压缩采样(Compressive sampling)陶哲轩

Robust Uncertainty Principles:

Exact Signal Reconstruction from Highly Incomplete

Frequency Information

Emmanuel Candes ?,Justin Romberg ?,and Terence Tao

?Applied and Computational Mathematics,Caltech,Pasadena,CA 91125 Department of Mathematics,University of California,Los Angeles,CA 90095

June 2004;Revised August 2005

Abstract

This paper considers the model problem of reconstructing an object from incomplete frequency samples.Consider a discrete-time signal f ∈C N and a randomly chosen set of frequencies ?.Is it possible to reconstruct f from the partial knowledge of its Fourier coe?cients on the set ??

A typical result of this paper is as follows.Suppose that f is a superposition of |T |spikes f (t )= τ∈T f (τ)δ(t ?τ)obeying

|T |≤C M ·(log N )?1·|?|,

for some constant C M >0.We do not know the locations of the spikes nor their amplitudes.Then with probability at least 1?O (N ?M ),f can be reconstructed exactly as the solution to the 1minimization problem

min g N ?1 t =0|g (t )|,s.t.?g (ω)=?f

(ω)for all ω∈?.In short,exact recovery may be obtained by solving a convex optimization problem.We give numerical values for C M which depend on the desired probability of success.Our result may be interpreted as a novel kind of nonlinear sampling theorem.In e?ect,it says that any signal made out of |T |spikes may be recovered by convex programming from almost every set of frequencies of size O (|T |·log N ).Moreover,this is nearly optimal in the sense that any method succeeding with probability 1?O (N ?M )would in general require a number of frequency samples at least proportional to |T |·log N .

The methodology extends to a variety of other situations and higher dimensions.For example,we show how one can reconstruct a piecewise constant (one-or two-dimensional)object from incomplete frequency samples—provided that the number of jumps (discontinuities)obeys the condition above—by minimizing other convex func-tionals such as the total variation of f .

Keywords.Random matrices,free probability,sparsity,trigonometric expansions,uncertainty principle,convex optimization,duality in optimization,total-variation minimization,image recon-struction,linear programming.

Acknowledgments. E.C.is partially supported by a National Science Foundation grant DMS 01-40698(FRG)and by an Alfred P.Sloan Fellowship.J.R.is supported by National Science Foundation grants DMS01-40698and ITR ACI-0204932.T.T.is a Clay Prize Fellow and is supported in part by grants from the Packard Foundation.E.C.and T.T.thank the Institute for Pure and Applied Mathematics at UCLA for their warm hospitality. E.C.would like to thank Amos Ron and David Donoho for stimulating conversations,and Po-Shen Loh for early numerical experiments on a related project.We would also like to thank Holger Rauhut for corrections on an earlier version and the anonymous referees for their comments and references.

1Introduction

In many applications of practical interest,we often wish to reconstruct an object(a discrete signal,a discrete image,etc.)from incomplete Fourier samples.In a discrete setting,we may pose the problem as follows;let?f be the Fourier transform of a discrete object f(t), t=(t1,...,t d)∈Z d

:={0,1,...,N?1}d,

N

?f(ω)=

f(t)e?2πi(ω1t1+...+ωd t d)/N.

t∈Z d

N

The problem is then to recover f from partial frequency information,namely,from?f(ω), whereω=(ω1,...,ωd)belongs to some set?of cardinality less than N d—the size of the discrete object.

In this paper,we show that we can recover f exactly from observations?f|?on small set of frequencies provided that f is sparse.The recovery consists of solving a straightforward optimization problem that?nds f of minimal complexity with?f (ω)=?f(ω),?ω∈?.

1.1A puzzling numerical experiment

This idea is best motivated by an experiment with surprisingly positive results.Consider a simpli?ed version of the classical tomography problem in medical imaging:we wish to reconstruct a2D image f(t1,t2)from samples?f|?of its discrete Fourier transform on a star-shaped domain?[4].Our choice of domain is not contrived;many real imaging devices collect high-resolution samples along radial lines at relatively few angles.Figure1(b) illustrates a typical case where one gathers512samples along each of22radial lines. Frequently discussed approaches in the literature of medical imaging for reconstructing an object from polar frequency samples are the so-called?ltered backprojection algorithms.In a nutshell,one assumes that the Fourier coe?cients at all of the unobserved frequencies are zero(thus reconstructing the image of“minimal energy”under the observation constraints). This strategy does not perform very well,and could hardly be used for medical diagnostics [24].The reconstructed image,shown in Figure1(c),has severe nonlocal artifacts caused by the angular undersampling.A good reconstruction algorithm,it seems,would have to guess the values of the missing Fourier coe?cients.In other words,one would need to interpolate?f(ω1,ω2).This seems highly problematic,however;predictions of Fourier coe?cients from their neighbors are very delicate,due to the global and highly oscillatory nature of the Fourier transform.Going back to the example in Figure1,we can see the

(c)(d)

Figure1:Example of a simple recovery problem.(a)The Logan-Shepp phantom test image.(b)Sampling domain?in the frequency plane;Fourier coe?cients are sampled along22approximately radial lines.(c)Minimum energy reconstruction obtained by setting unobserved Fourier coe?cients to zero.(d)Reconstruction obtained by minimizing the total variation,as in(1.1).The reconstruction is an exact replica of the image in(a).

problem immediately.To recover frequency information near(2πω1/N,2πω2/N),where 2πω1/N is near±π,we would need to interpolate?f at the Nyquist rate2π/N.However, we only have samples at rate aboutπ/22;the sampling rate is almost50times smaller than the Nyquist rate!

We propose instead a strategy based on convex optimization.Let g T V be the total-variation norm of a two-dimensional object g.For discrete data g(t1,t2),0≤t1,t2≤N?1,

g T V=

t1,t2

|D1g(t1,t2)|2+|D2g(t1,t2)|2,

where D1is the?nite di?erence D1g=g(t1,t2)?g(t1?1,t2)and D2g=g(t1,t2)?g(t1,t2?1).To recover f from partial Fourier samples,we?nd a solution f to the optimization problem

min g T V subject to?g(ω)=?f(ω)for allω∈?.(1.1) In a nutshell,given partial observation?f|?,we seek a solution f with minimum complexity—here Total Variation(TV)—and whose“visible”coe?cients match those of the unknown object f.Our hope here is to partially erase some of the artifacts that classical reconstruc-tion methods exhibit(which tend to have large TV norm)while maintaining?delity to the observed data via the constraints on the Fourier coe?cients of the reconstruction.

When we use(1.1)for the recovery problem illustrated in Figure1(with the popular Logan-Shepp phantom as a test image),the results are surprising.The reconstruction is exact; that is,f =f!This numerical result is also not special to this phantom.In fact,we performed a series of experiments of this type and obtained perfect reconstruction on many similar test phantoms.

1.2Main results

This paper is about a quantitative understanding of this very special phenomenon.For which classes of signals/images can we expect perfect reconstruction?What are the trade-o?s between complexity and number of samples?In order to answer these questions,we ?rst develop a fundamental mathematical understanding of a special one-dimensional model problem.We then exhibit reconstruction strategies which are shown to exactly reconstruct certain unknown signals,and can be extended for use in a variety of related and sophisti-cated reconstruction applications.

For a signal f∈C N,we de?ne the classical discrete Fourier transform F f=?f:C N→C N by

?f(ω):=N?1

t=0

f(t)e?2πiωt/N,ω=0,1,...,N?1.(1.2)

If we are given the value of the Fourier coe?cients?f(ω)for all frequenciesω∈Z N,then one can obviously reconstruct f exactly via the Fourier inversion formula

f(t)=1

N

N?1

ω=0

?f(ω)e2πiωt/N.

Now suppose that we are only given the Fourier coe?cients?f|?sampled on some partial subset? Z N of all frequencies.Of course,this is not enough information to reconstruct

f exactly in general;f has N degrees of freedom and we are only specifying|?|

Suppose,however,that we also specify that f is supported on a small(but a priori unknown) subset T of Z N;that is,we assume that f can be written as a sparse superposition of spikes

f(t)=

τ∈T

f(τ)δ(t?τ),δ(t)=1{t=0}.

In the case where N is prime,the following theorem tells us that it is possible to recover f exactly if|T|is small enough.

Theorem1.1Suppose that the signal length N is a prime integer.Let?be a subset of {0,...,N?1},and let f be a vector supported on T such that

|T|≤1

2

|?|.(1.3)

Then f can be reconstructed uniquely from?and?f|?.Conversely,if?is not the set of all N frequencies,then there exist distinct vectors f,g such that|supp(f)|,|supp(g)|≤1

2

|?|+1 and such that?f|?=?g|?.

Proof We will need the following lemma[29],from which we see that with knowledge of T,we can reconstruct f uniquely(using linear algebra)from?f|?:

Lemma1.2([29],Corollary1.4)Let N be a prime integer and T,?be subsets of Z N. Put 2(T)(resp. 2(?))to be the space of signals that are zero outside of T(resp.?).The restricted Fourier transform F T→?: 2(T)→ 2(?)is de?ned as

F T→?f:=?f|?for all f∈ 2(T),

If|T|=|?|,then F T→?is a bijection;as a consequence,we thus see that F T→?is injective for|T|≤|?|and surjective for|T|≥|?|.Clearly,the same claims hold if the Fourier transform F is replaced by the inverse Fourier transform F?1.

To prove Theorem1.1,assume that|T|≤1

2

|?|.Suppose for contradiction that there were

two objects f,g such that?f|?=?g|?and|supp(f)|,|supp(g)|≤1

2|?|.Then the Fourier

transform of f?g vanishes on?,and|supp(f?g)|≤|?|.By Lemma1.2we see that F supp(f?g)→?is injective,and thus f?g=0.The uniqueness claim follows.

We now examine the converse claim.Since|?|

does not lie in?.Applying Lemma1.2,we have that F T∪S→?∪{ω

0}is a bijection,and thus

we can?nd a vector h supported on T∪S whose Fourier transform vanishes on?but is nonzero onω0;in particular,h is not identically zero.The claim now follows by taking f:=h|T and g:=?h|S.

Note that if N is not prime,the lemma(and hence the theorem)fails,essentially because of the presence of non-trivial subgroups of Z N with addition modulo N;see Sections1.3

and1.4for concrete counter examples,and[7],[29]for further discussion.However,it is plausible to think that Lemma1.2continues to hold for non-prime N if T and?are assumed to be generic—in particular,they are not subgroups of Z N,or cosets of subgroups. If T and?are selected uniformly at random,then it is expected that the theorem holds with probability very close to one;one can indeed presumably quantify this statement by adapting the arguments given above but we will not do so here.However,we refer the reader to Section1.7for a rapid presentation of informal arguments pointing in this direction.

