- #1
yungman
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This is a derivation of radiating power of dipole antenna, I cannot follow the derivation of the notes. Here is my derivation starting from (1) or the notes.
[tex]I_{int}=\frac 1 2 \int_{-1}^1\frac{[1+\cos(klu)+1+\cos(kl)]}{(1+u)}du\;-\;\int_{-1}^1\frac{\cos\left (\frac{kl}{2}(1+u)\right)+\cos\left (\frac{kl}{2}(1-u)\right) }{(1+u)}du[/tex]
[tex]\hbox {Let }\;(1+u)kl=v\;\Rightarrow\;v=(kl+klu),\;u=(\frac {v}{kl}-1),\;(1+u)=\frac {v}{kl}, \;\hbox { and } \;du=\frac {dv}{kl}[/tex]
[tex]\Rightarrow\; I_{int}=\frac 1 2 \int_0^{2kl} \frac {[2 + \cos (v-kl)+\cos(kl)]}{v} dv-\int_0^{2kl}\frac{\cos\left(\frac v 2 \right)}{v}dv-\int_{-1}^1 \frac{\cos\left[\frac{kl}{2}(1-u)\right]}{1+u}du[/tex]
[tex]\hbox {For the second term, let }\; w=\frac v 2 \;\hbox { and then Let } v=w \hbox { later}[/tex]
[tex]\Rightarrow\; I_{int}=\frac 1 2 \int_0^{2kl} \frac {[2 + \cos (v-kl)+\cos(kl)]}{v} dv-\int_0^{kl}\frac{\cos(v)}{v}dv-\int_{-1}^1 \frac{\cos\left[\frac{kl}{2}(1-u)\right]}{1+u}du[/tex]
[tex]\Rightarrow\; I_{int}=\frac 1 2 \int_0^{2kl} \frac {[2 +\cos(kl)]}{v} dv+ \frac 1 2 \int_0^{2kl} \frac { \cos (v-kl)}{v} dv -\int_0^{kl}\frac{\cos(v)}{v}dv-\int_{-1}^1 \frac{\cos\left[\frac{kl}{2}(1-u)\right]}{1+u}du[/tex]
Now compare the line (2) in the scanned notes:
[tex]\Rightarrow\; I_{int}=\frac 1 2 \int_0^{2kl} \frac {[2 + \cos (klv)+\cos(kl)]}{v} dv-\int_0^{kl}\frac{\cos(v)}{v}dv-\int_{-1}^1 \frac{\cos\left[\frac{kl}{2}(1-u)\right]}{1+u}du[/tex]
[tex]\Rightarrow\; I_{int}=\frac 1 2 \int_0^{2kl} \frac {[2 +\cos(kl)]}{v} dv+ \frac 1 2 \int_0^{2kl} \frac { \cos (klv)}{v} dv -\int_0^{kl}\frac{\cos(v)}{v}dv-\int_{-1}^1 \frac{\cos\left[\frac{kl}{2}(1-u)\right]}{1+u}du[/tex]
If you compare the equation I derive with (2) of the notes, the only terms that is different is:
[tex]\frac 1 2 \int_0^{2kl} \frac { \cos (v-kl)}{v} dv\;\hbox { from my equation to }\;\frac 1 2 \int_0^{2kl} \frac { \cos (klv)}{v} dv\;\hbox { from the notes.}[/tex]
I just don't see how I can get the answer from the notes. As you see, I separate the longer integration into two parts. I am pretty sure I did this right this time...hopefully. Please help.
I attach the whole page of the notes as I might have more question coming and I might have typos. This is a difficult derivation that involve sine and cosine integral and I am rusty in math! I am not math major, I do this as part of the derivation of radiating power of antenna. I am not like you guys that are in "mid season condition"!. Everything is rusty for me and need some kick start! What I posted is from equation (1) to (2) in the notes.
[tex]I_{int}=\frac 1 2 \int_{-1}^1\frac{[1+\cos(klu)+1+\cos(kl)]}{(1+u)}du\;-\;\int_{-1}^1\frac{\cos\left (\frac{kl}{2}(1+u)\right)+\cos\left (\frac{kl}{2}(1-u)\right) }{(1+u)}du[/tex]
[tex]\hbox {Let }\;(1+u)kl=v\;\Rightarrow\;v=(kl+klu),\;u=(\frac {v}{kl}-1),\;(1+u)=\frac {v}{kl}, \;\hbox { and } \;du=\frac {dv}{kl}[/tex]
[tex]\Rightarrow\; I_{int}=\frac 1 2 \int_0^{2kl} \frac {[2 + \cos (v-kl)+\cos(kl)]}{v} dv-\int_0^{2kl}\frac{\cos\left(\frac v 2 \right)}{v}dv-\int_{-1}^1 \frac{\cos\left[\frac{kl}{2}(1-u)\right]}{1+u}du[/tex]
[tex]\hbox {For the second term, let }\; w=\frac v 2 \;\hbox { and then Let } v=w \hbox { later}[/tex]
[tex]\Rightarrow\; I_{int}=\frac 1 2 \int_0^{2kl} \frac {[2 + \cos (v-kl)+\cos(kl)]}{v} dv-\int_0^{kl}\frac{\cos(v)}{v}dv-\int_{-1}^1 \frac{\cos\left[\frac{kl}{2}(1-u)\right]}{1+u}du[/tex]
[tex]\Rightarrow\; I_{int}=\frac 1 2 \int_0^{2kl} \frac {[2 +\cos(kl)]}{v} dv+ \frac 1 2 \int_0^{2kl} \frac { \cos (v-kl)}{v} dv -\int_0^{kl}\frac{\cos(v)}{v}dv-\int_{-1}^1 \frac{\cos\left[\frac{kl}{2}(1-u)\right]}{1+u}du[/tex]
Now compare the line (2) in the scanned notes:
[tex]\Rightarrow\; I_{int}=\frac 1 2 \int_0^{2kl} \frac {[2 + \cos (klv)+\cos(kl)]}{v} dv-\int_0^{kl}\frac{\cos(v)}{v}dv-\int_{-1}^1 \frac{\cos\left[\frac{kl}{2}(1-u)\right]}{1+u}du[/tex]
[tex]\Rightarrow\; I_{int}=\frac 1 2 \int_0^{2kl} \frac {[2 +\cos(kl)]}{v} dv+ \frac 1 2 \int_0^{2kl} \frac { \cos (klv)}{v} dv -\int_0^{kl}\frac{\cos(v)}{v}dv-\int_{-1}^1 \frac{\cos\left[\frac{kl}{2}(1-u)\right]}{1+u}du[/tex]
If you compare the equation I derive with (2) of the notes, the only terms that is different is:
[tex]\frac 1 2 \int_0^{2kl} \frac { \cos (v-kl)}{v} dv\;\hbox { from my equation to }\;\frac 1 2 \int_0^{2kl} \frac { \cos (klv)}{v} dv\;\hbox { from the notes.}[/tex]
I just don't see how I can get the answer from the notes. As you see, I separate the longer integration into two parts. I am pretty sure I did this right this time...hopefully. Please help.
I attach the whole page of the notes as I might have more question coming and I might have typos. This is a difficult derivation that involve sine and cosine integral and I am rusty in math! I am not math major, I do this as part of the derivation of radiating power of antenna. I am not like you guys that are in "mid season condition"!. Everything is rusty for me and need some kick start! What I posted is from equation (1) to (2) in the notes.