generally expressed as $[mm/hr]$ versus $[hr]$. Hydrographs are $[m^3/s]$ versus $[hr]$. The relationship is purely empirical. On one hand you measure the rainflow with an instrument, on the other hand you measure the discharge from the river. You obtain the unit hydrograph using multiple datasets, through some calculations. Using this unit hydrograph, you can extrapolate the behavior of the basin for different storms with different rainfall intensities. – osolmaz Mar 25 '13 at 14:54
  • I see, thank you! so the only relationship we should expect based on the physical nature of your question between the hyetographs and the hydrographs would be that if the last non-zero value on the hyetograph (for rainfall) is at time $t$, then the last non-zero value for the hydrograph (for water runoff) should be slightly later - is this right? – Vincent Tjeng Mar 26 '13 at 00:00
  • Exactly, since the basin is a system that collects water, and routes it through exit points (rivers). Water travels from higher elevations to lower elevations, and that takes some time. The basin will not return to its original state immediately after the rainfall ceases. – osolmaz Mar 26 '13 at 11:35
  • By the way, I'll give the bounty because it's about to expire. – osolmaz Mar 26 '13 at 13:34
  • okay. but if I didn't fully answer your question, or if something is still not working, please tell me. I'd also suggest that you read up about LTI system theory; if your system is not LTI, then you will not be able to generate arbitrary hydrographs from hyetographs simply from the unit hyetograph and hydrograph. We can take this to chat if you'd like. – Vincent Tjeng Mar 26 '13 at 13:50
  • 1
    I've finally had a chance to look at the code, and I think we may have a misunderstanding. You are convolving the arbitrary hyetograph with the unit hyetograph right? What I did in my calculations was to convolve the unit hydrograph with the arbitrary hyetograph, which I now realize that assumes $i_u$ to be equal to the Dirac Delta function. I've updated the question with a figure and more information. – osolmaz Mar 27 '13 at 20:32
  • @I see! Actually, by comparing the unit hyetograph and hydrograph, I attempted to find the function $g(x)$ that, when convolved with the unit hyetograph, gives the unit hydrograph. I then convolved the same function $g(x)$ to the arbitrary hyetograph to give the hydrograph. – Vincent Tjeng Mar 28 '13 at 02:48
  • Either way, my problem is solved. Thank you so much for your efforts! – osolmaz Mar 28 '13 at 11:54
0

You probably want the $\frac 1n$ inside the limit. This makes $n$ into a dummy variable. Then $\lim_{n \to \infty}\frac 1n \sum f(t+dt)=f(t+dt)$. Unless you have a term subtracted from this, you can just let the $dt$ be zero and get $f(t)$

Ross Millikan
  • 374,822