Evaluate this integral - HW1: Difference between revisions
Jump to navigation
Jump to search
Max.Woesner (talk | contribs) (New page: == Max Woesner == Back to my Home Page === Homework #1 - Evaluate this integral === Evaluate the integral <math>\int_{-\frac{T}{2}}^{\frac{T}{2}}e^{j2 \pi (n-m)t/T}dt\!</...) |
Max.Woesner (talk | contribs) No edit summary |
||
| Line 7: | Line 7: | ||
For <math> n=m, \int_{-\frac{T}{2}}^{\frac{T}{2}}e^{j2 \pi (n-m)t/T}dt = \int_{-\frac{T}{2}}^{\frac{T}{2}}1 dt = T \Bigg|_{\frac{-T}{2}}^{\frac{T}{2}} = \frac{T}{2} - \frac{-T}{2} = T \!</math><br> |
For <math> n=m, \int_{-\frac{T}{2}}^{\frac{T}{2}}e^{j2 \pi (n-m)t/T}dt = \int_{-\frac{T}{2}}^{\frac{T}{2}}1 dt = T \Bigg|_{\frac{-T}{2}}^{\frac{T}{2}} = \frac{T}{2} - \frac{-T}{2} = T \!</math><br> |
||
For <math> n\neq m,\int_{-\frac{T}{2}}^{\frac{T}{2}}e^{j2 \pi (n-m)t/T}dt = \frac{e^{j2 \pi (n-m)t/T}}{\frac{j2 \pi (n-m)}{T}} \Bigg|_{\frac{-T}{2}}^{\frac{T}{2}} = \frac{e^{j \pi (n-m)} - e^{-j \pi (n-m)}}{\frac{j2 \pi (n-m)}{T}} = 0 \!</math><br> |
For <math> n\neq m,\int_{-\frac{T}{2}}^{\frac{T}{2}}e^{j2 \pi (n-m)t/T}dt = \frac{e^{j2 \pi (n-m)t/T}}{\frac{j2 \pi (n-m)}{T}} \Bigg|_{\frac{-T}{2}}^{\frac{T}{2}} = \frac{e^{j \pi (n-m)} - e^{-j \pi (n-m)}}{\frac{j2 \pi (n-m)}{T}} = 0 \!</math><br> |
||
So, <math>\int_{-\frac{T}{2}}^{\frac{T}{2}}e^{j2 \pi (n-m)t/T}dt = \begin{cases} T, & \mbox{for }n=m \\ 0, & \mbox{for }n\neq m \end{cases}\!</math><br><br> |
|||
Alternate method: <br> |
|||
<math>\int_{-\frac{T}{2}}^{\frac{T}{2}}e^{j2 \pi (n-m)t/T}dt = \int_{-\frac{T}{2}}^{\frac{T}{2}}[cos(2 \pi (n-m)t/T)+jsin(2 \pi (n-m)t/T)]dt\!</math><br> |
|||
For <math> n=m, \int_{-\frac{T}{2}}^{\frac{T}{2}}[cos(2 \pi (n-m)t/T)+jsin(2 \pi (n-m)t/T)]dt = \int_{-\frac{T}{2}}^{\frac{T}{2}}1 dt = T \Bigg|_{\frac{-T}{2}}^{\frac{T}{2}} = \frac{T}{2} - \frac{-T}{2} = T \!</math><br> |
|||
For <math> n\neq m,\int_{-\frac{T}{2}}^{\frac{T}{2}}[cos(2 \pi (n-m)t/T)+jsin(2 \pi (n-m)t/T)]dt = 0\!</math><br> |
|||
So, <math>\int_{-\frac{T}{2}}^{\frac{T}{2}}e^{j2 \pi (n-m)t/T}dt = \begin{cases} T, & \mbox{for }n=m \\ 0, & \mbox{for }n\neq m \end{cases}\!</math> |
So, <math>\int_{-\frac{T}{2}}^{\frac{T}{2}}e^{j2 \pi (n-m)t/T}dt = \begin{cases} T, & \mbox{for }n=m \\ 0, & \mbox{for }n\neq m \end{cases}\!</math> |
||
Revision as of 21:27, 18 October 2009
Max Woesner
Homework #1 - Evaluate this integral
Evaluate the integral
For
For
So,
Alternate method:
For
For
So,
![{\displaystyle \int _{-{\frac {T}{2}}}^{\frac {T}{2}}e^{j2\pi (n-m)t/T}dt=\int _{-{\frac {T}{2}}}^{\frac {T}{2}}[cos(2\pi (n-m)t/T)+jsin(2\pi (n-m)t/T)]dt\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ab48349f12ac1961cfe09cb4762a9f3191adcf33)
![{\displaystyle n=m,\int _{-{\frac {T}{2}}}^{\frac {T}{2}}[cos(2\pi (n-m)t/T)+jsin(2\pi (n-m)t/T)]dt=\int _{-{\frac {T}{2}}}^{\frac {T}{2}}1dt=T{\Bigg |}_{\frac {-T}{2}}^{\frac {T}{2}}={\frac {T}{2}}-{\frac {-T}{2}}=T\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d93da93452a60462aa0c83502042ae62ffe3e016)
![{\displaystyle n\neq m,\int _{-{\frac {T}{2}}}^{\frac {T}{2}}[cos(2\pi (n-m)t/T)+jsin(2\pi (n-m)t/T)]dt=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ba342f9ebf1b555c1f0bbeb0e9951bb6cb1be12)