# Can I write this integral in "closed-form" expression?

1. Aug 6, 2014

### EngWiPy

Hi all,

I have this integral, and I wish to write it in a closed-form in order to be able to program it. The integral is:

$$\int_{-\infty}^{\infty}e^{j2\pi f_c[a^{(u)}-a^{(m)}]}g(t[1+a^{(m)}]-kT_s-\tau_p^{(m,n)}) g(t[1+a^{(u)}]-lT_s-\tau_q^{(u,n)})\,dt$$

where

$$g(t)=\left\{\begin{array}{cc}1&t\in[0,T_s)\\0&\mbox{Otherwise}\end{array}\right.$$

Basically, I can write two different cases, namely, when:

$$\frac{kT_s+\tau_p^{(m,n)}}{1+a^{(m)}}\leq \frac{lT_s+\tau_q^{(u,n)}}{1+a^{(u)}}<\frac{(k+1)T_s+\tau_p^{(m,n)}}{1+a^{(m)}}$$

and when

$$\frac{lT_s+\tau_q^{(u,n)}}{1+a^{(u)}}\leq \frac{kT_s+\tau_p^{(m,n)}}{1+a^{(m)}}<\frac{(l+1)T_s+\tau_q^{(u,n)}}{1+a^{(u)}}$$

and solve the integral for these two separate cases. However, this requires me to check for every possible case separately which is taking too long time. I was hoping if I can write it as one expression which eases the programming part.

Thanks

2. Aug 12, 2014

### Greg Bernhardt

I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?

3. Aug 12, 2014

### D H

Staff Emeritus
Simplifying that a bit, you have $\int_{-\infty}^{\infty} c g_1(t) g_2(t)\, dt$ where
\begin{aligned} c &= e^{j2\pi f_c[a^{(u)}-a^{(m)}]} \\ g_1(t) &= g(t[1+a^{(m)}]-kT_s-\tau_p^{(m,n)}) \\ g_2(t) &= g(t[1+a^{(u)}]-lT_s-\tau_q^{(u,n)}) \end{aligned}
That constant $c=e^{j2\pi f_c[a^{(u)}-a^{(m)}]}[/tex] can be pulled out of the integral, and you're left with the integral of the product of two indicator functions. Each of those indicator functions is fairly simple (but messy): They describe a finite interval on the real number line. So the integral is just the constant c times the length of the intersection of these two intervals. I'll use [itex]a$ to denote the interval corresponding to $g_1(t)$ and $b$ to denote the nterval corresponding to $g_2(t)$:

\begin{aligned} a &= \left(\frac{kT_s+\tau_p^{(m,n)}}{1+a^{(m)}}, \frac{(k+1)T_s+\tau_p^{(m,n)}}{1+a^{(m)}}\right) \\ b &= \left(\frac{lT_s+\tau_q^{(u,n)}}{1+a^{(u)}}, \frac{(l+1)T_s+\tau_q^{(u,n)}}{1+a^{(u)}}\right) \end{aligned}

That's a bit messy yet. I'll define four items to represent those endpoints:
\begin{aligned} a_1 &= \frac{kT_s+\tau_p^{(m,n)}}{1+a^{(m)}} \\ a_2 &= \frac{(k+1)T_s+\tau_p^{(m,n)}}{1+a^{(m)}} \\ b_1 &= \frac{lT_s+\tau_q^{(u,n)}}{1+a^{(u)}} \\ b_2 &= \frac{(l+1)T_s+\tau_q^{(u,n)}}{1+a^{(u)}}\end{aligned}
With this, the intervals simplify to
\begin{aligned} a &= (a_1, a_2) \\ b &= (b_1, b_2) \end{aligned}

You missed four cases. Here are all six, depicted in ASCII art. In all cases, I'm portraying interval a above and interval b below the real number line, and the intersection between intervals a and b below the line displaying interval b.
Code (Text):

(    a    )
Case 1:  ------------------------------------>
(    b   )
Intersection is null

(     a    )
Case 2:  ------------------------------------>
(     b     )
( a∩b )

(  a  )
Case 3:  ------------------------------------>
(    b     )
( a∩b )

(    a     )
Case 4:  ------------------------------------>
(  b  )
( a∩b )

(     a     )
Case 5:  ------------------------------------>
(     b    )
( a∩b )

(    a   )
Case 6:  ------------------------------------>
(    b    )
Intersection is null

All you need is the length of that intersection. This is actually quite easy: it's $d=\max(0, \min(a_2,b_2) - \max(a_1,b_1))$. That outer max protects against a negative length, which is what would otherwise in the case of a null intersection.

Last edited: Aug 12, 2014
4. Aug 12, 2014

### EngWiPy

I am really thankful of your effort, but I have a typo in my post where the integral is actually:

$$\int_{-\infty}^{\infty}e^{j2\pi f_c[a^{(u)}-a^{(m)}]t}g(t[1+a^{(m)}]-kT_s-\tau_p^{(m,n)}) g(t[1+a^{(u)}]-lT_s-\tau_q^{(u,n)})\,dt$$

where the exponential is a function of t, otherwise I wouldn't have put it inside the integral. This is actually what complicates the integration. Now, in its presence can I find something similar to what you had without it?

5. Aug 12, 2014

### D H

Staff Emeritus
It doesn't complicate it by much. Those g(αt+β) are either zero or one, so instead of a length times a constant you get

$$\int_{\max(a_1, b_1)}^{\min(a_2,b_2)} \exp(j\omega t)\,dt$$

Here I've collapsed that big messy constant of yours into a single constant, $\omega$. That's an easy integral. There is one tricky issue now: You have to watch out for the two null cases. The above is correct only if there is a non-null intersection.