Dear DSP Forum viewers, I have a question about minimum phase systems, which I ask under "Question:" in what follows. Thanks in advance for your effort, Dani I am coming from a mathematics background, and am interested in DSP. In the book of Papolis I opened a page to Spectral theory, and it begins discussing minimum phase systems. It defines this as is a linear system that is causal and has finite energy impulse response. First, I guess a signal here depends on time, as well as the output; and causality means convolution kernel zero outside the negative half- line; and impulse response is the convolution kernel. Finite energy means bounded L2 norm? Then he writes that this means L(s) and 1/L(s) are analytic in the right hand plane Re(s)>0. Question: What is s? What is L(s)? Is it the Fourier transform of the convolution kernel? And, why is it analytic in the right half-plane? I guess he is not referring to FT since FT would map the signal to a complex valued function on the real line, so already that is ruled out.
minimum phase system - poles
Started by ●April 27, 2011
Reply by ●April 27, 20112011-04-27
On Apr 27, 1:04�pm, DAN <dyan...@gmail.com> wrote:> Dear DSP Forum viewers, > > I have a question about minimum phase systems, which I ask under > "Question:" in what follows. > Thanks in advance for your effort, > Dani > > I am coming from a mathematics background, and am interested in DSP. > In the book of Papolis I opened a page to Spectral theory, and it > begins discussing minimum phase systems. It defines this as is a > linear system that is causal and has finite energy impulse response. > > First, I guess a signal here depends on time, as well as the output; > and causality means convolution kernel zero outside the negative half- > line; and impulse response is the convolution kernel. Finite energy > means bounded L2 norm? > > Then he writes that this means L(s) and 1/L(s) are analytic in the > right hand plane Re(s)>0. > > Question: > What is s? What is L(s)? Is it the Fourier transform of the > convolution kernel? And, why is it analytic in the right half-plane? I > guess he is not referring to FT since FT would map the signal to a > complex valued function on the real line, so already that is ruled > out.L(s) is typically used to denote the Laplace transform of the time domain signal l(t). With regard to minimum phase system then l(t) is the impulse response of the system in question. A bit out of my area of expertise - but the impulse response is similar to a green's function. Given an input x(t) to a system with an impulse response l(t) the output y(t) is given by the convolution of x(t) and l(t). Additionally, the Fourier transform would given by evaluating F(jw) i.e. let s=jw, where j=sqrt(-1). Hope that helps. Dave
Reply by ●April 27, 20112011-04-27
On 04/27/2011 10:04 AM, DAN wrote:> Dear DSP Forum viewers, > > I have a question about minimum phase systems, which I ask under > "Question:" in what follows. > Thanks in advance for your effort, > Dani > > I am coming from a mathematics background, and am interested in DSP. > In the book of Papolis I opened a page to Spectral theory, and it > begins discussing minimum phase systems. It defines this as is a > linear system that is causal and has finite energy impulse response. > > First, I guess a signal here depends on time, as well as the output; > and causality means convolution kernel zero outside the negative half- > line; and impulse response is the convolution kernel. Finite energy > means bounded L2 norm? > > Then he writes that this means L(s) and 1/L(s) are analytic in the > right hand plane Re(s)>0. > > Question: > What is s? What is L(s)? Is it the Fourier transform of the > convolution kernel? And, why is it analytic in the right half-plane? I > guess he is not referring to FT since FT would map the signal to a > complex valued function on the real line, so already that is ruled > out.'s' is the Laplace frequency operator. L(s) is the Laplace-domain transfer function of the system l. L(s) is the the Laplace transform of the convolution kernel (or, in terms more familiar to me, the system impulse response). It is analytic in the right-half plane (inclusive of the imaginary axis) because L(s) remains continuous and finite for all s such that Re(s) >= 0. He is not referring to the Fourier transform. Note: Either his definition is incomplete, or one of us is misunderstanding it. Consider a system with transfer function L(s) = (s-3)/((s+2)*(s+1)). This system is most certainly not minimum phase, because it has a zero in the right half plane and poles in the left. Yet it describes a system whose impulse response is x(t) = (5 * e^-(2*t) - 4 * e^-t) * u(t), which is both of finite energy and causal. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●April 27, 20112011-04-27
