Suppose a length-16 sequence x[n] needs to be reordered in bit-reversed order ,then it is used as an input to a FFT algorithm.The sample number of the sequence in bit-reversed order is:
A: [img=583x69]1803a2f846ecc71.png[/img]
B: [img=581x71]1803a2f853645c7.png[/img]
C: [img=583x77]1803a2f85f65a12.png[/img]
D: [img=598x77]1803a2f86c48c3f.png[/img]
A: [img=583x69]1803a2f846ecc71.png[/img]
B: [img=581x71]1803a2f853645c7.png[/img]
C: [img=583x77]1803a2f85f65a12.png[/img]
D: [img=598x77]1803a2f86c48c3f.png[/img]
举一反三
- 17da426f4cb2265.jpg,计算[img=23x22]17da426f58ddf0c.jpg[/img]实验命令为( ). A: f=diff(log(x),3)f=2/x^3 B: syms x; f=diff(log(x),3)f=2/x^3 C: syms x;f=diff(logx,3)f=2/x^3
- 17e0a756f3d6e2a.jpg,计算[img=23x22]17e0b849ab0b36c.jpg[/img]实验命令为( ). A: f=diff(log(x),3)f=2/x^3 B: syms x; f=diff(log(x),3)f=2/x^3 C: syms x;f=diff(logx,3)f=2/x^3
- 函数f(x)=[img=15x18]17e0a6f63edb84a.jpg[/img]-5x-4, 则f′(x)=, f′(2)=.
- 在下列命题中:如果f(x)=[img=28x44]17e0bf9914bb2f1.png[/img],那么[img=27x29]17e0bf97582597b.png[/img]f(x)=0;如果f(x)=[img=28x44]17e0bf992111a1c.png[/img],那么[img=27x29]17e0bf97582597b.png[/img]f(x)=0;如果f(x)=[img=55x44]17e0bf992d8de0a.png[/img],那么[img=29x29]17e0bf9939482bb.png[/img]f(x)不存在;如果f(x)=[img=87x53]17e0bf99450fa82.png[/img],那么[img=27x29]17e0bf97582597b.png[/img]f(x)=0。其中错误命题的个数是( A: 0 B: 1 C: 2 D: 3
- 7、二次函数[img=120x26]18031ef2ad3cc4e.png[/img]的对称轴x=2,则f(3)＜f(2).