linear vs constant phase response of LTI system

Hello All,

It is mentioned that ideal phase response of a LTI filter is linear. The
reason quoted for this is, for such a system, an sinusoidal input would be
pass out without any shape distortion (unless the specifci frequency is
being filtered)

I am not able to wrap my head around this.

I take a signal x(n) = sin1(n) + sin2 (n);

Both sin1 and sin2 are of different frequencies but fall in the pass band
of a LTI filter.

Now, assuming a constant phase for the filter, when I pass x(n) through the
filter then both sin1 and sin2 undergo equal phase shifts. Combining them
back (conceptually, of course), would give me exact same shape as x(n). 

However, if filter had linear phase, then sin1 and sin2 would undergo
unequal phase shift and hence when combined back would result in a
difference shape.

Can experts weigh in please?	 

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On 16.08.14 09.08, SBR123 wrote:
> It is mentioned that ideal phase response of a LTI filter is linear. The
> reason quoted for this is, for such a system, an sinusoidal input would be
> pass out without any shape distortion (unless the specifci frequency is
> being filtered)
>
> I am not able to wrap my head around this.
>
> I take a signal x(n) = sin1(n) + sin2 (n);
>
> Both sin1 and sin2 are of different frequencies but fall in the pass band
> of a LTI filter.

You should have written more precisely
   x(t) = cos(w1*t + p1) + cos(w2*t + p2);
t := time
w := angular frequency = 2 pi f
p := phase
(Be aware of the units.)

> Now, assuming a constant phase for the filter, when I pass x(n) through the
> filter then both sin1 and sin2 undergo equal phase shifts. Combining them
> back (conceptually, of course), would give me exact same shape as x(n).

No. The phase shift is defined relative to the frequency. Shifting by 2 
pi = 360� is a no-op. But shifting different frequencies by the same 
phase changes everything. Think of a 90� phase shift, i.e. replacing sin 
by cos.

To get the time delay you need to divide the phase by the frequency. Or 
in other words a constant time shift equals a phase shift that increases 
linearly with the frequency.

It is also quite obvious from the units. A time shift must carry seconds 
or some other time unit while the argument of sin must not have any 
units and hence not the summand p.


Marcel

On Saturday, August 16, 2014 3:08:17 AM UTC-4, SBR123 wrote:
> Hello All,
> 
> 
> 
> It is mentioned that ideal phase response of a LTI filter is linear. The
> 
> reason quoted for this is, for such a system, an sinusoidal input would be
> 
> pass out without any shape distortion (unless the specifci frequency is
> 
> being filtered)
> 
> 
> 
> I am not able to wrap my head around this.
> 
> 
> 
> I take a signal x(n) = sin1(n) + sin2 (n);
> 
> 
> 
> Both sin1 and sin2 are of different frequencies but fall in the pass band
> 
> of a LTI filter.
> 
> 
> 
> Now, assuming a constant phase for the filter, when I pass x(n) through the
> 
> filter then both sin1 and sin2 undergo equal phase shifts. Combining them
> 
> back (conceptually, of course), would give me exact same shape as x(n). 
> 
> 
> 
> However, if filter had linear phase, then sin1 and sin2 would undergo
> 
> unequal phase shift and hence when combined back would result in a
> 
> difference shape.
> 
> 
> 
> Can experts weigh in please?	 
> 
> 

Delay = convolve with \delta(t-T) in time domain, equivalent to multiply by \exp(1i \omega T) in freq. This gives it linear phase. 

>No. The phase shift is defined relative to the frequency. Shifting by 2 
>pi = 360� is a no-op. But shifting different frequencies by the same 
>phase changes everything. Think of a 90� phase shift, i.e. replacing sin

>by cos.
>
>To get the time delay you need to divide the phase by the frequency. Or 
>in other words a constant time shift equals a phase shift that increases 
>linearly with the frequency.

Mr. Mueller,

Thank you. I was confusing time shift for different frequencies with phase
shift for each frequency
	 

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On Sat, 16 Aug 2014 02:08:17 -0500, SBR123 wrote:

> Hello All,
> 
> It is mentioned that ideal phase response of a LTI filter is linear. The
> reason quoted for this is, for such a system, an sinusoidal input would
> be pass out without any shape distortion (unless the specifci frequency
> is being filtered)
> 
> I am not able to wrap my head around this.
> 
> I take a signal x(n) = sin1(n) + sin2 (n);
> 
> Both sin1 and sin2 are of different frequencies but fall in the pass
> band of a LTI filter.
> 
> Now, assuming a constant phase for the filter, when I pass x(n) through
> the filter then both sin1 and sin2 undergo equal phase shifts. Combining
> them back (conceptually, of course), would give me exact same shape as
> x(n).
> 
> However, if filter had linear phase, then sin1 and sin2 would undergo
> unequal phase shift and hence when combined back would result in a
> difference shape.
> 
> Can experts weigh in please?

