continuous and discrete time systems

[PG1995]

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.

 

[steveB]

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

 

[PG1995]

Thanks, a lot.

Could you please help me with this? Thank you.

Regards
PG

 

[steveB]

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

 

[PG1995]

Thank you.

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

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

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

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

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

Regards
PG

 

[steveB]

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.

 

[PG1995]

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

 

[steveB]

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.

 

[PG1995]

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.

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

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.

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

Regards
PG

 

[steveB]

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

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.

 

[PG1995]

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

 

[steveB]

Don't you think using Tayloy exapansion is the easiest way to find limit in this case?

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

 

[PG1995]

Hi

Could you please help me with this query? Thank you.

Regards
PG

 

[steveB]

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.