A re?nement of the argument in Theorem1.1shows that for?xed subsets T,S of time domain an?in the frequency domain,the space of vectors f,g supported on T,S such that?f|?=?g|?has dimension|T∪S|?|?|when|T∪S|≥|?|,and has dimension|T∩S| otherwise.In particular,if we letΣ(N t)denote those vectors whose support has size at most N t,then set of the vectors inΣ(N t)which cannot be reconstructed uniquely in this class from the Fourier coe?cients sampled at?,is contained in a?nite union of linear spaces of dimension at most2N t?|?|.SinceΣ(N t)itself is a?nite union of linear spaces of dimension N t,we thus see that recovery of f from?f|?is in principle possible generically whenever|supp(f)|=N t<|?|;once N t≥|?|,however,it is clear from simple degrees-of-freedom arguments that unique recovery is no longer possible.While our methods do not quite attain this theoretical upper bound for correct recovery,our numerical experiements suggest that they do come within a constant factor of this bound(see Figure2). Theorem1.1asserts that one can reconstruct f from2|T|frequency samples(and that, in general,there is no hope to do so from fewer samples).In principle,we can recover f exactly by solving the combinatorial optimization problem

(P0)min

g∈C N g

,?g|?=?f|?,(1.4)

where g

is the number of nonzero terms#{t,g(t)=0}.This is a combinatorial optimization problem,and solving(1.4)directly is infeasible even for modest-sized signals. To the best of our knowledge,one would essentially need to let T vary over all subsets

T?{0,...,N?1}of cardinality|T|≤1

2|?|,checking for each one whether f is in the

range of F T→?or not,and then invert the relevant minor of the Fourier matrix to recover f once T is determined.Clearly,this is computationally very expensive since there are exponentially many subsets to check;for instance,if|?|～N/2,then the number of subsets scales like4N·3?3N/4!As an aside comment,note that it is also not clear how to make this algorithm robust,especially since the results in[29]do not provide any e?ective lower bound on the determinant of the minors of the Fourier matrix,see Section6for a discussion of this point.

A more computationally e?cient strategy for recovering f from?and?f|?is to solve the convex problem

(P1)min

g∈C N g

1

:=

t∈Z N

|g(t)|,?g|?=?f|?.(1.5)

The key result in this paper is that the solutions to(P0)and(P1)are equivalent for an overwhelming percentage of the choices for T and?with|T|≤α·|?|/log N(α>0is a constant):in these cases,solving the convex problem(P1)recovers f exactly.

To establish this upper bound,we will assume that the observed Fourier coe?cients are randomly sampled.Given the number Nωof samples to take in the Fourier domain,we

choose the subset ?uniformly at random from all sets of this size;i.e.each of the N N ω possible subsets are equally likely.Our main theorem can now be stated as follows.Theorem 1.3Let f ∈C N be a discrete signal supported on an unknown set T ,and choose ?of size |?|=N ωuniformly at random.For a given accuracy parameter M ,if

|T |≤C M ·(log N )?1·|?|,(1.6)

then with probability at least 1?O (N ?M ),the minimizer to the problem (1.5)is unique and is equal to f .

Notice that (1.6)essentially says that |T |is of size |?|,modulo a constant and a logarithmic factor.Our proof gives an explicit value of C M ,namely,C M 1/[23(M +1)](valid for |?|≤N/4,M ≥2and N ≥20,say)although we have not pursued the question of exactly what the optimal value might be.

In Section 5,we present numerical results which suggest that in practice,we can expect to recover most signals f more than 50%of the time if the size of the support obeys |T |≤|?|/4.By most signals,we mean that we empirically study the success rate for randomly selected signals,and do not search for the worst case signal f —that which needs the most frequency samples.For |T |≤|?|/8,the recovery rate is above 90%.Empirically,the constants 1/4and 1/8do not seem to vary for N in the range of a few hundred to a few thousand.

1.3For almost every ?

As the theorem allows,there exist sets ?and functions f for which the 1-minimization procedure does not recover f correctly,even if |supp(f )|is much smaller than |?|.We sketch two counter examples:

?A discrete Dirac comb.Suppose that N is a perfect square and consider the picket-fence signal which consists of spikes of unit height and with uniform spacing equal to √N .This signal is often used as an extremal point for uncertainty principles [7,8]as one of its remarkable properties is its invariance through the Fourier transform.Hence suppose that ?is the set of all frequencies but the multiples of √N ,namely,|?|=N ?√N .Then ?f

|?=0and obviously the reconstruction is identically zero.Note that the problem here does not really have anything to do with 1-minimization per se;f cannot be reconstructed from its Fourier samples on ?thereby showing that Theorem 1.1does not work “as is”for arbitrary sample sizes.

?Boxcar signals .The example above suggests that in some sense |T |must not be greater than about |?|.In fact,there exist more extreme examples.Assume the sample size N is large and consider for example the indicator function f of the interval T :={t :?N ?0.01

is a nonnegative bump function adapted to the interval {ω:?N/6<ωc for some absolute constant c >0.Because of

this,the signal f?ε|h|2will have smaller 1-norm than f forε>0su?ciently small (and N su?ciently large),while still having the same Fourier coe?cients as f on?.

Thus in this case f is not the minimizer to the problem(P1),despite the fact that the support of f is much smaller than that of?.

The above counter examples relied heavily on the special choice of?(and to a lesser extent of supp(f));in particular,it needed the fact that the complement of?contained a large interval(or more generally,a long arithmetic progression).But for most sets?,large arithmetic progressions in the complement do not exist,and the problem largely disappears. In short,Theorem1.3essentially says that for most sets of T of size about|?|,there is no loss of information.

1.4Optimality

Theorem1.3states that for any signal f supported on an arbitrary set T in the time domain, (P1)recovers f exactly—with high probability—from a number of frequency samples that is within a constant of M·|T|log N.It is natural to wonder whether this is a fundamental limit.In other words,is there an algorithm that can recover an arbitrary signal from far fewer random observations,and with the same probability of success?

It is clear that the number of samples needs to be at least proportional to|T|,otherwise F T→?will not be injective.We argue here that it must also be proportional to M log N to guarantee recovery of certain signals from the vast majority of sets?of a certain size. Suppose f is the Dirac comb signal discussed in the previous section.If we want to have a chance of recovering f,then at the very least,the observation set?and the frequency support W=supp?f must overlap at one location;otherwise,all of the observations are zero,and nothing can be done.Choosing?uniformly at random,the probability that it includes none of the members of W is

P(?∩W=?)= N?√N

|?|

N

|?|

≥

1?

2|?|

N

√N

,

where we have used the assumption that|?|>|T|=√

N.Then for P(?∩W=?)to be

smaller than N?M,it must be true that

√N·log

1?

2|?|

N

≤?M log N,

and if we make the restriction that|?|cannot be as large as N/2,meaning that log(1?

2|?| N )≈?2|?|

N

,we have

|?|≥Const·M·

√

·log N.

For the Dirac comb then,any algorithm must have|?|～|T|M log N observations for the identi?ed probability of success.

Examples for larger supports T exist as well.If N is an even power of two,we can su-perimpose2m Dirac combs at dyadic shifts to construct signals with time-domain support

|T |=2m √N and frequency-domain support |W |=

2?m √N for m =0,...,log 2√N .The same argument as above would then dictate that

|?|≥Const ·M ·N |W |

·log N =Const ·M ·|T |·log N.In short,Theorem 1.3identi?es a fundamental limit.No recovery can be successful for all signals using signi?cantly fewer observations.

1.5Extensions

As mentioned earlier,results for our model problem extend easily to higher dimensions and alternate recovery scenarios.To be concrete,consider the problem of recovering a one-dimensional piecewise constant signal via

min g t ∈Z N

|g (t )?g (t ?1)|?g |?=?f |?,(1.7)where we adopt the convention that g (?1)=g (N ?1).In a nutshell,model (1.5)is obtained from (1.7)after di?erentiation.Indeed,let δbe the vector of ?rst di?erence δ(t )=g (t )?g (t ?1),and note that δ(t )=0.Obviously,

?δ(ω)=(1?e ?2πiω/N )?g (ω),for all ω=0

and,therefore,with υ(ω)=(1?e ?2πiω/N )?1,the problem is identical to

min δ δ 1?δ

|?\{0}=(υ?f )|?\{0},?δ(0)=0,which is precisely what we have been studying.

Corollary 1.4Put T ={t,f (t )=f (t ?1)}.Under the assumptions of Theorem 1.3,the minimizer to the problem (1.7)is unique and is equal f with probability at least 1?

O (N ?M )—provided that f be adjusted so that f (t )=?f

(0).We now explore versions of Theorem 1.3in higher dimensions.To be concrete,consider the two-dimensional situation (statements in arbitrary dimensions are exactly of the same ?avor):

Theorem 1.5Put N =n 2.We let f (t 1,t 2),1≤t 1,t 2≤n be a discrete real-valued image and ?of a certain size be chosen uniformly at random.Assume that for a given accuracy parameter M ,f is supported on T obeying (1.6).Then with probability at least 1?O (N ?M ),the minimizer to the problem (1.5)is unique and is equal to f .

We will not prove this result as the strategy is exactly parallel to that of Theorem 1.3.Letting D 1f be the horizontal ?nite di?erences D 1f (t 1,t 2)=f (t 1,t 2)?f (t 1?1,t 2)and D 2f be the vertical analog,we have just seen that we can think about the data as the properly renormalized Fourier coe?cients of D 1f and D 2f .Now put d =D 1f +iD 2f ,where i 2=?1.Then the minimum total-variation problem may be expressed as

min δ 1subject to F ?δ=F ?d,

(1.8)

where F?is a partial Fourier transform.One then obtains a statement for piecewise constant 2D functions,which is similar to that for sparse1D signals provided that the support of f be replaced by{(t1,t2):|D1f(t1,t2)|2+|D2f(t1,t2)|2=0}.We omit the details.

The main point here is that there actually are a variety of results similar to Theorem 1.3.Theorem1.5serves as another recovery example,and provides a precise quantitative understanding of the“surprising result”discussed at the beginning of this paper.

To be complete,we would like to mention that for complex valued signals,the minimum 1problem(1.5)and,therefore,the minimum TV problem(1.1)can be recast as special convex programs known as second order cone programs(SOCP’s).For example,(1.8)is equivalent to

min

t u(t)subject to

|δ1(t)|2+|δ2(t)|2≤u(t),(1.9)

F?(δ1+iδ2)=F?d,

with variables u,δ1andδ2in R N(δ1andδ2are the real and imaginary parts ofδ).If in addition,δis real valued,then this is a linear program.Much progress has been made in the past decade on algorithms to solve both linear and second-order cone programs[25], and many o?-the-shelf software packages exist for solving problems such as(P1)and(1.9).

