# Inverse DTFT of left-sided sequence

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):

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

• Hey! No its not a typo. Sep 15 '17 at 18:11
• set $z = e^{j \omega}$ and see what you get. Sep 15 '17 at 18:30
• robert, if I use z instead, may I use properties from z transform without any problems? Sep 15 '17 at 18:40
• 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. Sep 16 '17 at 19:19

There seems to be a mistake in your derivations, the following code shows the correct one for $|\alpha| < 1$ , as indicated by @Matt.L $: N = 128; % simulation length (choose large enough) n = -N : 2; % simulation time span C = 1.5; % set some values D = 2.78; a = 0.33; % a is for alpha % x = -(n-2).*C.*(a.^(-n+1)).*stepfun(-n+2,0) + ... % -(n-1).*D.*(a.^(-n)).*stepfun(-n+1,0); x = zeros(1,130); % use the explicit code as stepfun() has problem with n. for n = -N:1 x(n+N+1) = -C*(n-2)*(a^(-n+1)) - D*(n-1)*(a^(-n)); end M=2000; % Select a FFT length for demonstration purposes F=fft(x,M); % compute the DFT of x[n] so obtained w = linspace(0,2*pi-2*pi/M,M); % choose DFT frequency samples H = (C*exp(-1j*w)+D)./((1-a*exp(1j*w)).^2); % Evaluate the given DTFT figure,plot(w,abs(H));title('H'); % compare the magnitudes of DTFTs figure,plot(w,abs(F));title('F'); % Phases will be different as this % is an acausal sequence.  Derivation of the result for the left-sided sequence$|\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: $$h[n] = C (-n+2) a^{-n+1} u[-n+2] + D (-n+1) a^{-n} u[-n+1]$$