Quantifying the Maximum Phase-Distortion Error

Jan Verspecht bvba Gertrudeveld 15 1840 Steenhuffel Belgium email: [email protected] web: http://www.janverspecht.com

Quantifying the Maximum Phase-Distortion Error Introduced by Signal Samplers

Jan Verspecht

IEEE Transactions on Instrumentation and Measurement, Vol. 46, No. 3, pp. 660-666

© 1997 IEEE. Personal use of this material is permitted. However, permission to reprint/republish this material for advertising or promotional purposes or for creating new collective works for resale or redistribution to servers or lists, or to reuse any copyrighted component of this work in other works must be obtained from the IEEE.

IEEE TRANSACTIONS ON INSTRUMENTATION AND MEASUREMENT, VOL. 46, NO. 3, JUNE 1997

Quantifying the Maximum Phase-Distortion Error Introduced by Signal Samplers Jan Verspecht, Member, IEEE

Abstract - Quantifying the errors introduced by signal sampler imperfections is of interest to people who are doing frequency domain measurements. The phase-distortion introduced by the sampler is hard to quantify and is usually neglected. In this article upper bounds for this phase-distortion error are derived which are based upon simple assumptions, namely that the sampler weighting function is strictly positive, that it is limited in time and that the function has only one local maximum. The theoretical limits are applied in order to specify the accuracy of a calibration procedure for broadband sampling oscilloscopes.

Introduction Signal samplers are a very important part of waveform digitizers. Quantifying the errors introduced by the signal sampler imperfections is necessary in order to specify these instruments. A measurement error which is hard to quantify and which is of particular interest to people who are doing frequency domain measurements is the phase-distortion introduced by the signal sampler. In this paper upper bounds for a signal sampler phase-distortion error are derived which are based upon simple assumptions. First, it is assumed that the sampler is linear such that it is completely characterized by a weighting function p(t). Note that p(t) describes the mathematical relationship between the input signal x(t) and the sampled value q at sampling instant T by means of the following relationship: ∞

q(T ) =

∫ p ( t – T )x ( t ) dt

.

(1)

–∞

In practice, it is often impossible to measure p(t) directly. Some assumptions concerning the shape of p(t) can be made, however, resulting in upper bounds for the phase-distortion error. First, an upper bound is calculated based upon the assumption that p(t) is strictly positive and is limited in

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time. Next, a smaller bound is derived by adding the constraint that p(t) has only one local maximum. After the mathematical analysis, the theory is applied in order to specify the accuracy of the “nose-to-nose” calibration procedure [2]-[5] for broadband sampling oscilloscopes (bandwidth exceeding 50 GHz).

Mathematical formulation of the problem To define the meaning of the phase-distortion error of a sampler, suppose one uses a sampler, with a weighting function p(t), in order to characterize a signal x(t). Using (1), one notes that Q(ω), the Fourier transform of the measured signal , is related to X(ω), the Fourier transform of x(t) by: +

Q ( ω ) = X ( ω )P ( ω ) ,

(2)

+

with P ( ω ) denoting the complex conjugate of the Fourier transform of p(t). The phase-distortion error of the signal sampler with weighting function p(t), Φ[ p ]( f ) , is +

defined as the phase of P ( ω ) with the linear part of the phase characteristic eliminated; this linear part is not considered as phase-distortion, since it corresponds to a pure delay in time domain. This is achieved by adding a phase component linear in f such that d ( Φ[ p ] ( f )) df

= 0.

(3)

f =0

Using the definition of the Fourier transform Φ[ p ] ( f ) is uniquely defined as  Ta   p ( t ) sin ( 2πft ) dt  ∫  Ta 0  - – ∫ p ( t )2πft dt . Φ[ p ]( f ) = arctan -------------------------------------------Ta    0  ∫ p ( t ) cos ( 2πft ) dt 0 

(4)

In order to quantify the maximum amount of phase-distortion possible, one needs to describe, in a mathematical way, the set of all possible sampler weighting functions. Within this set one has then to find the function which maximizes the phase-distortion. Note that this process

