discrete exponentials

[PG1995]

Hi

Could you please help me with **broken link removed** queries? Thank you.

Regards
PG

 

[steveB]

Q1: The stems are above and below the axis because exponentials of complex and imaginary numbers generate oscillatory functions. In continuous time, you would have g(t)=A*exp(z(t))=A*exp(x(t)+j*y(t))=A*exp(x)*exp(j*y). The exp(j*y) is equal to cox(y)+j*sin(y) and is oscillatory. It's the same with the discrete time exponential.

Q2: If you use the Argand Diagram, then you need to animate it to see how the vector moves in time. But this is acceptable if you have the software to do it, and it is a very good way to think about it even if you only animate it in your mind. Generally, oscillatory functions rotate around in a circular or spiral-like path. When |z|<1, the vector will spiral inwards. When |z|=1, you get a circular or elliptical shape. When |z|>1 the vector will spiral outwards.

 

[PG1995]

Thank you.

But in the Q1 we have real z's, not complex. Or, perhaps I still have it wrong. What do you say? Kindly let me know. Thanks.

Regards
PG

 

[steveB]

Thank you.

But in the Q1 we have real z's, not complex. Or, perhaps I still have it wrong. What do you say? Kindly let me know. Thanks.

Regards
PG

Did you read the figure captions for Fig. 3.9?

 

[PG1995]

Thanks.

Did you read the figure captions for Fig. 3.9?

Yes. But the Q1 is about the figure 3.8 and the caption says for different real values of z.

 

[steveB]

Thanks.



Yes. But the Q1 is about the figure 3.8 and the caption says for different real values of z.

Oh, yes correct.

For that case you have |z|<0. Now think about what happens when you raise a negative number to an even power (i.e. n=2, 4, 6 ...). Now, what happens when you raise a negative number to an odd power (i.e. n=1, 3, 5 ...).

 

[PG1995]

Thanks a lot.

 

[PG1995]

average signal power for periodic discrete-time function

Hi

Kindly help me with **broken link removed**. Many, many thanks.

Best wishes
PG

 

[steveB]

I think it's just a choice that was made. Notice that the average is taken by summing over 2N points and then dividing by 2N. If the summation goes from -N to +N, there will be 2N+1 points. I suppose one could just then divide by 2N+1 and take the limit. There may also be some other benefits to the given definition, but if so, I don't see them right now.

 

[MrAl]

Hello,


The diagram drawn for this question might be a little misleading because the difference between successive amplitudes is quite large and this might not be typical in a real life system. In a real system usually the difference between successive samples is small. Although not mandatory this helps ensure accuracy.
The reason for the N-1 is probably because you can not use N points when calculating area. If you use N points you get an area that is too big.
For example, if you had N=3 points x1,x2,x3, that only gives you two areas not three, so that's N-1 area segments.
You'll notice that with successive amplitudes that are nearly the same this works out quite well, even if the time increments are quite large. Doing it the other way, with N samples, the error would rise.

I should also mention however that i've seen a definition summing from n1 to n2 and dividing by n2-n1, n1 and n2 inclusive and for a finite interval.