On 27 אפריל, 20:26, Tim Wescott <t...@seemywebsite.com> wrote:> On 04/27/2011 10:04 AM, DAN wrote: > > > > > > > Dear DSP Forum viewers, > > > I have a question about minimum phase systems, which I ask under > > "Question:" in what follows. > > Thanks in advance for your effort, > > Dani > > > I am coming from a mathematics background, and am interested in DSP. > > In the book of Papolis I opened a page to Spectral theory, and it > > begins discussing minimum phase systems. It defines this as is a > > linear system that is causal and has finite energy impulse response. > > > First, I guess a signal here depends on time, as well as the output; > > and causality means convolution kernel zero outside the negative half- > > line; and impulse response is the convolution kernel. Finite energy > > means bounded L2 norm? > > > Then he writes that this means L(s) and 1/L(s) are analytic in the > > right hand plane Re(s)>0. > > > Question: > > What is s? What is L(s)? Is it the Fourier transform of the > > convolution kernel? And, why is it analytic in the right half-plane? I > > guess he is not referring to FT since FT would map the signal to a > > complex valued function on the real line, so already that is ruled > > out. > > 's' is the Laplace frequency operator. L(s) is the Laplace-domain > transfer function of the system l. L(s) is the the Laplace transform of > the convolution kernel (or, in terms more familiar to me, the system > impulse response). It is analytic in the right-half plane (inclusive of > the imaginary axis) because L(s) remains continuous and finite for all s > such that Re(s) >= 0. He is not referring to the Fourier transform. > > Note: Either his definition is incomplete, or one of us is > misunderstanding it. Consider a system with transfer function > > L(s) = (s-3)/((s+2)*(s+1)). > > This system is most certainly not minimum phase, because it has a zero > in the right half plane and poles in the left. Yet it describes a > system whose impulse response is > > x(t) = (5 * e^-(2*t) - 4 * e^-t) * u(t), >I hope this enters the discussion tree- Thank you Tim and Dave. regarding Tim's note, yes it's my mistake - the causal & finite energy condition he assumes for 1/L(s) as well. I did not see the laplace transform in math major courses except mentioned once in harmonic analysis course. But looking at the definition, it looks precisely the extension of the fourier transform to complex values, although it's applied to functions of one real variable?. I guess the use is to extend the fourier decomposition to functions that are not of class L1 on the real line?> which is both of finite energy and causal. > > -- > > Tim Wescott > Wescott Design Serviceshttp://www.wescottdesign.com > > Do you need to implement control loops in software? > "Applied Control Theory for Embedded Systems" was written for you. > See details athttp://www.wescottdesign.com/actfes/actfes.html-הסתר טקסט מצוטט- > > -הראה טקסט מצוטט-
Reply by ●April 27, 20112011-04-27
On 04/27/2011 01:22 PM, DAN wrote:> On 27 אפריל, 20:26, Tim Wescott<t...@seemywebsite.com> wrote: >> On 04/27/2011 10:04 AM, DAN wrote: >> >> >> >> >> >>> Dear DSP Forum viewers, >> >>> I have a question about minimum phase systems, which I ask under >>> "Question:" in what follows. >>> Thanks in advance for your effort, >>> Dani >> >>> I am coming from a mathematics background, and am interested in DSP. >>> In the book of Papolis I opened a page to Spectral theory, and it >>> begins discussing minimum phase systems. It defines this as is a >>> linear system that is causal and has finite energy impulse response. >> >>> First, I guess a signal here depends on time, as well as the output; >>> and causality means convolution kernel zero outside the negative half- >>> line; and impulse response is the convolution kernel. Finite energy >>> means bounded L2 norm? >> >>> Then he writes that this means L(s) and 1/L(s) are analytic in the >>> right hand plane Re(s)>0. >> >>> Question: >>> What is s? What is L(s)? Is it the Fourier transform of the >>> convolution kernel? And, why is it analytic in the right half-plane? I >>> guess he is not referring to FT since FT would map the signal to a >>> complex valued function on the real line, so already