1.6Relationship to uncertainty principles

From a certain point of view,our results are connected to the so-called uncertainty principles [7,8]which say that it is di?cult to localize a signal f∈C N both in time and frequency at the same time.Indeed,classical arguments show that f is the unique minimizer of(P1)

if and only if

t∈Z N |f(t)+h(t)|>

t∈Z N

|f(t)|,?h=0,?h|?=0

Put T=supp(f)and apply the triangle inequality

Z N |f(t)+h(t)|=

T

|f(t)+h(t)|+

T c

|h(t)|≥

T

|f(t)|?|h(t)|+

T c

|h(t)|.

Hence,a su?cient condition to establish that f is our unique solution would be to show

that

T |h(t)|<

T c

|h(t)|?h=0,?h|?=0.

or equivalently

T

|h(t)|<1

2

h

1

.The connection with the uncertainty principle is now

explicit;f is the unique minimizer if it is impossible to concentrate half of the 1norm of a signal that is missing frequency components in?on a“small”set T.For example,[7] guarantees exact reconstruction if

2|T|·(N?|?|)

Take|?|

By re?ning these uncertainty principles,[6]shows that a much stronger recovery result is possible.The central results of[6]imply that a signal consisting of|T|spikes which are spread out in a somewhat even manner in the time domain can be recovered from C·|T| lowpass observations.Theorem1.3is di?erent in that it applies to all signals with a certain support size,and does not rely on a special choice of?(almost any?which is large enough will work).The price for this additional power is that we require a factor of log N more observations.

In truth,this paper does not follow this classical approach of deriving a recovery condition directly from an uncertainty principle.Instead,we will use duality theory to study the solution of(P1).However,a byproduct of our analysis will be a novel uncertainty principle that holds for generic sets T,?.

1.7Robust uncertainty principles

Underlying our results is a new notion of uncertainty principle which holds for almost any pair(supp(f),supp(?f)).With T=supp(f)and?=supp(?f),the classical discrete uncertainty principle[7]says that

|T|+|?|≥2√

(1.10)

with equality obtained for signals such as the Dirac comb.As we mentioned above,such extremal signals correspond to very special pairs(T,?).However,for most choices of T and?,the analysis presented in this paper shows that it is impossible to?nd f such that T=supp(f)and?=supp(?f)unless

|T|+|?|≥γ(M)·(log N)?1/2·N,(1.11) which is considerably stronger than(1.10).Here,the statement“most pairs”says again that the probability of selecting a random pair(T,?)violating(1.11)is at most O(N?M). In some sense,(1.11)is the typical uncertainty relation one can generally expect(as opposed to(1.10)),hence,justifying the title of this paper.Because of space limitation,we are unable to elaborate on this fact and its implications further,but will do so in a companion paper.

1.8Connections with existing work

The idea of relaxing a combinatorial problem into a convex problem is not new and goes back a long way.For example,[5,27]used the idea of minimizing 1norms to recover spike trains.The motivation is that this makes available a host of computationally feasible procedures.For example,a convex problem of the type(1.5)can be practically solved using techniques of linear programming such as interior point methods[3].

Using an 1minimization program to recover sparse signals has been proposed in several di?erent contexts.Early work in geophysics[23,26,27]centered on super-resolving spike trains from bandlimited observations,i.e.the case where?consists of low-pass https://www.sodocs.net/doc/6517627593.html,ter works[6,7]provided a uni?ed framework in which to interpret these results by demonstrating that the e?ectiveness of recovery via minimizing 1was linked to discrete uncertainty principles.As mentioned in Section1.6,these papers derived explicit bounds

on the number of frequency samples needed to reconstruct a sparse signal.The earlier reference[7]also contains a conjecture that more powerful uncertainty principles may exist if one of T,?is chosen at random,which is essentially the content of Section1.7here. More recently,there exists a series of beautiful papers[8–10,13,18]concerned with prob-lem of?nding the sparsest decomposition of a signal f using waveforms from a highly overcomplete dictionary D.One seeks the sparsestαsuch that

f=Dα,(1.12)

where the number of columns M from D is greater than the sample size N.Consider the solution which minimizes the 0norm ofαsubject to the constraint(1.12)and that which minimizes the 1norm.A typical result of this body of work is as follows:suppose that s can be synthesized out of very few elements from D,then the solution to both problems are unique and are equal.We also refer to[30,31]for very recent results along these lines. This literature certainly in?uenced our thinking in the sense it made us suspect that results such as Theorem1.3were actually possible.However,we would like to emphasize that the claims presented in this paper are of a substantially di?erent nature.We give essentially two reasons:

1.Our model problem is di?erent since we need to“guess”a signal from incomplete

data,as opposed to?nding the sparsest expansion of a fully speci?ed signal.

2.Our approach is decidedly probabilistic—as opposed to deterministic—and thus calls

for very di?erent techniques.For example,underlying our analysis are delicate esti-mates for the norms of certain types of random matrices,which may be of independent interest.

Apart from the wonderful properties of 1,several novel sampling theorems have in intro-duced in recent years.In[11,12]the authors study universal sampling patters that allow the exact reconstruction of signals supported on a small set.In[32],ideas from spectral analysis are leveraged to show that a sequence of N t spikes can be recovered exactly from 2N t+1consecutive Fourier samples(in[32]for example,the recovery requires solving a system of equations and factoring a polynomial).Our results,namely,Theorems1.1and 1.3require slightly more samples to be taken(C·N t log N versus C·N t),but are again more general in that they address the radically di?erent situation in which we do not have the freedom to choose the sample locations at our convenience.

Finally,it is interesting to note that our results and the references above are also related to recent work[15]in?nding near-best B-term Fourier approximations(which is in some sense the dual to our recovery problem).The algorithm in[15,16],which operates by estimating the frequencies present in the signal from a small number of randomly placed samples, produces with high probability an approximation in sublinear time with error within a constant of the best B-term approximation.First,in[16]the samples are again selected to be equispaced whereas we are not at liberty to choose the frequency samples at all since they are speci?ed a priori.And second,we wish to produce as a result an entire signal or image of size N,so a sublinear algorithm is an impossibility.

1.9Random sensing

Against this background,the main contribution of this paper is the idea that one can use randomness as a sensing mechanism;that is,as a way of extracting information about an object of interest from a small number of randomly selected observations.For example, we have seen that if an object has a sparse gradient,then we can“image”this object by measuring a few Fourier samples at random locations,rather than by acquiring a large number of pixels.

This point of view is very broad.Suppose we wish to reconstruct a signal f assumed to be sparse in a?xed basis,e.g.a wavelet basis.Then by applying random sensing—taking a small number of random measurements—the number of measurement we need depends far more upon the structural content of the signal(the number of signi?cant terms in the wavelet expansion)than the resolution N.From a quantitative viewpoint,our methodology should certainly be amenable to such general situations,as we will discuss further in Section6.3. 2Strategy

There exists at least one minimizer to(P1)but it is not clear why this minimizer should be unique,and why it should equal f.In this section,we outline our strategy for answering these questions.In Section2.1,we use duality theory to show that f is the unique solution to(P1)if and only if a trigonometric polynomial with certain properties exists(a similar duality approach was independently developed in[14]for?nding sparse approximations from general dictionaries).We construct a special polynomial in Section2.2and the re-mainder of the paper is devoted to showing that if(1.6)holds,then our polynomial obeys the required properties.

2.1Duality

Suppose that f is supported on T,and we observe?f on a set?.The following lemma shows that a necessary and su?cient condition for the solution f to be the solution to(P1) is the existence of a trigonometric polynomial P whose Fourier transform is supported on ?,matches sgn(f)on T,and has magnitude strictly less than1elsewhere.

Lemma2.1Let??Z N.For a vector f∈C N with T:=supp(f),de?ne the sign vector sgn(f)(t):=f(t)/|f(t)|when t∈T and sgn(f)=0otherwise.Suppose there exists a vector P whose Fourier transform?P is supported in?such that

P(t)=sgn(f)(t)for all t∈T(2.1) and

|P(t)|<1for all t∈T.(2.2) Then if F T→?is injective,the minimizer f to the problem(P1)is unique and is equal to f.Conversely,if f is the unique minimizer of(P1),then there exists a vector P with the above properties.

This is a result in convex optimization whose proof is given in the Appendix.

Since the space of functions with Fourier transform supported in?has|?|degrees of freedom,and the condition that P match sgn(f)on T requires|T|degrees of freedom,one now expects heuristically(if one ignores the open conditions that P has magnitude strictly less than1outside of T)that f should be unique and be equal to f whenever|T| |?|; in particular this gives an explicit procedure for recovering f from?and?f|?.

2.2Architecture of the Argument

We will show that we can recover f supported on T from observations on almost all sets ?obeying(1.6)by constructing a particular polynomial P(that depends on T and?) which automatically satis?es the equality constraints(2.1)on T,and then showing the the inequality constraints(2.2)on T c hold with high probability.

With|?|>|T|,and if F T→?is injective(has full column rank),there are many trigono-metric polynomials supported on?in the Fourier domain which satisfy(2.1).We choose, with the hope that its magnitude on T c is small,the one with minimum energy:

P:=F??F T→?(F?T→?F T→?)?1ι?sgn(f)(2.3)

where F?=F Z

N→?is the Fourier transform followed by a restriction to the set?;the

embedding operatorι: 2(T)→ 2(Z N)extends a vector on T to a vector on Z N by placing zeros outside of T;andι?is the dual restriction mapι?f=f|T.It is easy to see that?P is

supported on?,and noting thatι?F?

?=F?

T→?

,P also satis?es(2.1)

ι?P=ι?sgn(f).

Fixing f and its support T,we will prove Theorem1.3by establishing that if the set?is chosen uniformly at random from all sets of size Nω≥C?1

M

·|T|·log N,then

1.Invertibility.The operator F T→?is injective,meaning that F?

T→?F T→?in(2.3)is

invertible,with probability1?O(N?M).

2.Magnitude on T c.The function P in(2.3)obeys|P(t)|<1for all t∈T c again with

probability1?O(N?M).

Making these arguments directly for the case where?of a certain size is chosen uniformly at random would be complicated,as the probability of a particular frequency being included in the set?would depend on whether or not each other frequency is included.To simplify the analysis,the next subsection introduces a Bernoulli probability model for selecting the set?,and shows how results using this model can be translated into results for the uniform probability model.

2.3The Bernoulli model

A set? of Fourier coe?cients is sampled using the Bernoulli model with parameter0<τ<1by?rst creating the sequence

Iω=

0with prob.1?τ,

1with prob.τ

(2.4)

and then setting

? :={ω:Iω=1}.(2.5) The size of the set|? |is also random,following a binomial distribution,and E(|? |)=τN. In fact,classical large deviations arguments tell us that as N gets large,|? |/N≈τwith high probability.