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has to be repeated for all frequencies, since the phase-distortion is a function of frequency, f, and one does not know whether the function maximizing the distortion at one frequency will also maximize the phase-distortion at another frequency. Two sets of admissible weighting functions will be considered in the following. A first set is the collection of all positive functions defined on a time interval with duration Ta, Ta being the so called “sampler aperture time”. A second set of admissible functions will be all functions belonging to the previous set but which have only one local maximum. This will result in a lower upper bound for the possible phase-distortion error. Since the phase-distortion as defined by (4) is invariant versus multiplication of the function with any strictly positive real and versus delaying the function, it will be sufficient to consider all functions supported by the open interval (0,Ta) with their integral value on the interval equal to 1. Note that an open interval is preferred since this implies that

p ( 0 ) = p ( T a ) = 0 , which

property will be used during the calculations. The first set of functions described, with the only constraint that the functions are positive, will be called Π , the second set, with the constraint added of only one local maximum, will be called Π L . Note that Dirac-delta functions are allowed. The problem can then be formulated as finding, for each f, the element in the set of functions ∆ , with ∆ equal to Π respectively Π L , called p MAX[ ∆ ] ( f ) , which maximizes the absolute value of Φ [ p ] ( f ) . An upper bound for the maximum phase-distortion error possible within ∆ will be noted Φ MAX[ ∆ ] ( f ) , and will be given by Φ MAX[ ∆ ] ( f ) = Φ [ p MAX[ ∆ ]( f ) ] ( f ) .

(5)

In order to simplify mathematics, the normalized variables θ and u are introduced. They are defined as follows: θ = 2πf T a , and

(6)

t u = ------ . Ta

(7)

With these variables the problem becomes the following. Define ∆ N as the set of all positive functions belonging to ∆ with the normalization (6) and

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(7) applied. For each θ , find the element in ∆ N , called p MAX[ ∆ N ]( θ ) , which maximizes the absolute value of Φ[ p ]( θ ) , with Φ[ p ] ( θ ) defined by 1

   ∫ p ( u ) sin ( θu ) du  1 0  Φ[ p ]( θ ) = arctan  --------------------------------------- – ∫ p ( u )θu du 1    ∫ p ( u ) cos ( θu ) du 0  

.

(8)

0

An upper bound for the maximum phase-distortion error possible, designated as Φ MAX[ ∆ N ] ( θ ) , will then be given by Φ MAX[ ∆ N ]( θ ) = Φ [ p MAX[ ∆ N ]( θ ) ] ( θ )

.

(9)

p MAX[ ∆ N ] ( θ ) will be called the extremal function.

Calculation of the extremal function in ΠN. In what follows

p MAX[ Π N ] ( θ ) will be calculated. Consider a fixed θ and define the function

x 3 Γ ( x, y, z ) = arctan  -- – z . This function maps the three dimensional real vector space, |R ,  y into the set of reals, |R . One notes that 1

1

1

  Φ[ p ]( θ ) = Γ  ∫ p ( u ) sin ( θu ) du, ∫ p ( u ) cos ( θu ) du, ∫ p ( u )θu du 0  0 0

.

(10)

3

Next one maps the set of functions Π N onto |R through the transformation:

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1

∫ p ( u ) sin ( θu ) du 0 1

p(u) →

∫ p ( u ) cos ( θu ) du

.

(11)

0 1

∫ p ( u )θu du 0

This transformation will map the infinite dimensional Π N onto a three-dimensional volume which will be noted V. Suppose one knows V, the problem of finding an extremal function is then transformed into finding the vector, called v MAX , element of V, which maximizes the absolute value of the function Γ . One can then write: Φ MAX ( θ ) = Γ ( v MAX ) .

(12)

Next V will be mathematically described. sin ( θu ) p ( u ) ∪ ∫ cos ( θu ) du . p ( u ) ∈ ΠN 0 θu 1

V =

(13)

Because of the definition of Π N as the set of all positive functions defined on (0,1) with an integral value equal to 1, (13) reveals that V is the collection of all possible weighted averages of all the vectors belonging to the one dimensional curve, called C, defined as follows:

C =

∪

0≤α≤1

sin ( θα ) cos ( θα ) . θα

(14)

In mathematical terms it says that V is the convex hull of C, given by V = co ( C ) . Carathéodory’s theorem [1] states that co(C) is the union of all possible tetrahedra that can be defined on C. In order to find v MAX we calculate the gradient of Γ :

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y – ---------------2 2 x +y grad ( Γ ) = . x ---------------2 2 x +y –1

(15)