that is ruled >>> out. >> >> 's' is the Laplace frequency operator. L(s) is the Laplace-domain >> transfer function of the system l. L(s) is the the Laplace transform of >> the convolution kernel (or, in terms more familiar to me, the system >> impulse response). It is analytic in the right-half plane (inclusive of >> the imaginary axis) because L(s) remains continuous and finite for all s >> such that Re(s)>= 0. He is not referring to the Fourier transform. >> >> Note: Either his definition is incomplete, or one of us is >> misunderstanding it. Consider a system with transfer function >> >> L(s) = (s-3)/((s+2)*(s+1)). >> >> This system is most certainly not minimum phase, because it has a zero >> in the right half plane and poles in the left. Yet it describes a >> system whose impulse response is >> >> x(t) = (5 * e^-(2*t) - 4 * e^-t) * u(t), >> > I hope this enters the discussion tree- > Thank you Tim and Dave. > regarding Tim's note, yes it's my mistake - the causal& finite energy > condition he assumes for 1/L(s) as well.I find that an interesting way to specify things. The 'engineer's definition' of minimum phase is that all poles and zeros are strictly in the left-half plane. That probably leaves out some corner cases (i.e. systems with delay), which is perhaps why he specifies it the way he does.> I did not see the laplace transform in math major courses except > mentioned once in harmonic analysis course.That's interesting. I know it's in one of my differential equations books, and that's a course that's fundamental enough that all the engineering students get it. But I don't recall us actually opening the book to that section in the differential equations course...> But looking at the > definition, it looks precisely the extension of the fourier transform > to complex values, although it's applied to functions of one real > variable?. I guess the use is to extend the fourier decomposition to > functions that are not of class L1 on the real line?I'm not enough of a mathematician to know what "class L1" means. I believe that there are cases where the Fourier transform does not apply and the Laplace transform does, however. Certainly with the Laplace transform you avoid the use of a lot of complex math -- a real-valued ordinary linear differential equation in the time domain turns into one or more real-valued polynomials in the Laplace domain; you don't need to mess with complex numbers unless you're getting frequency responses or are factoring the polynomials. It's worth it just for that. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●April 28, 20112011-04-28
On Apr 27, 7:04�pm, DAN <dyan...@gmail.com> wrote:> Dear DSP Forum viewers, > > I have a question about minimum phase systems, which I ask under > "Question:" in what follows. > Thanks in advance for your effort, > Dani > > I am coming from a mathematics background, and am interested in DSP. > In the book of Papolis I opened a page to Spectral theory, and it > begins discussing minimum phase systems. It defines this as is a > linear system that is causal and has finite energy impulse response. > > First, I guess a signal here depends on time, as well as the output;You might want to review this in some detail: The *system* has constant coefficients (they don't change over time), and is thus 'invariant'. The *signal* is free to vary with time, in the sense 'exhibit transient behaviour'. However, the theory is based on the assumption that the spectrum of a transient signal is known, and constant for all time.> and causality means convolution kernel zero outside the negative half- > line;Maybe yes, maybe no. Depends on what part of th etheory you read. For 'analog' systems (that is, continuous time), you are right. The Laplace Transform is stable if the poles are located in this or that half plane. For *discrete* (time) system, one uses the Z Transform instead of the LT. The ZT is stable if the poles are inside the unit circle.> and impulse response is the convolution kernel.Yep.> Finite energy > means bounded L2 norm?Yep.> Then he writes that this means L(s) and 1/L(s) are analytic in the > right hand plane Re(s)>0.If so, he is discussing the LT, not the ZT.> Question: > What is s? What is L(s)? Is it the Fourier transform of the > convolution kernel? And, why is it analytic in the right half-plane? I > guess he is not referring to FT since FT would map the signal to a > complex valued function on the real line, so already that is ruled > out.No, this is the Laplace Transform. Stanard material in control theory, the same technique you will use to solve linear differential equations with constant coefficients. Don't know if 'Laplace Transform' is a term much used by mathematicians, but a mathematician would certainly already know the method. Rune
