Answer on Question #39182 – Math – Integral Calculus
1. Evaluate the integral:
1 T ∫ 0 T / 2 sin ( 2 π T t ) e − j n 2 π T t d t . \frac {1}{T} \int_ {0} ^ {T / 2} \sin \left(\frac {2 \pi}{T} t\right) e ^ {- j n \frac {2 \pi}{T} t} d t. T 1 ∫ 0 T /2 sin ( T 2 π t ) e − jn T 2 π t d t .
Solution.
Let us use the formula sin a = e j a − e − j a 2 j \sin a = \frac{e^{ja} - e^{-ja}}{2j} sin a = 2 j e ja − e − ja
I = 1 T ∫ 0 T / 2 sin ( 2 π T t ) e − j n 2 π T t d t = 1 T ∫ 0 T / 2 e j 2 π T t − e − j 2 π T t 2 j e − j n 2 π T t d t = 1 2 T j ∫ 0 T / 2 [ e − j ( n − 1 ) 2 π T t − e − j ( n + 1 ) 2 π T t ] d t = = 1 2 T j ⋅ [ − e − j ( n − 1 ) 2 π T t j ( n − 1 ) 2 π T + e − j ( n + 1 ) 2 π T t j ( n + 1 ) 2 π T ] ∣ 0 T / 2 = 1 2 T j ⋅ [ − e − j ( n − 1 ) π − 1 j ( n − 1 ) 2 π T + e − j ( n + 1 ) π − 1 j ( n + 1 ) 2 π T ] = = 1 4 π [ e − j ( n − 1 ) π − 1 n − 1 − e − j ( n + 1 ) π − 1 n + 1 ] . \begin{array}{l} I = \frac {1}{T} \int_ {0} ^ {T / 2} \sin \left(\frac {2 \pi}{T} t\right) e ^ {- j n \frac {2 \pi}{T} t} d t = \frac {1}{T} \int_ {0} ^ {T / 2} \frac {e ^ {j \frac {2 \pi}{T} t} - e ^ {- j \frac {2 \pi}{T} t}}{2 j} e ^ {- j n \frac {2 \pi}{T} t} d t = \frac {1}{2 T j} \int_ {0} ^ {T / 2} \left[ e ^ {- j (n - 1) \frac {2 \pi}{T} t} - e ^ {- j (n + 1) \frac {2 \pi}{T} t} \right] d t = \\ = \frac {1}{2 T j} \cdot \left[ - \frac {e ^ {- j (n - 1) \frac {2 \pi}{T} t}}{j (n - 1) \frac {2 \pi}{T}} + \frac {e ^ {- j (n + 1) \frac {2 \pi}{T} t}}{j (n + 1) \frac {2 \pi}{T}} \right] \Bigg | _ {0} ^ {T / 2} = \frac {1}{2 T j} \cdot \left[ - \frac {e ^ {- j (n - 1) \pi} - 1}{j (n - 1) \frac {2 \pi}{T}} + \frac {e ^ {- j (n + 1) \pi} - 1}{j (n + 1) \frac {2 \pi}{T}} \right] = \\ = \frac {1}{4 \pi} \left[ \frac {e ^ {- j (n - 1) \pi} - 1}{n - 1} - \frac {e ^ {- j (n + 1) \pi} - 1}{n + 1} \right]. \\ \end{array} I = T 1 ∫ 0 T /2 sin ( T 2 π t ) e − jn T 2 π t d t = T 1 ∫ 0 T /2 2 j e j T 2 π t − e − j T 2 π t e − jn T 2 π t d t = 2 T j 1 ∫ 0 T /2 [ e − j ( n − 1 ) T 2 π t − e − j ( n + 1 ) T 2 π t ] d t = = 2 T j 1 ⋅ [ − j ( n − 1 ) T 2 π e − j ( n − 1 ) T 2 π t + j ( n + 1 ) T 2 π e − j ( n + 1 ) T 2 π t ] ∣ ∣ 0 T /2 = 2 T j 1 ⋅ [ − j ( n − 1 ) T 2 π e − j ( n − 1 ) π − 1 + j ( n + 1 ) T 2 π e − j ( n + 1 ) π − 1 ] = = 4 π 1 [ n − 1 e − j ( n − 1 ) π − 1 − n + 1 e − j ( n + 1 ) π − 1 ] .
If n ∈ Z n \in \mathbb{Z} n ∈ Z e − j ( n − 1 ) π = ( − 1 ) n − 1 e^{-j(n - 1)\pi} = (-1)^{n - 1} e − j ( n − 1 ) π = ( − 1 ) n − 1 e − j ( n + 1 ) π = ( − 1 ) n + 1 = ( − 1 ) n − 1 e^{-j(n + 1)\pi} = (-1)^{n + 1} = (-1)^{n - 1} e − j ( n + 1 ) π = ( − 1 ) n + 1 = ( − 1 ) n − 1
I = 1 4 π [ ( − 1 ) n − 1 − 1 n − 1 − ( − 1 ) n − 1 − 1 n + 1 ] = ( − 1 ) n − 1 − 1 4 π ( 1 n − 1 − 1 n + 1 ) = ( − 1 ) n − 1 − 1 2 ( n 2 − 1 ) π = { 0 , n = 2 k − 1 − 1 ( n 2 − 1 ) π , n = 2 k I = \frac {1}{4 \pi} \left[ \frac {(- 1) ^ {n - 1} - 1}{n - 1} - \frac {(- 1) ^ {n - 1} - 1}{n + 1} \right] = \frac {(- 1) ^ {n - 1} - 1}{4 \pi} \left(\frac {1}{n - 1} - \frac {1}{n + 1}\right) = \frac {(- 1) ^ {n - 1} - 1}{2 (n ^ {2} - 1) \pi} = \left\{ \begin{array}{l l} 0, & n = 2 k - 1 \\ - \frac {1}{(n ^ {2} - 1) \pi}, & n = 2 k \end{array} \right. I = 4 π 1 [ n − 1 ( − 1 ) n − 1 − 1 − n + 1 ( − 1 ) n − 1 − 1 ] = 4 π ( − 1 ) n − 1 − 1 ( n − 1 1 − n + 1 1 ) = 2 ( n 2 − 1 ) π ( − 1 ) n − 1 − 1 = { 0 , − ( n 2 − 1 ) π 1 , n = 2 k − 1 n = 2 k
Answer: 1 4 π [ e − j ( n − 1 ) π − 1 n − 1 − e − j ( n + 1 ) π − 1 n + 1 ] . \frac{1}{4\pi}\left[\frac{e^{-j(n - 1)\pi} - 1}{n - 1} -\frac{e^{-j(n + 1)\pi} - 1}{n + 1}\right]. 4 π 1 [ n − 1 e − j ( n − 1 ) π − 1 − n + 1 e − j ( n + 1 ) π − 1 ] .
#339914 on Dec 2023
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