I am having trouble understanding the relationship between a frequency function and it's Inverse Fourier transform.

The Frequency function is

$$\frac{1+0.8(e^{-j 2\pi f}+e^{j 2\pi f})+0.64}{1+1.4\cos(2\pi f)+0.49}$$

And the inverse Forier transform is $$\frac{(-0.7)^{|n|}}{1-0.49}+\frac{0.64(-0.7)^{|n|}}{1-0.49}+\frac{0.8(-0.7)^{|n-1|}}{1-0.49}+\frac{0.8(-0.7)^{|n+1|}}{1-0.49}$$

  • $\begingroup$ Where did you get these equations from? Did you calculate them yourself, or are they from a book, or somewhere else? What is it exactly that you aren't understanding? $\endgroup$ – lxop Dec 21 '16 at 21:03
  • $\begingroup$ What is $\tau$ in the IFT? $\endgroup$ – MBaz Dec 22 '16 at 0:59

Consider the signal $$s[n]=\frac{a^{|n|}}{1-a^2},\, |a|<1$$

Assuming $$x[n]=a^n u[n]\stackrel{\text{DTFT}}\longleftrightarrow\frac{1}{1-a e^{-j\omega}},$$ one can show that


The proof is simple. For example, for $n>0$ we have $$\sum_{k=n}^{\infty}x[k]x[k-n]=\sum_{k=n}^{\infty}a^ka^{k-n}=\sum_{k=n}^{\infty}a^{2k-n}=a^n(1+a^2+a^4+\cdots)=\frac{a^n}{1-a^2}$$

Hence, the DTFT of the signal $s[n]$ is given by $$S(e^{j\omega})=X(e^{j\omega})X(e^{-j\omega})=\frac{1}{1-a e^{-j\omega}}\frac{1}{1-a e^{j\omega}}=\frac{1}{1-2a\cos(\omega)+a^2}$$ In summary, $$\boxed{\frac{a^{|n|}}{1-a^2}\stackrel{\text{DTFT}}\longleftrightarrow\frac{1}{1-2a\cos(\omega)+a^2},\, |a|<1}$$

Now consider

$$G(f)=\frac{1}{1+1.4\cos(2\pi f)+0.49}=\frac{1}{(1+0.7e^{-j2\pi f})(1+0.7e^{j2\pi f})}$$

whose IDTFT can be calculated using the above formula for $a=-0.7$ as


Since $$\frac{1+0.8(e^{-j 2\pi f}+e^{j 2\pi f})+0.64}{1+1.4\cos(2\pi f)+0.49}=G(f)+0.8e^{-j 2\pi f}G(f)+0.8e^{j 2\pi f}G(f)+0.64G(f)$$ The IDTFT is $$g[n]+0.8g[n-1]+0.8g[n+1]+0.64g[n]$$ But obviously, the variable $\tau$ is used instead of $n$ in the give equation.


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