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  • z^{n} =e^{-2\pi fn\Delta} \\ =\cos(2\pi fn\Delta)\\
    5 KB (682 words) - 10:16, 16 July 2019
  • ...t(f\right)=\frac{{0.5}e^{i\left({2}\pi fn\Delta f+\pi \right)}}{e^{i{2}\pi fn\Delta t}}=0.5e^{i\pi }. {{\rm Input}}=\frac{b_0e^{i{2}\pi fn\Delta t}}{e^{j{2}\pi f\Delta t}}=b_0,
    11 KB (1,819 words) - 13:19, 3 May 2021
  • ...}+ \dots\ + b_{{\rm l}}e^{i2\pi f\left(n-{\rm l}\right)\Delta t}}{e^{i2\pi fn}}, e^{i2\pi f\Delta t}+\dots +b_Ne^{i2\pi f\Delta fN}.
    11 KB (1,656 words) - 13:27, 3 May 2021
  • ...t(f\right)=\frac{{0.5}e^{i\left({2}\pi fn\Delta f+\pi \right)}}{e^{i{2}\pi fn\Delta t}}=0.5e^{i\pi }. {{\rm Input}}=\frac{b_0e^{i{2}\pi fn\Delta t}}{e^{j{2}\pi f\Delta t}}=b_0,
    10 KB (1,707 words) - 13:21, 3 May 2021
  • ...t(f\right)=\frac{{0.5}e^{i\left({2}\pi fn\Delta f+\pi \right)}}{e^{i{2}\pi fn\Delta t}}=0.5e^{i\pi }. {{\rm Input}}=\frac{b_0e^{i{2}\pi fn\Delta t}}{e^{j{2}\pi f\Delta t}}=b_0,
    10 KB (1,716 words) - 15:44, 19 May 2021
  • ...}+ \dots\ + b_{{\rm l}}e^{i2\pi f\left(n-{\rm l}\right)\Delta t}}{e^{i2\pi fn}}, e^{i2\pi f\Delta t}+\dots +b_Ne^{i2\pi f\Delta fN}.
    10 KB (1,542 words) - 13:29, 3 May 2021
  • ...}+ \dots\ + b_{{\rm l}}e^{i2\pi f\left(n-{\rm l}\right)\Delta t}}{e^{i2\pi fn}}, e^{i2\pi f\Delta t}+\dots +b_Ne^{i2\pi f\Delta fN}.
    10 KB (1,554 words) - 15:45, 19 May 2021