Because circular time shift is just simple time shift mod N, i.e., whenever the value of n-n0 exceeds or becomes equal to N it goes back to zero. The proof for why it happens is because the value of DFT is repeated after every N samples and, that whole proof is not required here.
Your smile at starting is worth tons of gold sir😍🤗
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Wait a sec sir
E^-j2Pi n°K /N
Ko kaise alag kar saktey hai
Usme K hai jo ki sigma ka part hai
Toh sigma solve kiye bina kaise alag hoga
Thanks a lot sir i've learnt DSP completely from your playlist its the best
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Thank you sir 🙏
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Very helpful sir 👏❤️
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in 3’33’ . K is in side as part of summation. Why k can be moved out of summation?
Very well explained..👏👏
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Your are just awesome
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Is this circular convolution??
thank you sir for teaching this hard subject
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Sir, at last proved
DFT{x(n-n0)} = .....
But we had to prove that
DFT{x((n-n0))N} = .......
Because circular time shift is just simple time shift mod N, i.e., whenever the value of n-n0 exceeds or becomes equal to N it goes back to zero. The proof for why it happens is because the value of DFT is repeated after every N samples and, that whole proof is not required here.
Thank u sir
Welcome
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Superbbbb explanation sir
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sir, is this vtu scheme??
Yes it's as per Vtu…
Tq sir well explined
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But where is circular convolution mentioned in the proof?
Sir why we need IDFT as it is a property of dft....please answer me,,
No… IDFT is not property of dft but IDFT is inverse Fourier transform…
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@@ECAcademy sir why Idft used in this time shifting property and in my material time shifting property shown x(n)=X(e power of jw. Please explain 🙏
@youtube i am ewatching videos at 2X at you are giving me 18 sec ad in normal speed :(
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