フィルターのクリア

Inverse Fourier Transformation with a function in the exponent

3 ビュー (過去 30 日間)
Shoaibur Rahman
Shoaibur Rahman 2018 年 8 月 1 日
編集済み: Shoaibur Rahman 2018 年 8 月 1 日
I have a signal x[n] with it's Fourier Transform X[k]. I want a frequency-warped signal (x_warp[n]) by taking the inverse Fourier of X[k], with a warping function fun. I know the fun will be somehow incorporated to the exponent of original inverse Fourier e^(j*2*pi*k*n/N)
So far, I have the following code:
x_hat = 1/N * X(1:N/2+1) * exp(1j*fun'/Fs*2*pi*(0:N-1))
x_warp = 2*real(x_hat)
Fs=Sampling frequency. Could you please tell me if the expression in the exponent is correct or not? I.E. what will be exp(???) in the above code?
Thanks in advance!

回答 (0 件)

カテゴリ

Help Center および File ExchangeDiscrete Fourier and Cosine Transforms についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by