One notes that the function has no stationary points. This means that v MAX is situated on the boundary of V. Since V is the union of all possible tetrahedra that can be defined on C, every point on the boundary of V belongs to a triangle with its corner points situated on the curve C (the boundary of one such a tetrahedron). It will thus be sufficient to look for v MAX within this collection of triangles. Consider now one such triangle. Take the intersection between this triangle and a plane that contains the z-axis (characterized by x/y is constant). This intersection will be a line segment. On this line segment Γ will be maximum at one of the end points (belonging to the borders of the triangle). This is true since Γ is just a linear function of z on this line segment (x/y is constant) such that the maximum Γ is reached for the point on the line segment with lowest or the highest z. The important conclusion is that v MAX is on the border of such a triangle, such that it is sufficient to look for v MAX within the set of all line segments with their two end points belonging to C. This set of line segments is the image of the transformation (11) applied to all elements of Π N which can be written as the sum of two Dirac-delta functions. One can thus conclude that the extremal function one is looking for is a weighted average of two Dirac-delta functions. Since Φ[ p ] ( θ ) is invariant with delaying the function p, it will be sufficient to look for the extremal functions within the set of functions that is described as βδ ( u ) + ( 1 – β )δ ( u – γ ) , where β and γ are two parameters ranging from 0 to 1, and δ ( u ) stands for a Dirac-delta function. One can then define the phase-distortion as a function of θ , β and γ : ( 1 – β ) sin ( γθ ) Φ ( β, γ , θ ) = arctan  ---------------------------------------------- – ( 1 – β )γθ  β + ( 1 – β ) cos ( γθ )

.

(16)

The problem is now reduced to the search of the point in the two-dimensional parameter

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space

β

and

γ,

both

ranging

from

0

to

1,

where

Φ

is

extremal.

Since

Φ ( 1 – β, γ , θ ) = – Φ ( β, γ , θ ) and since one is only interested in the absolute value of Φ , it will be sufficient to confine the search to values of β ranging from 0 to 0.5. Investigating the dependency of Φ on γ , it is clear that Φ ( β, 0, θ ) = 0 . One then calculates the partial derivative of Φ versus γ : γθ 1 4θ sin2  ----- β  --- – β ( 1 – β )     2 2 ∂ Φ ( β, γ , θ ) = -------------------------------------------------------------------- . 1 – 2 ( 1 – cos ( γθ ) )β ( 1 – β ) ∂γ

(17)

Both the numerator and the denominator of (17) are always positive for β ranging from 0 to 0.5, so the partial derivative of Φ versus γ is always positive. From this one can conclude that, for γ ranging from 0 to 1, the absolute value of Φ will be maximum for γ equal to 1. Investigating the dependency of Φ ( β, 1, θ ) on β , since Φ ( 0, 1, θ ) = Φ ( 0.5, 1, θ ) = 0 , there exists a value between 0 and 0.5, called β MAX ( θ ) , which maximizes the absolute value of Φ ( β, 1, θ ) . β MAX ( θ ) is found by solving the equation: ∂ Φ ( β, 1, θ ) ∂β

β MAX

sin θ = θ – ---------------------------------------------------------------------------------- = 0 . 1 – 2 ( 1 – cos θ )β MAX ( 1 – β MAX )

(18)

This equation has two roots, one chooses the one that is in the interval [0,0.5]: sin θ 1 – ----------1 1 θ β MAX ( θ ) = --- – --- – ----------------------------- . 2 4 2 ( 1 – cos θ )

(19)

Finally, one can write: Φ MAX[ Π N ]( θ ) = Φ ( β MAX ( θ ), 1, θ ) ( 1 – β MAX ( θ ) ) sin θ = arctan  ------------------------------------------------------------------------------ – ( 1 – β MAX ( θ ) )θ  β MAX ( θ ) + ( 1 – β MAX ( θ ) ) cos θ

(20)

.

(21)

For ease of interpretation Φ MAX[ Π N ] will be expressed in degrees and as a function of the θ variable ------ , which is equal to f T a (cf. (6)). Φ MAX[ Π N ] as a function of f T a is illustrated in 2π

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Table 1.