Reply by ●April 28, 20112011-04-28
On Apr 28, 4:11�pm, Rune Allnor <all...@tele.ntnu.no> wrote:> On Apr 27, 7:04�pm, DAN <dyan...@gmail.com> wrote: > > > Dear DSP Forum viewers, > > > I have a question about minimum phase systems, which I ask under > > "Question:" in what follows. > > Thanks in advance for your effort, > > Dani > > > I am coming from a mathematics background, and am interested in DSP. > > In the book of Papolis I opened a page to Spectral theory, and it > > begins discussing minimum phase systems. It defines this as is a > > linear system that is causal and has finite energy impulse response. > > > First, I guess a signal here depends on time, as well as the output; > > You might want to review this in some detail: The *system* has > constant coefficients (they don't change over time), and is thus > 'invariant'. The *signal* is free to vary with time, in the sense > 'exhibit transient behaviour'. However, the theory is based > on the assumption that the spectrum of a transient signal is known, > and constant for all time. > > > and causality means convolution kernel zero outside the negative half- > > line; > > Maybe yes, maybe no. Depends on what part of th etheory you read. > For 'analog' systems (that is, continuous time), you are right. > The Laplace Transform is stable if the poles are located in this > or that half plane. > > For *discrete* (time) system, one uses the Z Transform instead > of the LT. The ZT is stable if the poles are inside the unit > circle. > > > and impulse response is the convolution kernel. > > Yep. > > > Finite energy > > means bounded L2 norm? > > Yep. > > > Then he writes that this means L(s) and 1/L(s) are analytic in the > > right hand plane Re(s)>0. > > If so, he is discussing the LT, not the ZT. > > > Question: > > What is s? What is L(s)? Is it the Fourier transform of the > > convolution kernel? And, why is it analytic in the right half-plane? I > > guess he is not referring to FT since FT would map the signal to a > > complex valued function on the real line, so already that is ruled > > out. > > No, this is the Laplace Transform. Stanard material in control > theory, the same technique you will use to solve linear differential > equations with constant coefficients. Don't know if 'Laplace > Transform' is a term much used by mathematicians, but a mathematician > would certainly already know the method. > > RuneYes, Laplace is Laplace nomatter where! Laplace was a French mathematician. The transfer function however was pioneered by English engineer Oliver Heaviside. In fact it is Engineers who have really taken the supject to the next level after Laplace. We cannot forget the work of another famous Frenchman of course - Cauchy. If he hadnot have bumped into Reiman that night down the pub, God only knows where complex theory would be today. Hardy
Reply by ●April 28, 20112011-04-28
On Apr 28, 8:48�am, HardySpicer <gyansor...@gmail.com> wrote:> On Apr 28, 4:11�pm, Rune Allnor <all...@tele.ntnu.no> wrote: > > > > > > > On Apr 27, 7:04�pm, DAN <dyan...@gmail.com> wrote: > > > > Dear DSP Forum viewers, > > > > I have a question about minimum phase systems, which I ask under > > > "Question:" in what follows. > > > Thanks in advance for your effort, > > > Dani > > > > I am coming from a mathematics background, and am interested in DSP. > > > In the book of Papolis I opened a page to Spectral theory, and it > > > begins discussing minimum phase systems. It defines this as is a > > > linear system that is causal and has finite energy impulse response. > > > > First, I guess a signal here depends on time, as well as the output; > > > You might want to review this in some detail: The *system* has > > constant coefficients (they don't change over time), and is thus > > 'invariant'. The *signal* is free to vary with time, in the sense > > 'exhibit transient behaviour'. However, the theory is based > > on the assumption that the spectrum of a transient signal is known, > > and constant for all time. > > > > and causality means convolution kernel zero outside the negative half- > > > line; > > > Maybe yes, maybe no. Depends on what part of th etheory you read. > > For 'analog' systems (that is, continuous time), you are right. > > The Laplace Transform is stable if the