With this pobability model,we establish two formal statements showing that P in(2.3) obeys the conditions of Lemma2.1.Both are proven in Section3.

Theorem2.2Let T be a?xed subset,and choose?using the Bernoulli model with pa-rameterτ.Suppose that

|T|≤C M·(log N)?1·τN,(2.6)

where C M is the same as in Theorem1.3.Then F?

T→?F T→?is invertible with probability

at least1?O(N?M).

Lemma2.3Under the assumptions of Theorem2.2,P in(2.3)obeys|P(t)|<1for all t∈T c with probability at least1?O(N?M).

We now explain why these two claims give Theorem1.3.De?ne Failure(?0)as the event where no dual polynomial P,supported on?0in the Fourier domain,exists that obeys the conditions(2.1)and(2.2)above.Let?of size Nωbe drawn using the uniform model,and let? be drawn from the Bernoulli model withτ=Nω/N.We have

P

Failure(? )

=

N

k=0

P

Failure(? )

|? |=k

·P

|? |=k

=

N

k=0

P(Failure(?k))·P

|? |=k

where?k is selected uniformly at random with|?k|=k.We make two observations:?P(Failure(?k))is a nonincreasing function of k.This follows directly from the fact that

?1??2?P(Failure(?2))≤P(Failure(?1)), (the larger?becomes,it only becomes easier to construct a valid P).

?SinceτN is an integer,it is the median of|? |:

P

|? |≤τN?1

<1/2

|? |≤τN

.

(See[19]for a proof.)

With the above in mind,we continue

P

Failure(? )

≥

Nω

k=1

P(Failure(?k))·P

|? |=k

≥P(Failure(?))·

Nω

k=1

P

|? |=k

≥

1

2

·P(Failure(?)).

Thus if we can bound the probability of failure for the Bernoulli model,we know that the failure rate for the uniform model will be no more than twice as large.

3Construction of the Dual Polynomial

The Bernoulli model holds throughout this section,and we carefully examine the minimum energy dual polynomial P de?ned in(2.3)and establish Theorem2.2and Lemma2.3.The main arguments hinge on delicate moment bounds for random matrices,which are presented in Section4.From here on forth,we will assume that|τN|>M log N since the claim is vacuous otherwise(as we will see,C M≤1/M and thus(1.6)will force f≡0,at which point it is clear that the solution to(P1)is equal to f=0).

We will?nd it convenient to rewrite(2.3)in terms of the auxiliary matrix

Hf(t):=?

ω∈?

t ∈T:t =t

e2πiω(t?t

)

N f(t ),(3.1)

and de?ne

H0=ι?H.

To see the relevance of the operators H and H0,observe that

ι?

1

|?|

H=

1

|?|

F??F T→?

I T?

1

|?|

H0=

1

|?|

F?T→?F T→?

where I T is the identity for 2(T)(note thatι?ι=I T).Then

P=

ι?

1

|?|

H

I T?

1

|?|

H0

?1

ι?sgn f.

The point here is to separate the constant diagonal of F?

T→?F T→?(which is|?|everywhere)

from the highly oscillatory o?-diagonal.We will see that choosing?at random makes H0 essentially a“noise”matrix,making I T?1

|?|

H0well conditioned.

3.1Invertibility

We would like to establish invertibility of the matrix I T?1

|?|

H0with high probability.One way to proceed would be to show that the operator norm(i.e.the largest eigenvalue)of H0

is less than|?|.A straightforward way to do this is to bound the operator norm H0 by the Frobenius norm H0 F:

H0 2≤ H0 2F:=Tr(H0H?0)=

t1,t2|(H0)t

1,t2

|2.(3.2)

where(H0)t

1,t2

is the matrix element at row t1and column t2.

Using relatively simple statistical arguments,we can show that with high probability

|(H0)t

1,t2|2～|?|.Applying(3.2)would then yield invertibility when|T|～

|?|.To

show that H0is“small”for larger sets T(recall that|T|～|?|·(log N)?1is the desired result),we use estimates of the Frobenius norm of a large power of H0,taking advantage of cancellations arising from the randomness of the matrix coe?cients of H0.

Our argument relies on a key estimate which we introduce now and shall be discussed in greater detail in Section3.2.Assume thatτ≤1/(1+e)and n≤τN/[4|T|(1?τ)].Then the2n th moment of H0obeys

E(Tr(H2n0))≤2

4

e(1?τ)

n

n n+1·|τN|n|T|n+1.(3.3)

Now this moment bound gives an estimate for the operator norm of H0.To see this,note that since H0is self-adjoint

H0 2n= H n0 2≤ H n0 2F=Tr(H2n0).

Lettingαbe a positive number0<α<1,it follows from the Markov inequality that

P( H n0 F≥αn·|τN|n)≤

E H n0 2

F α2n|τN|2n

.

We then apply inequality(3.3)(recall H n0 2

F

=Tr(H2n0))and obtain

P( H n0 F≥αn·|τN|n)≤2n e?n

4n

α2(1?τ)

n

|T|

|τN|

n

|T|.(3.4)

We remark that the last inequality holds for any sample size|T|(with the proviso that n≤τN/[4|T|(1?τ)])and we now specialize(3.4)to selected values of|T|.

Theorem3.1Assume thatτ≤(1+e)?1and suppose that T obeys

|T|≤α2

M

(1?τ)

4

|τN|

n

,for someαM≤α≤1.(3.5)

Then

P( H n0 F≥αn·|τN|n)≤1

2

α2e?n|τN|.(3.6)

Select n=(M+1)log N which corresponds to the assumptions of Theorem2.2.Then the operator I T?1

|?|

H0is invertible with probability at least1?1.25N?M.

Proof The ?rst part of the theorem follows from (3.4).For the second part,we begin by observing that a typical application of the large deviation theorem gives

P (|?|

(3.7)Slightly more precise estimates are possible,see [1].It then follows that P (|?|<(1? M )|τN |)≤N ?M , M := 2M log N |τN |.(3.8)We will denote by B M the event {|?|<(1? M )|τN |}.

We now take n =(M +1)log N and α=1/√2and assume that T obeys (3.5)(note that

|T |obeys the assumptions of Theorem 2.2).Put A M :={ H 0 ≥|τN |/√2}.Then

P (A M )≤14·|τN |·N ?(M +1)≤14

N ?M ,and on the complement of A M ∪B M ,we have H 0 ≤τN/√2≤|?|/[√2(1? M )].Hence I T ?1|?|H 0is invertible with the desired probability.

We have thus established Theorem 2.2,and thus P is well de?ned with high probability.To conclude this section,we would like to emphasize that our analysis gives a rather precise estimate of the norm of H 0.

Corollary 3.2Assume for example that |T |log |T |≤τN/(4(1?τ))and set γ= 4/(1?τ).

For any >0,we have P H 0 >(1+ )γ log |T | |T ||τN | →0as |T |,|τN |→∞.

Proof Put λ=γ log |T | |T ||τN |.The Markov inequality gives

P ( H 0 ≥(1+ )λ)≤E [Tr(H 2n 0](1+ )2n λ2n

.Select n = log |T | so that

e ?n n n |T |≤ log |T | n .

For this n ,E [Tr(H 2n 0]≤2nλ2n (3.3).Therefore,the probability is bounded by 2n (1+ )?2n which goes to zero as n = log |T | goes to in?nity.

3.2The key estimate

Our key estimate (3.3)is stated below.The proof is technical and deferred to Section 4.Theorem 3.3Let τ≤1/(1+e ).Then with the Bernoulli model E (Tr(H 2n 0)≤ 2 4e (1?τ) n n n +1|τN |n |T |n +1n ≤τN 4(1?τ)|T |,n 1?τ(4n )

2n ?1|τN ||T |2n ,otherwise .(3.9)In other words,when n ≤τN 4|T |(1?τ),the 2n th moment obeys (3.3).

3.3Magnitude of the polynomial on the complement of T

In the remainder of Section3,we argue that max t/∈T|P(t)|<1with high probability and prove Lemma2.3.We?rst develop an expression for P(t)by making use of the algebraic identity

(1?M)?1=(1?M n)?1(1+M+...+M n?1).

Indeed,we can write

(I T?

1

|?|n

H n0)?1=I T+R where R=

∞

p=1

1

|?|pn

H pn0,

so that the inverse is given by the truncated Neumann series

(I T?

1

|?|

H0)?1=(I T+R)

n?1

m=0

1

|?|m

H m0.(3.10)

The point is that the remainder term R is quite small in the Frobenius norm:suppose that

ι?H F≤α·|?|,then

R F≤

αn

1?αn

.

In particular,the matrix coe?cients of R are all individually less thanαn/(1?αn).Intro-duce the ∞-norm of a matrix as M ∞=sup x

∞≤1

Mx ∞which is also given by

M ∞=sup

i

j

|M(i,j)|.

It follows from the Cauchy-Schwarz inequality that

M 2∞≤sup

i #col(M)

j

|M(i,j)|2≤#col(M)· M 2F,

where by#col(M)we mean the number of columns of M.This observation gives the crude

estimate

R ∞≤|T|1/2·

αn

1?αn

.(3.11)

As we shall soon see,the bound(3.11)allows us to e?ectively neglect the R term in this formula;the only remaining di?culty will be to establish good bounds on the truncated

Neumann series1

|?|H

n?1

m=0

1

|?|m

H m0.

3.4Estimating the truncated Neumann series From(2.3)we observe that on the complement of T

P=

1

|?|

H(I T?

1

|?|

H0)?1ι?sgn(f),

since theιcomponent in(2.3)vanishes outside of T.Applying(3.10),we may rewrite P as

P(t)=P0(t)+P1(t),?t∈T c,

where

P0=S n sgn(f),P1=

1

|?|

HRι?(I+S n?1)sgn(f)

and

S n=

n

m=1

|?|?m(Hι?)m.

Let a0,a1>0be two numbers with a0+a1=1.Then

P

sup

t∈T c

|P(t)|>1

≤P( P0 ∞>a0)+P( P1 ∞>a1),

and the idea is to bound each term individually.Put Q0=S n?1sgn(f)so that P1= 1

|?|

HRι?(sgn(f)+Q0).With these notations,observe that

P1 ∞≤

1

|?|

HR ∞(1+ ι?Q0 ∞).

Hence,bounds on the magnitude of P1will follow from bounds on HR ∞together with bounds on the magnitude ofι?Q0.It will be su?cient to derive bounds on Q0 ∞(since ι?Q0 ∞≤ Q0 ∞)which will follow from those on P0since Q0is nearly equal to P0(they di?er by only one very small term).