Adding the constraint of only one local maximum A smaller upper bound can be found if the constraint is added that p(u) has only one local maximum. Since p(u) has an integral value on the interval (0,1) equal to 1, one can say that p(u) is constrained to all possible unimodal distributions supported by (0,1). Note that we hereby include the cases where p(u) reaches this maximum value on an interval of the u-axis (p(u) being equal to a constant function on this interval). The formulation of the problem is identical to the case of the previous paragraph, except that the search for the extremal function is now limited to a subset of Π N , called Π LN , which contains all elements of Π N with only one local maximum. The 3

mathematics is however more subtle. First, one will look for the image in |R associated with Π LN through transformation (11), this image will be called VL. In order to find VL it will be shown that: 1

Π LN

   = ∪  ∪  ∫ W ( α )K ( α, γ , u ) dα  , 0 ≤ γ ≤ 1  W ( α ) ∈ ΠN   0

(22)

1 with K ( α, γ , u ) equal to --------------- for α < u < γ and for γ < u < α and equal to 0 in all other α–γ cases. (22) is proven in two steps. First one shows that: 1

   Π LN ⊂ ∪  ∪  ∫ W ( α )K ( α, γ , u ) dα  . 0 ≤ γ ≤ 1  W ( α ) ∈ ΠN   0

(23)

Consider therefore an element p ( u ) ∈ Π LN . Choose W ( α ) as follows: W ( α ) = p' ( α ) ( α – γ ) , with γ equal to the modus of p(u).

(24)

One can then perform the following calculation:

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1

∫ W ( α )K ( α, γ , u ) dα

=

0 γ

.

1

(25)

∫ p' ( α ) ( γ – α )K ( α, γ , u ) dα – ∫ p' ( α ) ( α – γ )K ( α, γ , u ) dα γ

0

For u < γ this becomes: u

∫ p' ( α ) dα

= p(u) – p(0) = p(u) ,

(26)

– ∫ p' ( α ) dα = p ( u ) – p ( 1 ) = p ( u ) .

(27)

0

and for u > γ this becomes: 1

u

Next will be shown that W ( α ) ∈ Π N . Two conditions need to be fulfilled. The first one is that W ( α ) ≥ 0 , which is trivial, the second one is that the integral value on the interval (0,1) equals 1. This integral is calculated as follows: γ

1

∫ p' ( α ) ( γ – α ) dα 0

=

1

∫ p' ( α ) ( γ – α ) dα – ∫ p' ( α ) ( α – γ ) dα

.

(28)

γ

0

Using partial integration this becomes: [ p(α)(γ –

γ α ) ]0

γ

1

+ ∫ p ( α ) dα + [ p ( α ) ( γ – 0

1 α ) ]γ

+ ∫ p ( α ) dα = 1

.

(29)

γ

(25) until (29) show that: 1

   p ( u ) ∈ ∪  ∪  ∫ W ( α )K ( α, γ , u ) dα  0 ≤ γ ≤ 1  W ( α ) ∈ ΠN   0

,

(30)

which is a prove of (23). Secondly one proves that:

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1

   Π LN ⊃ ∪  ∪  ∫ W ( α )K ( α, γ , u ) dα  . 0 ≤ γ ≤ 1  W ( α ) ∈ ΠN   0

(31)

Consider therefore one γ and one W ( α ) ∈ Π N . Define p(u) as: 1

∫ W ( α )K ( α, γ , u ) dα

.

(32)

0

One then has to prove that the integral value of p(u) on the interval (0,1) equals 1 and that p(u) is unimodal. The first conjecture is considered trivial. The second conjecture, namely that p(u) is unimodal is proven as follows. Consider two values u 1 and u 2 such that 0 ≤ u 1 < u 2 ≤ γ . One then finds: 1

p ( u1 ) =

∫ W ( α )K ( α, γ , u1 ) dα 0 u1

=

W (α)

- dα ∫ -------------α–γ

0 1

p ( u2 ) =

∫ W ( α )K ( α, γ , u2 ) dα 0 u2

=

.

(33)

W (α)

- dα ∫ -------------α–γ

0 u1

W (α) = ∫ --------------- dα + α–γ 0

u2

W (α)

- dα ∫ -------------α–γ

u1

= p ( u1 ) + ε Since ε ≥ 0 it follows that p ( u 2 ) ≥ p ( u 1 ) . In an analog way it can also be shown that if γ ≤ u 1 < u 2 ≤ 1 then p ( u 1 ) ≥ p ( u 2 ) . This proves that p(u) is an unimodal distribution with modus γ such that one can conclude that p ( u ) ∈ Π LN which proves (31). To find VL transformation (11) is applied to (22). One can then write:

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VL

      = ∪  ∪ 0 ≤ γ ≤ 1  W ( α ) ∈ ΠN     

    ∫  ∫ W ( α )K ( α, γ , u ) dα sin ( θu ) du   0 0  1 1     W ( α )K ( α , γ , u ) d α cos ( θu ) d u   ∫ ∫   0 0  1 1     ∫  ∫ W ( α )K ( α, γ , u ) dα θu du   0 0 1 1

.