poles are located in this > > or that half plane. > > > For *discrete* (time) system, one uses the Z Transform instead > > of the LT. The ZT is stable if the poles are inside the unit > > circle. > > > > and impulse response is the convolution kernel. > > > Yep. > > > > Finite energy > > > means bounded L2 norm? > > > Yep. > > > > Then he writes that this means L(s) and 1/L(s) are analytic in the > > > right hand plane Re(s)>0. > > > If so, he is discussing the LT, not the ZT. > > > > Question: > > > What is s? What is L(s)? Is it the Fourier transform of the > > > convolution kernel? And, why is it analytic in the right half-plane? I > > > guess he is not referring to FT since FT would map the signal to a > > > complex valued function on the real line, so already that is ruled > > > out. > > > No, this is the Laplace Transform. Stanard material in control > > theory, the same technique you will use to solve linear differential > > equations with constant coefficients. Don't know if 'Laplace > > Transform' is a term much used by mathematicians, but a mathematician > > would certainly already know the method. > > > Rune > > Yes, Laplace is Laplace nomatter where!- The *person* Laplace: Yes, he's the same no matter what. - The *technique* he is credited for: Yes, that's the same no matter what However, the term 'Laplace Transform' needs not be universally used about the techniqe. I have a vague recollection having heard or read somewhere that what is known in the English-speaking world as 'Snell's law' is known in France as 'Descarte's law'. Who knows what confusion might arise if the Russians start using their terms about such mundane matters. Rune
Reply by ●April 28, 20112011-04-28
>On Apr 28, 8:48=A0am, HardySpicer <gyansor...@gmail.com> wrote: >> On Apr 28, 4:11=A0pm, Rune Allnor <all...@tele.ntnu.no> wrote: >> >> >> >> >> >> > On Apr 27, 7:04=A0pm, DAN <dyan...@gmail.com> wrote: >> >> > > Dear DSP Forum viewers, >> >> > > I have a question about minimum phase systems, which I ask under >> > > "Question:" in what follows. >> > > Thanks in advance for your effort, >> > > Dani >> >> > > I am coming from a mathematics background, and am interested inDSP.>> > > In the book of Papolis I opened a page to Spectral theory, and it >> > > begins discussing minimum phase systems. It defines this as is a >> > > linear system that is causal and has finite energy impulseresponse.>> >> > > First, I guess a signal here depends on time, as well as theoutput;>> >> > You might want to review this in some detail: The *system* has >> > constant coefficients (they don't change over time), and is thus >> > 'invariant'. The *signal* is free to vary with time, in the sense >> > 'exhibit transient behaviour'. However, the theory is based >> > on the assumption that the spectrum of a transient signal is known, >> > and constant for all time. >> >> > > and causality means convolution kernel zero outside the negativehalf=>- >> > > line; >> >> > Maybe yes, maybe no. Depends on what part of th etheory you read. >> > For 'analog' systems (that is, continuous time), you are right. >> > The Laplace Transform is stable if the poles are located in this >> > or that half plane. >> >> > For *discrete* (time) system, one uses the Z Transform instead >> > of the LT. The ZT is stable if the poles are inside the unit >> > circle. >> >> > > and impulse response is the convolution kernel. >> >> > Yep. >> >> > > Finite energy >> > > means bounded L2 norm? >> >> > Yep. >> >> > > Then he writes that this means L(s) and 1/L(s) are analytic in the >> > > right hand plane Re(s)>0. >> >> > If so, he is discussing the LT, not the ZT. >> >> > > Question: >> > > What is s? What is L(s)? Is it the Fourier transform of the >> > > convolution kernel? And, why is it analytic in the right half-plane?=>I >> > > guess he is not referring to FT since FT would map the signal to a >> > > complex valued function on the real line, so already that is ruled >> > > out. >> >> > No, this is the Laplace Transform. Stanard material in control >> > theory, the same technique you will use to solve linear differential >> > equations with constant coefficients. Don't know if 'Laplace >> > Transform' is a term much used by mathematicians, but a mathematician >> > would certainly already know the method. >> >> > Rune >> >> Yes, Laplace is Laplace nomatter where! > >- The *person* Laplace: Yes, he's the same no matter what. >- The *technique* he is credited for: Yes, that's the same > no matter what > >However, the term 'Laplace Transform' needs not be universally >used about the techniqe. I have a vague recollection having heard >or read somewhere that what is known in the English-speaking world >as 'Snell's law' is known in France as 'Descarte's law'. Who knows >what confusion might arise if the Russians start using their terms >about such mundane matters. > >RuneTry looking up the refraction law and you will find it substantially predates either of the above. That's so well know, even Wikipedia gets it right. :-) Steve
