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

[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

 

[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?

 

[D H]

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.

Simplifying that a bit, you have \int_{-\infty}^{\infty} c g_1(t) g_2(t)\, dt where
\begin{aligned}<br /> c &amp;= e^{j2\pi f_c[a^{(u)}-a^{(m)}]} \\<br /> g_1(t) &amp;= g(t[1+a^{(m)}]-kT_s-\tau_p^{(m,n)}) \\<br /> g_2(t) &amp;= g(t[1+a^{(u)}]-lT_s-\tau_q^{(u,n)})<br /> \end{aligned}
That constant c=e^{j2\pi f_c[a^{(u)}-a^{(m)}]}[/tex] can be pulled out of the integral, and you&#039;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 <i>c</i> times the length of the intersection of these two intervals. I&#039;ll use a to denote the interval corresponding to g_1(t) and b to denote the nterval corresponding to g_2(t):<br /> <br /> \begin{aligned}&lt;br /&gt; a &amp;amp;= \left(\frac{kT_s+\tau_p^{(m,n)}}{1+a^{(m)}},&lt;br /&gt; \frac{(k+1)T_s+\tau_p^{(m,n)}}{1+a^{(m)}}\right) \\&lt;br /&gt; b &amp;amp;= \left(\frac{lT_s+\tau_q^{(u,n)}}{1+a^{(u)}},&lt;br /&gt; \frac{(l+1)T_s+\tau_q^{(u,n)}}{1+a^{(u)}}\right)&lt;br /&gt; \end{aligned}<br /> <br /> That&#039;s a bit messy yet. I&#039;ll define four items to represent those endpoints:<br /> \begin{aligned}&lt;br /&gt; a_1 &amp;amp;= \frac{kT_s+\tau_p^{(m,n)}}{1+a^{(m)}} \\&lt;br /&gt; a_2 &amp;amp;= \frac{(k+1)T_s+\tau_p^{(m,n)}}{1+a^{(m)}} \\&lt;br /&gt; b_1 &amp;amp;= \frac{lT_s+\tau_q^{(u,n)}}{1+a^{(u)}} \\&lt;br /&gt; b_2 &amp;amp;= \frac{(l+1)T_s+\tau_q^{(u,n)}}{1+a^{(u)}}\end{aligned}<br /> With this, the intervals simplify to<br /> \begin{aligned}&lt;br /&gt; a &amp;amp;= (a_1, a_2) \\&lt;br /&gt; b &amp;amp;= (b_1, b_2)&lt;br /&gt; \end{aligned}<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Basically, I can write two different cases, namely, when:<br /> <br /> \frac{kT_s+\tau_p^{(m,n)}}{1+a^{(m)}}\leq \frac{lT_s+\tau_q^{(u,n)}}{1+a^{(u)}}&amp;lt;\frac{(k+1)T_s+\tau_p^{(m,n)}}{1+a^{(m)}}<br /> <br /> and when<br /> <br /> \frac{lT_s+\tau_q^{(u,n)}}{1+a^{(u)}}\leq \frac{kT_s+\tau_p^{(m,n)}}{1+a^{(m)}}&amp;lt;\frac{(l+1)T_s+\tau_q^{(u,n)}}{1+a^{(u)}} </div> </div> </blockquote><br /> You missed four cases. Here are all six, depicted in ASCII art. In all cases, I&#039;m portraying interval <i>a</i> above and interval <i>b</i> below the real number line, and the intersection between intervals <i>a</i> and <i>b</i> below the line displaying interval <i>b</i>.<br /> <div class="bbCodeBlock bbCodeBlock--screenLimited bbCodeBlock--code"> <div class="bbCodeBlock-title"> <i class="fa--xf fal fa-code "><svg xmlns="http://www.w3.org/2000/svg" role="img" aria-hidden="true" ><use href="/data/local/icons/light.svg?v=1756418028#code"></use></svg></i> Code: </div> <div class="bbCodeBlock-content" dir="ltr"> <pre class="bbCodeCode" dir="ltr" data-xf-init="code-block" data-lang=""><code> ( a ) Case 1: ------------------------------------&gt; ( b ) Intersection is null ( a ) Case 2: ------------------------------------&gt; ( b ) ( a∩b ) ( a ) Case 3: ------------------------------------&gt; ( b ) ( a∩b ) ( a ) Case 4: ------------------------------------&gt; ( b ) ( a∩b ) ( a ) Case 5: ------------------------------------&gt; ( b ) ( a∩b ) ( a ) Case 6: ------------------------------------&gt; ( b ) Intersection is null</code></pre> </div> </div><br /> All you need is the length of that intersection. This is actually quite easy: it&#039;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.

