[SOLVED] What will be the output when the input frequency is equal to the pole?

Consider a system whose transfer function is given by

T(s)= 1 / (s+ja)(s-ja)

It's very clear that we have a pole at the frequency w=a; Assume that this 'a' is very near to 0. Then you can say this as low pass filter.

Now my question is this-

What will happen when you give a input sine wave with frequency equal to a?

From the low pass filter characteristics you will say that it is approximately equal to the passband gain which is 1 here.

But if you substitute s=ja in the above equation you will get it as infinite.


I am a bit confused with this.

Will you please clear my doubt?


thanks in advance.

T(s)=Vout(s)/Vin(s).
Think this is sufficient to answer your query.

checkmate said:

T(s)=Vout(s)/Vin(s).
Think this is sufficient to answer your query.

Click to expand...

I mean

Vout(s)= Vin (s) * T(s)

if T(s) becomes infinite when the input frequency is equal to 'a' then the output becomes infinite. But from the bode plot we can say that it's near the passband gain and it's not infinite. That's the reason why I got confused.

That's the point. Your Vin(s) will have a term that cancels the pole!
Try replacing the Vin(s) with the laplace transform of a sinusoid.

checkmate said:

That's the point. Your Vin(s) will have a term that cancels the pole!
Try replacing the Vin(s) with the laplace transform of a sinusoid.

Click to expand...

Ha ha. Nice explanation. Thanks a lot. Now I have one more question which I will ask in my next thread

---------- Post added at 10:01 ---------- Previous post was at 09:55 ----------

Now I got another doubt.

What will happen if we have two poles at the same frequency?

eg: T(s) = 1/ (s+ja)(s+ja)(s-ja)(s-ja)

Now we get infinite if the input is at the pole frequency

---------- Post added at 11:28 ---------- Previous post was at 10:01 ----------

checkmate said:

That's the point. Your Vin(s) will have a term that cancels the pole!
Try replacing the Vin(s) with the laplace transform of a sinusoid.

Click to expand...

If you try replacing the transform of the sinusoid it doesn't cancel that term. Please help me.

---------- Post added at 11:32 ---------- Previous post was at 11:28 ----------

Hey I got it. It's indeed infinite. It's because you have real term to be zero in other words the poles lie on the jw axis and hence becomes unstable.

Hi iVenky,
I guess you are talking abt the closed loop transfer function. If so, then you have 2 poles, but one is LHP & the other is RHP. So, as such you cannot find any information from mag plot (eventhough there would be 40dB rolloff), but I guess the phase plot would be weird, ie, kind of without any phase shift. So, it is not exactly a Low pass characteristics. Correct me If I'm wrong.