We have a periodic input signal with the period of T=12. Can we have a periodic output signal with the period of T=3 in a LTI system?
Your question is equivalent to asking if it is possible for an LTI system to have an output at four times the frequency of the forcing function. The order of the LTI is important and you have not specified it in this context. The following is from the wikipedia article Because sinusoids are a sum of complex exponentials with complex-conjugate frequencies, if the input to the system is a sinusoid, then the output of the system will also be a sinusoid, perhaps with a different amplitude and a different phase, but always with the same frequency upon reaching steady-state. LTI systems cannot produce frequency components that are not in the input. Systems with non-linear elements do create new frequencies that are not in the input. For example a multiplier will create new frequencies at the sum and difference of the two input frequencies.
Hello there, I am not sure that is entirely correct. If the input is a single sinusoidal, then yes, that's a very good explanation. But the question did not specify that the input had to be a single sinusoidal: Given that it only has to be periodic, we could imagine choosing two frequencies one 100Hz and the other 400Hz where the 400Hz signal is of lower amplitude than the 100Hz. This kind of input would look like it had a period of 1/100 seconds. We could then imagine a high pass filter or maybe a band pass filter that attenuates the 100Hz sine a lot but doesnt affect the 400Hz sine too much. We'd then see something that looked more like it had a period of 1/400 seconds. It's good to remember that a single sinusoidal passing through a linear system can only change in amplitude and phase, but for two or more sines we can end up with some very interesting outputs. A square wave has lots of frequencies, but if we filter it right we get a single sine wave at the fundamental frequency out for the most part, with only very small remnants of the other frequencies. If we use a different filter, we get a sine that has frequency three times the fundamental, a different one yet and five times the fundamental, etc.
Well if it is a linear system then the principle of superposition applies, and the output will contain the same frequencies but there won't be any new frequencies. That was the substance of the original question. Creating new frequencies REQUIRES a non-linear element. Challenge: Show me an LTI with any combination of input sinusoids that produces new frequencies.
Hi, Is that a challenge in response to my previous post? If so, the point was not to be able to create new frequencies from a single sine wave it was to be able to create a signal with a different (shorter) period. Please note in the original post the word 'frequency' was never mentioned, only the 'period', and also 'sinusoid' was never mentioned either. So we dont have to create new frequencies from a single sine we just have to show a different (shorter) period. Even though a different period means a different frequency, it's not the same because it does not have to be a sinusoid. Quote from original post: I already agreed that any sinusoidal signal as input will only produce the same frequency. Combinations of sinusoids could be different because we are looking only at the period of a signal that is not sinusoidal. So for pure sine waves of a single frequency your post was accurate, but because the question was about the period of the signal that could mean we have more than one sinusoid at a time and we may be filtering out any one or more of them which results in a period that is not the same as the input wave. More basically, 'period' is a measure in the time domain and does not necessarily imply that there has to be a single sine wave present. If the wave is non sinusoidal, the period could change once it is filtered. The individual frequencies do not change, but the overall period does and we might then call that the frequency. Example: Vin(t)=10*sin(2*pi*f*t)+sin(6*pi*f*t) The period is T=1/f. Now filter out the fundamental then we have: v2(t)=sin(6*pi*f*t) so now the period is T=1/(3*f) which is 1/3 of what it was before. Note we have not 'created' any new frequencies in the same way we can do with a non linear filter, but the period itself did change.
I don't think that your construction meets the original requirement since the final signal with the shorter period is present in the input. I agree that if you start with a superposition of signals with different periods then you can get one or more of those periods in the output. We'll have to let the OP offer a ruling.
In this formulation we have T as a constant for all possible values of k, belonging to Z, the set of all integers. If the value of T is constant for all members of the input set x(t), with non-zero values of a_sub_k then all members of the output set y(t) will have the same value of T for all the non-zero members of b_sub_k. No new values of T are possible because it is a constant. BTW filtering the output is equivalent to setting one of the non-zero b_sub_k to 0. All the members of the remaining output have the same value of T, but different values of k and b_sub_k.
For some reason you want to over specify the original question where we have to impose some previously unmentioned criterion on the question result. Are we not allowed to reply to the EXACT question being asked without adding more criteria? You seem to be worrying about creating 'new' frequencies as we do with a non linear circuit, yet to satisfy the question all we have to do is show that the period changed. If the question was not asked correctly, that's not my problem So it is not a matter of a 'ruling' because the question was answered correctly. If a different question comes up then perhaps we can address that next. Note also that there is only one 'period' to any given waveform. There are not several periods, just one. That means for example if we had a square wave of frequency 1Hz with a low amplitude 1kHz sine riding on it, the period would be 1, not 1 and 0.001. Yes, we could describe the 1kHz sine as having a period of 0.001, but that's not the period of the square wave, which is the dominate wave. The original question: Part 1: "We have a periodic input signal with the period of T=12" Ok, here it is with period T=12: 10*sin(2*pi*(1/12)*t)+sin(2*pi*(1/3)*t) Part 2: "Can we have a periodic output signal with the period of T=3 in a LTI system?" Answer: Yes. After filtering we have as output: sin(2*pi*(1/3)*t) and this output has period T=3. Note to OP: If you would like to narrow the question down with more constraints, now would be the time to do so If you like we can look at the mathematical response to a filter and prove that it is possible to change the period. It's also important to note what Papa is saying here too, about creating new frequencies, or not.
You can draw any conclusion you want from a false premise. The period of the input waveform is incorrect according to the definition given by the OP
Hi my friends, to MrAl: you have a more practical view and what I want is a theoretical proof. BTW is that really a LTI system, I mean the Filtering system you mentioned? to papa: I agree with you somehow and your first Answer gave me the motivation to change my mindset about the problem. Well What I think is when we have X (t) that is sum of some exponentials with determined coefficients and then we put it in a LTI system that produces Y(t) which is the sum of the same exponentials with different coefficients we might lose some exponentials because of the different coefficients but can't add new exponentials . What I conclude is because the exponentials are periodic and sum of the periodic signals is periodic with a period that is only related to exponentials with nonzero coefficients so there can be no new frequency (new exponential with a new period) in the output signal. I still don't know the exact mathematical proof though ... Excuse me for my late answers my english is poor.sorry for my mistakes
The input signal can be any periodic signals with T=12 that satisfies the convergence conditions of Fourier series.
As I look at the definition of period T, I notice that it refers to the period of the fundamental component. That is, it is the period of the exponential with k = 1. Therefore if T=12 for k=1 then it is not possible to have any signal in the output with T=3. We can only have T=12 and k belongs to the set of all integers. There is no room for a T=3 in that formulation.