Expressing DTFTs in terms of one another with similar time domain signals

I cam across a question in my DSP book asking this:

1). Express $$X_2(e^{j\omega})$$ in terms of $$X_1(e^j\omega)$$ without explicitly computing $$X_1(e^{j\omega})$$. ($$X_1(e^j\omega)$$ represents the DTFT of $$x_1[n]$$)

\begin{align} x_1[n] &= \left[1 \ 2 \ 2 \ 1\right]\\ x_2[n] &= \begin{cases} x_1[n] & 0 \leq n \leq 3\\ x_1[n-4] & 4 \leq n \leq 7\\ 0 & \text{otherwise}\end{cases} \end{align}

Here is my code and answer, but Im dont think my answer is correct. What should I be looking for in the time domain graphs that can give clues to the DTFT? ( i know my dtft function needs some work but I will do that next)

Thanks

%% P3.2
% 1). The magnitude plot of X2(jw) should have a similar shape to X1(jw) because
%     they contain the similar frequency components, but part of X2 is just phase
%     shifted, therefore the angle plots might be a bit different.
% 2).

n1 = 0:3;
x1 = [1 2 2 1];
stem(n1,x1); axis([0 7 -4 4]);

n2 = 0:7;
x2 = [1 2 2 1 1 2 2 1];
%    ( x1[n])(x1[n-4])
figure;
stem(n2,x2); axis([0 7 -4 4]);

k = 0:200;

X1 = dtft(n1,x1,k);

X2 = dtft(n2,x2,k);

function X = dtft(n,x,k)
% DESCRIPTION
%  Performs an approximate Discrete Time Fourier Transform
%  the approximation is due to the fact that the output is
%  not continous. ex. X = dtft(n,x,k);
%
% NOTE : Only for finite duration signals.
%
% INPUT VARIABLES
%  n = number of samples
%  x = array containing time domain signal samples
%  k = evenly spaced divisions of omega (frequency)
%
% OUTPUT VARIABLES
%  X = the transformed output
%  Various graphs of signal aspects
%
% REFERENCES
%  Adapted from "Digital Signal Processing Using MATLAB 3rd ed." - Ingle, V.,
%  Proakis, J. (pg. 65).
%
% DOCUMENTATION
%  ver 1.0 by Dominic Meads  5/8/2020
%  filename: dtft.m
%
%  The authors reccommend to use this more as an
%  excersize rather than a full function. As stated above, use the DFT for
%  better results.
%

w = (pi/100)*k;  % calculates the evenly spaced frequencies
X = x * (exp(-j*pi/100)) .^(n'*k); % calculates the DTFT

magX = abs(X);  % for graphs
angX = angle(X);
realX = real(X);  % divide into real and imaginary parts
imagX = imag(X);

figure('Color', [1 1 1]);
subplot(2,2,1); plot(w/pi,magX); grid off;
xlabel('frequency in units of pi'); title('Magnitude part');
subplot(2,2,2); plot(w/pi,realX); grid off;
xlabel('frequency in units of pi'); title('Real part');
subplot(2,2,3); plot(w/pi,angX); grid off;
xlabel('frequency in units of pi'); title('Angle part');
subplot(2,2,4); plot(w/pi,imagX); grid off;
xlabel('frequency in units of pi'); title('Imaginary part');
end

The problem is asking you to find an analytical expression given $$x_1[n]$$ and $$x_2[n]$$. There is no need to compute anything, let alone code it up!

The problem wants you to identify that given the discrete-time domain functions, you use properties of the DTFT to write $$X_2(e^{j\omega})$$ in terms of $$X_1(e^{j\omega})$$. You've already identified that $$x_2[n]$$ contains a shifted version of $$x_1[n]$$.

You can use the time-shift property which says that if the DTFT of $$x[n]$$ is $$X(e^{j\omega})$$ then

$$\mathcal{F}(x[n - m]) = e^{-j\omega m}X(e^{j\omega})$$

Keeping in mind how the domain of $$x_2[n]$$ is defined, you can use the DTFT properties to express $$X_2(e^{j\omega})$$ in terms of $$X_1(e^{j\omega})$$.

• Gotcha. Thanks. Lesson learned: dont jump straight into coding, sometimes the math is all you need.
– Dom
Jul 16 '20 at 20:09