Folks, I am very new to the area of DSP and just getting a hang of it. Please have patience with me since my questions are going to be very basic. Question - it is given that if a sinusoid input is applied to an LTI system then the output is another sinusoid (albeit scaled and delayed version). Similarly, is this the case if a square input (or a unit value) is applied to such a LTI system? I know this is not the case but wondering where is the difference in LTI system. Can anyone provide a very simple example of a LTI system to help me analyze? Many thanks ...
sinusoidal input to LTI system
Started by ●July 19, 2011
Reply by ●July 19, 20112011-07-19
On Jul 19, 8:55�am, "manishp" <manishp.p18@n_o_s_p_a_m.gmail.com> wrote:> Folks, > > I am very new to the area of DSP and just getting a hang of it. > Please have patience with me since my questions are going to be very > basic. > > Question - > > it is given that if a sinusoid input is applied to an LTI system then the > output is another sinusoid (albeit scaled and delayed version).yes, it's true. no, it's not given, but is a result that one can derive. implied with the "albeit scaled and delayed", the output sinusoid is exactly the same frequency as the input sinusoid.> Similarly, is this the case if a square input (or a unit value) is applied > to such a LTI system?no. not in general.> I know this is not the case but wondering where is the difference in LTI > system.sinusoids can be expressed as exponentials with "i" or "j" in the exponent. exponentials are what we call "eigenfunctions" of LTI systems. if an eigenfunction goes in, then the same function comes out, but scaled by a complex scaler (the angle of that complex scaler is what accounts for the phase shift or the apparent delay). this complex scaler is a function of the frequency of the sinusoid. the way to think about the square wave is that it is the sum of a bunch of sinusoids of different frequencies, that all get passed through the LTI system. but they are not all passed through with the same delay or scaling. so when they are added up in the output, they add up to something different than what went in.> Can anyone provide a very simple example of a LTI system to help me > analyze?this is probably too trivial: y(t) = 0 this one is also probably too trivial: y(t) = x(t) this gets a little less trivial: y(t) = x(t-1) but it still doesn't change the wave shape. this one *will* affect the wave shape for non-sinusoidal input: y(t) = x(t) + x(t-1) have fun. r b-j
Reply by ●July 19, 20112011-07-19
On Jul 19, 11:18�am, robert bristow-johnson <r...@audioimagination.com> wrote:> On Jul 19, 8:55�am, "manishp" <manishp.p18@n_o_s_p_a_m.gmail.com> > wrote: > > > Folks, > > > I am very new to the area of DSP and just getting a hang of it. > > Please have patience with me since my questions are going to be very > > basic. > > > Question - > > > it is given that if a sinusoid input is applied to an LTI system then the > > output is another sinusoid (albeit scaled and delayed version). > > yes, it's true. �no, it's not given, but is a result that one can > derive. > > implied with the "albeit scaled and delayed", the output sinusoid is > exactly the same frequency as the input sinusoid. > > > Similarly, is this the case if a square input (or a unit value) is applied > > to such a LTI system? > > no. �not in general. > > > I know this is not the case but wondering where is the difference in LTI > > system. > > sinusoids can be expressed as exponentials with "i" or "j" in the > exponent. �exponentials are what we call "eigenfunctions" of LTI > systems. �if an eigenfunction goes in, then the same function comes > out, but scaled by a complex scaler (the angle of that complex scaler > is what accounts for the phase shift or the apparent delay). �this > complex scaler is a function of the frequency of the sinusoid. > > the way to think about the square wave is that it is the sum of a > bunch of sinusoids of different frequencies, that all get passed > through the LTI system. �but they are not all passed through with the > same delay or scaling. �so when they are added up in the output, they > add up to something different than what went in. > > > Can anyone provide a very simple example of a LTI system to help me > > analyze? > > this is probably too trivial: > > � �y(t) = 0 > > this one is also probably too trivial: > > � �y(t) = x(t) > > this gets a little less trivial: > > � �y(t) = x(t-1) > > but it still doesn't change the wave shape. > > this one *will* affect the wave shape for non-sinusoidal input: > > � �y(t) = x(t) + x(t-1) > > have fun.Manship, Robert knows much more that I do and I love him dearly. In this case, however, he knows too much. In particular, he thinks in math. I can answer your question more simply. Every waveform can be decomposed into sinusoidal components. (I presume that you know the Fourier series (as the collection of harmonics is called) of a square waveform. Each of those harmonics will pass through an LTI system unchanged in frequency but (except for a wire) changed in amplitude and/or delay. The sum of the emerging sinusoids will not in general have the same shape the sum of the original ones. Jerry -- Engineering is the art of making what you want from things you can get.