 

[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?

Simplifying that a bit, you have \int_{-\infty}^{\infty} c g_1(t) g_2(t)\, dt where
\begin{aligned}<br /> c &amp;= e^{j2\pi f_c[a^{(u)}-a^{(m)}]} \\<br /> g_1(t) &amp;= g(t[1+a^{(m)}]-kT_s-\tau_p^{(m,n)}) \\<br /> g_2(t) &amp;= g(t[1+a^{(u)}]-lT_s-\tau_q^{(u,n)})<br /> \end{aligned}
That constant c=e^{j2\pi f_c[a^{(u)}-a^{(m)}]}[/tex] can be pulled out of the integral, and you&#039;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 <i>c</i> times the length of the intersection of these two intervals. I&#039;ll use a to denote the interval corresponding to g_1(t) and b to denote the nterval corresponding to g_2(t):<br /> <br /> \begin{aligned}&lt;br /&gt; a &amp;amp;= \left(\frac{kT_s+\tau_p^{(m,n)}}{1+a^{(m)}},&lt;br /&gt; \frac{(k+1)T_s+\tau_p^{(m,n)}}{1+a^{(m)}}\right) \\&lt;br /&gt; b &amp;amp;= \left(\frac{lT_s+\tau_q^{(u,n)}}{1+a^{(u)}},&lt;br /&gt; \frac{(l+1)T_s+\tau_q^{(u,n)}}{1+a^{(u)}}\right)&lt;br /&gt; \end{aligned}<br /> <br /> That&#039;s a bit messy yet. I&#039;ll define four items to represent those endpoints:<br /> \begin{aligned}&lt;br /&gt; a_1 &amp;amp;= \frac{kT_s+\tau_p^{(m,n)}}{1+a^{(m)}} \\&lt;br /&gt; a_2 &amp;amp;= \frac{(k+1)T_s+\tau_p^{(m,n)}}{1+a^{(m)}} \\&lt;br /&gt; b_1 &amp;amp;= \frac{lT_s+\tau_q^{(u,n)}}{1+a^{(u)}} \\&lt;br /&gt; b_2 &amp;amp;= \frac{(l+1)T_s+\tau_q^{(u,n)}}{1+a^{(u)}}\end{aligned}<br /> With this, the intervals simplify to<br /> \begin{aligned}&lt;br /&gt; a &amp;amp;= (a_1, a_2) \\&lt;br /&gt; b &amp;amp;= (b_1, b_2)&lt;br /&gt; \end{aligned}<br /> <br /> <br /> <br /> You missed four cases. Here are all six, depicted in ASCII art. In all cases, I&#039;m portraying interval <i>a</i> above and interval <i>b</i> below the real number line, and the intersection between intervals <i>a</i> and <i>b</i> below the line displaying interval <i>b</i>.<br /> <div class="bbCodeBlock bbCodeBlock--screenLimited bbCodeBlock--code"> <div class="bbCodeBlock-title"> <i class="fa--xf fal fa-code "><svg xmlns="http://www.w3.org/2000/svg" role="img" aria-hidden="true" ><use href="/data/local/icons/light.svg?v=1756418028#code"></use></svg></i> Code: </div> <div class="bbCodeBlock-content" dir="ltr"> <pre class="bbCodeCode" dir="ltr" data-xf-init="code-block" data-lang=""><code> ( a ) Case 1: ------------------------------------&gt; ( b ) Intersection is null ( a ) Case 2: ------------------------------------&gt; ( b ) ( a∩b ) ( a ) Case 3: ------------------------------------&gt; ( b ) ( a∩b ) ( a ) Case 4: ------------------------------------&gt; ( b ) ( a∩b ) ( a ) Case 5: ------------------------------------&gt; ( b ) ( a∩b ) ( a ) Case 6: ------------------------------------&gt; ( b ) Intersection is null</code></pre> </div> </div><br /> All you need is the length of that intersection. This is actually quite easy: it&#039;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.

 

[D H]

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?

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.