Answer:
C = 292 Mbps
Explanation:
Given:
- Signal Transmitted Power P = 250mW
- The noise in channel N = 10 uW
- The signal bandwidth W = 20 MHz
Find:
what is the maximum capacity of the channel?
Solution:
-The capacity of the channel is given by Shannon's Formula:
C = W*log_2 ( 1 + P/N)
- Plug the values in:
C = (20*10^6)*log_2 ( 1 + 250*10^-3/10)
C = (20*10^6)*log_2 (25001)
C = (20*10^6)*14.6096
C = 292 Mbps
Answer:
Rubber-like solids with elastic properties are called elastomers. Polymer chains are held together in these materials by relatively weak intermolecular bonds, which permit the polymers to stretch in response to macroscopic stresses. Natural rubber, neoprene rubber, buna-s and buna-n are all examples of such elastomers.
Answer:
Damping ratio 
Explanation:
Given that
m=4.2 kg,K=85.9 N/m,C=1.3 N.s/m
We need to find damping ratio
We know that critical damping co-efficient
N.s/m
Damping ratio(
) is the ratio of damping co-efficient to the critical damping co-efficient
So 
So damping ratio
Answer:
a.
y[n] = x[n] x[n-1] x[n+1]
(i) Memory-less - It is not memory-less because the given system is depend on past or future values.
(ii) Causal - It is non-casual because the present value of output depend on the future value of input.
(iii) Invertible - It is invertible and the inverse of the given system is ![\frac{1}{x[n] . x[n-1] x[n+1]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%5Bn%5D%20.%20x%5Bn-1%5D%20x%5Bn%2B1%5D%7D)
(iv) Stable - It is stable because for all the bounded input, output is bounded.
(v) Time invariant - It is not time invariant because the system is multiplying with a time varying function.
b.
y[n] = cos(x[n])
(i) Memory-less - It is memory-less because the given system is not depend on past or future values.
(ii) Causal - It is casual because the present value of output does not depend on the future value of input.
(iii) Invertible - It is not invertible because two or more than two input values can generate same output values .
For example - for x[n] = 0 , y[n] = cos(0) = 1
for x[n] = 2
, y[n] = cos(2) = 1
(iv) Stable - It is stable because for all the bounded input, output is bounded.
(v) Time invariant - It is time invariant because the system is not multiplying with a time varying function.
Determine whether w is in the span of the given vectors v1; v2; : : : vn
. If your answer is yes, write w as a linear combination of the vectors v1; v2; : : : vn and enter the coefficients as entries of the matrix as instructed is given below
Explanation:
1.Vector to be in the span means means that it contain every element of said vector space it spans. So if a set of vectors A spans the vector space B, you can use linear combinations of the vectors in A to generate any vector in B because every vector in B is within the span of the vectors in A.
2.And thus v3 is in Span{v1, v2}. On the other hand, IF all solutions have c3 = 0, then for the same reason we may never write v3 as a sum of v1, v2 with weights. Thus, v3 is NOT in Span{v1, v2}.
3.In the theory of vector spaces, a set of vectors is said to be linearly dependent if at least one of the vectors in the set can be defined as a linear combination of the others; if no vector in the set can be written in this way, then the vectors are said to be linearly independent.
4.Given a set of vectors, you can determine if they are linearly independent by writing the vectors as the columns of the matrix A, and solving Ax = 0. If there are any non-zero solutions, then the vectors are linearly dependent. If the only solution is x = 0, then they are linearly independent.