Reply by ●July 19, 20112011-07-19
On Jul 20, 12:55�am, "manishp" <manishp.p18@n_o_s_p_a_m.gmail.com> wrote:> Folks, > > I am very new to the area of DSP and just getting a hang of it. > Please have patience with me since my questions are going to be very > basic. > > Question - > > it is given that if a sinusoid input is applied to an LTI system then the > output is another sinusoid (albeit scaled and delayed version). > > Similarly, is this the case if a square input (or a unit value) is applied > to such a LTI system? > > I know this is not the case but wondering where is the difference in LTI > system. > > Can anyone provide a very simple example of a LTI system to help me > analyze? > > Many thanks ...1/(s+1) Hardy
Reply by ●July 19, 20112011-07-19
On Jul 20, 6:05�am, Jerry Avins <j...@ieee.org> wrote:> On Jul 19, 11:18�am, robert bristow-johnson > > > > <r...@audioimagination.com> wrote: > > On Jul 19, 8:55�am, "manishp" <manishp.p18@n_o_s_p_a_m.gmail.com> > > wrote: > > > > Folks, > > > > I am very new to the area of DSP and just getting a hang of it. > > > Please have patience with me since my questions are going to be very > > > basic. > > > > Question - > > > > it is given that if a sinusoid input is applied to an LTI system then the > > > output is another sinusoid (albeit scaled and delayed version). > > > yes, it's true. �no, it's not given, but is a result that one can > > derive. > > > implied with the "albeit scaled and delayed", the output sinusoid is > > exactly the same frequency as the input sinusoid. > > > > Similarly, is this the case if a square input (or a unit value) is applied > > > to such a LTI system? > > > no. �not in general. > > > > I know this is not the case but wondering where is the difference in LTI > > > system. > > > sinusoids can be expressed as exponentials with "i" or "j" in the > > exponent. �exponentials are what we call "eigenfunctions" of LTI > > systems. �if an eigenfunction goes in, then the same function comes > > out, but scaled by a complex scaler (the angle of that complex scaler > > is what accounts for the phase shift or the apparent delay). �this > > complex scaler is a function of the frequency of the sinusoid. > > > the way to think about the square wave is that it is the sum of a > > bunch of sinusoids of different frequencies, that all get passed > > through the LTI system. �but they are not all passed through with the > > same delay or scaling. �so when they are added up in the output, they > > add up to something different than what went in. > > > > Can anyone provide a very simple example of a LTI system to help me > > > analyze? > > > this is probably too trivial: > > > � �y(t) = 0 > > > this one is also probably too trivial: > > > � �y(t) = x(t) > > > this gets a little less trivial: > > > � �y(t) = x(t-1) > > > but it still doesn't change the wave shape. > > > this one *will* affect the wave shape for non-sinusoidal input: > > > � �y(t) = x(t) + x(t-1) > > > have fun. > > Manship, > > Robert knows much more that I do and I love him dearly. In this case, > however, he knows too much. In particular, he thinks in math. I can > answer your question more simply. > > Every waveform can be decomposed into sinusoidal components. (I > presume that you know the Fourier series (as the collection of > harmonics is called) of a square waveform. > > Each of those harmonics will pass through an LTI system unchanged in > frequency but (except for a wire) changed in amplitude and/or delay. > > The sum of the emerging sinusoids will not in general have the same > shape the sum of the original ones. > > Jerry > -- > Engineering is the art of making what you want from things you can > get.Yes, but don't forget that there is also a transient component and the op never specified steady-state or full output. hardy
Reply by ●July 19, 20112011-07-19
On Tue, 19 Jul 2011 07:55:50 -0500, manishp wrote:> Folks, > > I am very new to the area of DSP and just getting a hang of it. Please > have patience with me since my questions are going to be very basic. > > Question - > > it is given that if a sinusoid input is applied to an LTI system then > the output is another sinusoid (albeit scaled and delayed version). > > Similarly, is this the case if a square input (or a unit value) is > applied to such a LTI system? > > I know this is not the case but wondering where is the difference in LTI > system.Like Jerry said, anything that is not itself a sinusoid can be thought of as being composed of as many sinusoids. An LTI system will, in general, treat different-frequency signals differently. They'll get munged*. The output will be the sum of the munged components, and will be munged itself.> Can anyone provide a very simple example of a LTI system to help me > analyze?In Math, or in English? It's kind of hard to do in English, because the 'system' in 'LTI system' is a mathematical concept. Dangle a rope or chain from your hand. Move your hand horizontally, and look at the motion of the end of the rope. When you define that rope as a system whose input is your hand motion and whose output is the horizontal (not vertical!!) motion of it's tip, then you have a LTI system. * "Munged" passes the spell checker in Pan, so it must be a real word... -- www.wescottdesign.com
