# continuous and discrete time systems

Discussion in 'Mathematics and Physics' started by PG1995, Nov 3, 2012.

1. ### PG1995Active Member

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Okay. Let's do it. I have said that x[n]=2u[n] where u[n] is a unit step function, which is unity for n>=0. Let's take a=5.

y[0]=5^2(1)=25
y[1]=5^2(1)=25
y[2]=5^2(1)=25
.
.
.

But you see it doesn't blow up. Where am I going wrong with my visualization? Please help me. Thanks.

2. ### steveBWell-Known MemberMost Helpful Member

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Yes, correct. Sorry I looked too fast and got confused. I was thinking of the exponential function a^n, not a^x.

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3. ### PG1995Active Member

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Thanks, a lot.

Regards
PG

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5. ### steveBWell-Known MemberMost Helpful Member

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the "a" coefficients are arbitrary, so you can define them either way. Just be consistent in your definitions.

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6. ### PG1995Active Member

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Thank you.

Though I was busy with other stuff, I thought I should clear these points.

The system, y[n]=e^x[n], is going to be bounded for a bounded input, I think. But the system,
for n>=0, is going to be unbounded for a bounded input such as a unit step function.

I don't really get it. Please have a look here. Thanks.

Regards
PG

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7. ### steveBWell-Known MemberMost Helpful Member

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So, the difference between 2 and 3 is the negative sign on the a_k coefficients. These definitions for a_k are different with one definition being the negative of the other definition. You can't say one is correct and the other is wrong. You can only decide on which definition you prefer, and once you decide, don't change definitions later. I just opened up one of my books which uses your definition #2. I can't recall if all books use this definition, but I believe this is the most common form. But, just be careful to check the definition used in your reference.

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8. ### PG1995Active Member

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Hi

Could you please check it to see if my working is completely correct? Thank you.

Could you please also help me with this query? Thanks.

Regards
PG

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9. ### steveBWell-Known MemberMost Helpful Member

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I think there is an error in part B. The phase should be 90 degrees for w=0 and 2pi. If you take the limit of sin(w)/(1-cos(w)), you get a x/x^2 form which goes to infinity as x goes to zero. The arctan of infinity is 90 degrees.

The other question I would say consult whatever book you are using for the definitions. You may find one notation used for the standard FT and the other for the DFT, but don't guess, - look at the context you are in or look at the definitions to be sure.

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10. ### PG1995Active Member

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Could you please tell me how you get x/x^2? Thanks.

Yes, you are right in saying that some books might use H(e^jw) for DTFT and H(Ω) for DFT. Thanks.

Regards
PG

11. ### steveBWell-Known MemberMost Helpful Member

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Well, there are many ways to take limits, but one way is by Taylor expansion.

If we expand sin(w) and take the lowest order term, we get x, and if we expand cos(w) to two lowest order terms we get 1-x^2/2.

So, this will actually give 2/x for the ratio sin(w)/(1-cos(w)) which becomes infinity as w goes to zero, and then arctan gives 90 degrees.

You may have been confused by the factor of 2 that I was missing before. I just did it quickly in my head and ignored the factor because infinity or 2 times infinity is the same thing.

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12. ### PG1995Active Member

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Thank you. I get it now.

Don't you think using Tayloy exapansion is the easiest way to find limit in this case? Perhaps, we can use some trigonometric identity(ies) which I can't think of at the moment. Thanks.

Regards
PG

13. ### steveBWell-Known MemberMost Helpful Member

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I do. I was able to do it in my head, so that is about as easy a way as you can get.

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Hi

Regards
PG

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15. ### steveBWell-Known MemberMost Helpful Member

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The sine of a multiple of pi is zero. Hence sin(pi(n-m))/(pi(n-m)) is zero for any m and n where n is not equal to m. When n equals m, we have a 0/0 limit to evaluate, but we know that the limit of sin(x)/x is 1 as x goes to zero. Hence, this function is an shifted impulse function as shown.

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