Fix t∈T c and write P0(t)as

P0(t)=

n

m=1

|?|?m X m(t),X m=(Hι?)m sgn(f).

The idea is to use moment estimates to control the size of each term X m(t).

Lemma3.4Set n=km.Then E|X m(t0)|2k obeys the same estimate as that in Theorem 3.3(up to a multiplicative factor|T|?1),namely,

E|X m(t0)|2k≤

1

|T|

B n,(3.12)

where B n is the right-hand side of(3.9).In particular,following(3.3)

E|X m(t0)|2k≤2e?n(4/(1?τ))n n n+1·|T|n|τN|n,(3.13) provided that n≤τN

4|T|(1?τ)

.

The proof of these moment estimates mimics that of Theorem3.3and may be found in the Appendix.

Lemma3.5Fix a0=.91.Suppose that|T|obeys(3.5)and let B M be the set where |?|<(1? M)·|τN|with M as in(3.8).For each t∈Z N,there is a set A t with the property

P(A t)>1? n, n=2(1? M)?2n·n2e?nα2n·(0.42)?2n,

and

|P0(t)|<.91,|Q0(t)|<.91on A t∩B c M.

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百度文库- 让每个人平等地提升自我 压缩空气系统确认方案 文件编号：JH-YZ-SB-025-R00 制定人： 制定日期： 审核人： 审核日期： 批准人： 批准日期： 实施日期： 四川利君精华制药股份有限公司

目录1：概述 压缩空气系统简介 压缩空气设备基本情况 净化压缩空气处理流程图 压缩空气系统的主要技术参数 2：目的 3：范围 4：依据 5：可接受标准 6：职责 7：培训 8：确认时间 9：确认内容 设计确认DQ 安装确认IQ 运行确认OQ 性能确认PQ 10：异常情况处理 11：偏差处理 12：变更控制 13：确认结果评定 14：拟定再确认周期 15：附表

1 概述 压缩空气系统简介 本压缩空气系统主要是作为制剂车间（固体制剂、提取车间和凝胶剂、栓剂车间）生产工艺的辅助设备，为车间提供符合生产工艺要求的压缩空气，压缩空气系统由压缩机、电动机、压力开关、单向阀、储气罐、压力表、自动排水器、安全阀、主管道过滤器等组成。 压缩空气设备基本情况 水润滑单螺杆空气压缩机 项目 栓剂、凝胶剂 固体制剂 产品型号 出厂编号 03214203 03214205 生产厂家 广东正力精密机械有限公司 净化压缩空气处理流程图 压缩空气系统的主要技术参数 序号 项目 主要技术参数 1 排气量（m 3/min ） 2 排气压力（MPa ） 3 螺杆润滑方式 水润滑 4 吸气状态 温度（℃） 2～40 压力 大气压 5 供气温度 环境温度+20 6 传动方式 弹性连轴器 7 冷却方式 分冷 空气 空气压缩机 冷干机 除油过滤器 精密过滤器 除菌过滤器 除臭过滤器 各使用点

压缩空气系统监测操作规程

压缩空气监测检验操作规程 目的：建立一个压缩空气监测检验操作程序，以便控制压缩空气给药品带来的污染。范围：直接接触药品生产的压缩空气。责任人：QC人员、QA人员依据: 《药品生产质量管理规范（2010年修订）》、《药品GMP指南》内容： 1 频率：每半年或压缩空气设备大修后对压缩空气进行检验。 2 采样检查人：经授权的取样人 3 采样工具：1000ml烧杯、经过灭菌处理的培养皿 4 压缩空气性状检查： 4.1 从设备上拔下压缩空气细管，调节压缩空气量，手感有微风即可。 4.2 将压缩空气通入装入1000ml纯化水的烧杯中，持续10分钟，水面不得有油花或其他 杂质。 5 微生物检查 5.1 采样： 5.1.1 静态取样，在空调系统正常运行30min后，洁净室内没有生产人员，测试人员不多 于2人情况下开始采样。 5.1.2 从设备上拔下压缩空气细管（每个设备特性细管数量不同），将其固定，调压缩 空气量，手感有微风即可。 5.1.3 用酒精棉消毒压缩空气细管的管口。 5.1.4 将已倾注胰酪大豆胨琼脂（TSA）培养基的平皿（φ90mm×15mm），平皿数量与压 缩空气细管数量相同，打开盖，置管口下5～10cm处收集压缩空气中的生物粒子于培养基平皿内，0.5h后盖上平皿。 5.1.5 用玻璃笔在培养皿盖上标注取样点，取样时间。 5.1.6 填写压缩空气取样、交接记录（附件I）。 5.2 采集样品后的平皿，立即送至化验室。 5.3 检验： 5.3.1 将采集样品后的培养皿置恒温培养箱中30～35℃培养48h。 5.3.2 菌落计数：用肉眼直接计数，然后用5～10倍放大镜检查有否遗漏。 5.3.3 结果计算：

压缩空气系统验证方案(1)

压缩空气系统验证方案 设备名称：压缩空气系统 设备型号： 设备编号：JD-0204-004 制造厂商： 安装位置： 验证方案编号：

目录 一、概述 (4) 二、目的 (4) 三、范围 (4) 四、压缩空气的组成及流程 (4) 五、验证依据和文件 (5) 六、人员职责及人员培训 (5) 七、风险评估 (6) 八、验证计量确认 (9) 九、性能确认 (9) 十、偏差处理 (11) 十一、变更控制 (11) 十二、验证结论 (12) 十三、再确认周期 (12) 十四、验证结论 (12)

验证方案起草审批方案起草 方案审核 方案批准 验证小组名单及职责

1.概述 本压缩空气系统是按照GMP要求设计、安装的压缩空气气源，由两台阿特拉斯·科普柯型固定式螺杆压缩机、一台冷冻式空气干燥机、一级P级精密过滤器、二级S级精密过滤器、一个的缓冲罐和无缝钢管输气管道组成。其基本流程是：将自然空气经固定式螺杆空气压缩机压缩，经缓冲罐、一级P级精密过滤器，再使用冷冻式干燥机将其除湿干燥，然后通过二级S级精过滤器得到无油、无水、无尘的压缩空气，经过无缝钢管输气管道，输送至车间各用气点，与药品直接接触各用气点再经μm过滤器过滤，压缩空气符合药品生产要求。 2、目的 确认系统生产的压缩空气性能达到使用标准 3.范围 对本厂区内接触药品内包材的压缩空气用气点进行性能确认。 4.压缩空气组成及流程 压缩空气系统设备一览表

净化区压缩空气用气点一览表： 5、验证依据及文件 药品生产质量管理规范（2010年修订） 空气压缩机标准操作规程 药品生产验证指南 6.人员培训确认 人员培训 确认目的：确认所有参与本次验证的人员是否接受了本次验证方案的培训。 合格标准：所有参与本次验证的人员均已接受了本次验证方案的培训。 确认记录：详见附件1，“验证方案培训记录”。

2014~2015年压缩空气系统质量回顾

1 压缩空气系统年度质量回顾 分析报告 （2014年01月～2014年12月）

2 1. 目的：对压缩空气系统运行、检查和检测情况进行回顾，确认压缩空气运行的稳定性和持续性，证明压缩空气系统的设计确认、安装确认、运行确认、性能确认符合设计要求，系统按现行的GMP 文件操作、维护保养等能够满足GMP 设计标准和实际生产工艺的需要，并评估是否启动压缩空气系统再验证。 2. 依据：年度回顾计划、压缩空气系统确认方案和报告、各压缩空气使用点数据检查 及检测记录。 3. 范围：适用于一车间、二车间、三车间压缩空气系统质量回顾。 4. 职责：总经办、设备管理部、质量保证部、质量控制部、一车间、二车间、三车间对本回顾内容起草、审核、批准负责。 5. 正文： 5.1. 概述： 5.1.1. 根据2010版《药品生产质量管理规范》对药品生产用压缩空气的要求，公司于2013年5月对压缩空气系统进行了再验证，证明压缩空气系统的设计确认、安装确认、运行确认、性能确认符合设计要求，系统按现行的GMP 文件操作、维护保养等能够满足GMP 设计标准和实际生产工艺的需要， 5.1.2. 总结压缩空气系统2014年的动态变化，确认目前压缩空气系统运行的现状，保证压缩空气系统运行质量安全、有效，且持续性运转。 5.1.3. 收集压缩空气系统相关所有记录；对压缩空气系统运行的关键过程控制点及各控制检测结果进行回顾；并检查仪器仪表是否定期校验并在有效期内。 5.2. 压缩空气系统简介 5.2.1. 我公司压缩空气系统有三台空压机组，两台德国进口空压机、一台国产无锡空压机厂的空压机交替工作，系统由空压机、储气罐、除油过滤器、冷冻干燥机、除水过滤器、精密除菌过滤器、终端过滤器等组成，具有除油、除水、除菌等功能。压缩空气经除油过滤器、除水过滤器、除菌过滤器、终端过滤器四级过滤后由不锈钢管道送至生产区房间使用点。 5.2.2. 压缩空气系统示意图：

压缩空气系统验证方案剖析

1 概述 1.1压缩空气系统描述 本压缩空气系统由预处理系统连接管路至车间各用气点构成。预处理系统位于制剂大楼二楼空调机房内，主要有LS10-30H固定式螺杆空压机、储气罐、主管路过滤器、冷冻式压缩空气干燥机、压缩空气精密过滤器等设施；连接管路及阀门全部采用304L不锈钢材质，并且双面抛光。系统为工艺生产气动设备及仪表的使用而提供无油无水的干燥空气，空压机排出的压缩空气，首先经过主管路过滤器，过滤粒径为1μm，然后经过冷冻式压缩空气干燥机除去水份，最后分别再经过二台0.01μm的压缩空气精密过滤器，保证了压缩空气质量满足GMP生产要求。 1.2设备基本情况 寿力空气压缩机组中一个重要部件是一单级容积式，油润滑螺杆压缩机。它提供稳定无脉动的压缩空气，并且无需保养和内部检查。 冷冻式压缩空气干燥机主要的功能是除去压缩空气所含的水份，是根据空气热交换原理，将压缩空气温度降至露点温度2℃～10℃，可凝结压缩空气所含的水份，再经油分离器分离空气和水滴，水滴经自动排水器排出系统外，即完成压缩空气干燥过程。 干燥的压缩空气经过压缩空气精密过滤器除油、除尘、除臭得到符合药品生产要求压缩空气。 压缩空气系统设备基本情况 序号名称规格型号编号供应商 1 固定式螺杆压缩机LS16-75H 01-008-01美国寿力公司 2 储气罐R11A2187 01-008-02台州中威空压机制造有限公司 3 冷冻式压缩空气 干燥机SLAD-10HTF 01-008-03 杭州山立净化 设备有限公司 4 主管路过滤器SLAF-10HT 01-008-03-F1杭州山立净化设备有限公司 5 微油雾过滤器SLAF-10HA 01-008-03-F2杭州山立净化设备有限公司 6 除油除臭超精过滤器SLAF-10HH 01-008-03-F3杭州山立净化设备有限公司 主要技术参数： 1.3压缩空气系统的流程示意图和各用气点分布图

