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#2 Re: Help Me ! » z transform » 2011-01-20 11:31:18

Excellent. Please share how you got the answer, I have been trying for hours

#3 Re: Help Me ! » Quadratic Equations » 2011-01-20 11:28:36

Shouldnt the answer to d just be a*p?

#4 Help Me ! » z transform » 2011-01-20 11:18:58

Onyx
Replies: 5

Hi, what is the inverse Z-transform of:

Where k is a constant.

I cannot find it in any Z-transform table. Only:

Thank you

#5 Re: Help Me ! » polynomial long division modulo 2 » 2011-01-07 07:29:43

Thanks, thats exactly what I was after, but would you mind explaining the process to me?

You start with the MSB (Most significant bit) of the divisor aligned with the second MSB of the dividend, why not the first?

Then I was expecting an XOR operation, since the arithmetic is modular in the finite field GF(2), so addition is equivalent to subtraction is equivalent to a bitwise XOR operation between digits of the operands, meaning there should no be carry or borrow bits in the calculation. However, the first stage of the division, you have:

1011011010
 11011
 ------
 100101

Shouldnt the result be 101101, not 100101, since 1XOR0=0XOR1=1, and 1XOR1=0XOR0=0.

#6 Help Me ! » polynomial long division modulo 2 » 2011-01-06 08:13:00

Onyx
Replies: 5

I'm stuck with the algorithm of polynomial long division modulo 2. Say we have

, and
, where
, the finite field with elements 0 and 1. Then P(x) and Q(x) can be represented as (n+1) and (m+1) bit binary numbers respectively.

I need to know how to make the division:

In order to find the remainder, R(x). The application is Cyclic Redundancy Checks (CRCs) used for error detection of digital packets, and the quotient polynomial (or binary number) is not of interest.

I know that long division is repeated subtraction, and that subtraction modulo 2 represents a bitwise XOR between the coefficients of the operands, and the CRC hardware used to compute these checks is a Shift Register containing XOR gates with feedback to the SR, reflecting the fact that the algorithm is based on repeated shifting and XORing of the divisor. I'm not sure exactly how this goes though, could someone help me out using an example?

Let P(x)=1011011010, Q(x)=11011, then how would the long division go (using binary numbers)?

#7 Re: Help Me ! » Inverse z-transform » 2011-01-01 04:55:24

Well most the tables I have looked at actually do define the inverse transforms in terms of the unit step, and the unit impulse (Kronecker delta) functions, which is expected, since it represents causality of the discrete time systems I'm working with. I think I mayu need partial fractions...

#8 Re: Help Me ! » solve for a » 2011-01-01 04:49:56

Why is it not possible to do this?

Then:

Then by the quadratic formula:

but,

#9 Re: Help Me ! » Inverse z-transform » 2011-01-01 04:42:23

Thanks, I know I need to use inverse z transform, but I can't find anything in a z-transform table, so I'm not sure how to do it.

#10 Help Me ! » Inverse z-transform » 2010-12-31 05:51:35

Onyx
Replies: 5

I have a transfer function:

I can't find it in any transform table. Is it possible to use a convolution sum to solve this from inverse z-transforms of its constituent factors? I can't find a way to do it. Thanks

#11 Re: Help Me ! » solve for a » 2010-12-31 05:45:46

Hi bobbym, actually as I said there are two approximate roots of the equation, given the constraint |a|<1. They are:

But only,

satisfies the constraint |a|<1, since,

Also, I know that these are only approximations, and for my engineering design application, they are extremely accurate approximations, and will suffice for the limited precision required.

I'm also familiar with iterative methods such as Newton or Newton-Raphson in numerical analysis, and am a proficient programmer, and would be able to implement such algorithms myself....Excel could also be used. I was just interested to see if the problem had a solution using algebraic manipulation, since my maths is a bit rusty in some areas, and this one really stumped me. I appreciate the offer though, but the solution is good enough for my problem.

