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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:


  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.

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You sum a constant? $f$ does not depend on $i$? – user63181 Mar 18 '13 at 13:15
Apologies I made a typo. Correcting it right now – hos Mar 18 '13 at 13:23
You mean $f(t+idt)$? – user63181 Mar 18 '13 at 13:25
God, yes thanks :D – hos 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
up vote 3 down vote accepted

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 = 
   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]"}]
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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? – hos 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. – hos 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 2:01
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 2:05
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. – hos Mar 27 '13 at 20:32

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 $f(t)$

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