压缩空气系统验证

哈尔滨新三勤制药有限公司再验证方案 类别：再验证方案编号：SQS—VTP—EN—2001—01 部门：工程部页码：共15页，第 1 页 头孢固体车间 压缩空气系统再验证方案 版次：□新订□替代： 起草：年月日 审阅会签： （验证小组） 批准：年月日 实施日期：年月日 授权：现授权下列部门拥有并执行本标准（复印数：） 复印序列号：

目录 1. 综述 --------------------------------------------------------------------------------------------------------------- 3 2.再验证目的 ------------------------------------------------------------------------------------------------------------- 3 3. 职责与成员------------------------------------------------------------------------------------------------------------ 3 3.1 验证委员会 ---------------------------------------------------------------------------------------------------- 3 3.2 工程部 ---------------------------------------------------------------------------------------------------------- 3 3.3 质量保证部 ---------------------------------------------------------------------------------------------------- 3 3.4车间 --------------------------------------------------------------------------------------------------------------- 4 3.5 成员 -------------------------------------------------------------------------------------------------------------- 4 3.6 验证实施的时间进度 ---------------------------------------------------------------------------------------- 4 4.相关性文件 -------------------------------------------------------------------------------------------------------------- 4 5.验证内容 ----------------------------------------------------------------------------------------------------------------- 5 5.1压缩空气系统安装确认 -------------------------------------------------------------------------------------- 5 5.2压缩空气系统运行确认 -------------------------------------------------------------------------------------- 6 5.3性能确认--------------------------------------------------------------------------------------------------------- 6 6. 拟订日常监测程序及验证周期：--------------------------------------------------------------------------------- 8 7. 验证结果评价与结论： --------------------------------------------------------------------------------------------- 8 8. 附录 ---------------------------------------------------------------------------------------------------------------------- 8

环境空气采样操作规程..

一、采样工作流程 1、接受任务 现场监测和采样承担部门的负责人在接到任务后提前通知有关科室配合，质量管理室填写任务传递单，将任务传递至现场监测人员。 2、对监测项目基本情况进行调查 现场监测人员认真了解监测对象的生产设备、工艺流程，清楚主要污染源、主要污染物及其排放规律，查看环保措施落实和环保设施运行情况，监控生产负荷，调查现场环境（如：气象、水文、污染源）有关参数和周边环境敏感点，检查监测点位符合性及安全性，搜集与编制技术（监测）报告有关的各种技术资料并做好相关的记录。 3、领取并检查采样所需仪器设备和辅助材料，进行采样前准备 现场监测人员根据任务传递单领取采样容器、滤膜，准备现场监测和采样所需的仪器设备、器具、样品标签、现场固定剂等，并完成仪器设备的运行检查。 （1）采样前准备的仪器设备和辅助材料 包括：采样器、风速风向仪、气温气压计、GPS；吸收瓶（内装配置好的吸收液，装箱，含空白、平行）、滤膜（含空白和备用膜）、镊子、凡士林、剪刀、手套、封口膜、电池、原始记录单、交接单、样品标签和笔等相关仪器物品。 （2）仪器设备的运行检查 在领用时，要检查并填写仪器的使用记录，尤其检查采样器流量是否需要校准，并对采样器进行气密性检查。 （3）现场采样前的准备 1）复核现场工况，是否适宜进行采样； 2）观测现场风速风向、局地流场、大气稳定度等气候条件，确定监测点位置； 3）按要求连接采样系统，并检查连接是否正确； 4）气密性检查，检查采样系统是否有漏气现象。 4、现场采样 （1）气态污染物采样 １）将气样捕集装置串联到采样系统中，核对样品编号，并将采样流量调至

所需的采样流量，开始采样。记录采样流量、开始采样时间、气样温度、压力等参数。气样温度和压力可分别用温度计和气压表进行同步现场测量。 ２）采样结束后，取下样品，将气体捕集装置进、出气口密封，记录采样流量、采样结束时间、气样温度、压力等参数。按相应项目的标准监测分析方法要求运送和保存待测样品。 （2）颗粒物采样 １）打开采样头顶盖，取出滤膜夹，用清洁干布擦掉采样头内滤膜夹及滤膜支持网表面上的灰尘，将采样滤膜毛面向上，平放在滤膜支持网上。同时核查滤膜编号，放上滤膜夹，安好采样头顶盖。启动采样器进行采样。记录采样流量、开始采样时间、温度和压力等参数。 ２）采样结束后，取下滤膜夹，用镊子轻轻夹住滤膜边缘，取下样品滤膜，并检查在采样过程中滤膜是否有破裂现象，或滤膜上尘的边缘轮廓不清晰的现象。若有，则该样品膜作废，需重新采样。确认无破裂后，将滤膜的采样面向里对折两次放入与样品膜编号相同的滤膜袋（盒）中。记录采样结束时间、采样流量、温度和压力等参数。 5、采样记录相关事项 环境空气质量采样记录包括：监测项目、样品批号、采样点位、采样日期、采样时间（开始、结束）、样品编号、气温、大气压、采样流量、采样体积、天气状况、风速、风向、采样人、审核人。 填写采样记录注意事项： 1）样品批号和样品种类一定要填写； 2）标况体积一定要计算正确； 3）发生异常情况，备注栏和附加说明处一定要填写清楚； 4）记录单上不能有涂改的痕迹，有错划掉，盖监测人印章。 6、样品转移、交接 工作结束后，现场监测人员应妥善保管原始记录，安全、规范运输样品，及时与样品管理员进行交接并填写交接记录。 二、采样工作中需要注意的事项 1、采样前检查气密性时要接干燥瓶，吸收瓶不能接以防倒吸。

压缩空气系统再确认方案

针剂车间压缩空气系统(运行、性能)再确认方案 编号Qua-01EM-005-R0-2019审批 程序 部门职务签名日期 起草验证工作小组组长 审核经理办公室生产副总经理办公室质量受权人设备科科长 生产技术科科长 针剂车间车间主任质管科 QA主管 QC主管 批准验证领导小组组长批复意见： 同意确认方案。 批准： 日期： 安徽金太阳生化药业有限公司

目录 1.概述 2.验证目的 3.范围 4.验证小组其责任 5.风险评估 6.确认内容 6.1确认所需文件 6.2确认用仪器仪表校验 6.3运行确认 6.4性能确认 7.偏差处理情况 8.验证结果数据汇总分析与评价 9.验证周期 10.批准

1.概述 公司小容量注射液车间压缩空气系统主要用于为洗瓶、灌封工序提供经除油、除水、除菌和净化处理的洁净工艺用气及为纯化水机组、蒸馏水机组、水浴式灭菌柜、纯蒸汽灭菌柜等设备上的气动元件提供气源。公司压缩空气系统主要由空气压缩机、空气储罐、冷冻式干燥机、多级别过滤器及使用点终端过滤器组成，生产出的洁净压缩空气通过不透钢管道，输送至车间各用气点。公司螺杆空气压缩机由上海德耐尔压缩机制造有限公司生产，型号为GA37P-7.5，排气量：7.3m 3/min ；最大工作压力：0.8Mpa ，压力露点为-40℃。冷冻式压缩空气干燥机为德耐尔压缩机制造有限公司生产，型号为DAD-15HTF ，处理量为18m 3/min 。微热再生吸附式压缩空气干燥机为德耐尔压缩机制造有限公司生产型号为DAD-15MXF 处理量18m 3/min 。管路系统由安徽仁和轻工机械有限公司完成，管路的材质均采用304不锈钢。洗瓶、灌封、起泡点试验用压缩空气均经过0.22μm 过滤器过滤。系统流程图如下： 洗瓶 HF7主管路过滤器 精度：1μm 灌封 0.22μm 过滤 0.22μm 过滤 水浴式灭菌柜 纯蒸汽灭菌柜 纯化水制备系统 空压机 7.3m 3/min 满载压力：0.8Mpa 压缩空气罐 容积2m 3 多效蒸馏水机 起泡点实验 0.22μm 除菌过滤 干燥机 18m 3/min 工作压力：1.0 冷干机 18m 3 /min 最大工作压力：1.3MPa 合成车间

压缩空气系统确认报告

报告编号：TS-70018-00 设备编码：4C008 项目负责人： 确认/验证领导小组审查汇签：

压缩空气系统确认报告 1．概述 从2015年08 月10日到2015 年08月20日对压缩空气系统（设备编码：4C008）进行了确认。在仔细总结、审核确认记录的基础上，根据确认方案的可接受标准，得出了确认报告。 4．批准 确认/验证领导小组已审阅上述所有检测结果及评价分析意见，确认该设备符合确认要求，同意投入使用。 5．建议 本确认方案建议两年后再进行确认。 批准人：日期：年月日

xx制药有限公司 压缩空气系统确认证书 证书编号：TS-70018-00 公司确认小组于2015年08 月10日到2015 年08月20日对压缩空气系统（设备编码：4C008）进行了确认，确认/验证领导小组已审阅该确认数据和结果，准予合格，特授予此证书，颁发执行，批准该设备投入使用。 质量受权人： xx制药有限公司 年月日

（附件）确认项目总结与分析 1.概述：螺杆式空气压缩机BOGE（上海）压缩机有限公司生产，安装于空调、压缩空气机房车间，用 于我公司固体车间、液体车间、前处理提取车间、搽剂车间药品生产及生产清洁吹扫。在确认/验证领导小组的统一组织和领导下，由质量部、生产技术部、工程设备部及其它相关部门的共同合作下，依据《压缩空气系统确认方案（TS-70017-00）》的要求，对《压缩空气确认方案（TS-70017-00）》中编制的确认项目进行了确认，现总结分析如下 2.确认/验证依据与参考文献 《药品生产质量管理规范（2010年修订）》 《药品生产质量管理规范实施指南（2010年修订）》 《药品生产确认指南（2003）》 《中华人民共和国药典（2010）》 《医药工业洁净室(区)沉降菌的测试方法（GBT16294-2010）》 《国际标准（ISO）生物污染控制（ISO14644）》 《国际制药工程协会基准指南(ISPE)》 《工业金属管道施工及验收规范（GB50235-97）》 《现场设备、工业管道焊接工程施工及验收规范（GB50236-98）》 《建筑给水排水及采暖工程施工质量验收规范（GB50242-2002）》 《压缩空气系统（固体、液体）确认方案（TS-700017-00）》 《压缩空气系统使用、维护保养及检修SOPSOP》 《压缩空气系统清洁SOP》 《螺杆式空气压缩机使用说明书》 3.确认目的： 压缩空气系统是生产中的重要设备，对其设计、安装、运行及性能予以确认，以确保药物质量，从而保证最终产品的质量。