Reply by ●July 19, 20112011-07-19
On 7/19/2011 5:55 AM, manishp wrote:> Folks, > > I am very new to the area of DSP and just getting a hang of it. > Please have patience with me since my questions are going to be very > basic. > > Question - > > it is given that if a sinusoid input is applied to an LTI system then the > output is another sinusoid (albeit scaled and delayed version). > > Similarly, is this the case if a square input (or a unit value) is applied > to such a LTI system? > > I know this is not the case but wondering where is the difference in LTI > system. > > Can anyone provide a very simple example of a LTI system to help me > analyze? > > Many thanks ... > >Well, you got several good answers. Just in case, I'll give another if similar: > it is given that if a sinusoid input is applied to an LTI system then the > output is another sinusoid (albeit scaled and delayed version). Only if the sinusoid was applied long enough prior to be able to assume / observe "steady state". I like r b-j's example: y(t) = x(t) - x(t-1) That's an LTI system that consists of one delay element and the output is the current input minus the output of the delay element. Or, for a discrete system, it consists of subtracting the previous sample from the current sample to generate the output (normalizing the sample interval to 1.0). As the frequency tends to zero, i.e. at very low frequencies, the output tends to zero. As the frequency increases, the result grows until the delay reaches one quarter period: pi/2. Then it decreases until the delay reaches one half period: pi. For a continuous system, this is a bandpass filter with peaks at pi/2, 3pi/2 etc. For a discrete system, this is a bandpass filter with peak at pi/2 and zero at zeroand fs - which happens to "just" violate the sampling theorem. As others have mentioned, a square wave input can be represented as a sum of sinusoids - so each will have its own steady state response through the LTI and, by superposition, the output will have a shape determined by their amplitudes and phase. Because the system above isn't linear phase (or constant delay), the sinusoidal components of a square wave input aren't aligned in phase. That plus the amplitude changes causes the output to not look very "squarewave-like). Another interesting example: y(t) = x(t+1) - 2x(t) + x(t-1) In this case the result is linear phase / constant delay. This means that all of the sinusoidal components of the square wave (up to the limit imposed by sampling) arrive perfectly aligned relative to the input but are bandpass filtered peaking at fs/4. So, the output might be a bit more "squarewave-like" - I'm not sure, haven't tried it. Fred
Reply by ●July 20, 20112011-07-20
On Jul 19, 2:59�pm, HardySpicer <gyansor...@gmail.com> wrote:> On Jul 20, 6:05�am, Jerry Avins <j...@ieee.org> wrote: > > > > > > > > > > > On Jul 19, 11:18�am, robert bristow-johnson > > > <r...@audioimagination.com> wrote: > > > On Jul 19, 8:55�am, "manishp" <manishp.p18@n_o_s_p_a_m.gmail.com> > > > wrote: > > > > > Folks, > > > > > I am very new to the area of DSP and just getting a hang of it. > > > > Please have patience with me since my questions are going to be very > > > > basic. > > > > > Question - > > > > > it is given that if a sinusoid input is applied to an LTI system then the > > > > output is another sinusoid (albeit scaled and delayed version). > > > > yes, it's true. �no, it's not given, but is a result that one can > > > derive. > > > > implied with the "albeit scaled and delayed", the output sinusoid is > > > exactly the same frequency as the input sinusoid. > > > > > Similarly, is this the case if a square input (or a unit value) is applied > > > > to such a LTI system? > > > > no. �not in general. > > > > > I know this is not the case but wondering where is the difference in LTI > > > > system. > > > > sinusoids can be expressed as exponentials with "i" or "j" in the > > > exponent. �exponentials are what we call "eigenfunctions" of LTI > > > systems. �if an eigenfunction goes in, then the same function comes > > > out, but scaled by a complex scaler (the angle of that complex scaler > > > is what accounts for the phase shift or the apparent