Thank you

#12 Re: Help Me ! » solve for a » 2010-12-31 00:17:49

haha thanks both! smile

I beleive you are both right, since I have |a|<1, so there are two approximate solutions....thats good enough for me.

(Application is design of a discrete time LTI filter for an audio DSP)

#13 Help Me ! » solve for a » 2010-12-30 05:18:43

Onyx
Replies: 12

Hi, can anybody help me solve this for a:

I'm really stuck

#14 Help Me ! » Can I simplify this fraction? » 2010-11-24 09:16:30

Onyx
Replies: 2

Can this fraction be simplified?

I'm not sure if this can actually be simplified or not.

#15 Re: Help Me ! » partial fraction » 2010-04-15 07:01:35

Thanks, it seems this partial fraction doesn't make anything simpler at all, but I found another way to solve my problem. I do however have another partial fraction:

I thought something like this was written as:

I haven't seen partial fractions written with a constant as a term in its own right (As the term 'A' is in your example) before, and also thought the form Bx+C in a numerator was only used for squared variables within the parenthesis such as:

Is the form I have provided not correct?

#16 Re: Help Me ! » partial fraction » 2010-04-13 07:43:38

Where does the A and Cx come from? I thought it was simply:

#17 Help Me ! » partial fraction » 2010-04-13 04:34:24

Onyx
Replies: 9

Hi, how can I reduce something like the following into partial fractions?

Thanks

#18 Re: Help Me ! » Taylor series and power series » 2009-08-30 09:13:25

So a Taylor series is just one type of power series?

#20 Help Me ! » Taylor series and power series » 2009-08-30 08:56:28

Onyx
Replies: 3

Hi, what's actually the difference between a Taylor series and a power series? I looked at the definitions on wikipedia and couldn't work out how they were different.

#21 Re: Help Me ! » tangent plane/normal vector » 2009-08-29 06:32:06

Thanks, do you have any idea about this?..

Onyx wrote:

Thanks, is there any theorem that says whenever we have a function,

, generating a surface in
, then when we express this function implicitly as
, the normal vector to the surface can be given by:

This is the tecnique you have used here, though I've never came across it.

#22 Re: Help Me ! » tangent plane/normal vector » 2009-08-26 00:14:08

Thanks, is there any theorem that says whenever we have a function,

, generating a surface in
, then when we express this function implicitly as
, the normal vector to the surface can be given by:

This is the tecnique you have used here, though I've never came across it.

Also can someone tell me what was wrong with my method of parameterization?

Thanks

#23 Re: Help Me ! » tangent plane/normal vector » 2009-08-20 01:10:47

Thanks, but I don't understand the method. Doesn't a function of three variables such as:

describe a surface in

, and not
?

Then the gradient of f,

would be a vector in
- I know this is what we are looking for but don't understand how it relates to the function of three variables, f, or even how you can simply rewrite the equation as you have and declare it as a function. I would have though parameterizing would have been the only way, but obviously this is wrong (I'm still not sure why though).

#24 Help Me ! » tangent plane/normal vector » 2009-08-17 06:53:46

Onyx
Replies: 7

Hi, I'm working through Schaums outlines linear algebra, and one of the excercises from the first chapter asks to find the normal vector

and tnagent plane H to the surface
at the point P(1,3,2).

Heres my working:


parameterizing:

at P(1,3,2):

at P(1,3,2):

at P(1,3,2):

The answer given is

, although no method is given.

I'm fairly sure vector calculus isn't assumed as a pre-requisite so I have two questions:

-Whats wrong with my method? (why has it given the wrong answer?)
-Whats the easier (or should I say correct?) way to do this? Since the first chapter only deals with basic vector properties and identities I assume it can be done using only this.

Thanks alot.

#25 Re: Help Me ! » moment of vector about point in 2D » 2009-04-23 02:43:56

Thanks, I had seen the cross product used in 3D examples, and know this isn't defined in 2D, but I see the formula you've given is that of a determinant. Basically we're finding the component of the force vector acting perpendicuar to the position vector of the point right? (assuming the moment is about the origin)

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