(34)

,

(35)

Exchanging the order of integration results into:

VL

      = ∪  ∪ 0 ≤ γ ≤ 1  W ( α ) ∈ ΠN     

 1    ∫ W ( α )  ∫ K ( α, γ , u ) sin ( θu ) du dα  0 0  1 1     d α W ( α ) K ( α , γ , u ) cos ( θu ) d u   ∫ ∫  0  0  1 1     ∫ W ( α )  ∫ K ( α, γ , u )θu du dα   0 0 1

and calculating the integral over u results into: 1

VL

  = ∪  ∪ ∫ W ( α )Q ( α, γ , θ ) dα , with 0 ≤ γ ≤ 1  W ( α ) ∈ ΠN  0

– cos ( θα ) + cos ( θγ ) ---------------------------------------------------θα – θγ 3 sin ( θα ) – sin ( θγ ) Q ( α, γ , θ ) = , element of |R . --------------------------------------------θα – θγ θα + θγ -------------------2

(36)

(37)

The problem of finding V L is now transformed into a problem very similar to finding V. Indeed, (36) reveals that V L is equal to the union over γ (ranging from 0 to 1) of all convex hulls associated with the one dimensional curves Q ( α, γ , θ ) , with parameter α ranging from 0 to 1. This can be written as:

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VL =

∪

0≤γ≤1

co (

∪

0≤α≤1

Q ( α, γ , θ ) ) .

(38)

Consider now one such a curve with a particular value for γ and θ . The same reasoning as done previously holds. The convex hull of Q ( α, γ , θ ) is the union of all tetrahedra that can be 3

defined on Q ( α, γ , θ ) . Since the gradient of Γ is never 0 in |R , Γ will reach its extremal value on the border of this convex hull, which implies that this extremal value occurs on the border of a tetrahedron defined on Q ( α, γ , θ ) (cf. theorem of Carathéodory [1]). This further implies that the extremal value occurs on a triangle defined on Q ( α, γ , θ ) . Taking the intersection between this triangle and a plane containing the z-axis, results in a line segment. On this line segment Γ is only function of z, such that the extremal value will be reached at one of the end points of these line segments (the point with the highest or lowest z-coordinate). The important conclusion is that the extremal value of Γ will be reached on a line segment which has its two end points belonging to Q ( α, γ , θ ) . This means that one has to consider only a limited set of W ( α ) ‘s, described as µδ ( α – η ) + ( 1 – µ )δ ( α – ν ) . This further implies that the extremal function in Π LN belongs to a subset which can be written as follows: 1

p(u) =

∫ ( µδ ( α – η ) + ( 1 – µ )δ ( α – ν ) )K ( α, γ , u ) dα

, or

(39)

0

p ( u ) = µK ( η, γ , u ) + ( 1 – µ ) K ( ν, γ , u ) ,

(40)

with parameters µ , γ and ε all ranging from 0 to 1. Within the set of functions described by (40) one has then to look for the element that maximizes Φ[ p ] ( θ ) . Considering the definition of K, this set of functions can easily be interpreted as consisting of two adjacent rectangles situated on the interval (0,1), with a total area of 1. Taking into account the fact that Φ[ p ] ( θ ) is invariant with delaying the function p(u), it will be sufficient to take into consideration all functions of the form illustrated in Figure1, with a, b and c parameters ranging from 0 to 1. One then writes Φ[ p ] ( θ ) as a function of a, b and c:

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a ( 1 – b ) + ( a – 1 )b cos ( cθ ) + ( b – a ) cos ( bcθ ) Φ ( a, b, c, θ ) = arctan  ------------------------------------------------------------------------------------------------------------------ –   – ( a – 1 )b sin ( cθ ) – ( b – a ) sin ( bcθ )

.

