# Question regarding DTFT of a complex signal

I have been doing DTFT practice problems for my DSP course, and I encountered this problem in the textbook that completely stumped me.

The question asks to find the DTFT of the shown signal and to plot its magnitude and phase. I am very comfortable with doing this with real-valued signals, but have no idea where to start when it comes to complex signals. Especially how the plotting would be done. I would appreciate any advice on how to tackle the problem.

• HINT: Complex numbers are numbers too. In exactly the same way that you say $u=2$ and $v = 4 \cdot u$, you can also say $u = 4 + 5j$ and $v = 8j \cdot u$. All operations are also defined over complex numbers. So, all that you have to do here is run your numbers through the sums. HINT2: This waveform has a particularly useful characteristic when it comes to the DFT. – A_A Feb 24 '18 at 7:48
• it's simpler than a "complex" number. it's purely imaginary. just factor out the $j$. – robert bristow-johnson Feb 24 '18 at 7:54
• Look up the definition of a linear operator. The DTFT is one. – hotpaw2 Feb 24 '18 at 12:52
• So my representation of the signal is the following: jdel[n] -jdel[n-4]. Is this the correct representation? From here, I will factor the j and apply the DTFT as it is a linear operator. – DSPer Feb 24 '18 at 18:03

I think you almost have the solution. As you suggest in your comment, if you express your signal as : $$x(n) = j\cdot \delta(n) - j\cdot \delta(n-4)$$ As DFT is linear, you will get : \begin{align} X(k) &= \sum_{n=0}^{N-1}x(n)\cdot e^{-2j\pi \frac{kn}{N}}\\ &= j\cdot \sum_{n=0}^{N-1}\delta (n)\cdot e^{-2j\pi \frac{kn}{N}} - j\cdot \sum_{n=0}^{N-1}\delta (n-4)\cdot e^{-2j\pi \frac{kn}{N}}\\ &=j-j\cdot e^{-8j\pi \frac{k}{N}}\\ &=\text{ }... \end{align}