First, this expert really hates to see statements like "it is mentioned 
that".  Mentioned by whom?  How credible are they?  Why should I take the 
follow-on statement as fact when the author doesn't have enough 
credibility to be mentioned by name?  When the "whom" is unknown, it 
becomes difficult to clarify an incorrect or overly-narrow statement of 
"fact".

At any rate, the ideal phase response of an LTI filter is often not linear 
-- and you may use my name in citing this fact.  For different reasons, 
both control systems and audio applications often demand minimum-phase 
filters, whose phase response is most certainly not linear.

Second, you err in taking linear phase to be constant phase.  Linear phase 
shift (if the unknown speakers mean what I think they do) means that the 
phase shift of the filter is proportional to frequency.  Such a phase 
shift means that the filter has a constant delay -- so there are no phase-
induced changes in signal shape (although there may still be changes in 
shape induced by changes in the amplitudes of the signal components).

-- 

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

On Sat, 16 Aug 2014 02:08:17 -0500, "SBR123" <100967@dsprelated>
wrote:

>Hello All,
>
>It is mentioned that ideal phase response of a LTI filter is linear. The
>reason quoted for this is, for such a system, an sinusoidal input would be
>pass out without any shape distortion (unless the specifci frequency is
>being filtered)
>
>I am not able to wrap my head around this.
>
>I take a signal x(n) = sin1(n) + sin2 (n);
>
>Both sin1 and sin2 are of different frequencies but fall in the pass band
>of a LTI filter.
>
>Now, assuming a constant phase for the filter, when I pass x(n) through the
>filter then both sin1 and sin2 undergo equal phase shifts. Combining them
>back (conceptually, of course), would give me exact same shape as x(n). 
>
>However, if filter had linear phase, then sin1 and sin2 would undergo
>unequal phase shift and hence when combined back would result in a
>difference shape.
>
>Can experts weigh in please?	 

Hi SBR123,
  When you see the acronym 'LTI" describing 
a digital filter, the 'L' stands for "linear" 
and the 'TI' stands for "time-invariant."

The "linear" phrase means that if I apply a 
sin#1 + sin#2 signal to the input of the filter, 
the filter's output will be the sum of *ONLY* two 
sine waves.  The freqs of those two output sine 
waves will be the same as the freqs of the two 
input sine waves.  That is, a "linear" filter 
will NOT generate any output sine waves whose 
freqs are different than the input sine waves' 
freqs.

The "time-invariant" phrase means that the filter's 
impulse response does not change over time.
That is, the filter's impulse response is the 
same at 12:00 noon on Tuesday as it was at 12:00 
noon on Monday.

[-Rick-]

Rick

It's important to note one exception to your LTI example. Never measure an impulse response after lunch on Friday. It's always different. Don't know why. 

Bob

On Fri, 22 Aug 2014 01:45:17 -0700, Rick Lyons wrote:

> [snip]

> The "linear" phrase means that if I apply a 
> sin#1 + sin#2 signal to the input of the filter, 
> the filter's output will be the sum of *ONLY* two 
> sine waves.  The freqs of those two output sine 
> waves will be the same as the freqs of the two 
> input sine waves.  That is, a "linear" filter 
> will NOT generate any output sine waves whose 
> freqs are different than the input sine waves' 
> freqs.
> 
> The "time-invariant" phrase means that the filter's 
> impulse response does not change over time.
> That is, the filter's impulse response is the 
> same at 12:00 noon on Tuesday as it was at 12:00 
> noon on Monday.
> 
> [-Rick-]

The key part of linearity is _superposition_ : if you
know the response of a linear system to signal 'A', and 
separately know the response to signal 'B', then the
response to any combination of these signals is:
    k1 * f(A) + k2 * f(B) = f(k1 * A + k2 * B)
where k1 and k2 are constant multipliers.

Things get much messier when the system is time-varying.

On Fri, 22 Aug 2014 03:45:09 -0700 (PDT), radams2000@gmail.com wrote:

>Rick
>
>It's important to note one exception to your LTI example. Never measure an impulse response after lunch on Friday. It's always different. Don't know why. 
>
>Bob

Hi Bob,
   Ha ha.  Funny how that happens!

[-Rick-]

Thanks a lot, everyone. I think I was confusing phase response directly
with time delay that a specific signal would undergo after passing through
an LTI system. I get this now.

>First, this expert really hates to see statements like "it is mentioned 
>that".  Mentioned by whom?  How credible are they?  Why should I take the

>follow-on statement as fact when the author doesn't have enough 
>credibility to be mentioned by name?  When the "whom" is unknown, it 
>becomes difficult to clarify an incorrect or overly-narrow statement of 
>"fact".

Hello Tim,

Since I had multiple reference before putting this question, I was vague. I
will keep that in mind. Note that my standard reference (so far) is DSP
book by Prof. Proakis and Prof. Manolakis.

Thanks a lot 
	 

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