若信噪比S N的值为100,则对应的分贝值为( )dB
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![若信噪比S N的值为100,则对应的分贝值为( )dB](/uploads/image/f/6970898-2-8.jpg?t=%E8%8B%A5%E4%BF%A1%E5%99%AA%E6%AF%94S+N%E7%9A%84%E5%80%BC%E4%B8%BA100%2C%E5%88%99%E5%AF%B9%E5%BA%94%E7%9A%84%E5%88%86%E8%B4%9D%E5%80%BC%E4%B8%BA%28+%29dB)
∵Sn-Sn-3=51(n>3),∴an+an-1+an-2=51(n>3),又数列{an}为等差数列,∴3an-1=51(n≥2),∴an-1=17.(n≥2),又a2=3,Sn=100,∴Sn=(
根据等差数列的性质有a1+a13=2a7a2+a12=2a7……S13=13/2(a1+a13)b1+b13=2b7所以S13/T13=(a1+a13)/(b1+b13)=a7/b7=(7×13+2)
(1)由已知有:2a1=4096得a1=2048,又an+sn=4096,an+1+Sn+1=4096,两式相减得an+1=an/2,所以an是以1/2为公比的等比数列,故an=2048*(1/2)^
设等比数列{an}的公比为q,前n项和为Sn,且Sn+1,Sn,Sn+2成等差数列,则2Sn=Sn+1+Sn+2.若q=1,则Sn=na1,式子显然不成立.若q≠1,则有2a1(1−qn)1−q=a1
S4=S8=>(a1+a4)*4/2=(a1+a8)*8/2=>a1+a4=2a1+2a8=>a4=a1+2a8=>a1+3d=a1+2(a1+7d)=>3d=2a1+14d=>a1=-5.5d>0=
(1)∵{An}为等比数列,则有An+1=An·q,又∵Sn+1,Sn,Sn+2成等差数列,∴Sn+1+Sn+2=2Sn∴Sn+An+Sn+An+An·q=2Sn∴可得2+q=0所以q=-2(2)这里
因为Sn+1,Sn,Sn+2成等差数列S(n+1)+S(n+2)=2*S(n)(q^(n+1)-1)*a1/(q-1)+(q^(n+2)-1)*a1/(q-1)=2*(q^(n)-1)*a1/(q-1
S(2m-1)=(A1+A(2m-1))×(2m-1)/2=(A1+A1+2(m-1)d)×(2m-1)/2=(A1+(m-1)d)×(2m-1)=Am×(2m-1)同理S(2n-1)=An×(2n-
因为等差数列的前9项的均值等于第五项a5:b5=S9:S9''n=9带入Sn:Sn''=5n+3:2n+7,得a5:b5=48:25
解题思路:利用函数的性质解题过程:
q=1,讨论一下就可以了,首先你写等比求和公式的时候,需要讨论的是q是否为1,假设q=1,你会发现这个结果是可以的,再讨论q不等于1,因为sn-s(n-1)=a1*q^n,对吧?因为sn为等差,那么a
设等差数列{an}的首项为a1,由S10=0,得10a1+10×(10−1)d2=10a1+45d=0,∴a1=−92d.由Sn≥-5,得:na1+n(n−1)d2=−9d2n+d2n2−d2n=d2
Sm=[2A1+(m-1)d]*m/2=2n.(1)Sn=[2A1+(n-1)d]*n/2=2m.(2)上方程组化简为:2A1+(m-1)d=4n/m.(3)2A1+(n-1)d=4m/n.(4)(3
∵Sn=2/3an-1/3∴3Sn=2an-1∴3Sn=2(Sn-Sn-1)-1∴Sn=﹣2Sn-1-1∴Sn+1/3=2(Sn-1+1/3)∴(Sn+1/3)/(Sn-1+1/3)=2∵Sn=2/3
数列{an}为等比数列,首项a1≠0.公比q=1时,Sn=nSn+1=n+1Sn+2=n+22Sn=2nSn+1+Sn+2=2n+32Sn≠Sn+1+Sn+2,不满足题意,因此公比q≠1Sn+1、Sn
S6=(a1+a6)*6/2=36得a1+a6=12记为1式Sn-S(n-6)=[a(n-5)+an]*6/2=324-144=180得a(n-5)+an=60记为2式又a6+a(n-5)=a1+an
(1)∵an+Sn=4096,∴a1+S1=4096,a1=2048.当n≥2时,an=Sn-Sn-1=(4096-an)-(4096-an-1)=an-1-an∴anan−1=12an=2048(1