difference between impulses in dt and ct cases

[purnapragna]

difference between Δ dt

hi i have a doubt like
1)Is δ[n] derived from δ(t). that is is DT impulse a discretized version of CT

impulse? and

2) How do u represent δ[3n] ?

thank u

purna

 

[steve10]

(1). δ[n] is a discretized version of δ(t), which is uniquely characterized by (for any f(t))

∫f(t)δ(t)=f(0).

Similarly, for any digital signal {f[n]}, δ[n] is the one that makes

∑f[n]δ[n]=f[0].

(2). All you need to do is to check (for any {f[n]}), what is ∑f[n]δ[3n]? We have

∑f[n]δ[3n]=f[0]=∑f[n]δ[n].

Therefore,

δ[3n]=δ[n].

 

[purnapragna]

my question is that is δ[n] derived from δ(t)

and also sory for again asking you then how to plot δ[3n-6] ? Is it δ[n-2] ?

thanx

sri hari

 

[steve10]

When I said that is a discretized version of δ(t), I meant it's NOT derived from δ(t). Rather, it develops in parallel. They work in completely different paradises, how could they be derived from each other?

Also, I told you how to decide δ[3n]. It woks for the two news functions, too.

 

[ranjeeth]

Hi ,
δ[n] is discrete time domain signal defined as follows :
δ[n] = 1 n = 0
δ[n] = 0 for all n ≠ 0

δ(t) is ct signal defined as follows :
∫ ( δ(t) , -∞ , ∞ ) = 1 ;

The amplitude of δ(t) is not defined. The area under the curve is 1.
δ(t) (unit impulse ) is a signal with ∞ amplitude at time t = 0 ; it's amplitude is otherwise 0.
δ[n] (unit discrete impulse ) is a signal with amplitude = 1 at time t = 0 .
it's amplitude at all n ≠ 0 is zero.

The signal δ(t) is called the Dirac-delta function wheras δ[n] is called as Kronecker's delta.

 

[claudiocamera]

my question is that is δ[n] derived from δ(t)

and also sory for again asking you then how to plot δ[3n-6] ? Is it δ[n-2] ?

thanx

sri hari

Answering the remaining question, ( I think it was not answered yet) δ[3n-6] has the same plot of δ[n-2]. The operations must be done in the correct order, first the time shifting and after time scaling, when you shift the discrete impulse in 6 steps and after you scale (compress) it by 3 you have the equivalent of shifiting the discrete impulse by 2.

Regarding that the plots are equal because the signals are discrete impulse, actualy these operations doesnt happen in this way because any property of discrete impulse, they are basic operations on signals.

In order to understand better, refer to time scaling and time shifting operations on signals, This subject is usualy presented as basic operations on signals in the first chapter of signals and systems books.

 

[purnapragna]

hi

i.e., we can say that δ[3n-2] is same as δ[kn -2] where k is some integer value?

thanx

purna

 

[claudiocamera]

hi

i.e., we can say that δ[3n-2] is same as δ[kn -2] where k is some integer value?

thanx

purna

No you can't , δ[3n-2] is a empty set, considering that the order of operations is first time shifiting and then time scaling, when you time shift in 2 step and after you scale by 3 you will not have a integer number (2/3) so the discrete impulse in 2 will be discharged after compress the time scale by 3.
The analysis of the signal δ[kn -2] is the same the signal exists only to k=1 or k=2 , any order value of k represents a compression that discharges the signal orinally shifted by 2.

As I wrote earlier it is not a matter of the signal be a discrete impulse but a matter of basic operations on signals.

 

[ranjeeth]

hi

i.e., we can say that δ[3n-2] is same as δ[kn -2] where k is some integer value?

thanx

purna

I agree with cladiocamera δ[3n-2] doesnot exist.
A simple technique would be to substitute 3n-2 = k
Now δ[k] = 1 for k=0
= 0 for all other k ε I { ...,-2,-1,0,1,2,3 .... }

Thus at k =0 we have 3n-2 =0 which gives n =2/3
δ[3n-2] = 1 at n = 2/3
For any digital signal x[k] the sample is not defined at n =2/3 (i.e. any non integral values)
Hence δ[3n-2] doesnot exist.

 

[Kral]

Another difference:
The discrete impulse can be physically implemented, while the continuous (dirac delta function) can not.