Reply by ●April 28, 20112011-04-28
On 28 אפריל, 11:18, "steveu" <steveu@n_o_s_p_a_m.coppice.org> wrote:> >On Apr 28, 8:48=A0am, HardySpicer <gyansor...@gmail.com> wrote: > >> On Apr 28, 4:11=A0pm, Rune Allnor <all...@tele.ntnu.no> wrote: > > >> > On Apr 27, 7:04=A0pm, DAN <dyan...@gmail.com> wrote: > > >> > > Dear DSP Forum viewers, > > >> > > I have a question about minimum phase systems, which I ask under > >> > > "Question:" in what follows. > >> > > Thanks in advance for your effort, > >> > > Dani > > >> > > I am coming from a mathematics background, and am interested in > DSP. > >> > > In the book of Papolis I opened a page to Spectral theory, and it > >> > > begins discussing minimum phase systems. It defines this as is a > >> > > linear system that is causal and has finite energy impulse > response. > > >> > > First, I guess a signal here depends on time, as well as the > output; > > >> > You might want to review this in some detail: The *system* has > >> > constant coefficients (they don't change over time), and is thus > >> > 'invariant'. The *signal* is free to vary with time, in the sense > >> > 'exhibit transient behaviour'. However, the theory is based > >> > on the assumption that the spectrum of a transient signal is known, > >> > and constant for all time. > > >> > > and causality means convolution kernel zero outside the negative > half= > >- > >> > > line; > > >> > Maybe yes, maybe no. Depends on what part of th etheory you read. > >> > For 'analog' systems (that is, continuous time), you are right. > >> > The Laplace Transform is stable if the poles are located in this > >> > or that half plane. > > >> > For *discrete* (time) system, one uses the Z Transform instead > >> > of the LT. The ZT is stable if the poles are inside the unit > >> > circle. > > >> > > and impulse response is the convolution kernel. > > >> > Yep. > > >> > > Finite energy > >> > > means bounded L2 norm? > > >> > Yep. > > >> > > Then he writes that this means L(s) and 1/L(s) are analytic in the > >> > > right hand plane Re(s)>0. > > >> > If so, he is discussing the LT, not the ZT. > > >> > > Question: > >> > > What is s? What is L(s)? Is it the Fourier transform of the > >> > > convolution kernel? And, why is it analytic in the right half-plane? > = > >I > >> > > guess he is not referring to FT since FT would map the signal to a > >> > > complex valued function on the real line, so already that is ruled > >> > > out. > > >> > No, this is the Laplace Transform. Stanard material in control > >> > theory, the same technique you will use to solve linear differential > >> > equations with constant coefficients. Don't know if 'Laplace > >> > Transform' is a term much used by mathematicians, but a mathematician > >> > would certainly already know the method. > > >> > Rune > > >> Yes, Laplace is Laplace nomatter where! > > >- The *person* Laplace: Yes, he's the same no matter what. > >- The *technique* he is credited for: Yes, that's the same > > no matter what > > >However, the term 'Laplace Transform' needs not be universally > >used about the techniqe. I have a vague recollection having heard > >or read somewhere that what is known in the English-speaking world > >as 'Snell's law' is known in France as 'Descarte's law'. Who knows > >what confusion might arise if the Russians start using their terms > >about such mundane matters. > > >Rune > > Try looking up the refraction law and you will find it substantially > predates either of the above. That's so well know, even Wikipedia gets it > right. :-) > > Steve-הסתר טקסט מצוטט- > > -הראה טקסט מצוטט-yes, I guess saw something about the discrete case, so I was refering here to the continuous signal case. In the ODE class I took, as I remember we solved linear constant coeficient systems by by transforming the system matrix to Jordan form, the change of basis defining the solution set and the Jordan form characterising the system's response in the phase plane. So I need to learn the Laplace transform point of view. Thank you for helping! Dani