(41)

cθ ------ ( 1 – a + b ) 2 The problem of finding the extremal function is then reduced to: for a given θ , find the values for a, b and c ranging from 0 to 1 maximizing Φ ( a, b, c, θ ) . To simplify the problem one will use the fact that Φ ( a, b, c, θ ) = Φ ( a, b, cθ, 1 ) , such that it will be sufficient to study the function a ( 1 – b ) + ( a – 1 )b cos θ' + ( b – a ) cos ( bθ' ) ψ ( a, b, θ' ) = arctan  --------------------------------------------------------------------------------------------------------- –   – ( a – 1 )b sin θ' – ( b – a ) sin ( bθ' ) θ' ---- ( 1 – a + b ) 2

,

(42)

with Φ ( a, b, c, θ ) = ψ ( a, b, cθ ) . This function ψ ( a, b, θ' ) has some interesting properties. ψ ( a, b, θ' ) will equal zero if a=b or if a=0 or if a=1. The proof of this is trivial since the corresponding functions are symmetric, so that

there

is

no

phase-distortion

present.

Another

property

is

that

ψ ( a, b, θ' ) = – ψ ( 1 – a, 1 – b, θ' ) since replacing a and b by 1-a and 1-b corresponds to time inversion of the function which results in changing the sign of the phase-distortion. Because one is only interested in the maximum of the absolute value of ψ , this property implies that it will be sufficient to look for this maximum in the triangle in the (a,b)-plane defined by the line segments a=1, a=b and b=0. Numerical evaluation [6] in the (a,b)-plane for several values of θ' reveals that ∂ ψ ( a, b, θ' ) ≥ 0 for all values of a and b between 0 and 1, and for all values of θ' between 0 ∂b and π . Combining the fact that the partial derivative versus b is positive with the knowledge that the function equals zero for a=b and for a=1 results in the conclusion that the absolute value of ψ ( a, b, θ' ) will reach its maximum at the line segment b=0 of the triangle described above. This fact further reduces the problem since one will now be able to work with the function ψ RED ( a, θ' ) defined as follows:

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    θ' 1 – cos θ' ψ RED ( a, θ' ) = lim ψ ( a, b, θ' ) = arctan  ---------------------------------------- –  ---- ( 1 – a ) .  2 b→0 a    ---------- θ' + sin θ'   1 – a 

Numerical evaluation shows that

(43)

∂ ψ ( a, θ' ) ≤ 0 for a ranging from 0 to 1 and for θ' ∂ θ' RED

ranging from 0 to π . ψ RED ( a, θ' ) will thus reach its maximum absolute value for the largest value of θ' . Since θ' = cθ and c ranges from 0 to 1, this largest value for a certain a will be equal to ψ RED ( a, θ ) . The function ψ RED ( a, θ ) has then to be maximized with respect to a. This can be done by solving the following equation: ∂ ψ ( a, θ ) ∂a

= 0 , which results in

(44)

a MAX

4 – 4 cos θ – 2θ sin θ a MAX ( θ ) = -----------------------------------------------------------2- , 2 – 2 cos θ – 2θ sin θ + θ

(45)

provided that the trivial solution a MAX = 0 has been omitted. The maximum phasedistortion in Π LN for a given θ is then finally given by: Φ MAX[ Π LN ] ( θ ) = ψ RED ( a MAX ( θ ), θ ) ,

(46)

which can also be written as θ   2 sin2 --  2 Φ MAX[ Π LN ] ( θ ) = – arctan  -------------------------------------------------------------------------- + θ ( 4 – 4 cos θ – 2θ sin θ )  sin θ + ------------------------------------------------------ – 2 + θ 2 + 2 cos θ 

.

(47)

θ ( – 2 + θ 2 + 2 cos θ ) ------------------------------------------------------------------2 ( 2 + θ 2 – 2 cos θ – 2θ sin θ ) The corresponding extremal functions are the sum of a Dirac-delta function at the origin and a rectangular function on the interval (0,1). Some values of Φ MAX[ Π LN ] ( θ ) are given in Table 2, note the use of the variable fT a for ease of physical interpretation (cf. (6)). One notes that the corresponding upper bound for Π LN is about three times lower than the one found for Π N .

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A graph of both upper bounds is given in Figure1, the associated extremal functions are depicted in Figure1.

