已知一基带信号m(t)=cos2πt+2cos4πt,对其进行理想抽样:(1)为了在接收端能不失真地从已抽样信号ms(t)中恢复m(t),试问抽样间隔应如何选择?(2)若抽样间隔取为0.2s,试画出已抽样信号的频谱图。 参考答案: 您可能感兴趣的试卷你可能感兴趣的试题 1.问答题设发送数字信息序列为01011000110100,分别画出相应的...
已知xa(t)=2cos(2πf0t)式中f0=100HZ,以采样频率fs=400Hz对xa(t)进行采样,得到采样信号 和时域离散信号x(n),求: (1)写出xa(t)的傅里叶变换表示式Xa(jΩ); (2)写出xa(t)和x(n)的表达式。 参考答案: 你可能感兴趣的试题 1.问答题 ...
Cos 2pi is the value of the cosine function cos x when x = 2π. The value of cos 2pi is equal to the value of cos 0 which is 1. So, cos 2pi is equal to 1.
7 cos2 x + sinx cosx − 3 = 0 View Solution ভাগ পদ্ধতিতে sqrt0.050625 -এর মান লেখো | View Solution Exams IIT JEE NEET UP Board Bihar Board CBSE Free Textbook Solutions KC Sinha Solutions for Maths ...
cos(2pi/257) à la GaussGauss's method for expressing cos(2pi/17) in nested square roots is implemented. The program can generate all cos(2pi/p) when p is a Fermat prime.Michael Trott
计算cos(2pi) 代数 示例 cos(2π)cos(2π) 减去2π2π的全角,直至角度大于等于00且小于2π2π。 cos(0)(0) cos(0)cos(0)的准确值为11。 11
cos(π2)cos(π2) ( ) | [ ] √ ≥ ° 7 8 9 ≤ θ 4 5 6 / ^ × > π 1 2 3 - + ÷ < , 0 . %
Step 1. Multiply and divide the given expression by ππ2sin2π15, we get cos2π15 cos4π15 cos8π15 cos16π15 ππππππππππππ=12sin2π15×2sin2π15×cos2π15×cos4π15×cos8π15×cos16π15 ππππππππππ=12sin2π15×sin4π15×cos4π15×cos8π15×co...
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如只考虑最近邻的相互作用,用紧束缚近似方法可以得到简立方晶体中s态电子能带为E(k)=E0-A-2J(cos2πkx+cos2πky+cos2πkz)试求:(1)能带宽度。 (2)能带底部和能带顶部的电子有效质量。