Essa derivação é complicada. A abordagem sugerida anteriormente tem uma falha. Deixe-me demonstrar isso primeiro; então eu darei a solução correta.
Desejamos relacionar a transformação do sinal de baixa amostragem, Y D ( z ) = Z { x [ M n ] } , à transformação Z do sinal original X ( z ) = Z { x [ n ] } .ZYD(z)=Z{x[Mn]}ZX(z)=Z{x[n]}
O caminho errado
Pode-se pensar em simplesmente conectar a expressão para o sinal de amostragem reduzida na expressão da transformação :Z
YD(z)=∑n=−∞+∞x[Mn]z−n
Uma mudança da variável parece óbvia:n′=Mn
YD(z)=∑n′∈MZx[n′]z−n′/M
No entanto, é importante perceber que, embora o novo índice de somatório ainda funcione de - ∞ an′−∞ ,a soma está agora acima de 1 dentre M números inteiros. Em outras palavras,∞
,n′∈MZ={...,−2M,−M,0,M,2M,...}
enquanto a definição da transformação requerZ
.n∈{...,−2,−1,0,1,2,...}
Como isso não é mais uma transformação , não podemos escrever:Z
YD(z)=X(z1/M)
O caminho certo
Vamos primeiro definir um sinal de trem de impulso 'auxiliar' como:tM[n]
tM[n]=∑k=−∞+∞δ[n−kM]={10:n∈MZ:n∉MZ
This function is 1 at one out of every M samples, and zero everywhere else.
Equivalently, the pulse train function can be written as:
tM[n]=1M∑k=0M−1ej2πkn/M
Proof: We need to consider separately the cases n∈MZ and n∉MZ:
tM[n]=1M∑k=0M−1ej2πkn/M=⎧⎩⎨⎪⎪⎪⎪1M∑k=0M−111M1−ej2πkn1−ej2πkn/M:n∈MZ:n∉MZ={1MM1M1−11−ej2πkn/M:n∈MZ:n∉MZ={10:n∈MZ:n∉MZ
In the case
n∉MZ, we used the expression for
the finite sum of a geometric series.
Now let's go back to our original problem of finding the Z-transform of a downsampler:
YD(z)=∑n=−∞+∞x[Mn]z−n
n′=Mn
YD(z)=∑n′∈MZx[n′]z−n′/M
n∈Z
YD(z)=∑n=−∞+∞tM[n]x[n]z−n/M
Using the above formulation for the impulse train function as a finite sum of exponentials, we get:
YD(z)=∑n=−∞+∞(1M∑k=0M−1ej2πkn/M)x[n]z−n/M=1M∑k=0M−1∑n=−∞+∞ej2πkn/Mx[n]z−n/M=1M∑k=0M−1∑n=−∞+∞x[n](e−j2πk/Mz1/M)−n
The summation on the right is a summation over all integers, and is therefore a valid Z-transform in terms of z′=e−j2πk/Mz1/M. Therefore, we can write:
YD(z)=1M∑k=0M−1X(e−j2πk/Mz1/M)
This is the formula for the Z-transform of a downsampler.
I've not seen this notation before. However, it does seem to make sense. TheM -downsampler is defined by the equation:
Itsz transform is defined by the equation:
Apply a change of variable, lettingn′=Mn . The ranges of the summation are unaffected by the change of variable since they extend to infinity.
This looks similar to thez transform of x[n] itself. Recall that it is defined as:
By inspection, we can therefore conclude the following relationship between thez transforms of x[n] and yD[n] :
Therefore, thez transform of the downsampler output is closely related to the z transform of the input signal, which is to be expected. In the frequency domain, this results in an M -fold stretching of the signal's frequency content.
But how do you go from the above equation to the one you referenced in the paper? It gives a definition ofYD(z) in terms of z only, while the expression we derived is a function of z1/M . So for a particular value of z that you would like to evaluate YD(z) at, you would first calculate z1/M (i.e. take the M -th root of z ) and then substitute that into X(z) . However, all nonzero z∈C have M distinct M -th roots:
whereWk is the DFT kernel value ej2πk/M referenced in your question, and rp is what I define to be the principal M -th root of the complex value z :
That is,z 's principal M -th root rp is obtained by converting z to polar form, taking the M -th root of z 's magnitude (which is a real number), and dividing z 's angle by M . The resulting values express rp in polar form.
Why go to all of this trouble? Because, as I noted before, the mapping fromYD(z) 's domain to the domain of X(z1/M) is not one-to-one. I'll now begin some handwaving. For any particular value of z that you would like to evaluate YD(z) for, there are M corresponding points in X(z1/M) that you could map to. Therefore, each of those M points in X(z1/M) contribute to the corresponding value of YD(z) . You then end up with a sum like that shown in the paper:
whererp(z) refers to the principal M -th root calculation I showed earlier. In reality, you could pick any of z 's M -th roots as the principal one; I picked this definition because it's the most straightforward. If you were to properly and rigorously derive this relationship, I believe the factor of 1M comes in because of a derivative of z1/M .
In mathematician-speak, I believe this would be referred to as a composition of functions;YD(z)=f(g(z)) , where f(z)=X(z) and g(z)=z1/M . In order to unroll the function composition and write YD(z) as a function of z only, you would chop the domain of YD(z) into chunks that are one-to-one, invert the function over those intervals, and then sum the results with appropriate scaling factors. I've used this technique before to calculate the probability distribution function of a function of a random variable given the original random variable's pdf (e.g. to derive the pdf of X−−√ given X 's pdf), but the name of the technique escapes me.
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