delay). �this > > > complex scaler is a function of the frequency of the sinusoid. > > > > the way to think about the square wave is that it is the sum of a > > > bunch of sinusoids of different frequencies, that all get passed > > > through the LTI system. �but they are not all passed through with the > > > same delay or scaling. �so when they are added up in the output, they > > > add up to something different than what went in. > > > > > Can anyone provide a very simple example of a LTI system to help me > > > > analyze? > > > > this is probably too trivial: > > > > � �y(t) = 0 > > > > this one is also probably too trivial: > > > > � �y(t) = x(t) > > > > this gets a little less trivial: > > > > � �y(t) = x(t-1) > > > > but it still doesn't change the wave shape. > > > > this one *will* affect the wave shape for non-sinusoidal input: > > > > � �y(t) = x(t) + x(t-1) > > > > have fun. > > > Manship, > > > Robert knows much more that I do and I love him dearly. In this case, > > however, he knows too much. In particular, he thinks in math. I can > > answer your question more simply. > > > Every waveform can be decomposed into sinusoidal components. (I > > presume that you know the Fourier series (as the collection of > > harmonics is called) of a square waveform. > > > Each of those harmonics will pass through an LTI system unchanged in > > frequency but (except for a wire) changed in amplitude and/or delay. > > > The sum of the emerging sinusoids will not in general have the same > > shape the sum of the original ones. > > > Jerry > > -- > > Engineering is the art of making what you want from things you can > > get. > > Yes, but don't forget that there is also a transient component and the > op never specified steady-state or full output. > > hardyHow does that affect what I wrote? Jerry -- Engineering is the art of making what you want from things you can get.
Reply by ●July 20, 20112011-07-20
On Jul 19, 10:14�pm, Fred Marshall <fmarshallxremove_th...@acm.org> wrote:> On 7/19/2011 5:55 AM, manishp wrote: > > > > > Folks, > > > I am very new to the area of DSP and just getting a hang of it. > > Please have patience with me since my questions are going to be very > > basic. > > > Question - > > > it is given that if a sinusoid input is applied to an LTI system then the > > output is another sinusoid (albeit scaled and delayed version). > > > Similarly, is this the case if a square input (or a unit value) is applied > > to such a LTI system? > > > I know this is not the case but wondering where is the difference in LTI > > system. > > > Can anyone provide a very simple example of a LTI system to help me > > analyze? > > > Many thanks ... > > Well, you got several good answers. �Just in case, I'll give another if > similar: > > �> it is given that if a sinusoid input is applied to an LTI system then the > �> output is another sinusoid (albeit scaled and delayed version). > > Only if the sinusoid was applied long enough prior to be able to assume > / observe "steady state". > > I like r b-j's example: > y(t) = x(t) - x(t-1) > That's an LTI system that consists of one delay element and the output > is the current input minus the output of the delay element.except i had a plus sign. still LTI and, with the plus sign, it's linear phase.> Another interesting example: > > y(t) = x(t+1) - 2x(t) + x(t-1) > > In this case the result is linear phase / constant delay.are you sure, Fred? i know this is: y(t) = x(t+1) + 2x(t) + x(t-1) maybe yours is. the other issue is that while this it's LTI, it ain't causal or, using the language my prof had in the 70s, "realizable". output happens before input. delaying it a little makes it causal and phase linear: y(t) = x(t) + 2x(t-1) + x(t-2) BTW, i wasn't meaning to normalize T to 1 nor even describe a discrete- time LTI. i just wanted to leave the unit time unspecified since that was not needed to elucidate the topic. but, strictly speaking (and at other times when i wanna be anal-retentive about details), t-1 and t-2 are not dimensionally consistent. also, normally when i wanna be discrete-time, i follow the convention that O&S and many other authors use and use brackets (and "n" for the sampling index value): y[n] = x[n] + 2x[n-1] + x[n-2] that's a kosher expression, even when i'm anal about math style. r b-j
Reply by ●July 20, 20112011-07-20
On Jul 19, 11:03�pm, Jerry Avins <j...@ieee.org> wrote:> On Jul 19, 2:59�pm, HardySpicer <gyansor...@gmail.com> wrote: >...> > > Yes, but don't forget that there is also a transient component and the > > op never specified steady-state or full output. > > > How does that affect what I wrote? >and, Hardy, can i have some of what you got? it's evidently much gooder than what i got to smoke. :-O :-) r b-j