压缩空气系统确认方案

压缩空气系统确认方案 目录 1验证方案审批 2概述 3目的 4范围 5 职责 6风险评估 7培训 8验证依据与标准 9验证内容 9.1设计确认（DQ） 9.2安装确认（IQ） 9.3运行确认（OQ） 9.4性能确认（PQ） 10偏差与处理 11结果评审 12再验证周期确认

13验证报告 14文件修订变更历史 15附件 江西庐山百草堂生物制药有限公司 确认/验证立项申请表 文件编号：-00 立项部门生产部申请日期年月日 立项题目压缩空气系统验证 新项目确认、确认/验证周期性确认、确认/验证变更确认、确认/验证非周期性确认/验证 前确认、确认/验证同步确认、确认/验证回顾性确认、确认/验证再确认、确认/验证

2. 概述 按照GMP的要求，设备在正式生产使用前，需要经过验证来证实所使用的设备能够达到设计要求及规定的技术指标，符合生产工艺要求，以便使所生产的产品符合预定的质量标准，从设备方面为产品的质量提供保证，因此需要对压缩空气系统进行设备验证。 本公司固体制剂生产所需的压缩空气系统为新购设备，本次确认为新购系统首次确认。该系统由一台WS-7508PV/PSV型（变频VSD）螺杆式空压机、SLAD-10NF常温风冷型冷冻式压缩空气干燥机、储气罐和过滤系统组成。本压缩机为无油型压缩机，其生产厂家为深圳寿力亚洲实业有限公司，流量可达12.5m3/min，压力0.8MPa，使用时压缩空气经过三级过滤器滤分别滤去水分、尘埃、杂质、尘油后，到各使用点使用. 本系统为全封闭结构，具有气量足压缩空气洁净，低噪音，振动小，重量轻，占地面积小，操作方便，易损件少，运行效率高，无需安装基础。压缩排出的气体进入贮气罐，然后气体分别经过主管路过滤器（HC级），冷冻式干燥机、油雾过滤器（HT级）、微油雾过滤器（HA级）对压缩空气进行干燥，除油，净化，使其达到干燥，无油、清洁、洁净度达到D 级的要求，净化后的压缩空气通过不锈钢分配管道输送到车间各用气点。保证其无油、无水、无菌、微粒数等指标符合规定，从而保证药品的质量。 2.1 螺杆压缩机压缩原理： 2.1.1第一步吸气过程：当电机驱动转子时，主、从转子的齿沟空间在转至进气端口时，其空间大，外界的空气充满其中，当转子的进气侧端面转离了机壳之进气口时，在齿沟间的空气被封闭在主、从转子与机壳之间，此为“吸气过程”完成。 2.1.2第二步封闭及输送过程：在吸气终了时，主、从转子齿峰与机壳形成的封闭容积随着转子角度的变化而减少并按螺旋状移动，此为“封闭与输送过程”

压缩空气系统再验证报告

压缩空气系统再验证报告再验证报告审批表 目录

1.验证组织系统 2.概述 3.验证目的 4.相关文件 5.验证范围 6.人员培训 7.验证内容 7.1压缩空气系统安装情况的稳定性检查 7.2运行确认 7.3性能确认 8 特殊情况处理 9再验证结果评定与结论 10文件执行 11文件归档 12附表 附表1：再验证方案变更申请表 附表2：压缩空气系统机组上仪器仪表校验记录附表3：压缩空气系统空调机组安装检查记录附件4：压缩空气系统运行确认检测记录 附表5：压缩空气系统油污检测记录 附表6：压缩空气系统尘埃粒子数检测报告 附表7：压缩空气系统微生物数检测记录 附表8：漏项、偏差处理表 附表9：压缩空气系统空气干燥检测记录 1验证组织系统

1.1.1验证委员会成员及其职责 1.1.2验证委员会职责 主任：负责验证方案、验证报告的批准；负责签发验证证书。 委员：审核验证方案、验证报告，制定验证计划。 1.2验证小组成员及其职责 1.2.1系统验证小组成员 1.2.2各成员职责 组长——负责验证实施全过程的组织协调工作； 组员——负责验证过程中的具体工作，并做好记录工作。 1.2.3验证过程中各相关部门职责 1.2.3.1质量管理部： 负责组织验证方案、报告与结果的会审会签；负责对验证全过程实施监控；负责核查、汇总验证数据；负责建立验证档案，及时将批准实施的验证资料收存

归档。 1.2.3.2生产技术部 负责指导车间相关人员做好验证记录。 1.2.3.3设备工程部 负责提供设备相关文件；负责编制设备使用标准操作规程、维护标准操作规程及清洁规程。 1.2.3.4化验室 负责验证过程的取样、检验及结果报告。 1.2.3.5综合制剂车间 负责设备所在操作间的清洁处理，保证运行环境符合设计要求； 负责协助验证小组保证验证工作顺利进行。 2概述 压缩空气系统是药品在生产过程中环境的控制点，所以压缩空气系统供给符合要求的空气是保证产品质量的重要方面。本次验证是再次确认车间使用一定时期后的压缩空气系统的各项性能是否符合规定，能否满足车间生产要求。 2.1工艺流程简述 车间使用的压缩空气是由空压机组供给，要求无水、无油、无菌。空压机组由空气压缩机、空气贮罐及冷干机、过滤器组成。 2.2压缩空气系统流程图 压缩空气 通过对螺杆式空气压缩机进行再验证，确认其经过长期运行后，能否稳定地为药品生产提供符合工艺要求的压缩空气。 4相关文件

压缩空气系统验证报告

福建太平洋制药有限公司GMP 文件 文件名称 压缩空气系统验证报告 文件编码 STP(R)-YZ-01-001-00 页 码 第1页，共10页 起草 人 审 核 人 QA 审核 起草日期 审核日期 审核日期 批 准 人 颁发部门 质量管理部 执行部门 各职能部门 批准日期 颁发日期 生效日期 分发部门 验证小组、质量管理部 1 概述 本压缩空气系统由预处理系统连接管路至车间各用气点构成。预处理系统位于制剂大楼二楼空调机房内，主要有LS16-75H 固定式螺杆空压机、储气罐、主管路过滤器、冷冻式压缩空气干燥机、压缩空气精密过滤器等设施，这些设施于2011年5月采购，2011年10月进行安装；连接管路及阀门全部采用304L 不锈钢材质，并且双面抛光。系统为工艺生产气动设备及仪表的使用而提供无油无水的干燥空气，空压机排出的压缩空气，首先经过主管路过滤器，过滤粒径为1μm ，然后经过冷冻式压缩空气干燥机除去水份，最后分别再经过二台0.01μm 的压缩空气精密过滤器，保证了压缩空气质量满足GMP 生产要求。 验证小组根据验证方案中法规要求和风险分析结论，对压缩空气系统进行了设计确认（DQ ）、安装确认（IQ ）运行确认（OQ ）、性能确认（PQ ）。现对验证过程中所取得的数据进行收集和整理报告如下： 2 验证范围 检查并确认验证范围包括 （1）压缩空气系统的设计确认（DQ ） （2）压缩空气系统的安装确认（IQ ） （3）压缩空气系统的运行确认（OQ ） （4）压缩空气系统的性能确认（PQ ） 3 验证目的 3.1 检查并确认该系统所有设备所用材质、设计、制造及制造厂家符合GMP 和本公司生产工艺的要求。 3.2 检查该系统的文件资料齐全且符合GMP 要求；检查并确认设备的安装符合生产

压缩空气系统再验证方案.

压缩空气系统再验证方案起草人 年月日 审核人年月日 批准人年月日 共 7 页 1主要内容与适用范围 1.1再验证概述 本次验证属再验证，该系统设备在使用期间未作任何改动，设备的操作规程未作修订，所以本次验证只对该系统做性能验证，并按《中国药典》2005年版压缩空气系统质量标准及GMP要求进行各项指标的验证。 1.2验证目的 ·检查并确认L-15压缩空气系统的各项性能指标符合设计要求。 ·检查并确认L-15压缩空气系统运行是否符合GMP尘埃粒子数和微生物限度的标准，油、水含量是否达到要求。 1.3验证范围 本方案适用于压缩空气系统验证。 2 引用标准

《中国药典》（2005） 《药品生产验证指南》（2003年修订） 《KGS10型液体自动计量灌装加塞机标准操作规程》 3术语 无 4 职责 4.1验证项目小组 组长：设备能源部部经理 成员：设备主管、QA人员、岗位操作人员、QC人员 4.2职责 4.2.1验证项目小组组长 负责本验证的全部工作，根据验证计划部署实施方案 检查督促、协调验证工作进度及完成质量 负责验证方案及验证报告的审核 4.2.2成员 设备主管负责验证方案的起草及具体实施工作，对验证过程的技术质量负责设备部负责保证在每次验证过程中严格执行批准生效的相关规程

QA负责按标准操作规程及验证方案规定的取样计划取样，并及时报告检验 结果 岗位操作人员负责本验证步骤的操作 QC负责样品的检验并出据相应的报告 5 验证程序及内容 5.1验证前的准备 ·人员培训 ·检查并确认操作人员的培训记录和对设备、SOP的操作熟练程度。 ·该项检查的具体内容和标准、结果见附表1 5.2验证的必要性 压缩空气系统在使用前应进行系统的验证工作，以确认设备符合生产要求，设备运行一段时间后，或者大修，或者其他原因往往会造成设备参数的飘移，影响设备的正常使用，因此有必要对设备进行回顾性验证。 5.3验证内容 性能确认 5.3.1性能确认的目的：检查并确认L-15压缩空气系统运行是否符合GMP尘埃粒子数和微生物限度的标准，油、水含量是否达到要求 5.3.2试验条件：在符合压缩空气系统的环境条件下试验。 5.3.3试验方法。

压缩空气系统验证方案(最终)

