I'm self studying signal and system. I've come across this problem: if $a_1 = 1, a_2 = j$, what are $a_3, a_4, a_5$ for a discrete odd signal x[n] with fundamental period of N=6?
Asked
Active
Viewed 33 times
1 Answers
1
Since the signal is odd we have $$a_n=-a_{-n}$$also $$a_n=a_{n+N}$$therefore $$a_5=a_{-1}=-a_1=-1\\a_4=a_{-2}=-a_2=-j$$and $$a_0=-a_0=0\\a_3=-a_{-3}=-a_{6-3}=-a_3\to a_3=0$$finally here are the coefficients $$(a_0,a_1,a_2,a_3,a_4,a_5)=(0,1,j,0,-j,-1)$$
Mostafa Ayaz
- 31,924
-
what's this property $a_N = a_{n+N}$? – drerD Nov 30 '18 at 23:03
-
Actually this is because of the discrete nature of the signal. The proof is easy $$a_{n+N}={1\over N}\sum x[k]e^{- k\Omega (n+N)}={1\over N}\sum x[k]e^{- k\Omega n}e^{- k\Omega N}=a_n$$since $e^{- k\Omega N}=1$ – Mostafa Ayaz Nov 30 '18 at 23:11
-
Thanks for your help. Why is $e^{k\Omega N}=1$? I am guessing it has to do with fundamental period? – drerD Dec 08 '18 at 01:06
-
You're welcome. Use the definition $\Omega ={2\pi \over N}$ – Mostafa Ayaz Dec 08 '18 at 20:47