Application: the “nose-to-nose” calibration procedure In the past this calibration procedure was introduced as an accurate method to determine the frequency response function of broadband sampling oscilloscopes like the HP54124T, with a bandwidth exceeding 50 GHz. The method is based on the principle that each sampling head acts as a pulse generator when sampling direct current and that this sampler so called “kick-out” waveform is proportional to the oscilloscope impulse response. During a “nose-to-nose” measurement the inputs of 2 sampling oscilloscopes are connected to each other and one of the oscilloscopes is measuring the “kick-outs” generated by the other oscilloscope. Applying the Fourier transform together with deconvolution techniques on the measured waveforms results in the frequency response function of one oscilloscope. Several articles on this subject are available in literature [2]-[5]. As explained in these articles, the method can not determine the phasedistortion of the oscilloscope frequency response characteristic introduced by the asymmetry of the sampler weighting function (also called sampling aperture waveform in the context of broadband sampling oscilloscopes). In the past it was always assumed that this error is very small and could be neglected. The error that is made by neglecting the phase contribution of the sampling aperture waveform was, however, never practically quantified. This is necessary if one wants to refer to the method as a “broadband time domain standard”. In this article a useful upper bound for this phase error can be calculated based upon the theory explained above. The sampling aperture waveform of a sampling oscilloscope of the type mentioned above, is different from zero only during a limited time span (called the aperture time). Because of the switching process of the sampling diodes (they are first turned on by increasing the conductance and are then turned off by decreasing the conductance), one can assume that the sampler weighting function has only one local maximum. For an HP54124T the sampler aperture time is estimated to be about 10 ps. This implies that fT a equals 0.2 for f equal to 20 GHz, and 0.5 for f equal to 50 GHz. Table 2 then reveals that the remaining phase-distortion error (after the “nose-to-nose” calibration) is smaller than 0.72° at 20 GHz and smaller than 13° at 50 GHz. Since one knows that, in practice, the aperture waveform is far from being the sum of a Dirac-delta distribution and a rectangle, the real

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error will be less than the theoretical worst case. However, in order to justify a smaller bound, it is necessary to describe the set of all possible sampling aperture waveforms more precisely. Unfortunately finding a description which is more stringent than the one used in this paper is far from trivial.

Conclusion It is possible to find an upper bound for the phase-distortion introduced by signal samplers which is based upon simple assumptions, namely that the sampler weighting function is limited in time, is positive and has only one local maximum. A particular application is found in the determination of the accuracy of the “nose-to-nose” calibration procedure for broadband sampling oscilloscopes.

References [1] Jan Van Tiel,”Convex analysis - An introductory text,” John Wiley & Sons Ltd., pp. 41, 1984. [2] Ken Rush, Steve Draving, John Kerley,”Characterizing high-speed oscilloscopes,” IEEE Spectrum, pp. 38-39, September 1990. [3] D. Henderson, A. G. Roddie, A. J. A. Smith,”Recent developments in the calibration of fast sampling oscilloscopes,” IEE Proceedings-A, Vol. 139, No. 5, pp. 254-260, September 1992. [4] Jan Verspecht and Ken Rush,”Individual Characterization of Broadband Sampling Oscilloscopes with a “Nose-to-Nose” Calibration Procedure,” IEEE Transactions on Instrumentation and Measurement, vol. IM-43, no. 2, pp. 347-354, 1994. [5] Jan Verspecht,”Broadband Sampling Oscilloscope Characterization with the “Nose-to-Nose” Calibration Procedure: A Theoretical and Practical Analysis,” IEEE Transactions on Instrumentation and Measurement, vol. IM-44, no. 6, pp. 991-997, December 1995. [6] Stephen Wolfram,”Mathematica - A system for doing mathematics by computer,” AddisonWesley, Second Edition.

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p(u)

a -----bc 1–a ------------------( 1 – b )c

0 Figure1

c

bc

1

u

Set of functions to which belongs the extremal function in ΠNL.

20 Upper bound ( ° )

17.5 15 12.5 10 7.5 5 2.5 0

0.1

0.2

0.3

0.4

0.5

fTa Figure2

Upper bounds for the phase-distortion versus normalized frequency (fTa). : Φ MAX [ Π N ]

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aMAX ( θ )

β MAX ( θ )

1 – βMAX ( θ )

0 Figure3

1

1 – aMAX ( θ )

0

1

Extremal functions in ΠN (left) and ΠNL (right), an arrow indicates a Dirac-delta function.

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Table 1 Upper bound for the phase-distortion error based upon ΠN fT a

Φ MAX[ Π N ]

0.0

0.00˚

0.1

0.23˚

0.2

1.98˚

0.3

7.54˚

0.4

22.1˚

0.5

90.0˚

Table 2 Upper bound for the phase-distortion error based upon ΠNL fT a

Φ MAX [ Π LN ]

0.0

0.00˚

0.1

0.09˚

0.2

0.72˚

0.3

2.52˚

0.4

6.26˚

0.5

13.0˚

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