压缩空气系统验证方案 验证方案编号： 设备（系统）名称：压缩空气系统 设备（系统）编号： ******制药

证小组成员名单项目主管： 小组成员：

目录 1 概述 2 目的 3 范围 4 职责 4.1 验证领导小组 4.2 验证工作小组 4.3 生产部 4.4 质监部 5 验证实施的步骤和要求 5.1 验证依据及标准 5.2 预确认 5.3 安装确认所需文件资料确认及仪器仪表确认5.4 安装确认 5.5 运行确认 5.6 性能确认 6 结果分析与评价 7 验证记录、验证项目有关记录表格

l 概述 洁净压缩空气系统为制剂车间公共设施。主要设备有FHOG75A型蜗杆式空气压缩机、气水分离器、2.5m3空气储罐、SAD冷冻式干燥机、LY型高效除油器、A级过滤器等组成洁净压缩空气系统的装置。该系统按车间工艺布置图安装于车间二楼空压机室内。为固体制剂车间湿法混合制粒机、沸腾制粒干燥器、高效沸腾干燥机、高效包衣机、铝塑包装机等设备提供洁净压缩空气。 FHOG75A型蜗杆式空气压缩机系统流程将空气经过空气滤清器滤去尘埃、杂质，由减荷阀控制进入压缩机工作腔，随着蜗杆与两侧星轮片的合运动，空气被压缩，并在压缩过程开始时与喷入的润滑油混合，经压缩后的混合气体进入油气分离器，利用旋风分离法和上返分离法粗分离油气后，经精分离器滤芯进行精分离、通过最小压力阀排出的气体是比较纯净压缩空气；然后经过板翘式冷却器，将压缩空气冷却，空气中水蒸汽饱和析出，与压缩空气一起排出。 空气压缩机排出的压缩空气还含有一些的水分、油气和杂质，所以增加了一套后处理设备，包括：气水分离器（除水）、冷冻干燥机（进一步处除水）、高效除油器（除油）、A级精密过滤器（除尘、进一步除油）等。经过处理后的压缩空气常压露点≤-23℃；含油量≤0.01ppm；固体尘≤0.01μm。 本系统采用自动控制系统，操作方便。 FHOG75A型蜗杆式空气压缩机： 容积流量（排气量）：10m3／min 排气压力（表压）：0.55～0.7MPa 电机转速： 2970r/min

压缩空气系统验证方案

压缩空气系统验证方案 Company number：【0089WT-8898YT-W8CCB-BUUT-202108】

压缩空气系统验证方案 设备名称：压缩空气系统 设备型号： 设备编号：JD-0204-004 制造厂商： 安装位置： 验证方案编号： 目录 一、概述 (4) 二、目的 (4) 三、范围 (4) 四、压缩空气的组成及流程 (4) 五、验证依据和文件 (5) 六、人员职责及人员培训 (5) 七、风险评估 (6) 八、验证计量确认 (9) 九、性能确认 (9) 十、偏差处理 (11) 十一、变更控制 (11) 十二、验证结论 (12) 十三、再确认周期 (12) 十四、验证结论 (12)

验证方案起草审批方案起草 方案审核 方案批准 验证小组名单及职责

1.概述 本压缩空气系统是按照GMP要求设计、安装的压缩空气气源，由两台阿特拉斯·科普柯型固定式螺杆压缩机、一台冷冻式空气干燥机、一级P级精密过滤器、二级S级精密过滤器、一个的缓冲罐和无缝钢管输气管道组成。其基本流程是：将自然空气经固定式螺杆空气压缩机压缩，经缓冲罐、一级P级精密过滤器，再使用冷冻式干燥机将其除湿干燥，然后通过二级S级精过滤器得到无油、无水、无尘的压缩空气，经过无缝钢管输气管道，输送至车间各用气点，与药品直接接触各用气点再经μm过滤器过滤，压缩空气符合药品生产要求。 2、目的 确认系统生产的压缩空气性能达到使用标准 3.范围 对本厂区内接触药品内包材的压缩空气用气点进行性能确认。 4.压缩空气组成及流程 压缩空气系统设备一览表 净化区压缩空气用气点一览表：

环境空气采样规范

欢迎阅读 环境空气采样作业指导书 1.采样工作流程 1.1监测项目调查 现场监测人员认真了解监测对象的生产设备、工艺流程，清楚主要污染源、主要污染物及其排放规律，查看环保措施落实和环保设施运行情况。监控生产负荷，调查现场环境（气象、水温、污染源）有关参数和周边环境敏感点，检查监测点位符合性及安全性，搜集与编制监测报告有关的各种技术资料并做好相关记录。 1.2实验室采样前准备 1.3 1.4. 1.5 1.5.1 1.5.2 1.6 结束）、 1.7样品转移、交接。 工作结束，现场监测人员应妥善保管原始记录，安全、规范运输样品，及时与样品管理员进行交接并填写交接记录。 2.采样工作中的注意事项 2.1采样前检查气密性时要接干燥瓶，吸收瓶不能接以防倒吸。 2.2采样结束后，取下样品，将气体吸收装置进、出口密封，按相应项目的标准监测分析方法要求运送和保存待测样品。 2.3用超细玻璃纤维滤膜采样时，应对光线检查滤膜是否有损坏，如有损坏，停止使用。 2.4采集气体样品时，注意吸收瓶溶液的颜色，如果未采样已变色，则该样品作废。

2.5现场空白样的放置：启动采样气路时，同时将空白样的吸收瓶封口膜打开，气路采样结束时，同时将空白样封口；准备空白滤膜装入切割头中放置在空气中，采样后结束后按照滤膜采样同样方法放入滤膜袋中，运回实验室检测，空白滤膜前后两次称量质量之差应远小于采样滤膜上的颗粒物负载量，否则次批次采样监测数据无效。 2.6向采样器中放置和取出滤膜时，应佩戴聚乙烯手套等实验室专用手套（和实验室人员称量滤膜所带的手套相同），使用无锯齿状镊子。 2.7采样进气口必须暴露在空气中（箱体盖子可以不盖，几乎不影响吸收液温度）。 2.8夏天仪器应尽量避免放置在太阳下暴晒，以防止吸收液蒸发，可将仪器放置于树荫处或适当遮盖。 2.9 2.10 2.11 2.12 3. 3.1 3.2 1 ? ?2 ? ?3 ? 3.2.2 除排放污染物的种类和排放速率（估计值）之外，还应重点调查被测无组织排放源的排出口形状、尺寸、高度及其处于建筑物的具体位置等，应有无组织排放口及其所在建筑物的照片。 3.2.3排放源所在区域的气象资料调查一般情况下，可向被测污染源所在地区的气象台（站）了解当地的“常年”气象资料，其内容应包括： 1）按月统计的主导风向和风向频率； 2）按月统计的平均风速和最大、最小风速； 3）按月统计的平均气温和气温变化情况等。? 如有可能，最好直接了解当地的逆温和大气稳定度等污染气象要素的变化规律。了解当地“常年”气象资料的目的，是为了对监测时段的选择作指导。

压缩空气验证

注射针制剂用压缩空气系统的验证方案 一. 压缩空气系统概述 1.1. 制备过程 由于注射针制剂用的压缩空气直接与物料接触, 因此必须控制压缩空气中的油、水、固体微粒和生物粒子(细菌) 等, 本系统设计中已考虑药品对压缩空气的要求, 其制备过程为选用水循环无油式空压机, 杜绝了润滑油带入空气, 经空压储罐、主管过滤器(1μm)对干燥之前压缩空气中的污染物(油、液态水、尘) 进行有效地过滤后, 进入无热再生压缩空气干燥器对气体进行干燥除水, 再经0.01μ精密过滤器除去99.99%游离状态的水份、油份、过滤粒径≦1μm, 再通过活性炭过滤器吸附除去气体中的臭气、异味, 最后通过0.2μ除菌过滤器以得到无菌的压缩空气供水针剂生产使用。为保证使用为无菌压缩空气, 对无菌过滤器及其至使用点的压空管道定期用蒸汽灭菌(121℃，30min), 对蒸汽的进入采用蒸汽过滤器过滤后使用。 1.2. 工艺流程方图框 1.3. 设备一览表 表1 空压系统设备一览表

二. 压缩空气系统的验证方案 1. 验证目的 通过对水循环无油式空压机为主的空压系统的验证, 从而确认其生产能力及其性能在任何条件下始终满足设计能力和无菌生产工艺的要求。 2. 安装确认 2.1.准备工作 在设备到位开箱后, 设备管理人员或验证检查人员必须检查设备厂家提供的设备图纸、手册、备件清单、操作说明书、维修、保养指南等技术资料。在熟悉这些资料的基础上进行质量检查。 2.2. 空气压缩机及其系统(含管道)安装质量检查 在设备按工艺流程要求安装后, 对照设计图纸及供货商提供的技术参数, 检查设备及其管道安装是否符合设计和《规范》。具体检查技术要求见表2 表2 空压系统设备及其管道安装质量检查表

洁净压缩空气的验证报告(模板)

洁净压缩空气的验证报告 文件编号: 版本号 : 实施部门：……部 审核： 批准： 验证时间：……年……月……日～……年……月……日

目录 1.概述 2.验证目的 3.验证所需的相关文件 4.验证的内容及过程 4.1预确认 4.2安装确认 4.3运行确认 4.4性能确认 5. 结果分析与评价 6.再验证周期的确定 7.验证时间的安排 8.验证结果及批准

1.概述 1.1洁净压缩空气系统采用空气压缩机产生压缩空气，经过冷冻干燥机去处水分，通过 三级空气过滤去除粒、油分，达到洁净空气净化，并在使用点终点根据需要安装除菌 过滤器。使用压缩空气的洁净度等合工艺用气的要求。 1.2系统工艺流程 空气压缩机储气罐冷冻干燥粗滤精滤除菌过滤各使用点 1.3本方案仅适用于洁净压缩空气系统的验证。 2.验证目的 2.1对空压系统的设计及本型号设备的可靠性进行评估。 2.2对空压系统的设备、管道安装能否达到生产工艺要求作出确认。 2.3通过对空压机所提供的压缩空气检测，以评价空压系统的产气量能否满足生产要求；通过对过滤装置过滤后的空气检测，以确定安装的合理性和适用性；确定过滤后 的压缩空气无油、无尘，微粒在规定范围内，空气洁净度达到相应级别净化要求；过 滤装置的过滤效果达到生产工艺所规定的要求。 3.验证所需的相关文件 序号文件名称 1 xxxx 空气压缩机使用说明书 2 xxxx 空气压缩机电脑控制器用户手册 3 压力容器产品质量保证书

4.验证的内容及过程 4.1确认 4.1.1工艺设计对设备的要求 能连续不断地为气动生产设备及通气检验提供稳定的洁净的气源；并能根据空气的使用情况自动调节产气量，保证工作气源压力稳定可靠；过滤后的空气符合相应洁净级别的要求。 4.1.2系统配置情况 检查空气压缩机、储蓄罐、过滤器管道等系统配置是否符合生产工艺要求？ 4.1.3售后维修服务 维修服务单位：xxxx 有限公司 详细地址： 联系人： 联系电话：