I am pretty new to inverse Fourier transforms and I would like to ask a question. Does anyone know how to bring back to the "sequence domain" this relation below? In other words, get the inverse DTFT of this expression? (Given that C, D and $a$ are constants and you know them):

$$ \frac{C \cdot e^{-j \omega}+D}{(1 - a \cdot e^{ j \omega })^2} $$

  • $\begingroup$ Hey! No its not a typo. $\endgroup$
    – Taiwaninja
    Sep 15, 2017 at 18:11
  • $\begingroup$ set $z = e^{j \omega}$ and see what you get. $\endgroup$ Sep 15, 2017 at 18:30
  • $\begingroup$ robert, if I use z instead, may I use properties from z transform without any problems? $\endgroup$
    – Taiwaninja
    Sep 15, 2017 at 18:40
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    $\begingroup$ Note that there's not only one correct solution but two. Depending on the value of $\alpha$ you get two different sequences. For $\alpha>1$ you get a right-sided sequence, and for $\alpha<1$ you get a left-sided sequence. $\endgroup$
    – Matt L.
    Sep 16, 2017 at 19:19

1 Answer 1


There seems to be a mistake in your derivations, the following code shows the correct one for $|\alpha| < 1$ , as indicated by @Matt.L $:

N = 128;         % simulation length (choose large enough)
n = -N : 2;      % simulation time span

C = 1.5;          % set some values
D = 2.78;
a = 0.33;         % a is for alpha

% x = -(n-2).*C.*(a.^(-n+1)).*stepfun(-n+2,0) + ...
%    -(n-1).*D.*(a.^(-n)).*stepfun(-n+1,0);  

x = zeros(1,130);  % use the explicit code as stepfun() has problem with n.
for n = -N:1
    x(n+N+1) = -C*(n-2)*(a^(-n+1)) - D*(n-1)*(a^(-n));

M=2000;          % Select a FFT length for demonstration purposes
F=fft(x,M);      % compute the DFT of x[n] so obtained

w = linspace(0,2*pi-2*pi/M,M);              % choose DFT frequency samples
H = (C*exp(-1j*w)+D)./((1-a*exp(1j*w)).^2); % Evaluate the given DTFT

figure,plot(w,abs(H));title('H'); % compare the magnitudes of DTFTs
figure,plot(w,abs(F));title('F'); % Phases will be different as this
                                  % is an acausal sequence.

Derivation of the result for the left-sided sequence $|\alpha|< 1$ case) is as follows:

Intuition plays a central for getting the answer as clean as possible which involves inverse Fourier transforms. In this example we shall use those three fundamental Fourier transform properties.

1- $x[n] \leftrightarrow X(\omega) \longrightarrow y[n]=x[-n] \leftrightarrow Y(\omega) = X(-\omega) $

2- $x[n] \leftrightarrow X(\omega) \longrightarrow y[n]=nx[n] \leftrightarrow Y(\omega) = j \frac{d X(\omega)}{d\omega} $

3- $x[n] \leftrightarrow X(\omega) \longrightarrow y[n]=x[n-d] \leftrightarrow Y(\omega) = e^{-j \omega d} X(\omega)$

Begin by $$x[n] = a^n u[n] \longleftrightarrow X(\omega) = \frac{1}{1 - a e^{-j \omega} }$$

Let $y[n]=x[-n]=a^{-n} u[-n]$ then $Y(\omega)= \frac{1}{1-a e^{j \omega}}$

Apply second rule: $z[n] = n y[n] = n a^{-n} u[-n]$ then $Z(\omega) = \frac{-a e^{j \omega}}{(1-a e^{j \omega})^2}$

Apply linearity: $ w[n] = -a^{-1} z[n] = - n a^{-1} a^{-n} u[-n]$ then $ W(\omega) = \frac{e^{j\omega}}{(1-a e^{j \omega})^2}$

Apply shift property on $w[n]$ to get rid of the numerator exponential. $v[n] = w[n-1] = - (n-1) a^{-1} a^{-(n-1)} u[-(n-1)]$ then $ V(\omega) =\frac{1}{(1-a e^{j \omega})^2}$

Now apply shift property and linearity to $v[n]$ such that the resulting DTFT will be $$H(\omega) = \frac{ C \cdot e^{-j \omega} + D} {(1-a e^{j \omega})^2} = \frac{ C \cdot e^{-j \omega}} {(1-a e^{j \omega})^2} + \frac{ D} {(1-a e^{j \omega})^2}$$

which implies that $$h[n] = C v[n-1] + D v[n] $$ hence we get $h[n]$ as:

$$h[n] = C \left( - ((n-1)-1) a^{-1} a^{-((n-1)-1)} u[-((n-1)-1)] \right) - D \left( (n-1) a^{-1} a^{-(n-1)} u[-(n-1)] \right) $$

simplify to get the result: $$ h[n] = C (-n+2) a^{-n+1} u[-n+2] + D (-n+1) a^{-n} u[-n+1] $$

  • 1
    $\begingroup$ \@Fat32 thank you very much. Now it makes sense, I could not duplicate the result as it was before but now I get it. Thank you so much! $\endgroup$
    – Taiwaninja
    Sep 15, 2017 at 22:53
  • $\begingroup$ @Taiwaninja now there is also a derivation... $\endgroup$
    – Fat32
    Sep 16, 2017 at 20:39
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    $\begingroup$ @Taiwaninja Nice to see that I can help! Please don't hesitate to upvote and select the answer if it helped you, as this will help other people trust the answer as well and moreover the question will be closed properly... $\endgroup$
    – Fat32
    Sep 18, 2017 at 9:50

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