I am using Digital Signal Processing Principles, Algorithms, and Applications 4th edition and by Proakis. Here is what I don't understand

say the signal $x_a(t)$ has its Fourier Transform $X_a(F)$ and digital signal $x(n)=x_a(nT)$ which is $x_a(t)$ after sampling with sampling frequency $F_s$ has its DTFT $X(f)$

since $$x_a(t)=\int^{\infty}_{-\infty}X_a(F)\,e^{j2\pi Ft}dF\tag{1}$$

because $x(n)=x_a(nT)$ and $nT=\dfrac{n}{F_s}$ $$x(n)=x_a(nT)=\int^{\infty}_{-\infty}X_a(F)e^{j2\pi nF/F_s}dF\tag{2}$$

Using inverse DTFT for $x(n)$ then

$$\int^{\frac12}_{-\frac12}X(f)e^{j2\pi nf}df=\int^{\infty}_{-\infty}X_a(F)e^{j2\pi nF/F_s}dF\tag{3}$$

for $f=\dfrac{F}{F_s}$ then

$$\int^{\frac12}_{-\frac12}X(f)e^{j2\pi nf}df=\frac{1}{F_s}\int^{\frac {F_s}{2}}_{-{\frac {F_s}{2}}}X(F)e^{j2\pi nF/F_s}dF\tag{4}$$

That is what the book says and it use that to explain the reconstruction process in theory. However isn't it will be like this according to substitution in integral

$$\int^{\frac12}_{-\frac12}X(f)e^{j2\pi nf}df = \frac{1}{F_s} \int^{\frac {F_s}{2}}_{-{\frac {F_s}{2}}}X\left(\frac{F}{F_s}\right)e^{j2\pi nF/F_s}dF\tag{5}$$

I think this should be in Math Section but since it is about Digital Signal so I post it here.

  • 1
    $\begingroup$ Please check if you have transcribed this correctly. The first two integrals are missing an integration variable (probably dF ?). The first equation shouldn't have an "n" in it. I think. The second probably should be x(nT) where T is 1/Fs and instead of x(nt). $\endgroup$
    – Hilmar
    Dec 14, 2015 at 4:12
  • $\begingroup$ yes, thank you. I wrote it when I was about to go to bed so there're many mistake $\endgroup$
    – aukxn
    Dec 14, 2015 at 19:44
  • $\begingroup$ I use 4th edition but I don't want to downvote any answers. Read both edition and making comparison is waste of time. It seems like it's an error or author intend to use that notation without any specific purpose $\endgroup$
    – aukxn
    Dec 16, 2015 at 16:14

2 Answers 2


You are absolutely right. With $f=F/F_s$ you obtain the equality

$$\int_{\frac12}^{\frac12}X(f)e^{j2\pi nf}df=\frac{1}{F_s}\int_{-F_s/2}^{F_s/2}X\left(\frac{F}{F_s}\right)e^{j2\pi nF/F_s}dF\tag{1}$$

which is also what I find in my third edition of the book (the right-hand side of $(1)$ equals the left-hand side of Eq. $(4.2.81)$ in my edition).


The idea is to get the spectrum of continuous time signal from DTFT co-efficients as $$X_a(F) = \dfrac{1}{F_s}X\left(\dfrac{F}{F_s}\right)$$


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