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

Question

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.

Answer 1

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).

Answer 2

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